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T«l. Rochdale 50514 


Anal/sis and 


Second Edition 

Wiley International Edition 


Analysis and 


Second Edition 

by Franklin F. Kuo 

Bell Telephone Laboratories, Inc. 

John Wiley & Sons, Inc., New York | London | Sydney 

Toppan Company. Ltd. , Tokyo, Japan 

Copyright © 1962, 1966 by John Wiley & Sons, Inc. 
All Rights Reserved 

This book or any part thereof 

must not be reproduced in any form 

without the written permission of the publisher 

Wiley International Edition 

This book is not to be sold outside the country 
to which it is consigned by the publisher. 

Libray of Congress Catalog Card Number : 66-16127 
Printed in Singapore by Toppan Printing Co. (S) Pte. Ltd. 

To My Father and Mother 


In the second edition, I have tried to keep the organization of the first 
edition. Most of the new material are additions aimed at strengthening 
the weaknesses of the original edition. Some specific changes deserve 
mention. The most important of these is a new chapter on computer 
applications (Chapter 15). In the past five years, digital computers have 
brought about many significant changes in the content of engineering 
subject matter concerned with both analysis and design. In analysis, 
computation has become an important adjunct to theory. Theory estate 
lishes the foundation of the subject matter; computation provides clarity, 
depth, and insight. In design, the computer has not only contributed 
precision and speed to existing procedures but has made practicable 
design methods that employ iteration and simulation. The importance of 
computer-aided design cannot be overemphasized. In Chapter 15 I have 
attempted to survey some digital computer applications in the areas of 
network analysis and design. I strongly encourage all students to read 
this chapter for cultural interest, if not for survival. 

Another new section contains a rigorous treatment of the unit impulse. 
It was difficult to decide whether to incorporate this material in Chapter 2 
in the discussion of signals or in a separate appendix. By putting general- 
ized functions in an appendix, I have left the decision of whether to teach 
the rigorous treatment up to the individual instructor. 

Other changes worth mentioning are: (1) two new sections on the 
Fourier integral in Chapter 3 ; (2) a section on initial and final conditions 
in Chapter 5; (3) a section on Bode plots in Chapter 8; (4) revised 
material on two-port parameters in Chapter 9; and (5) new sections on 
frequency and transient responses of filters in Chapter 13. Major or 
minor changes may be found in every chapter, with the exception of 
Chapters 11 and 12. In addition, many new problems are included at the 
end of each chapter. 


viii Preface 

A brief description of the subject matter follows. Chapters 1 and 2 
deal with signal representation and certain general characteristics of 
linear systems. Chapter 3 deals with Fourier analysis, and includes the 
impulse method for evaluating Fourier coefficients. Chapters 4 and 5 
discuss solutions of network differential equations in the time domain. 
In Chapters 6 and 7, the goals are the same as those of the two preceding 
chapters, except that the viewpoint here is that of the frequency domain. 
Chapter 8 deals with the amplitude, phase, and delay of a system function. 
The final seven chapters are concerned with network synthesis. Chapter 
9 deals with two-ports. In Chapter 10, the elements of realizability theory 
are presented. Chapters 11 and 12 are concerned with elementary driving- 
point and transfer function synthesis procedures. In Chapter 13, some 
fundamental concepts in modern filter design are introduced. Chapter 14 
deals with the use of scattering matrices in network analysis and design. 
And, as mentioned earlier, Chapter 15 contains a brief survey of digital 
computer techniques in system analysis. In addition, there are five 
appendices covering the rudiments of matrix algebra, generalized functions, 
complex variables, proofs of Brune's theorems, and a visual aid to filter 

The book is intended for a two-semester course in network theory. 
Chapters 1 through 8 may be used in a one-semester undergraduate or 
beginning graduate course in transient analysis or linear system analysis. 
Chapters 9 and 15 are to be used in a subsequent course on network 

The second edition is largely a result of the feedback from the professors 
who have used this book and from their students, Who have discovered 
errors and weaknesses in the original edition. I wish to express my sincere 
appreciation to those who provided this feedback. 

Special thanks are due to the following people: Robert Barnard of 
Wayne State, Charles Belove and Peter Dorato of the Polytechnic 
Institute of Brooklyn, James Kaiser and Philip Sherman of the Bell 
Telephone Laboratories, Evan Moustakas of San Jose State College, 
A. J. Welch of the University of Texas, and David Landgrebe of Purdue. 
I am particularly indebted to Mac Van Valkenburg of Princeton University 
and Robert Tracey of Illinois for editorial advice and to Donald Ford 
of Wiley for help and encouragement. 

In addition, I wish to thank Elizabeth Jenkins, Lynn Zicchino, and 
Joanne Mangione of the Bell Telephone Laboratories for their efficient 
and careful typing of the manuscript. 

» • i tj ■ u. F. F. Kuo 

Berkeley Heights, 

New Jersey, 

December, 1965 

Preface to the First Edition 

This book is an introduction to the study of electric networks based upon 
a system theoretic approach. In contrast to many present textbooks, 
the emphasis is not on the form and structure of a network but rather on 
its excitation-response properties. In other words, the major theme is 
concerned with how a linear network behaves as a signal processor. 
Special emphasis is given to the descriptions of a linear network by its 
system function in the frequency domain and its impulse response in the 
time domain. With the use of the system function as a unifying link, the 
transition from network analysis to synthesis can be accomplished with 
relative ease. 

The book was originally conceived as a set of notes for a second course 
in network analysis at the Polytechnic Institute of Brooklyn. It assumes 
that the student has already had a course in steady-state circuit analysis. 
He should be familiar with Kirchhoff's laws, mesh and node equations, 
standard network theorems, and, preferably, he should have an elementary 
understanding of network topology. 

A brief description of the subject matter follows. Chapters 1 and 2 
deal with signal representation and certain characteristics of linear net- 
works. Chapters 3, 4, 5, and 6 discuss transient analysis from both a time 
domain viewpoint, i.e., in terms of differential equations and the impulse 
response, and a frequency domain viewpoint using Fourier and Laplace 
transforms. Chapter 7 is concerned with the use of poles and zeros in 
both transient and steady-state analysis. Chapter 8 contains a classical 
treatment of network functions. 

The final five chapters deal with network synthesis. In Chapter 9, the 
elements of realizability theory are presented. Chapters 10 and 1 1 are 
concerned with elementary driving-point and transfer function synthesis 
procedures. In Chapter 12, some fundamental concepts in modern filter 
design are introduced. Chapter 13 deals with the use of scattering matrices 


x Preface to the First Edition 

in network analysis and synthesis. In addition, there are three appendices 
covering the rudiments of matrix algebra, complex variables, and proofs 
of Brune's realizability theorems. 

The book is intended for a two-semester course in network theory. 
Chapters 1 through 7 can be used in a one-semester undergraduate or 
beginning graduate course in transient analysis or linear system analysis. 
Chapters 8 through 13 are to be used in a subsequent course on network 


It was my very good fortune to have studied under Professor M. E. Van 
Valkenburg at the University of Illinois. I have been profoundly influenced 
by his philosophy of teaching and writing, which places strong emphasis 
upon clarity of exposition. In keeping with this philosophy, I have tried 
to present complicated material from a simple viewpoint, and I have 
included a large number of illustrative examples and exercises. In addition, 
I have tried to take a middle ground between mathematical rigor and 
intuitive understanding. Unless a proof contributes materially to the 
understanding of a theorem, it is omitted in favor of an intuitive argument. 
For example, in the treatment of unit impulses, a development in terms 
of a generalized function is first introduced. It is stressed that the unit 
impulse is not really a function but actually a sequence of functions whose 
limit point is undefined. Then, the less rigorous, intuitive notion of an 
impulse "function" is presented. The treatment then proceeds along the 
nonrigorous path. 

There are a number of topics which have been omitted. One of these 
is network topology, which seems to be in vogue at present. I have pur- 
posely omitted topology because it seems out of place in a book that 
de-emphasizes the form and structure approach to network analysis. 

In an expository book of this nature, it is almost impossible to reference 
adequately all the original contributors in the vast and fertile field of 
network theory. I apologize to those whose names were omitted either 
through oversight or ignorance. At the end of the book, some supple- 
mentary textbooks are listed for the student who either wishes to fill in 
some gaps in his training or wants to obtain a different point of view. 

I acknowledge with gratitude the help and advice given to me by my 
colleagues at the Bell Telephone Laboratories and by my former colleagues 
at the Polytechnic Institute of Brooklyn. I wish to express my sincere 
appreciation to the many reviewers whose advice and criticism were 
invaluable in revising preliminary drafts of the manuscript. Professors 
R. D. Barnard of Wayne State University and R. W. Newcomb of Stanford 
University deserve specific thanks for their critical reading of the entire 
manuscript and numerous helpful suggestions and comments. 

In addition, I wish to thank Mrs. Elizabeth Jenkins and Miss Elizabeth 

Preface to the First Edition xi 

La Jeunesse of the Bell Telephone Laboratories for their efficient and 
careful typing of the manuscript. 

Finally, to my wife Dora, I owe a special debt of gratitude. Her 
encouragement and cooperation made the writing of this book an 
enjoyable undertaking. 

Murray Hill, New Jersey, F. F. KUO 

January, 1962 


Chapter I: Signals and Systems 


1.1 Signal Analysis 

1.2 Complex Frequency 

1.3 Network Analysis 

1.4 Network Synthesis 





Chapter 2: Signals and Waveforms 


2.1 General Characteristics of Signals 20 

2.2 General Descriptions of Signals 24 

2.3 The Step Function and Associated Waveforms 28 

2.4 The Unit Impulse 33 

Chapter 3: The Frequency Domain: Fourier Analysis 


3.1 Introduction 

3.2 Orthogonal Functions 

3.3 Approximation Using Orthogonal Functions 

3.4 Fourier Series 

3.5 Evaluation of Fourier Coefficients 

3.6 Evaluation of Fourier Coefficients Using Unit 

3.7 The Fourier Integral 

3.8 Properties of Fourier Transforms 




xiv Contents 

Chapter 4: Differential Equations 

4.1 Introduction 

4.2 Homogeneous Linear Differential Equations 

4.3 Nonhomogeneous Equations 

4.4 Step and Impulse Response 

4.5 Integrodifferential Equations 

4.6 Simultaneous Differential Equations 



Chapter 5: Network Analysis: I 

5.1 Introduction 

5.2 Network Elements 

5.3 Initial and Final Conditions 

5.4 Step and Impulse Response 

5.5 Solution of Network Equations 

5.6 Analysis of Transformers 



Chapter 6: The Laplace Transform 

6.1 The Philosophy of Transform Methods 

6.2 The Laplace Transform 

6.3 Properties of Laplace Transforms 

6.4 Uses of Laplace Transforms 

6.5 Partial-Fraction Expansions 

6.6 Poles and Zeros 

6.7 Evaluation of Residues 

6.8 The Initial and Final Value Theorems 



Chapter 7: Transform Methods in Network Analysis 175 

7.1 The Transformed Circuit 175 

7.2 Thevenin's and Norton's Theorems 180 

7.3 The System Function 187 

7.4 The Step and Impulse Responses 194 

7.5 The Convolution Integral 197 

7.6 The Duhamel Superposition Integral 201 

Contents xv 

Chapter 8: Amplitude, Phase, and Delay 



Amplitude and Phase Response 

Bode Plots 

Single-Tuned Circuits 

Double-Tuned Circuits 

On Poles and Zeros and Time Delay 


Chapter 9: Network Analysis: II 



Network Functions 

Relationships Between Two-Port Parameters 

Transfer Functions Using Two-Port Parameters 

Interconnection of Two-Ports 

Incidental Dissipation 

Analysis of Ladder Networks 


Chapter 10: Elements of Readability Theory 


10.1 Causality and Stability 

10.2 Hurwitz Polynomials 

10.3 Positive Real Functions 

10.4 Elementary Synthesis Procedures 


Chapter 1 1 : Synthesis of One-Port Networks with Two Kinds 
of Elements 



Properties of L-C Immittance Functions 315 

Synthesis of L-C Driving-Point Immittances 319 

Properties of R-C Driving-Point Impedances 325 

Synthesis of R-C Impedances or R-L. Admittances 329 

Properties of R-L Impedances and R-C Admittances 331 

Synthesis of Certain R-L-C Functions 333 

xvi Contents 

Chapter 12: Elements of Transfer Function Synthesis 341 

12.1 Properties of Transfer Functions 341 

12.2 Zeros of Transmission 345 

12.3 Synthesis of Y n and Z S1 with a 1-Q Termination 347 

12.4 Synthesis of Constant-Resistance Networks 352 

Chapter 13: Topics in Filter Design 365 

13.1 The Filter Design Problem 365 

13.2 The Approximation Problem in Network Theory 365 

13.3 The Maximally Flat Low-Pass Filter Approximation 368 

13.4 Other Low-Pass Filter Approximations 373 

13.5 Transient Response of Low-Pass Filters 388 

13.6 A Method to Reduce Overshoot in Filters 392 

13.7 A Maximally Flat Delay and Controllable Magnitude 
Approximation 395 

13.8 Synthesis of Low-Pass Filters 397 

13.9 Magnitude and Frequency Normalization 402 

13.10 Frequency Transformations 404 

Chapter 14: The Scattering Matrix 413 

14.1 Incident and Reflected Power Flow 413 

14.2 The Scattering Parameters for a One-Port Network 415 

14.3 The Scattering Matrix for a Two-Port Network 419 

14.4 Properties of the Scattering Matrix 426 

14.5 Insertion Loss 429 

14.6 Darlington's Insertion Loss Filter Synthesis 431 

Chapter 15: Computer Techniques in Circuit Analysis 438 

15.1 The Uses of Digital Computers in Circuit Analysis 438 

15.2 Amplitude and Phase Subroutine 450 

15.3 A Fortran Program for the Analysis of Ladder 
Networks 453 

15.4 Programs that Aid in Darlington Filter Synthesis 457 

Contents xvii 

Appendix A: Introduction to Matrix Algebra 461 

A.1 Fundamental Operations 461 

A.2 Elementary Concepts 462 

A.3 Operations on Matrices 464 

A.4 Solutions of Linear Equations 468 

A. 5 References on Matrix Algebra 469 

Appendix B: Generalized Functions and the Unit Impulse 470 

B.l Generalized Functions 470 

B.2 Properties of the Unit Impulse 476 

Appendix C: Elements of Complex Variables 481 

C.l Elementary Definitions and Operations 481 

C.2 Analysis 483 

C.3 Singularities and Residues 486 

C.4 Contour Integration 487 

Appendix D: Proofs of Some Theorems on Positive Real 

Functions 490 

Appendix E: An Aid to the Improvement of Filter Approxi- 
mation 493 

E.l Introduction 493 

E.2 Constant Logarithmic Gain Contours 494 

E.3 Constant Phase Contours 495 

E.4 Contour Drawings 496 

E.5 Correction Procedure 498 

E.6 Correction Network Design 502 

E.7 Conclusion 504 

Bibliography £05 

Name Index $09 

Subject Index si I 

chapter I 

Signals and systems 

This book is an introduction to electric network theory. The first half 
of the book is devoted to network analysis and the remainder to network 
synthesis and design. What are network analysis and synthesis? In a 
generally accepted definition of network analysis and synthesis, there are 
three key words: the excitation, the network, and the response as depicted 
in Fig. 1.1. Network analysis is concerned with determining the response, 
given the excitation and the network. In network synthesis, the problem 
is to design the network given the excitation and the desired response. 
In this chapter we will outline some of the problems to be encountered 
in this book without going into the actual details of the problems. We 
will also discuss some basic definitions. 


For electric networks, the excitation and response are given in terms of 
voltages and currents which are functions of time, t. In general, these 
functions of time are called signals. In describing signals, we use the two 
universal languages of electrical engineering— time and frequency. Strictly 
speaking, a signal is a function of time. However, the signal can be 
described equally well in terms of spectral or frequency information. As 
between any two languages, such as French and German, translation is 
needed to render information given in one language comprehensible in the 






FIG. I.I. The objects of our concern. 

Network analysis and synthesis 



FIG. 1.2. Sinusoidal signal. 

other. Between time and frequency, the translation is effected by the 
Fourier series, the Fourier integral, and the Laplace transform. We shall 
have ample opportunity to define and study these terms later in the book. 
At the moment, let us examine how a signal can be described in terms of 
both frequency and time. Consider the sinusoidal signal 

s(t) = A sin (a> t + O ) 


where A is the amplitude, O is the phase shift, and <w is the angular 
frequency as given by the equation 

coo = Y < L2 > 

where T is the period of the sinusoid. The signal is plotted aguinst time 
in Fig. 1.2. An equally complete description of the signal is obtained if we 

D Ao 




b>0 W 

Angular frequency 
FIG. 1.3a. Plot of amplitude A versus angular frequency <o. 

o>o to 

Angular frequency 

FIG. 1.3b. Plot of phase 6 versus angular frequency <o. 

— «a —6>4 —6)3 — a>2 — 6>i 

Signals and systems 


&>0 «">1 »2 "3 ft>4 +<<> 

FIG. 1.4a. Discrete amplitude spectrum. 

—ft) — fc>4 — W3 — ft)2 — ft>i 

b)i 0)2 0>3 ft>4 



FIG. 1.4b. Discrete phase spectrum. 

let the angular frequency co be the independent variable. In this case, the 
signal is described in terms of ^4 > w o> an£ l ^0. as shown in Fig. 1.3a, where 
amplitude is plotted against frequency, and in Fig. 1.36, where phase shift 
is plotted. 
Now suppose that the signal is made up of 2n + 1 sinusoidal components 

s(t) =%A i sin (<v + fy) 

<— n 


The spectral description of the signal would then contain 2n + l lines at 

±<*>i, ±«>2 ±(»„ as given in Figs. 1.4a and b. These discrete spectra 

of amplitude A versus co and phase shift 6 versus co are sometimes called 
line spectra. Consider the case when the number of these spectral lines 
become infinite and the intervals oo i+1 — w f between the lines approach 
zero. Then there is no longer any discrimination between one frequency 
and another, so that the discrete line spectra fuse into a continuous spectra, 
as shown by the example in Figs. 1.5a and b. In the continuous case, the 
sum in Eq. 1.3 becomes an integral 



A(o>) sin [mt + 0(to)] dm 


where A{m) is known as the amplitude spectrum and 0(co) as the phase 

As we shall see later, periodic signals such as the sine wave in Fig. 1.2 
can be described in terms of discrete spectra through the use of Fourier 
series. On the other hand, a nonperiodic signal such as the triangular 

4 Network anal/sis and synthesis 


—a <a 

FIG. 1.5a. Continuous amplitude spectrum. 

FIG. 1.5b. Continuous phase spectrum. 

pulse in Fig. 1.6 can only be described in terms of continuous spectra 
through the Fourier integral transform. 


In this section, we will consider the concept of complex frequency. As 
we shall see, the complex frequency variable 

s = a + ja> 


is a generalized frequency variable whose real part a describes growth and 
decay of the amplitudes of signals, and whose imaginary part/cu is angular 
frequency in the usual sense. The idea of complex frequency is developed 
by examining the cisoidal signal 

= aJ»* 

S(f) = Ae 


FIG. 1.6. Triangular signal. 

Signals and systems 5 


FIG. 1.7. Rotating phasor. 

when S(0 is represented as a rotating phasor, 1 as shown in Fig. 1.7. 
The angular frequency m of the phasor can then be thought of as a velocity 
at the end of the phasor. In particular the velocity to is always at right 
angles to the phasor, as shown in Fig. 1.7. However, consider the general 
case when the velocity is inclined at any arbitrary angle y> as given in 
Figs. 1.8a and 1.86. In this case, if the velocity is given by the symbol s, 
we see that s is composed of a component m at right angle to the phasor S 
as well as a component a, which is parallel to S. In Fig. 1.8a, s has a 
component — a toward the origin. As the phasor S spins in a counter- 
clockwise fashion, the phasor decreases in amplitude. The resulting wave 
for the real and imaginary parts of S(r) are damped sinusoids as given by 

ReS(f) = Ae~" cos cot 
ImS(0 = X<r*'sina>f 


M ]<b) 

FIG. 1.8. (a) Rotating phasor with exponentially decreasing amplitude, (b) Rotating 
phasor with exponentially increasing amplitude. 

1 A phasor S is a complex number characterized by a magnitude and a phase angle 
(see Appendix C). 

6 Network analysis and synthesis 

/Envelope = Ae~ 

FIG. 1.9. Damped sinusoids. 

/ ^~ Envelope = Ae** 

FIG. 1. 10. Exponentially increasing sinusoid. 

Signals and systems 7 


FIG. I.I I. Exponential signals. 

which are shown in Fig. 1.9. Note that the damped sinusoid has an 
exponential envelope decay, Ae~" % . In Fig. 1 .84, the phasor is shown with 
a positive real component of velocity +o. Therefore, as the phasor spins, 
the amplitudes of the real and imaginary parts increase exponentially 
with an envelope Ae +at , as shown by Im S(f) in Fig. 1.10. 
From this discussion, it is apparent that the generalized cisoidal signal 

S(0 = Ae" = Ae {a+Mt 


describes the growth and decay of the amplitudes in addition to angular 
frequency in the usual sense. When a = 0, the sinusoid is undamped, and 
wheny'a> = 0, the signal is an exponential signal 

S(0 = Ae*' 


as shown in Fig. 1.11. Finally, if a = ja> = 0, then the signal is a constant 
A. Thus we see the versatility of a complex frequency description. 


As mentioned before, the characterization of the excitation and response 
signals in time and frequency makes up only part of the analysis problem. 
The other part consists of characterizing the network itself in terms of 
time and frequency, and determining how the network behaves as a signal 
processer. Let us turn our attention now to a brief study of the properties 
of linear networks and the general characteristics of signal processing by 
a linear system. 

8 Network analysis and synthesis 



A system (network) is linear if (a) the principle of superposition and 
(b) the principle of proportionality hold. 

By the superposition principle, if, for a given network, [e x (t), r^t)] and 
[e g (0, r 2 (f)l are excitation-response pairs, then if the excitation were 
e[t) = e x (t) + e 2 (t), the response would be r(t) = r,(f) + r£i). By the 
proportionality principle, if the excitation were C^it), where C x is a 
constant, then the response would be C^r^t), i.e., the constant of propor- 
tionality C x is preserved by the linear network. The two conditions of 
linearity are summarized in Fig. 1.12. 

Another definition of a linear network is that the excitation and response 
of the network are related by a linear differential equation. We shall 
discuss this definition in Chapter 4 on differential equations. 


A linear network is passive* if (a) the energy delivered to the network is 
nonnegative for any arbitrary excitation, and (b) if no voltages or currents 
appear between any two terminals before an excitation is applied. 


A network is said to be reciprocal if when the points of excitation and 
measurement of response are interchanged, the relationship between 

Cieift) .. 


CuiM . 

C 2 e2(t) 


C 2 r^t) _ 



Cm(t) * CW<) 

FIG. 1.12. linear system. 

1 G. Raisbeck, "A Definition of Passive Linear Networks in Terms of Time and 
Energy," /. Appl. Phys., 25 (Dec. 1954), 1510-1514. 

Signals and systems 9 



- — — .M. 



>- B 



<• t 





^ B 


\ Ti 

hTi t 



FIG. 1.13. Time-invariant system. 

excitation and response remains the same, 
choice of points of excitation and response. 

Thus must be true for any 


We say a system is causal if its response is nonanticipatory, i.e., if 

«(0 = t < T 
then KO = ° ' < r 


In other words, a system is causal if before an excitation is applied at 
/ = T, the response is zero for — oo < t < T. 

Time invariant 

A system is time invariant ife(t) -*■ rtf) implies that e(t ± T)-*-r(t± T), 
where the symbol -»■ means "gives rise to." To understand the concept 
of time invariance in a linear system, let us suppose that initially the 
excitation is introduced at f = 0, which gives rise to a response r(t). If 
the excitation were introduced at / = T, and if the shape of the response 
waveform were the same as in the first case, but delayed by a time T 
(Fig. 1.13), then we could say the system is time invariant. Another way 
of looking at this concept is through the fact that time-invariant systems 
contain only elements that do not vary with time. It should be mentioned 
here that linear systems need not be time invariant. 

Derivative property 

From the time-invariant property we can show that, if e{f) at the input 
gives rise to r(i) at the output (Fig. 1.14), then, if the input were e'(f), 

10 Network analysis and synthesis 




> dr(t) 


* dt 



, d*Ht) 


' dt' 

e(T)dr — 


■k C'.{,\ 

> 1 l'(T) 

FIG. 1.14. Some implications of linear time-invariant systems. 

i.e., the derivative of e(t), the response would be r'(t). The proof is quite 
simple. Consider an excitation e(* + e) where e is a real quantity. By the 
time-invariant property, the response would be r(t + e). Now suppose the 
excitation were 

e 1 (r)--W* + «)-««)] (1-11) 


then according to the linearity and time-invariant properties, the response 
would be 

ri (0 = - [r(I + e) - K0] (112) 

Taking the limit as e ->- 0, we see that 

limc 1 (0 = 7-c(0 

e-»0 at 

lim r x (0 = ^ K0 
«-»o at 


We can extend this idea to higher derivatives as well as for the integrals of 
e(t) and r(f), as shown in Fig. 1.14. 

Ideal models 

Let us now examine some idealized models of linear systems. The 
systems given in the following all have properties which make them very 
useful in signal processing. 

Signals and systems 1 1 


■ Amplifier 



1.15. Amplifier. 


K d 

K dt 

. Kdfltt 

1 dt 


.16. Differentiator. 


>K i f f VaT 


1.17. Integ 






FIG. 1.18. Time-delay network. 

1 2 t 

FIG. 1.19. Excitation function. 

1. Amplifier: An amplifier scales up the magnitude of the input, i.e., 
lit) = Ke(t), where Kis a constant (Fig. 1.15). 

2. Differentiator: The input signal is differentiated and possibly scaled 
up or down (Fig. 1.16). 

3. Integrator: The output is the integral of the input, as shown in 
Fig. 1.17. 

4. Time delayer: The output is delayed by an amount T, but retains the 
same wave shape as the input (Fig. 1.18). 

Suppose we take the triangular pulse in Fig. 1.19 as the input signal. 
Then the outputs for each of the four systems just described are shown 
in Figs. 1.200-1 .20(2. 

12 Network analysis and synthesis 





2 t 




Ti Ti + 1 Ti + 2 

(c) (d) 

FIG. I JO. la) Amplifier output, (ft) Differentiator output, (c) Integrator output, (rf) 
Delayed output. 

Ideal elements 

In the analysis of electric networks, we use idealized linear mathematical 
models of physical circuit elements. The elements most often encountered 
are the resistor R, given in ohms, the capacitor C, given in farads, and 
the inductor L, expressed in henrys. The endpoints of the elements are 
called terminals. A port is defined as any pair of two terminals into which 
energy is supplied or withdrawn or where network variables may be 
measured or observed. In Fig. 1.21 we have an example of a two-port 

The energy sources that make up the excitation functions are ideal 
current or voltage sources, as shown in Figs. 1.22a and b. The polarities 


*a Response 
'/ measurement 

FIG. 1.21. Two-port network. 




Signals and systems 13 



FIG. 1.22a. Voltage source. FIG. 1.22b. Current source. 

indicated for the voltage source and the direction of flow for the current 
source are arbitrarily assumed for reference purposes only. An ideal 
voltage source is an energy source that provides, at a given port, a voltage 
signal that is independent of the current at that port. If we interchange the 
words "current" and "voltage" in the last definition, we then define an 
ideal current source. 

In network analysis, the principal problem is to find the relationships 
that exist between the currents and voltages at the ports of the network. 
Certain simple voltage-current relationships for the network elements also 
serve as defining equations for the elements themselves. For example, 
when the currents and voltages are expressed as functions of time, then the 
R, L, and C elements, shown in Fig. 1.23, are defined by the equations 

or i(t) = -v(t) 

t>(0 = L -J9 or iQ) = i J v(x) dx + j(0) (1.14) 
at L *o 

»(0 = 7 I' «(*) dx + v(0) or i(t) = C ^ 
CJo dt 

where the constants of integration i'(0) and »(0) are initial conditions to be 
discussed in detail later. 
Expressed as a function of the complex frequency variable s, the equations 


— *- 



— *— 





M (b) (e) 

FIG. 1.23. (a) Resistor, (b) Inductor, (c) Capacitor. 

14 Network analysis and synthesis 





— »— 



]sL V(s) 


(a) (b) (c) 

FIG. 1.24. (a) Resistor, (b) Inductor, (c) Capacitor. 

defining the R, L, and C elements, shown in Fig. 1.24, are (ignoring 
initial conditions for the moment) 

V(s) = RI(s) or I(s) = ^V(s) 


V(s) = sLI(s) or I(s) = — V(s) 



7(s) = -^/(s) 

or /(s) = sCV(s) 

We see that in the time domain, i.e., where the independent variable is t, 
the voltage-current relationships are given in terms of differential equations. 
On the other hand, in the complex-frequency domain, the voltage-current 
relationships for the elements are expressed in algebraic equations. 
Algebraic equations are, in most cases, more easily solved than differential 
equations. Herein lies the ration d'itre for describing signals and networks 
in the frequency domain as well as in the time domain. 

When a network is made up of an interconnection of linear circuit 
elements, the network is described by its system or transfer function H(s). 
The response R(s) and the excitation E(s) are related by the equation 

R(s) = H(s) E(s). 


In network analysis, we are given E(s), and we can obtain H(s) directly 
from the network. Our task is to determine R(s). 


We will now briefly introduce some of the problems germane to network 
synthesis. In network synthesis, we are given the response R(s) and the 
excitation E(s), and we are required to synthesize the network from the 

Signals and systems 15 




FIG. 1.25. Driving-point impedance 
Z{s) = R. 

FIG. I.M. Black box. 

system function 





Since R(s) and £(5) are voltages or currents, then H(s) is denoted generally 
as an immittance if R(s) is a voltage and E(s) is a current, or vice versa. 
A driving-point immittance 9 is defined to be a function for which the 
variables are measured at the same port. Thus a driving-point impedance 
Z(s) at a given port is the function 



where the excitation is a current I(s) and the response is a voltage V(s), as 

shown in Fig. 1.25. When we interchange the words "current" and 

"voltage" in the last definition, we then have a driving-point admittance. 

An example of a driving-point impedance is the network in Fig. 1.25, 


Z(s) = ^i = R 


Now suppose the resistor in Fig. 1.25 were enclosed in a "black box." 
We have no access to this black box, except at the terminals 1-1' in Fig. 
1.26. Our task is to determine the network in the black box. Suppose we 
are given the information that, for a given excitation /(*), the voltage 
response V(s) is proportional to I(s) by the equation 

V(s) = KI(s) (1.20) 

An obvious solution, though not unique, is that the network consists of a 
resistor of value R = KCl. Suppose next that the excitation is a voltage 
V(s), the response is a current I(s), and that 

ns) _M. 3 + 4j 

•IRE Standards on Circuits "Linear Passive Networks," 
(Sept. 1960), 1608-1610. 


Proe. IRE, 48, No. 9 

16 Network analysis and synthesis 

1 '■ hM 




=r 4 >i 





— < o 


FIG. 1.27. Network realiza- 
tion for Y(s). 

FIG. 1.28. Two-port network. 

Our task is to synthesize a network equivalent to the network in the black 
box. From a close scrutiny of the driving-point admittance Y(s), we see 
that a possible solution might consist of a resistor of J O in parallel with 
a capacitor of 4 farads, as seen in Fig. 1.27. 

The problem of driving-point synthesis, as shown from the examples 
just given, consists of decomposing a given immittance function into 
basic recognizable parts (such as 3 + 4s). Before we proceed with the 
mechanics of decomposition, we must first determine whether the function 
is realizable, i.e., can it be synthesized in terms of positive resistances, 
inductances, and capacitances? It will be shown that realizable driving- 
point immittances belong to a class of functions known as positive real or, 
simply, p.r. functions. From the properties of p.r. functions, we can test a 
given driving-point function for realizability. (The Appendices present a 
short introduction to complex variables as well as the proofs of some 
theorems on positive real functions.) With a knowledge of p.r. functions, 
we then go on to examine special driving-point functions. These include 
functions which can be realized with two kinds of elements only — the L-C, 
R-C, and R-L immittances. 

Next we proceed to the synthesis of transfer functions. According to 
the IRE Standards on passive linear networks, 4 a transfer function or 
transmittance is a system function for which the variables are measured at 
different ports. There are many different forms which a transfer function 
might take. For example, consider the two-port network in Fig. 1.28. 
If the excitation is I^s) and the response Vjs), the transfer function is a 
transfer impedance 

Z tt (s) = 


On the other hand, if V l (s) were the excitation and V^s) the response, then 
we would have a voltage-ratio transfer function 

tf( S )=™ 


4 Loe. eit. 

Signals and systems 17 


FIG. 1.2?. Ideal amplitude spectrum for low-pass filter. 

As for driving-point functions, there are certain properties which a 
transfer function must satisfy in order to be realizable. We shall study 
these realizability conditions and then proceed to the synthesis of some 
simple transfer functions. 

The most important aspect of transfer function synthesis is filter design. 
A filter is defined as a network which passes a certain portion of a fre- 
quency spectrum and blocks the remainder of the spectrum. By the term 
"blocking," we imply that the magnitude response \H(ja>)\ of the filter is 
approximately zero for that frequency range. Thus, an ideal low-pass 
filter is a network which passes all frequencies up to a cutoff frequency 
m c , and blocks all frequencies above w c , as shown in Fig. 1.29. 

One aspect of filter design is to synthesize the network from the transfer 
function H(s). The other aspect deals with the problem of obtaining a 
realizable transmittance H(s) given the specification of, for example, the 
magnitude characteristic in Fig. 1.29. This part of the synthesis is generally 
referred to as the approximation problem. Why the word "approxi- 
mation?" Because frequency response characteristics of the R, L, and C 
elements are continuous (with the exception of isolated points called 
resonance points), a network containing these elements cannot be made to 
cut off abruptly at <o c in Fig. 1.29. Instead, we can realize low-pass filters 
which have the magnitude characteristics of Fig. 1.30. In connection with 


FIG. 1.30. Realizable low-pass filter characteristics. 

18 Network analysis and synthesis 

the filter design problems, we will discuss certain problems in magnitude 
and frequency normalization so that, in designing a filter, we deal with 
element values such as R — 0.5 ohm and C = 2 farads instead of "practical" 
element values of, for example, R — 500,000 ohms and C = 2 picofarads 
(pico = 10 -12 ). Also we will study a method whereby low-pass filter 
designs might be transformed into high-pass, band-pass, and band- 
elimination filters. The mathematical basis of this method is called 
frequency transformation. 

We next discuss some aspects of analysis and synthesis in which the 
excitation and response functions are given in terms of power rather than 
of voltage and current. We will examine the power-transfer properties of 
linear networks, using scattering parameters, which describe the incident 
and reflected power of the network at its ports. 

Finally, in Chapter 15, we will examine some of the many uses of high- 
speed digital computers in circuit analysis and design. In addition to a 
general survey of the field, we will also study some specific computer 
programs in circuit analysis. 


1.1 Draw the line spectra for the signal 

s(t) - 3 sin (t + ^1 + 4sin (it - 1| 

+ 6 sin it 

1.2 Find the response to the excitation sin / into a sampler that closes every 
JtTir/4 seconds where K »= 0, 1, 2, Draw the response for ^ / ^ 2*. 

13 Find the response to the excitation shown in the figure when the network 
is (a) an ideal differentiator; (ft) an ideal integrator. 



1 1.5 2 

PROS. 1.3 

1.4 If the system function of a network is given as 



(s + 2X* + 3) 

Signals and systems 19 
find the response R(s) if the excitation is 


1.5 Given the driving-point functions find their simplest network realiz- 



1.6 For the network shown, write the mesh equation in terms of (a) differential 
equations and (b) the complex-frequency variable s. 

R l 

— — wv— npfir* — 




= 2s H — 

s + 2 


-3 1 * 

i+ s*+2 


_ 1 2, 

3s + 2 s* + 4 


PROB. 1.6 

1.7 For the network shown, write the node equation in terms of (a) differential 
equations and (6) complex-frequency form. 

PROB. 1.7 

1.8 Suppose the response of a linear system to an excitation «(/) were 
r(t) = 3e-*'. What would the response be to an excitation of t(t — 2)1 

chapter 2 

Signals and waveforms 

Our main concern in this chapter is the characterization of signals as 
functions of time. In previous studies we have dealt with d-c signals 
that were constant with time, or a-c signals which were sinusoids of 
constant amplitude, such as s(t) = A sin (cot + 0). In engineering practice, 
the class of signals encountered is substantially broader in scope than 
simple a-c or d-c signals. To attempt to characterize each member of 
the class is foolhardy in view of the almost infinite variety of signals 
encountered. Instead, we will deal only with those signals that can be 
characterized in simple mathematical terms and which serve as building 
blocks for a large number of other signals. We will concentrate on formu- 
lating analytical tools to aid us in describing signals, rather than deal with 
the representation of specific signals. Because of time and space limita- 
tions, we will cover only signals which do not exhibit random behavior, 
i.e., signals which can be explicitly characterized as functions of time. 
These signals are often referred to as deterministic signals. Let us first 
discuss certain qualitative aspects of signals in general. 


In this section we will examine certain behavior patterns of signals. 
Once these patterns are established, signals can be classified accordingly, 
and some simplifications result. The adjectives which give a general 
qualitative description of a signal axe periodic, symmetrical, and continuous. 
Let us discuss these terms in the given order. 

First, signals are either periodic or aperiodic. If a signal is periodic, then 
it is described by the equation 

s(t) = s(t±kT) k = 0,1,2,... (2.1) 


Signals and waveforms 21 










FIG. 2.1. Square wave. 

where T is the period of the signal. The sine wave, sin /, is periodic with 
period T = lit. Another example of a periodic signal is the square wane 
given in Fig. 2.1. On the other hand, the signals given in Fig. 2.2 are 
aperiodic, because the pulse patterns do not repeat after a certain finite 
interval T. Alternatively, these signals may be considered "periodic" with 
an infinite period. 

Next, consider the symmetry properties of a signal. The key adjectives 
here aie even and odd. A signal function can be even or odd or neither. 
An even function obeys the relation 



For an odd function 

s(i) = -s{-t) 


For example, the function sin t is odd, whereas cos t is even. The square 
pulse in Fig. 2.2a is even, whereas the triangular pulse is odd (Fig. 2.26). 
Observe that a signal need not be even or odd. Two examples of signals 
of this type are shown in Figs. 2.3a and 2.4a. It is significant to note, 
however, that any signal s(t) can be resolved into an even component 5,(0 
and an odd component s 9 (t) such that 

*(0 = *.(0 + *o(0 


For example, the signals in Figs. 2.3a and 2.4a can be decomposed into 
odd and even components, as indicated in Figs. 2.3b, 2.3c, 2.4b, and 2.4c. 


o t 


M (b) 

FIG. 12. (a) Even function, (b) Odd function. 

22 Network analysis and synthesis 






-1 o 1 

i t 








FIG. 2.3. Decomposition into 
odd and even components, 
(a) Original function, (6) 
Even part, (c) Odd part. 




FIG. 2.4. Decomposition into 
even and odd components, 
(a) Unit step function. (6) 
Even part of unit step, (c) 
Odd part of unit step. 


From Eq. 2.4 we observe that 

s(-t) = s t (-t) + * (-0 
- sXO - *o(0 
Consequently, the odd and even parts of the signal can be expressed as 

J#)- MK0 + *(-')] (2 .6; 

*(o = two - K-m 

Consider the signal s(t), shown in Fig. 2.5a. The function s(-t) is equal 
to s(t) reflected about the t = axis and is given in Fig. 2.5ft. We then 
obtain s t (t) and j„(0 as shown in Figs. 2.5c and d, respectively. 


Signals and waveforms 





1 t 
















I t 

(e) (d) 

FIG. US. Decomposition into odd and even components from s(t) and s(—t). 

Now let us turn our attention to the continuity property of signals. 
Consider the signal shown in Fig. 2.6. At t = T, the signal is discontinuous. 
The height of the discontinuity is 

f(T+) -f(T- ) = A (2.7) 


/(T+) = lim/(T+c) 


f(T-) - lim/(T- «) 


and € is a real positive quantity. In particular, we are concerned with 
discontinuities in the neighborhood of f = 0. From Eq. 2.8, the points 



T t 

FIG. 24. Signal with discontinuity. 


Network analysis and synthesis 



FIG. 2.7. Signal with two discontinuities. 

/(0+) and/(0-) are 

/(0+) - lim/( e ) 


/(0-) = lim/(-e) 


For example, the square pulse in Fig. 2.7 has two discontinuities, at T x and 
T t . The height of the discontinuity at T x is 

s(T x +) - s(T x -) = K (2.10) 

Similarly, the height of the discontinuity at T a is — K. 


In this section we consider various time domain descriptions of signals. 
In particular, we examine the meanings of the following terms: time 
constant, rms value, d-c value, duty cycle, and crest factor. The term, time 
constant, refers only to exponential waveforms; the remaining four 
terms describe only periodic waveforms. 

Time constant 

In many physical problems, it is important to know how quickly a 
waveform decays. A useful measure of the decay of an exponential is the 
time constant T. Consider the exponential waveform described by 

r(t) = Ke~* IT u(t) 
From a plot of r(t) in Fig. 2.8, we see that when t = T, 

r(T) = 0.37KO) 
Also r(4T) = 0.02r(0) 



Signals and waveforms 25 








FIG. 2.8. Normalized curve for time constant T ■ 

■ 1. 

Observe that the larger the time constant, the longer it requires for the 
waveform to reach 37% of its peak value. In circuit analysis, common 
time constants are the factors RC and RjL. 

RMS Value 

The rms or root mean square value of a periodic waveform e(t) is defined 



- i?f>*r 


where T is the period. If the waveform is not periodic, the term rms does 
not apply. As an example, let us calculate the rms voltage for the periodic 
waveform in Fig. 2.9. 



= J2J3Av 


D-C Value 

The d-c value of a waveform has meaning only when the waveform is 
periodic. It is the average value of the waveform over one period 

1 f T 

«oc = - e{t)dt 

TJ p 


26 Network analysis and synthesis 




FIG. 2.9. Periodic waveform. 

The square wave in Fig. 2.1 has zero d-c value, whereas the waveform in 
Fig. 2.9 has a d-c value of 

1 [AT ATI A 


Duty cycle 

The term duty cycle, D, is defined as the ratio of the time duration of the 
positive cycle t 9 of a periodic waveform to the period, T, that is, 



The duty cycle of a pulse train becomes important in dealing with wave- 
forms of the type shown in Fig. 2.10, where most of the energy is con- 
centrated in a narrow pulse of width f . The rms voltage of the waveform 
in Fig. 2.10 is 



= AJ7JT 
An A 



FIG. 2.10. Periodic waveform with small duty cycle. 

Signals and waveforms 27 




FIG. 2.1 1. Periodic waveform with zero d-c and small duty cycle. 

Wc see that the smaller the duty cycle, the smaller the rms voltage. The 
square wave in Fig. 2.1 has a 50% duty cycle. 

Crest factor 

Crest factor 1 is defined as the ratio of the peak voltage of a periodic 
waveform to the rms value (with the d-c component removed). Explicitly, 
for any waveform with zero d-c such as the one shown in Fig. 2.1 1 — crest 
factor, CF, is denned as 

CF = -^- or -^- 




whichever is greater. For the waveform in Fig. 2.11, the peak-to-peak 
voltage is defined as 

e, p = *„ + <?» 

Since the waveform has zero d-c value 

eJ* = e t (T-t ) 
Also, e b = e„D 

and e a = e PP (l - D) 

The rms value of the waveform is 

e mg = ( e »( l ~ D ? f ° + enWr- t ) \ * 




- e ppy /D(l - D) 

1 G. Justice, 'The Significance of Crest Factor," Hewlett-Packard Journal, 15, No. 5 
(Jan., 1964), 4-5. 

28 Network analysis and synthesis 
Since crest factor CF = ejerm*, we have 

e„(\ - D) 

CF = 

e p jD(l - D) 



For example, if D = — , 


= ^100 - 1 


UD = 


CF=VlO,000- 1~100 



A voltmeter with high crest factor is able to read accurately rms values of 
signals whose waveforms differ from sinusoids, in particular, signals with 
low duty factor. Note that the smallest value of crest factor occurs for 
the maximum value of D, that is, X>m»x = 0.5, 

CFnto = Vl/iW - 1 
= 1 




The unit step function w(f) shown in Fig. 2.12 is defined as 

„(/) = t < 

= 1 f£0 

The physical analogy of a unit step excitation corresponds to a switch S, 
which closes at t = and connects a d-c battery of 1 volt to a given circuit, 
as shown in Fig. 2.13. Note that the unit step is zero whenever the 







FIG. 2.12. Unit step function. FIG. 2.13. Network analog of unit step. 

Signals and waveforms 29 



*<t-a) 4 

a ( 

FIG. 2.14. Shifted step function. 

1 2 

FIG. 2.15. Square pulse. 

argument (/) within the parentheses is negative, and is unity when the 
argument (t) is greater than zero. Thus the function u(t — a), where 
a > 0, is defined by 

vit - a) = / < a 

= 1 t^a {Zil) 

and is shown in Fig. 2.14. Note that the jump discontinuity of the step 
occurs when the argument within the parentheses is zero. This forms the 
basis of the shifting property of the step function. Also, the height of the 
jump discontinuity of the step can be scaled up or down by the multiplica- 
tion of a constant K. 

With the use of the change of amplitude and the shifting properties of 
the step function, we can proceed to construct a family of pulse waveforms. 
For example, the square pulse in Fig. 2.15 can be constructed by the sum 
of two step functions 

sif) = 4u(t - 1) + (-4) u(t - 2) 


as given in Fig. 2.16. The "staircase" function, shown in Fig. 2.17, is 
characterized by the equation 

s(t) =>J,u(t - kT) (2.33) 




FIG. 2.16. Construction of square pulse by step function. 


Network analysis and synthesis 





FIG. 2.17. Staircase function. 


Finally, let us construct the square wave in Fig. 2.1. Using the shifting 
property, we see that the square wave is given by (for t ^ 0) 

s(t) - m(0 - 2u(t - T) + 2u(t - 2T) - 2k(* - 3D + • ' • (2.34) 

A simpler way to represent the square wave is by using the property 
that the step function is zero whenever its argument is negative. Restricting 
ourselves to the interval f ;> 0, the function 




is zero whenever sin (irtIT) is negative, as seen by the waveform in Fig. 
2.18. It is now apparent that the square wave in Fig. 2. 1 can be represented 

S (0 = «(sin^)-u(-sin^) (2.36) 

Another method of describing the square wave is to consider a generaliz- 
ation of the step function known as the sgn function (pronounced signum). 
The sgn function is defined as 

sgn [/(f)] =1 

fit) > o 
fit) - 

fit) < o 



T 2T 3T AT 

FIG. 2. IS. The signal u(sin7rr/r). 

Signals and waveforms 31 


ZT 3T 

FIG. 2.19. Sine pulse. 

Thus the square wave in Fig. 2.1 is simply expressed as 

s(0 = sgn I sin — 1 


Returning to the shifting property of the step function, we see that the 
single sine pulse in Fig. 2.19 can be represented as 

s(r) - sin ^ [u(t - IT) - u(t - 3T)] 


The step function is also extremely useful in representing the shifted or 
delayed version of any given signal. For example, consider the unit ramp 

p(t) = tu(t) 


shown in Fig. 2.20. Suppose the ramp is delayed by an amount t = a, as 
shown in Fig. 2.21. How do we represent the delayed version of ramp? 

l t 

FIG.2JM. Ramp function with zero time shift. 

32 Network analysis and synthesis 


FIG. 2.21. Ramp function with time shift = a. 

First, let us replace the variable t by a new variable t' — t — a. Then 

p(r') = f u(t') (2.41) 

When p(t') is plotted against t', the resulting curve is identical to the plot 
of 5(0 versus t in Fig. 2.20. If, however, we substitute t — a = t ' in p(t'), 

we then have 

p(f) = (t - a) u(t - a) (2.42) 

When we plot p(t') against t, we have the delayed version of pit) shown in 
Fig. 2.21. 

From the preceding discussion, it is clear that if any signal /(/) u(t) is 
delayed by a time T, the delayed or shifted signal is given by 

f{t')=f{t-T)u{t-T) (2.43) 

For example, let us delay the function (sin irtfT) u(i) by a period T. Then 
the delayed function s(t'), shown in Fig. 2.22, is 

s(t') = [sin - (t - T)] u(t - T) (2.44) 

FIG. 2.22. Shifted sine wave. 

Signals and waveforms 33 



1 2 

FIG. 2.23. Triangular pulse. 

As a final example, consider the waveform in Fig. 2.23, whose com- 
ponent parts are given in Fig. 2.24. For increasing t, the first nonzero 
component is the function 2{t — 1) u(t — 1), which represents the straight 
line of slope 2 at / = 1. At ? = 2, the rise of the straight line is to be 
arrested, so we add to the first component a term equal to — 2(t — 2)u(t — 2) 
with a slope of —2. The sum is then a constant equal to 2. We then add a 
term — 2u(t — 2) to bring the level down to zero. Thus, 

s(t) = 2{t - 1) u(t - 1) - 2(t - 2) u(t - 2) - 2u(t - 2) (2.45) 


The unit impulse, or delta function, is a mathematical anomaly. P. A. M. 
Dirac first used it in his writings on quantum mechanics. 2 He defined the 


FIG. 2.24. Decomposition of the triangular pulse in Fig. 2.23. 

• P. A. M. Dirac, The Principles of Quantum Mechanics Oxford University Press, 1930. 

34 Network anal/sis and synthesis 
delta function d\t) by the equations 


6\t) dt = 1 

<J(0 = for t * 
Its most important property is the sifting properly, expressed by 


J— i 




In this section we will examine the unit impulse from a nonrigorous 

approach. Those who prefer a rigorous treatment should refer to 

Appendix B for development of this discussion. The material in that 

appendix is based on the theory of generalized functions originated by 

G. Temple. 8 In Appendix B it is shown that the unit impulse is the 

derivative of the unit step .. - „ ,. /0 ao>l 

r o(0 = if (0 (2.49) 

At first glance this statement is doubtful. After all, the derivative of the 
unit step is zero everywhere except at the jump discontinuity, and it does 
not even exist at that point! However, consider the function g t (i) in Fig. 
2.25. It is clear that as e goes to zero, g ( (t) approaches a unit step, that is, 

lim g e (t) 



Taking the derivative of g t (t), we obtain £',(0. which is defined by the 

g'«(0 - - ; 


= 0; t<0, t>€ 

as shown in Fig. 2.26. Now let « take on a sequence of values e ( such that 
«« > e 4+1 . Consider the sequence of functions {g' c< (0} for decreasing 

e t (t) 


FIG. US. Unit step when « — 0. 

FIG. 1M. Derivative of g ( (t) in 
Fig. 2.25. 

* G. Temple, "The Theory of Generalized Functions," Proe. Royal Society, A, 228, 
1955, 175-190. 

Signals and waveforms 35 





8i t m 

n «s «s n * 

FIG. 127. The sequence {g fi (t)}. 

values of c { , as shown in Fig. 2.27. The sequence has the following 

g' (i (t)dt = l 


where t x and /, are arbitrary real numbers. For every nonzero value of e, 
there corresponds a well-behaved function (i.e., it does not "blow up") 
g't t (t). As €f approaches zero, 





so that the limit of the sequence is not defined in the classical sense. 
Another sequence of functions which obeys the property given in Eq. 
2.52 is the sequence {/«//)} m Fig- 2.28. We now define the unit impulse 
6{t) as the class of all sequences of functions which obey Eq. 2.52. In 
particular, we define 

J*«i>o r*t>o 

«t)A-lim g' t( (t)dt 

«i<0 e<-»0«/«i<0 

a r*» >o 

4hm UQdt 


36 Network analysis and synthesis 

FIG. 2.28. The sequence {/ £ ,(0> 

It should be stressed that this is not a rigorous definition (which, as 
stated previously, is found in Appendix B) but merely a heuristic one. 
From the previous definition we can think of the delta "function" as 
having the additional properties, 

<5(0) = oo 

<5(f) = for t^O 


Signals and waveforms 37 

Continuing with this heuristic treatment, we say that the area "under" 
the impulse is unity, and, since the impulse is zero for t # 0, we have 

d(t)dt = \ 6 

J— <x> Jo— 

6{i) dt = 1 



Thus the entire area of the impulse is "concentrated" at t = 0. Con- 
sequently, any integral that does not integrate through t = is zero, 

/*o- r+oo 

as seen by S(t) dt = (5(f) dt = 

J-oo J0+ 


The change of scale and time shift properties 
discussed earlier also apply for the impulse 
function. The derivative of a step function 

s(t) = Au(t — a) 


s(t) = Au(t- a) 



s'(t) = Ad(t-a) 
FIG. 2.29 

yields an impulse function 

s\t) = A6(t- a) (2.59) 

which is shown in Fig. 2.29. Graphically, we 
represent an impulse function by an arrow- 
head pointing upward, with the constant 
multiplier A written next to the arrowhead. 
Note that A is the area under the impulse 
A d(t - a). 

Consider the implications of Eqs. 2.58 and 2.59. From these equations 
we see that the derivative of the step at the jump discontinuity of height A 
yields an impulse of area A at that same point t = T. Generalizing on this 
argument, consider any function /(0 with a jump discontinuity at / = T. 
Then the derivative, /'(0 must have an impulse at t = T. As an example, 
consider /(0 in Fig. 2.30. At t = T,f(t) has a discontinuity of height A, 


T t 

FIG. 2.30. Function with discontinuity at T. 

38 Network analysis and synthesis 
which is given as 

A =/(r+) -f(T-) 


Let us define f x (t) as being equal to f(t) for t < T, and having the 
same shape as/(/), but without the discontinuity for t > T, that is, 

The derivative A0 is then 

A0 -AC) + ^ <K* - T) 



The following example illustrates this point more clearly. In Fig. 2.31a, 
the function /(f) is 

fit) = Au(t-a)-A u(t - b) (2.63) 

Its derivative is fit) = Ad(t - a) - Ad(t - b) 


and is shown in Fig. 2.316. Since/(/) has two discontinuities, at t = a and 
t — b, its derivative must have impulses at those points. The coefficient of 
the impulse at t = b is negative because 

f(b+) -fQ>-) - -A 







a b t 













e pulse, 


FIG. 2.31. (a) Squar 
(b) Derivative of squai 





.2.32. (a) Signal, 


Signals and waveforms 39 

As a second example, consider the function g(i) shown in Fig. 2.32a. 
We obtain #'(0 by inspection, and note that the discontinuity at / = 1 
produces the impulse in #'(0 of area 

£0+)-£(l-) = 2, (2.66) 

as given in Fig. 2.32*. 

Another interesting property of the impulse function is expressed by the 

f + 7(0 S(t-T)dt= f(T) (2.67) 

This integral is easily evaluated if we consider that d(t — T) = for all 
1 5* T. Therefore, the product 

f{t)d(t-T) = aXLtitT (2.68) 

If/(r) is single-valued at t = T,f(T) can be factored from the integral so 
that we obtain 

/(T) | d(t -T)dt= f(T) (2.69) 

J— O0 

Figure 2.33 shows /(r) and d(t - T), where /(/) is continuous at i = T. 
If/(0 has a discontinuity at / = T, the integral 

J +00 



is not denned because the value of f(T) is not uniquely given. Consider 
the following examples. 

Example 2.1 

Example Z2 

J +0O 

/(0 - 1*«* 
~&*Hf - T)dt - e*» r 

/(/) = sin t 

£>''('- 5)*-^ 



40 Network analysis and synthesis 


FIG. 2.34. Impulse scanning. 

Consider next the case where/ (0 is continuous for - oo < t < oo. Let 
us direct our attention to the integral 




which holds for all / in this case. If T were varied from -oo to +oo, 
then/(0 would be reproduced in its entirety. An operation of this sort 
corresponds to scanning the function/(f) by moving a sheet of paper with 
a thin slit across a plot of the function, as shown in Fig. 2.34. 

Let us now examine higher order derivatives of the unit step function. 
Here we represent the unit impulse by the function f t (i) in Fig. 2.28, 
which, as e -* 0, becomes the unit impulse. The derivative of/ £ (0 is given 
in Fig. 2.35. As e approaches zero,/' £ (0 approaches the derivative of the 
unit impulse d'(t), which consists of a pair of impulses as seen in Fig. 2.36. 
The area under d'(t), which is sometimes called a doublet, is equal to zero. 


6'(t) dt = 
The other significant property of the doublet is 

/(») d'(t -T)dt= -f\T) 





f 2 

r t (t) 


— € 




FIG. 2.35. Unit doubles as <= -* 0. 

FIG. 2.36. The doublet 

Signals and waveforms 41 

where f'(T) is the derivative of/(0 evaluated at t = T where, again, we 
assume that/(/) is continuous. Equation 2.74 can be proved by integration 
by parts. Thus, 

*/ — ( 



-00 «/— oo 

= -f'(T). 
It can be shown in general that 


f°° /(O <5 <B, (< -T)dt = (-l)y ( »'(r) (2.76) 

•r— oo 

where (5< n > and/ <"> denote nth derivatives. The higher order derivatives of 
d(t) can be evaluated in similar fashion. 


2.1 Resolve the waveforms in the figure into odd and even components. 






i t 


+1 t 




















PRO B. 2.1 

42 Network analysis and synthesis 

2.2 Write the equation for the waveforms in the figure using shifted step 










2 - 


1 2 



PROB. 2.2 


23 Find the derivative of the waveforms in Prob. 2.2 and write the equations 
for the derivatives, using shifted step and/or impulse functions. 
2.4 For the waveform /(r) given in the figure, plot carefully 

for a value of t > T. 







PROB. 2.4 

PROB. 2.5 

Signals and waveforms 43 

2^ For the waveform /(r), shown in the figure, determine what value K 
must be so that 

(«) P/(0<//-0 

J— 00 

<*) j"7<f)<fr-o 


2.6 For the waveforms shown in the figure, express in terms of elementary 
time functions, that is, /",«(/ - Q, d(t - /,), („)/(/) (*)/'(/) (c) f * + /( T ) A- 

CU.1.1. *u_ .. r * rL\ j * x .. J— 00 

Sketch the waveforms for (6) and (c) neatly. 

«f— 00 






. * 

2 At 

o i ; 

! 3 1 t 
(ii) J" 5 


PROS. 2.4 

2.7 Prove that 
(a) <*'(*) --<}'(- 



-<»(a;) = x d'(x) 



J— 00 

;*) «& - o 

2.8 The waveform /(0 in the figure is defined as 

/W-^(/-e)*, o<;/£« 

■= 0, elsewhere 
Show that as e — 0,/(/) becomes a unit impulse. 

44 Network analysis and synthesis 

2.9 Plot 

(a) %°s 

(ft) t sgn (cos 0; ' £t £2n 

2.10 Evaluate the following integrals 

( fl ) r <*t - TO u(t - T 2 ) dt; T i <T^ 

J— CO 


a,) I <5(a> — a> ) cos o>f rfo> 

J— 00 

(c) f OO [<5(0-/4<5(r-7' 1 )+2/l<5(f-7 , 2 )]e-^«"A 

J— 00 

2.11 Evaluate the following integral 

f " sin tU (t - l\ + S' (' " f ) + *'<' ~ »>] * 

2.12 The response from an impulse sampler is given by the equation 

r(t) = P sin n l s(t -Kj\dt; K = 0, 1, 2, 3, . . . 

Plot KO for <, t <. IT. 

2.13 If the step response of a linear, time-invariant system is r s (0 = 2e ' u(t), 
determine the impulse response h(f), and plot. 

2.14 For the system in Prob. 2.13 determine the response due to a staircase 



Plot both excitation and response functions. 

2.15 If the impulse response of a time-invariant system is W) = «~* "(')» 
determine the response due to an excitation 

Plot e(t). 

e(t) = 2S(t - 1) - 2d(t - 2) 

PROB. 2.16 

Signals and waveforms 45 
2.16 The unit step response of a linear system is 
<x(r) = (2e- 2t - 1) «(/) 

(a) Find the response r(f) to the input /(f). 

(b) Make a reasonably accurate sketch of the response. Show all pertinent 

chapter 3 

The frequency domain: 
Fourier analysis 


One of the most common classes of signals encountered are periodic 
signals. If T is the period of the signal, then 

s(t) = s(t±nT) 7i = 0,l,2,... (3.1) 

In addition to being periodic, if s(t) has only a finite number of discon- 
tinuities in any finite period and if the integral 


" T \s(t)\ dt 

is finite (where a is an arbitrary real number), then s(t) can be expanded 
into the infinite trigonometric series 

s(t) = ^ + a x cos tot + a t cos 2<ot + • • • „ 2 \ 

+ b x sin cot + b t sin "hot + • • • 
where m = 2w/r. This trigonometric series is generally referred to as the 
Fourier series. In compact form, the Fourier series is 

s(t) = ?* + 2 («» cos not + b n sin ntot) (3.3) 

2 «— i 

It is apparent from Eqs. 3.2 and 3.3 that, when s(t) is expanded in a 
Fourier series, we can describe s(t) completely in terms of the coefficients 


The frequency domain: Fourier analysis 47 

of its harmonic terms, a* a lt a„ . . . , b x , b These coefficients con- 
stitute a frequency domain description of the signal. Our task now is to 
derive the equations for the coefficients a ( , b t in terms of the given signal 
function s(t). Let us first discuss the mathematical basis of Fourier series, 
the theory of orthogonal sets. 


Consider any two functions / x (f) and/»(/) that are not identically zero. 
Then if 

f r, /i(0/*(0dt = O (3.4) 


we say that/i(r) and/^/) are orthogonal over the interval [T lt TJ. For 
example, the functions sin t and cos t are orthogonal over the interval 
nlit ^ r <; (« + l)2n. Consider next a set of real functions {&(*), 
<f>&) <j> H (t)}. If the functions obey the condition 

(*. 4>i) s I V<(0 4>M dt = 0, i*j (3.5) 

then the set {<f> t } forms an orthogonal set over the interval [7\, TJ. In 
Eq. 3.5 the integral is denoted by the inner product (^ 4 , <f> t ). For conven- 
ience here, we use the inner product notation in our discussions. 
The set {^J is orihonormal over [T u 7",] if 

= 1 t-.j 
The norm of an element <f> k in the set {^J is defined as 

l&l - (4, &)* - (£%»\0 *)* (3-7) 

We can normalize any orthogonal set {<f> x , <f>t, . . . , fa} by dividing each 
term <f> k by its norm ||^J|. 


' 3.1. The Laguerre set, 1 which has been shown to be very useful in 
time domain approximation, is orthogonal over [0, <»]. The first four terms 

1 W. H. Kautz, "Transient Synthesis in the Time Domain," Trans. IRE on Circuit 
Theory, CT-1, No. 3 (Sept. 1954), 29-39. 

48 Network analysis and synthesis 
of the Laguerre set are 

*i(0 - e~** 

Ut) - e-*[l - 2(fl01 

*,(/) = e-o'H - MfO) + Hatf\ (3.8) 

4> t (t) = e-°'[l - far + 6(flf)* - Ka/) 8 ] 

To show that the set is orthogonal, let us consider the integral 

f " Ut) hiO * = f V*"[l - MpQ + 2(«0*] dt (3.9) 

Letting t = at, we have 

«i,«-;;J'V ,, a-4T+2,*>rfr 


The norms of &(/) and &(/) are 

,«_(J[-^.*) M -^ (3,., 

- ( f "^[l ~ 4(«0 + 4< a ') a ] *} 

Uo J (3.12) 




It is not difficult to verify that the norms of all the elements in the set are also 
equal to l/V2aT Therefore, to render the Laguerre set orthonormal, we divide 
each element fa by 1/ V2a. 


In this section we explore some of the uses of orthogonal functions in 
the linear approximation of functions. The principal problem is that of 
approximating a function/(0 by a sequence of functions/ B (f) such that the 
mean squared error 

e = lim f *W) - /»(0] 2 dt - (3.13) 

n-»oo JTi 

The frequency domain: Fourier analysis 49 

When Eq. 3.13 is satisfied, we say that {/„(f)} converges in the mean to 

To examine the concept of convergence in the mean more closely, we 
must first consider the following definitions: 

Definition 3.1 Given a function/(f) and constant/; > for which 

( T, \f(t)\*dt<ao, 


we say that /(f) is integrable L* in [7\, TJ, and we write /(/) e L» in 
[T» T t ]. 

Definition 12 If/(f) eL'in [T lt T t ], and {/„(f)> is a sequence of func- 
tions integrable L* in [T lt TJ, we say that if 


then {/«.(*)} converges in the mean of order/? to /(f). Specifically, when 
p — 2 we say that {/„(f)} converges in the mean to /(f). 

The principle of least squares 

Now let us consider the case when/„(f ) consists of a linear combination 
of orthonormal functions 4>i,<l> <f> n - 

/„(0 = I *<&(<) (3.14) 


Our problem is to determine the constants a t such that the integral 
squared error 

11/ - All* - f *W) - M9f dt (3.15) 


is a minimu m. The principle of least squares states that in order to attain 
m in im um squared error, the constants a ( must have the values 

Ci = ]Kt)Mt)dt (3.16) 

Proof. We shall show that in order for \\f — /„|| s to be minimum , we 
must set a { = c t for every i = 1, 2, . . . , n. 

H/-/.II* = </,/) - 2{f,fJ + (/„,/„) 

= 11/11* -22a,(/,&)+2a,W 

50 Network analysis and synthesis 

Since the set {&} is orthonormal, |«\| a = 1, and by definition, c t = 
(/,&). We thus have 

I/-/-I' - l/l' - 22«A + i '* (318) 

Adding and subtracting 2 c«* gives 

i/-/.!" = ii/n 8 - 2i«A + i^ + i«*' - i*" 

<_1 <-l <-l » _1 



We see that in order to attain minimum integral squared error, we must 
set a t = c<. The coefficients c t , defined in Eq. 3.16, are called the Fourier 
coefficients of /(*) with respect to the orthonormal set {+&)}. 

Parseval's equality 

Consider f n (t) given in Eq. 3.14. We see that 


ft L/j(or*-i«' < 320 > 

since & are orthonormal functions. This result is known as Parseval's 
equality, and is important in determining the energy of a periodic signal. 


Let us return to the Fourier series as denned earlier in this chapter, 

aft) _ £s + V ( a , cos nmt + b n sin nmt) (3.21) 

2 £Ti 

From our discussion of approximation by orthonormal functions, we can 
see that the periodic function s(t) with period T can be approximated by a 
Fourier series s n (t) such that s n (t) converges in the mean to s(t), that is, 


im r^MO-s-O)]*^ — ( 322 > 

-»oo J a 

where a is any real number. We know, moreover, that if n is finite, the 
mean squared error ||j(0 - *»(0II* is minimized when the constants a { , b t 
are the Fourier coefficients of s(t) with respect to the orthonormal set 

/cos koat sin k<ot \ .__«,« 

The frequency domain: Fourier analysis 51 



-3x -2x -x x 2x 3x t 
FIG. 3.1. Rectified sine wave. 

in explicit form the Fourier coefficients, according to the definition given 
earlier, are obtained from the equations 

fl0=s rJ« * )A (323) 

2 r 


s(t) cos kmt dt (3.24) 

s(0 sin tor «/f (3.2S) 

We should note that because the Fourier series s n (t) only converges to 
s(t) in the mean, when s(t) contains a jump discontinuity, for example, 

S .0,) = ^±I±J(£o=) ( 3. 26) 

At any point t t that s(t) is differentiable (thus naturally continuous) 
*»(*i) converges to a(/J.* 

As an example, let us determine the Fourier coefficients of the fully 
rectified sine wave in Fig. 3.1. As we observe, the period is T= n so 
that the fundamental frequency is m = 2. The signal is given as 

s{t) = A |sin t\ (3.27) 

Let us take a = and evaluate between and ir. Using the formula just 
derived, we have . 

K = ~ | «(0 sin 2nt dt = (3.28) 

TT Jo 

2A C* &A 

«, = — \ sin tdt = — (3.29) 

IT Jo W 


IT Jo 

s(0 cos 2nt dt 
1 AA 

1 -An* 


• For a proof see H. F. Davis Fourier Series and Orthogonal Functions, Allyn and 
Bacon, Boston, 1963, pp. 92-95. 

52 Network analysis and synthesis 

Thus the Fourier series of the rectified sine wave is 

5(/) - £i(l + f —±— cos 2nt) (3.31) 


In this section we will consider two other useful forms of Fourier series. 
In addition, we will discuss a number of methods to simplify the evaluation 
of Fourier coefficients. First, let us examine how the evaluation of 
coefficients is simplified by symmetry considerations. From Eqs. 3.23-3.25 
which give the general formulas for the Fourier coefficients, let us take 
a = —T/2 and represent the integrals as the sum of two separate parts, 
that is, 



rr/t ro -i 

s(0 cos nmt dt + s(t) cos mot dt 


2 r f T,t f ° i 

ss — I s(t) sin nmt dt + I s(f) sin nmt dt 
TUo J-T/i J 

T/a " (3.32) 

Since the variable (0 in the above integrals is a dummy variable, let us 
substitute x = t in the integrals with limits (0 ; r/2), and let x = — t in the 
integrals with limits (—T/2; 0). Then we have 

2 C T/ * 
a n = — I [s(x) + s(—x)] cos nmx dx 

TJ ° (3.33) 

2 C T/ * 
b n = — I [s(a:) — s(— *)] sin nmx dx 
T Jo 

Suppose now the function is odd, that is, s(x) = —s(—x), then we see that 
a„ = for all n, and 


s{x) sin nmx dx (3.34) 


This implies that, if a function is odd, its Fourier series will contain only 
sine terms. On the other hand, suppose the function is even, that is, 
s(x) = s(—x), then b n = and 

a n = — I s(x) cos nmx dx (3.35) 


Consequently, the Fourier series of an even function will contain only 
cosine terms. 

The frequency domain: Fourier anal/sis 53 

FIG. 3.2 

Suppose next, the function s(t) obeys the condition 
,(, ± l)--<0 


as given by the example in Fig. 3.2. Then we can show that s(t) contains 
only odd harmonic terms, that is, 

«„ = b n = 0; 




T Jo 

s(t) cos nmt dt 

5(0 sin nmt dt, n odd 


With this knowledge of symmetry conditions, let us examine how we 
can approximate an arbitrary time function s(t) by a Fourier series within 
an interval [0, 71. Outside this interval, the Fourier series s n (t) is not 
required to fit s(t). Consider the signal s(t) in Fig. 3.3. We can approxi- 
mate s(t) by any of the periodic functions shown in Fig. 3.4. Observe 
that each periodic waveform exhibits some sort of symmetry. 

Now let us consider two other useful forms of Fourier series. The first 
is the Fourier cosine series, which is based upon the trigonometric identity, 

C„ cos (na>i + 8 J = C„ cos mot cos 0„ — C„ sin nmt sin 0„ (3.38) 

^| T t 

FIG. 3 J. Signal, to be approximated. 

54 Network anal/sis and synthesis 




^ ^ 



^ t 






"~«^ ° 


f~"-~. t 






I**" * 




FIG. 3.4. (a) Even function cosine terms only, (ft) Odd function sine terms only, (c) 
Odd harmonics only with both sine and cosine terms. 

We can derive the form of the Fourier cosine series by setting 

a„ = C„ cos 0„ 
and b n = — C„ sin 0„ 

We then obtain C n and 0„ in terms of a n and b n , as 

c,-(fl.' + M 


C a = 





If we combine the cosine and sine terms of each harmonic in the original 
series, we readily obtain from Eqs. 3.38-3.41 the Fourier cosine series 

f(t) = C + C t cos (mi + 00 + C, cos (2o>f + 0.) 

+ C.cos (3fl>f + 0,) + • • • + C n cos(nmt + J + • • • (3.42) 

The frequency domain: Fourier anal/sis 55 

It should be noted that the coefficients C„ are usually taken to be positive. 
If however, a term such as —3 cos 2cot carries a negative sign, then we can 
use the equivalent form 

— 3cos2a>f = 3 cos (2<u/ + tt) (3.43) 

For example, the Fourier series of the fully rectified sine wave in Fig. 3.1 
was shown to be 

s(0 = ^(l + f -^— , cos 2nt) (3.44) 

2ir\ n=i 1 — An I 

Expressed as a Fourier cosine series, s(t) is 

s(0 = ^[l + 1 —£— cos (2nt + *)] (3.45) 

2wL n-i4n — 1 J 

Next we consider the complex form of a Fourier series. If we express 
cos ncot and sin mot in terms of complex exponentials, then the Fourier 
series can be written as 

= — + 2 ( "* ~ J& " **"* + fl " + ^* *-'"■') 

2 n-l\ 2 2 / 



If we define 

P» = 2 ' "~ n 2 ' 2 

then the complex form of the Fourier series is 

5(0 = A + i(/ff„e*"" + /*-„«-'"•") 

„ "^ ( 34 «) 

- 2 PS- 

n— oo 

We can readily express the coefficient fi„ as a function of s(t), since 
/? a « ~ J b » 

Pn — 

l C T 

= — I s(0(cos nott — j sin na»0 dt (3.49) 

T Jo 

= i f T s(t)e- i ' mt dt 
T Jo 

Equation 3.49 is sometimes called the discrete Fourier transform of «(/) 
and Eq. 3.48 is the inverse transform of P„(na)) = /?„. 

56 Network analysis and synthesis 



C 2 



-Au -3« -2w -w 





w 2(i> 3d) 4a> 

FIG. 3.5. Amplitude spectrum. 








FIG. 3.6. Square wave. 

The frequency domain: Fourier anal/sis 57 

Observe that /?„ is usually complex and can be represented as 

/* n = Re/8 B +yIm£, (3.50) 

The real part of fi„ Re /?„, is obtained from Eq. 3.49 as 

1 C T 
Re Pn - - «(0 cos tuot dt (3.51) 

T Jo 

and the imaginary part of /?„ is 

JImfl,-^f s(0sinn«>r<fc (3.52) 

T Jo 

It is clear that Re /?„ is an even function in r, whereas Im /S„ is an odd 
function in n. The amplitude spectrum of the Fourier series is defined as 

|/JJ-(Re"fc.+ ImP«M (3.53) 

and the />/uue spectrum is denned as 

6, = arctan^ (3.54) 

It is easily seen that the amplitude spectrum is an even function and the 
phase spectrum is an odd function in n. The amplitude spectrum provides 
us with valuable insight as to where to truncate the infinite series and still 
maintain a good approximation to the original waveform. From a plot of 
the amplitude spectrum, we can almost pick out by inspection the non- 
trivial terms in the series. For the amplitude spectrum in Fig. 3.5, we see 
that a good approximation can be obtained if we disregard any harmonic 
above the third. 

As an example, let us obtain the complex Fourier coefficients for the 
square wave in Fig. 3.6. Let us also find the amplitude and phase spectra 
of the square wave. From Fig. 3.6, we note that s(t) is an odd function. 
Moreover, since s(t — T/2) = — s(t), the series has only odd harmonics. 
From Eq. 3.49 we obtain the coefficients of the complex Fourier series as 

n - - Ae-' Mi dt--\ Ae' ,nat dt 
TJo TJt/» 

(I — 2e~ linmT/i) -)- e~ ,tuaT ) 


Since ncoT = nlir, /?„ can be simplified to 

/». - rr- a - 2*"*" + «"*"*) ( 3 - 56 > 

58 Network analysis and synthesis 










I 3 

5 u 




-5 -3 



FIG. 3.7. Discrete spectra of square wave, (a) Amplitude. (*) Phase. 

Simplifying /5» one step further, we obtain 

n odd 



= n even 

The amplitude and phase spectra of the square wave are given in Fig. 3.7. 


In this section we make use of a basic property of impulse functions to 
simplify the calculation of complex Fourier coefficients. This method is 
restricted to functions which are made up of straight-line components only. 
Thus the method applies for the square wave in Fig. 3.6. The method is 
based on the relation 


f" /(*)«(«- TO* -/(TO 

J— 00 

Let us use this equation to evaluate the complex Fourier coefficients for 
the impulse train in Fig. 3.8. Using Eq. 3.58 with/(f) - e~ ,fmt , we have 

'• = T fH' - 2 V'"" d% '~T e ~ < *~ 2V,, (359> 


The frequency domain: Fourier analysis 59 



A . 


1 r 3r 2T sr 
2 T T 

FIG. 3.8. Impulse train. 

We see that the complex Fourier coefficients for impulse functions are 
obtained by simply substituting the time at which the impulses occur into 
the expression, e~ ,m " t . 

In the evaluation of Fourier coefficients, we must remember that the 
limits for the P„ integral are taken over one period only, i.e., we consider 
only a single period of the signal in the analysis. Consider, as an example, 
the square wave in Fig. 3.6. To evaluate /?„, we consider only a single 
period of the square wave, say, from t = to 
t = T, as shown in Fig. 3.9a. Since the square 
wave is not made up of impulses, let us 
differentiate the single period of the square 
wave to give s'(t), as shown in Fig. 3.96. We 
can now evaluate the complex Fourier coeffi- 
cients for the derivative s(t), which clearly is 
made up of impulses alone. Analytically, if 
s(t) is given as 









s(0 - 1 P«e 

»— — 00 

then the derivative of 5(0 is 




Here, we define a new complex coefficient 






If the. derivative s'(i) is a function which 
consists of impulse components alone, then 
we simply evaluate y n first and then obtain 
/?„ from Eq. 3.63. For example, the derivative 
of the square wave yields the impulse train 


FIG. 3.9. (a) Square wave 
over period [0, 71. (b) Deriv- 
ative of square wave over 
period [0, T\. 

60 Network analysis and synthesis 

in Fig. 3.96. In the interval [0, T], the signal s'(f ) is given as 

s'(t) = A6(t)- 2^4 d(t - |) + A d(t - T) (3.64) 

Then the complex coefficients are 




= —(\— 2e- linmT/ * } + e~ inmT ) 

The Fourier coefficients of the square wave are 

a _ y» 


— ^ (i — 2e~ linaTI * ) + e~ inmT ) 

which checks with the solution obtained in the standard way in Eq. 3.55. 
If the first derivative, /(f), does not contain impulses, then we must 
differentiate again to yield 

s »(f) = f X n e inmt (3.67) 

n— — oo 

where K =jno>Yn = (P 1 ™)* P« ( 3 - 68 ) 

For the triangular pulse in Fig. 3.10, the second derivative over the 
period [0, T] is 

s »(0 = ^Uf) - 2d(f - |) + ** - d] ( 3 - 69 > 

The coefficients A n are now obtained as 

X n = - ( T s"(i)e- inat dt 

2A ,j _ 2 e -OfK»27*> i e -fnmT\ 


which simplifies to give 

&A ., 

X =s — n odd 
" T* (3.71) 

= n even 

The frequency domain: Fourier analysis 61 

















2A [>"<'> 



FIG. 3.10. The triangular wave and its derivatives. 

From A_ we obtain 



= 2A 




A slight difficulty arises if the expression for s'(t) contains an impulse 
in addition to other straight-line terms. Because of these straight-line 
terms we must differentiate once more. However, from this additional 
differentiation, we obtain the derivative of the impulse as well. This 
presents no difficulty, however, because we know that 

I °%(r) d'(t - T) - -s\T) (3.73) 

J— GO 

so that f " d'(t - T)e- im "dt = jna>e-" u ' T (3.74) 

J— 00 

62 Network analysis and synthesis 

We can therefore tolerate doublets or even higher derivatives of impulses 
in the analysis. Consider the signal s(f) given in Fig. 3.1 la. Its derivative 
*'(<), shown in Fig. 3.116, can be expressed as 

S '(0 = |[ M (0 - u(t - f )] + «0 - 2«(l - f) 0.75) 

The second derivative s"(t) consists of a pair of impulses and a pair of 
doublets as given by 

m - |[aco - *(t - f )] + *W - 2d'(t - 1) 

as shown in Fig. 3.11c. We therefore evaluate A„ as 





— 2. (i _ e -(*~»r/») + 1™± (i _ 2e- UnaTin ) 



2 - ., 
1 ' — 



: t t 




(W t-M(l-f) 



r ?«(*-i) 


FIG. 3.11 

The frequency domain: Fourier analysis 63 
The complex coefficients /S„ are now obtained as 



= ±— (i _ e -o»»Z7«) + _!_ n _ 2e -(^«»r/a)) 

(jconT)* janT ' 

Simplifying, we have 

1 3 

Pn=-— t + — "odd 
mr j2irn 

j (3.79) 

= n even 


In conclusion, it must be pointed out that the method of using impulses 
to evaluate Fourier coefficients does not give the d-c coefficient, a /2 or /3 . 
We obtain this coefficient through standard methods as given by Eq. 3.23. 


In this section we extend our analysis of signals to the aperiodic case. 
We show through a plausibility argument that generally, aperiodic signals 
have continuous amplitude and phase spectra. In our discussion of Fourier 
series, the complex coefficient (i n for periodic signals was also called the 
discrete Fourier transform 

s(i)e- M "'*dt (3.80) 

1 C T 


and the inverse (discrete) transform was 

*(0 = 2 Knfo)e ln, ' M (3.81) 

n=— oo 

From the discrete Fourier transform we obtain amplitude and phase 
spectra which consist of discrete lines. The spacing between adjacent lines 
in the spectrum is 

A/ - (n + l)/„ - «/„ = i (3.82) 

As the period T becomes larger, the spacing between the harmonic lines in 
the spectrum becomes smaller. For aperiodic signals, we let T approach 
infinity so that, in the limit, the discrete spectrum becomes continuous. 

64 Network anal/sis and synthesis 

We now define the Fourier integral or transform as 

s(/) = lim £W£ = r s(()c -«" « it (3.83) 

T—x> Jo J— oo 


The inverse transform is 


<0-f" S(J)e iU,t df (3.84) 

J— 00 

Equations 3.83 and 3.84 are sometimes called the Fourier transform pair. 
If we let 7 denote the operation of Fourier transformation and 5~ x denote 
inverse transformation, then 

*</>-*-<0 (3 . 85) 

s(t) = ?-i-S(f) 

In general, the Fourier transform S(f) is complex and can be denoted as 

S(f) = ReS(f)+ j Im S(f) (3.86) 

The real part of S(f) is obtained through the formula 

ReS(/) = J[S(/) + S(-/)] 

= |" s(t) cos 2-nft dt 

J— CO 

and the imaginary part through 


Im S(f) = £ [SCO - S(-/)] 


= _ j °° s (0 sin 2tt/( dt 

J— CO 

The amplitude spectrum of S(f) is defined as 

A(f) = [Re SC/) 2 + Im £(/)*]* (3.89) 

and the phase spectrum is 

«"-"— Sin **» 

Using the amplitude and phase definition of the Fourier transform, the 
inverse transform can be expressed as 

s(t) = f °° A(f) cos [lirft - <Kf)] df (3.91) 

J— 00 

Let us examine some examples. 



The frequency domain: Fourier anal/sis 


+ / 

(a) ((,) 

FIG. 3.12. Amplitude and phase spectrum of A d(t — /,). 

Example 3.2.* 

sit) = A6\t-td 
S(f)~ (" A^t- t 9 )er^t* dt 

J— 00 

- Ae-i**"> 

Its amplitude spectrum is 
while its phase spectrum is 

<Kf) - -2^//o 


as shown in Fig. 3.12. 

Example 33. Next consider the rectangular function plotted in Fig. 3.13. 
Formally, we define the function as the rect function. 



1 HSy 


\x\ > - 


The inverse transform of rect/is defined as sine t (pronounced sink), 
3 r - 1 [rect/] ■» sine t 


e>*"*df (3.96) 




* It should be noted here that the Fourier transform of a generalized function is also 
a generalized function. In other words, if <f> e C, (^ • <f>) £ C. For example, J" • <5(f ) = 
1, where 1 is described by a generalized function 1„(/). We will not go into the formal 
details of Fourier transforms of generalized functions here. For an excellent treatment 
of the subject see M. J. Lighthill, Fourier Analysis and Generalized Functions, England, 
Cambridge University Press, 19SS. 

66 Network analysis and synthesis 



-* w 

" 2 

+ 2 

FIG. 3.13. Plot of rect function. 

sine t 






£. ± 3. 1. -L 

w ~w ~w ~w ~w 

i. 2. 3. 4. 5. 

W W W W W 

FIG. 3.14. The sine / curve. 



Narrow pulse 


Wide band 


Narrow pulse 






10 Ti 

10 Tx 
Wide pulse Narrower band Wide pulse 

FIG. 3.15. Illustration of the reciprocity relationships between time duration and 

The frequency domain: Fourier anal/sis 67 

From the plot of sine t in Fig. 3.14 we see that sine t falls as does |r| _1 , 
with zeros at t = n\W, n = 1, 2, 3, . . . We also note that most of the 
energy of the signal is concentrated between the points — \\W < t < \\W. 
Let us define the time duration of a signal as that point, t , beyond which 
the amplitude is never greater than a specified value, for example, e . We 
can effectively regard the time duration of the sine function as t = ± 1/ W. 
The value W, as we see from Fig. 3.13, is the spectral bandwidth of the 
rect function. We see that if W increases, t decreases. The preceding 
example illustrates the reciprocal relationship between the time duration 
of a signal and the spectral bandwidth of its Fourier transform. This 
concept is quite fundamental. It illustrates why in pulse transmission, 
narrow pulses, i.e., those with small time durations, can only be trans- 
mitted through filters with large bandwidths; whereas pulses with longer 
time durations do not require such wide bandwidths, as illustrated in 
Fig. 3.15. 


In this section we consider some important properties of Fourier 


The linearity property of Fourier transforms states that the Fourier 
transform of a sum of two signals is the sum of their individual Fourier 
transforms, that is, 

Hc x *i(0 + c t sjflj] = Cl S x (/) + c 2 S t (f) (3.97) 


This property states that the Fourier transform of the derivative of a 
signal isjlnf times the Fourier transform of the signal itself: 

J-s\i)=j2-nfS{f) (3.98) 

or more generally, 

?- s ™(t) = (j2*frS(f) (3.99) 

The proof is obtained by taking the derivative of both sides of the inverse 
transform definition, 

s'(0 = 7;f 00 S(J)e»*<*df 

at J -oo 

-f j2nfS(f)e»*"df 

J— CO 

68 Network analysis and synthesis 

Similarly, it is easily shown that the transform of the integral of s(t) is 

? [ f ' 5(t) dr\ = ^- S(J) (3.101) 

LJ-oo J j2nf 

Consider the following example 

j(0 = «-°'«(0 ( 3 - 102 > 

Its Fourier transform is 

s(/) = f" r^vHtyr^dt 

J— CO 


a + j2irf 
The derivative of s(t) is 

s '(t) . 3(0 - aer** «(0 (3.104) 

Its Fourier transform is 


?[''«)] - 1 - 

a+j2wf a+j2irf (3.105) 

= j2nfS(f) 


The symmetry property of Fourier transforms states that if 

3" • x(t) = X(f) (3.106) 

then ^ • X(t) = *(-/) (3.107) 

This property follows directly from the symmetrical nature of the Fourier 
transform pair in Eqs. 3.83 and 3.84. 

Example 3.4. From the preceding section, we know that 

^ • sine / = rect/ (3.108) 

It is then simple to show that 

& • rect t = sine ( -/) = sine/ (3.109) 

which conforms to the statement of the symmetry property. Consider next 
the Fourier transform of the unit impulse, & • 8(t) = 1. From the symmetry 
property we can show that 

?• 1 = 8(f) (3.110) 

as shown in Fig. 3.16. The foregoing example is also an extreme illustration 
of the time-duration and bandwidth reciprocity relationship. It says that zero 
time duration, 8(t), gives rise to infinite bandwidth in the frequency domain; 
while zero bandwidth, 8(f) corresponds to infinite time duration. 

The frequency domain: Fourier analysis 69 





o f 

FIG. 3.1*. Fourier transform of /(f) = 1.0. 

Scale change 

The scale-change property describes the time-duration and bandwidth 
reciprocity relationship. It states that 

[.(*)] -M*«0 


Proof. We prove this property most easily through the inverse trans- 

J-\\a\ S{af)] = \a\ f" S(af)e™* df 

J— oo 

Let/' = af; then 

*^l«l S(f')] = \a\ f " S(/V*"' ( ' /o, — 
J— co a 

As an example, consider 





Sy 1 



j2iraf + a 








The folding property states that 


J-W-01 = s(-f) 


70 Network anal/sis and synthesis 

The proof follows directly from the definition of the Fourier transform. 
An example is 

3=V u(-t)] = ^— ■ (3.117) 



If a signal is delayed by an amount t in the time domain, the corre- 
sponding effect in the frequency domain is to multiply the transform of the 
undelayed signal by er iz " ft <>, that is, 

For example, 


&[e- aU -'° ) u(t-t )] = 


a + j2nf 

The modulation or frequency shift property of Fourier transforms 
states that if a Fourier transform is shifted in frequency by an amount f , 
the corresponding effect in time is described by multiplying the original 
signal by e iufot , that is, 

3^[S(f - /<,)] = e mM s(t) (3.120) 

Example 3.5. Given S(f) in Fig. 3.17a, let us find the inverse transform of 
Si(f) in Fig. 3.176 in terms of s(t) = 5"" 1 S(f). We know that 

Sxif) = S(f -f ) + 5(/ + /„) (3.121) 




h *f 


FIG.' 3.17. Demonstration of amplitude modulation. 

The frequency domain: Fourier analysis 71 
Then 3^ 5,(/) - *>*".« s{t) + er'*'U* g(t) = 2s(t) cos 2nfy (3.122) 

Thus we see that multiplying a signal by a cosine or sine wave in the time domain 
corresponds to shifting its spectrum by an amount ±f . In transmission termi- 
nology /„ is the carrier frequency, and the process of multiplying s(t) by cos 2itfy 
is called amplitude modulation. 

Parseval's theorem 

An important theorem which relates energy in the time and frequency 
domains is Parseval's theorem, which states that 

I " «i(0 St(0 dt = f " S x (/) S,(-/) df (3.123) 

J— 00 J— no 

The proof is obtained very simply as follows: 

f " si(0 »i(0 dt = f " s„(t) dt f " SiC/V""* d/ 

-f" Si(/)4rf" s&V u "dt (3.124) 

•*— oo J—ao 

-f" sMSA-fldf 

J— 00 

In particular, when ^(r) = *j(f), we have a corollary of Parseval's theorem 
known as PlancheraVs theorem. 

f " aty) * = f " ISC/)! 1 4T (3.125) 

J— 00 J— 00 

If j(/) is equal to the current through, or the voltage across a 1-ohm 
resistor, the total energy is 


s 2 (0 dt 

We see from Eq. 3.125 that the total energy is also equal to the area under 
the curve of \S(f)\*. Thus \S(f)\* is sometimes called an energy density 
or energy spectrum. 


3.1 Show that the set {1, sin nnt/T, cos nnt/T}, n = 1,2, 3, . . . , forms an 
orthogonal set over an interval [a, a + 2T], where a is any real number. Find 
the norms for the members of the set and normalize the set. 

72 Network analysis and synthesis 
3.2 Given the functions/^) and/i(0 expressed in terms of complex Fourier 


/l(/)=2«»^ B,, " 

n— — oo 


where both/i(0 and/j(0 have the same period T, and 

«„ = Kl e**\ A. - lft»l e*» 

- «oA> + 22 l«g»»l cos(fl B - 4>J 

show that 

Note that 

\0, m * -n 

33 For the periodic signals in the figure, determine the Fourier coefficients 










T T 



The frequency domain: Fourier analysis 73 

3.4 For the waveforms in Prob. 3.3, find the discrete amplitude and phase 
spectra and plot. 

3.5 For the waveforms in Prob. 3.3 determine the complex Fourier coefficients 
using the impulse function method. 

3.6 Find the complex Fourier coefficients for the function shown in the 

A A A 

ST, 3T. Tj T, 3T, 51 

~r — r -t T ~T ~i 


PROB. 3.6 

3.7 Find the Fourier transform for the functions shown in the figure. 









*T t 












V t 



PROB. 3.7 

3.8 Find the Fourier transform for 
(a) /« - A <3(0 


/(f) — A sin atf 

74 Network anal/sis and synthesis 

3.9 Prove that (a) if /(f) is even, its Fourier transform F(ja>) is also an even 
function; (6) if /(/) is odd, its Fourier transform is odd and pure imaginary. 




— (i) +«0 

PROS. 3.10 

3.10 Find the inverse transform of 

F(jm) = P 6(fo - Wo) + P 8{(o + eo ) 

as shown in the figure. What can you say about line spectra in the frequency 

chapter 4 

Differential equations 


This chapter is devoted to a brief study of ordinary linear differential 
equations. We will concentrate on the mathematical aspects of differential 
equations and leave the physical applications for Chapter 5. The differen- 
tial equations considered herein have the general form 

/MO, *'('), • • • , *< B >(0, t] = (4.1) 

where t is the independent variable and x(t) is a function dependent upon 
t. The superscripted terms ««>(/) indicate the rth derivative of x(t) with 
respect to t, namely, 

«>/a d (i) x(t) 
x(t) = ~df~ (42) 

The solution of F = in Eq. 4.1 is x(t) and must be obtained as an 
explicit function of /. When we substitute the explicit solution x(t) into F, 
the equation must equal zero. If fin Eq. 4.1 is an ordinary linear differ- 
ential equation, it is given by the general equation 

«» * (B> (0 + a^x^Xt) + ••• + «, x\t) + oo x(t) =f(t) (4.3) 

The order of the equation is n, the order of the highest derivative term. 
The term/(0 on the right-hand side of the equation is Xht forcing function 
or driver, and is independent of x(i). When /(f) is identically zero, the 
equation is said to be homogeneous; otherwise, the equation is non- 

In this chapter we will restrict our study to ordinary, linear differential 
equations with constant coefficients. Let us now examine the meanings of 
these terms. 


76 Network anal/sis and synthesis 

Ordinary. An ordinary differential equation is one in which there is only 
one independent variable (in our case, t). As a result there is no need for 
partial derivatives. 

Constant coefficients. The coefficients a n , o^, ...,a„ a lt a„ are con- 
stant, independent of the variable t. 

Linear. A differential equation is linear if it contains only terms of the 
first degree in x(t) and all its higher derivatives, as given by Eq. 4.3. For 
example, the equation 

3x'(t) + 2z(0 = sin t (4.4) 

is a linear differential equation. On the other hand, 

3[* W + 2*(f) x\t) + 4x(0 = 5t (4.5) 

is nonlinear, because the terms [x'(t)]* and x(t) x'(t) are nonlinear by the 
definition just given. 

An important implication of the linearity property is the superposition 
property. According to the superposition property, if x x (0 and x^t) are 
solutions of a given differential equation for forcing functions/^) and/,(0, 
respectively, then, if the forcing function were any linear combination of 

Ut) and/,(0 such as 

J AO-afM + btff) (4-6) 

the solution would be 

x(r)-a«i(0 + &*40 ( 4 - 7 ) 

where a and b are arbitrary constants. It should be emphasized that the 
superposition property is extremely important and should be kept in 
mind in any discussion of linear differential equations. 


This section deals with some methods for the solution of homogeneous, 
linear differential equations with constant coefficients. First, let us find 
the solution to the equation 

x'(t) - 2*(<) = (4.8) 

Now, with a little prestidigitation, we assume the solution to be of the form 

*(0 = Ce" (4.9) 

where C is any arbitrary constant. Let us check to see whether x(t) = Ce 2 * 
is truly a solution of Eq. 4.8. Substituting the assumed solution in Eq. 4.8, 

we obtain tA tn . 

2Ce* f - 2Ce»* = (4.10) 

Differential equations 77 

It can be shown, in general, that the solutions of homogeneous, linear 
differential equations consist of exponential terms of the form C<e p <*. To 
obtain the solution of any differential equation, we substitute Ce vt for x(t) 
in the equation and determine those values ofp for which the equation is 
zero. In other words, given the general equation 

a n *<">(*) +••• + «! x'(t) + a, x(f) = (4.1 1) 

we let x(t) = Ce", so that Eq. 4.11 becomes 

Ce'Ka n p" + a^tf*- 1 + • • • + a# + a,) = (4.12) 

Since e pt cannot be zero except at/? = — oo, the only nontrivial solutions 
for Eq. 4.12 occur when the polynomial 

H(p) - <W" + a-tf" -1 + • ' * + W + «o - (4.13) 

Equation 4.13 is often referred to as the characteristic equation, and is 
denoted symbolically in this discussion as H(p). The characteristic equation 
is zero only at its roots. Therefore, let us factor H(j>) to give 

H(j>) = a n (p- P Mp- Pi )---(p- p^ (4.14) 

From Eq. 4.14, we note that C^e"**, C x e*** t .... C w _ 1 c*«-»' are all solutions 
of Eq. 4.11. By the superposition principle, the total solution is a linear 
combination of all the individual solutions. Therefore, the total solution 
of the differential equation is 

x(t) = CV' + Q**' + • • • + <:_!«»-»« (4.15) 

where C , C x , . . . , C B _i are generally complex. The solution x(t) in Eq. 
4. IS is not unique unless the constants C„, C ls . . . , C n _x are uniquely 
specified. In order to determine the constants C it we need n additional 
pieces of information about the equation. These pieces of information are 
usually specified in terms of values of x(t) and its derivatives at t — 0+, 
and are therefore referred to as initial conditions. To obtain ft coefficients, 
we must be given the values <r(0+), x'(0+), . . . , a; ( "- 1 >(0+). In a number 
of special cases, the values at t = 0— are not equal to the values at 
/ = 0+. If the initial specifications are given in terms of *(0— ), 
x'(0— ), . . . , ^"-^(O— ), we must determine the values at f = 0+ In order 
to solve for the constants C { . This problem arises when the forcing 
function /(f) is an impulse function or any of its derivatives. We will 
discuss this problem in detail in Section 4.4. 

For example, in Eq. 4.9 if we are given that *(0+) = 4, then we obtain 
the constant from the equation 

x(0+) = Ce*=C (4.16) 

78 Network analysis and synthesis 
so that x(t) is uniquely determined to be 

x(t) = 4e* 
Example 4.1 Find the solution for 

*'(/) + 5x\t) + 4a</) = (4.17) 

given the initial conditions 

a<0+)-2 x'(0+) = -1 
Solution. From the given equation, we first obtain the characteristic equation 
H(p) = p* + 5p + 4 = (4.18) 

which factors into (/> + 4X/> + 1) = (4.19) 

The roots of the characteristic equation (referred to here as characteristic values) 
step = -1 ; p = —4. Then x(t) takes the form 

x{t) = Qe-* + C^e-** (4.20) 

From the initial condition «(0+) = 2, we obtain the equation 

x(0+) =2 = Q + C a (4.21) 

In order to solve for C x and C, explicitly, we need the additional initial condition 
x'(0+) = -1. Taking the derivative of x(i) in Eq. 4.20, we have 

X '(f) = -Qe-' - 4Q*- 4 * (4.22) 

Atf =0+,a;'(0is 

a:'(0+) = -1 = -Q -4C a (4.23) 

Solving Eqs. 4.21 and 4.23 simultaneously, we find that 

Ci = i C % = —J 

Thus the final solution is a*/) = %r* - \e~** (4.24) 

Next, we examine the case when the characteristic equation H(p) has 
multiple roots. Specifically, let us consider the case where H(p) has a 
root p = p of multiplicity k as given by 

H(p) = a n (p - p ) k (p -pd---(p-p n ) (4-25) 

It will be left to the reader to show that the solution must then contain k 
terms involving e" * of the form 

x(i) = CV"»« + C 01 te*>* + Cnt'e*** + • • • + C^t*-^"* 

+ C^*** + Cje 1 "' + • • • + C n e*'* (4.26) 

where the double-scripted terms in Eq. 4.26 denote the terms in the solution 
due to the multiple root, (p — #,)*• 

Differential equations 79 

Example 4.2 Solve the equation 

*»(/) - Sx'(t) + 16*(r) = (4.27) 

with a<0+)«2 and *'(<>+) "=4 

Solution. The characteristic equation is 

H{p) ~p* - ip + 16 = (p - 4)» (4.28) 

Since H(p) has a double root at/> =4, the solution must take the form 

x(t) « C^* + Cy*" (4.29) 

In order to determine Q and C„ we evaluate x(f) and »'(') at r = 0+ to give 

«(0+) = C 1 =2 

*'(<>+) =4Q +C 2 =4 

Thus the final solution is a<r) = 2c*' - 4te*' (4.31) 

Another interesting case arises when H(p) has complex conjugate roots. 
Consider the equation 

H{p) - a 2 (p -pMp-pJ (4.32) 

where />! and />, are complex conjugate roots, that is, 

Px,Pi = o±jo> (4.33) 

The solution x(t) then takes the form 

x(t) = Ci«<*W">* + <V—'»>' (4.34) 

Expanding the term e*" f by Euler's equation, x(t) can be expressed as 

x(t) = C t e*(cos cot +j sin cat) + Cje^cos cor —y sin cat) (4.35) 

which reduces to 

x(t) = (Q + C,y* cos cor +j(C 1 — CJe'* sin cor (4.36) 

Let us introduce two new constants, M x and Af„ so that x(r) may be 
expressed in the more convenient form 

x(t) = Mje'* cos cor + Aftf"' sin cat (4.37) 

where 3/ x and M t are related to the constants C x and C 2 by the equations 

M! = d + C, 

^. =y(C! - C 2 ) 

The constants Afj and M t are determined in the usual manner from 
initial conditions. 

80 Network analysis and synthesis 

Another convenient form for the solution x(t) can be obtained if we 
introduce still another pair of constants, M and <f>, defined by the equations 

Ml = MsiD * (4.39) 

M t = M cos <f> 

With the constants M and <f> we obtain another form of x(i), namely, 

x{i) - MC* sin (a>t + <f>) (4.40) 

Example 4.3 Solve the equation 

x\t) + 2*'(/) + 5x(t) = (4.41) 

with the initial conditions 

a<0+) = l *'(<>+)= 

Solution. The characteristic equation H(p) is 

H(p) =/>* + 2/7 + 5 = (p + 1 +j2Kp + 1 -j2) (4.42) 

so that, assuming the form of solution in Eq. 4.37, we have a = -1 and a> = 2. 

Then x\t) is 

x(t) = M &-* cos 2t + Mtf-* sin It (4.43) 

At t = 0+ *(0+) - 1 - M t (4.44) 

The derivative of x(i) is 

*'(*) = M x ( -«"' cos 2/ - 2e"' sin 20 + M£ -e~* sin 2/ + 2c"« cos 2t) (4.45) 
At / = +, we obtain the equation 

«'(<)+) = - -Afj + 2M S (4.46) 

Solving Eqs. 4.44 and 4.46 simultaneously, we find M x = 1 and M, = +$. 
Thus the final solution is 

x(t) = <r*(cos 2/ + £ sin 2t) (4.47) 

If we had used the form of x(t) given in Eq. 4.40, we would have obtained the 

x(t) - Vfe-* sin [2t + tan" 1 (2)] (4.48) 

Now let us consider a differential equation that illustrates everything we 
have discussed concerning characteristic values. 

Example 4.4 The differential equation is 

*<«(/) + 9* (4 >(f) + 32x<»(0 + 5Sx'(t) + 56<r'(0 + 24a</) = (4.49) 

The initial conditions are 

a! U)(0+)=0 5c (S, (0+) = l 

x'(0+) = -1 x'(0+) =• a<0+) - 1 

Differential equations 81 
Solution. The characteristic equation is 

H(p) = p* + 9/ + 32/>» + SSp* + 56p + 24 - (4.50) 

which factors into 

H(p) - (p + 1 +yi)0» + 1 -;1)0> + 2)ty + 3) = (4.51) 

From H(p) we immediately write x(t) as 

a<0 - Mi* - * cos / + M^r* sin f + Qc - ** + Qte"** + Qe -3 * (4.52) 

Since there are five coefficients, we need a corresponding number of equations 
to evaluate the unknowns. These are 

a<0+) - M x + C + C 8 = 1 
a;'(0+) - -M x + M t - 2C - 3C t + Q - 
x'(0+) - -2M, + 4C - 4Q + 9C t = -1 (4.53) 

a;«)(0+) = 2Mj, + 2M t - 8C + 12Q - 27C, = 1 
*<«((> +) = -4M 1 + 16C + 81C, - 32C X = 

Solving these five equations simultaneously, we obtain 
Mi - M t - 1 C = 1 Q - i 
so that the final solution is 

a</) - $e~' sin /+«-*+ Jte - ** 

C. =0 


We have seen that the solution of a homogeneous, differential equation 
may take different forms depending upon the roots of its characteristic 
equation. Table 4.1 should be useful in determining the particular form of 

TABLE 4.1 

Roots of H(p) 

Forms of Solution 


Single real root, p =p 

e p,t 


Root of multiplicity, k, (p 



+ Cje'i* + • 

• + CfcV'-W 


Complex roots at/>, , = a 


M X C* 

cos mt + Mge"* sin ait 

Me°* sin (mt + +) 


Complex roots of multiplicity k 


cos mt + Mjte"* cos »/ + ■•• 

*tP*.6 = " ±j<>> 

+ Aft-if* -1 **' cos mt 

+ Nite?* sin mt H 

+ N k _ 1 t k - V sin mt 

+ NtfP* sin mt 

82 Network analysis and synthesis 


As we mentioned in the introduction to this chapter, a nonhomogeneous 
differential equation is one in which the forcing function f(t) is not 
identically zero for all t. In this section, we will discuss methods for 
obtaining the solution x(t) of an equation with constant coefficients 

a n *<»>(*) + a^ x»-«(0 + • • • + a x(t) =f(t) (4.55) 

Let x p (f) be a particular solution for Eq. 4.55, and let x e (t) be the solution 
of the homogeneous equation obtained by letting f(t) — in Eq. 4.55. 
It is readily seen that 

<t) = xjf) + x e (0 (4-56) 

is also a solution of Eq. 4.55. According to the uniqueness theorem, the 
solution x(t) in Eq. 4.56 is the unique solution for the nonhomogeneous 
differential equation if it satisfies the specified initial conditions at 
t = 0+. 1 In Eq. 4.56, x p (t) is the particular integral; x a (t) is the comple- 
mentary function; and x(t) is the total solution. 

Since we already know how to find the complementary function x e (t), 
we now have to find the particular integral x p (t). In solving for x p (t), a 
very reliable rule of thumb is that x p (t) usually takes the same form as the 
forcing function if/(0 can be expressed as a sum of exponential functions. 
Specifically, x p {t) assumes the form of /(/) plus all its derivatives. For 
example, if /(f) = a sin cot, then xjf) takes the form 

x P (t) = A sin cot + B cos cot 

The only unknowns that must be determined are the coefficients A and B 
of the terms in xjf). The method for obtaining x p (t) is appropriately 
called the method of undetermined coefficients or unknown coefficients. 
In illustrating the method of unknown coefficients, let us take/(0 to be 

/(0=<*e" (4.57) 

where a and /S are arbitrary constants. We then assume x p (t) to have a 
similar form, that is, 

xjf) = Ae" (4.58) 

and A is the unknown coefficient. To determine A, we simply substitute 
the assumed solution x p (t) into the differential equation. Thus, 

AefiXaJ" + a,,.^"- 1 + • • • + aj + a ) = oe" (4.59) 

•See, for example, C. R. Wylie, Advanced Engineering Mathematics (2nd ed.), 
McGraw-Hill Book Company, New York, 1960, pp. 83-84. 

Differential equations 83 

We sec that the polynomial within the parentheses is the characteristic 
equation H(p) with /» = /?. Consequently, the unknown coefficient is 
obtained as 

A = -^- (4.60) 

provided that H(fi) jt 0. 
Example 4.5 Determine the solution of the equation 

**(r) + 3x'(t) + Ttff) - 4e* (4.61) 

with the initial conditions, a<0+) = 1, x'(0+) = — 1. 
Solution. The characteristic equation is 

Hip) -/>» + 3/> + 2 = (p + 2Xp + 1) 

so that the complementary function is 

*JLi) = c ie -* + c*-* 

For the forcing function /(f) — 4e«, the constants in Eq. 4.60 are a = 4, => 1. 

Then ^ = _i_ = ? 

H(l) 3 

Thus we obtain x p (t) = |e* 

The total solution is 

*(') - *c(0 + *„(0 = C x e-« + Cir" + |e« (4.62) 

To evaluate the constants Q and C„ we substitute the given initial conditions, 

40+) = 1 "Q+Cg+f 

*'(0+) = -1 - -Ci - 2C, + f (4 ' 63) 

Solving Eq. 4.63, we find that Q = — 1, C, = $. Consequently, 

a</) . _ e -' + £«-» + fe* (4.64) 

It should be pointed out that we solve for the constants C x and C t from the 
initial conditions for the total solution. This is because initial conditions are 
not given for x e (t) or *„(/), but for the total solution. 

Next, let us consider an example of a constant forcing function/(f) = a. 
We may use Eq. 4.60 if we resort to the artifice 

/(r) = a = ae*° (4.65) 

84 Network analysis and synthesis 

that is, j8 = 0. For the differential equation in Example 4.5 with/(/) = 4, 

we see that 

and *(0 = C lt r* + C*r« + 2 (4.67) 

When the forcing function is a sine or cosine function, we can still 
consider the forcing function to be of exponential form and make use of 
the method of undetermined coefficients and Eq. 4.60. Suppose 

fit) = oe* * = a(cos <ot + j sin oat) (4.68) 

then the particular integral x vl it) can be written as 

x Pl it) - Re MO + J Im MO (4 ' 69) 

From the superposition principle, we can show that 

if fit) = a. cos o>t then x„(0 = Re MO 

if fit) = a sin tot then *,(0 = Im * rt (f) 

Consequently, whether the excitation is a cosine function a cos mt or a 
sine function a sin or, we can use an exponential driver fit) = <xe iat ; 
then we take the real or imaginary part of the resulting particular integral. 

Example 4.6 Find the particular integral for the equation 

x'it) + 5*'(0 + 4a<<) = 2 sin 3* (4.70) 

Solution. First, let us take the excitation to be 

/i(0 = 2e»* (4.71) 

so that the particular integral *«.(/) takes the form 

» rt (0 - Ae* (4.72) 

From the characteristic equation 

H(p) =•/>* +5p +4 
we determine the coefficient A to be 

_ =, —L^tan-Hs)--] (4.73) 

H(j2) -5 +jlS 5 VlO 


Then MO is *«« - 1^ o^ 1 ^*- 1 < 474 > 

Differential equations 85 
and the particular integral x„(t) for the original driver /(f) = 2 sin 3/ is 

xjf) - Im x pl (t) = — -= sin [3f + tan" 1 (3) - *] (4.75) 


There are certain limitations to the applicability of the method of 
undetermined coefficients. If /(f) were, for example, a Bessel function 
7 (f )> we could not assume x v (t) to be a Bessel function of the same form 
(if it is a Bessel function at all). However, we may apply the method to 
forcing functions of the following types : 

!• /(0 = A; constant. 

2. /(f) = A(t n + Vi' M + • • • + V ' + b ); n, integer. 

3. /(f) = e pt ; p real or complex. 

4. Any function formed by multiplying terms of type 1, 2, or 3. 

For the purposes of linear network analysis, the method is more than 
Suppose the forcing function were 

/(f) = At*e pt p = a+jm 

The particular integral can be written as 

x p (t) = (AJ* + 4^ + ... + A 1 t + A )e* f (4.76) 

where the coefficients A& A±_ lf . . . , A lt A are to be determined. 


In this section we will discuss solutions of differential equations with 
step or impulse forcing functions. In physical applications these solutions 
are called, respectively, step responses and impulse responses. As physical 
quantities, the step and impulse responses of a linear system are highly 
significant measures of system performance. In Chapter 7 it will be shown 
that a precise mathematical description of a linear system is given by 
its impulse response. Moreover, a reliable measure of the transient 
behavior of the system is given by its step and impulse response. In this 
section, we will be concerned with the mathematical problem of solving 
for the impulse and step response, given a linear differential equation with 
initial conditions at f = 0—. 

From Chapter 2, recall that the definition of the unit step function was 

«(f) = 1 f ^ 
= f < 

86 Network analysts and synthesis 

and the unit impulse was shown to have the properties: 

d(t) = oo t = 
= tj*0 



and <3(0 df = 1 

in addition, we have the relationship 

d(t) = i*@ 

As the definitions of d(t) and u(t) indicate, both functions have dis- 
continuities ait = 0. In dealing with initial conditions for step and impulse 
drivers, we must then recognize that the solution x{t) and its derivatives 
x'(t), x"(f), etc., may not be continuous at t = 0. In other words, it may be 

*<«>(()-) ji z<«>(0+) 

x (»-i)(0_) ^ a;<"-"(0+) 

x(0-) 5* *(0+) 

In many physical problems, the initial conditions are given at t = 0— . 
However, to evaluate the unknown constants of the total solution, we must 
have the initial conditions at t = 0+. Our task, then, is to determine the 
conditions at t = 0+, given the initial conditions at t = 0— . The method 
discussed here is borrowed from electromagnetic theory and is often 
referred to as "integrating through a Green's function."* 

Consider the differential equation with an impulse forcing function 

a n *<»>(t) + a*.,. xt«-\t) +-- + a„x(f) = A d\t) (4.77) 

To insure that the right-hand side of Eq. 4.77 will equal the left-hand 
side, one of the terms z (B, (f), * (n_1) (0> • • • » x 0) must contain an impulse. 
The question is, "Which term contains the impulse ?" A close examination 
shows that the highest derivative term x ln) (t) must contain the impulse, 
because if * <n-1) (f) contained the impulse, x (B) (f) would contain a doublet 
C d'(t). This argument holds, similarly, for all lower derivative terms of 
x(i). If the term 3 (n> (f ) contains the impulse, then s ( *- u (0 would contain a 
step and x {n ~* , (0> a ramp. We conclude therefore that, for an impulse 
forcing function, the two highest derivative terms are discontinuous at i = 0. 

'The Green's function is another name for impulse response; see, for example, 
Morse and Feshbach, Methods of Theoretical Physics, McGraw-Hill Book Company, 
New York, 1952, Chapter 7. 

Differential equations 87 

For a step forcing function, only the highest derivative term is discontinuous 
at t = 0. 

Since initial conditions are usually given at t = 0— , our task is to 
determine the values a; (B) (0+) and * <n - 1, (0+) for an impulse forcing 
function. Referring to Eq. 4.77, let us integrate the equation between 
t = 0— and t == 0+, namely, 

a n )x M (t) dt + a^ f ^'-"(O dt + • • • + a ]x{t)dt = A f %*)<*' 
Jo- Jo- Jo- Jo— 

AP . (4-78) 

After integrating, we obtain 

a B [*(»-»(0+) - *<»-i>(0-)] + a„-i[* (B - a, (0+) - *<-»(0-)] + • • • = A 


We know that all derivative terms below (n — 1) are continuous at t = 0. 
Consequently, Eq. 4.79 simplifies to 

a n [x<»- 1 >(0+) - ^—"(O-)] = A (4.80) 

so that x l ^ v (0+) = — + * ( *-"(0-) (4.81) 

We must next determine * (n> (0+). At t = 0+, the differential equation 
in Eq. 4.77 is 

a n *<">(0+) + a^ *<"-»(0+) + •• • + Oo *(0+) = (4.82) 

Since all derivative terms below (« — 1) are continuous, and since we 
have already solved for z (B-1) (0+), we find that 

*<">(<)+) - - -±- [0^x^X0+) + • • • + ai x'(0+) + flo *(0+)] 



For a step forcing function A u(t), all derivative terms except x<">(f), are 
continuous at t = 0. To determine a; (B) (0+), we derive in a manner 
similar to Eq. 4.83, the expression 

*<»>(0+) = - - -± [a^ 1 x (B - 1> (0+) + • • • + ao*(0+)] (4.84) 

The process of determining initial conditions when the forcing function 
is an impulse or one of its higher derivatives can be simplified by the visual 

88 Network analysis and synthesis 

process shown in Eqs. 4.85 and 4.86. Above each derivative term we draw 
its associated highest-order singularity. Note that we need only go as low 
as a step in this visual aid. 

OnX^Kt) + o B _ 1 a: (B - 1 »W + a^^-'KO + ••■+ ao^O = <*'(') (4.85) 

a„x<«\t) + OaV- 1 ^) + (Wf^O + ■ • ■ + flo^W = *0 (4.86) 

It should be noted that if a derivative term contains a certain singularity — 
for example, a doublet— it also contains all lower derivative terms. For 
example, in the equation 

*'(/) + 3»'(0 + 2a</) = 4<5'(/) (4.87) 

we assume the following forms for the derivative terms at t = 0: 

*"(0 = A S'(t) + B 8{f) + C «(0 
x'(f) - A 8(t) + B u(t) (4.88) 

x(t) = A u(t) 
Substituting Eq. 4.88 into Eq. 4.87, we obtain 

A d'(t) + B (5(0 + Cu(t) + I A 8(t) + 3B «(r) + 2A u(f) - 4d'(t) (4.89) 
Or in a more convenient form, we have 

A d'(t) + (B + 3A) <J(0 + (C + 3B + 2A) u(0 - 4d'(0 (4.90) 
Equating like coefficients on both sides of the Eq. 3.90 gives 

A = 4 
B + 3A = (4.91) 

C + 3B + 2A = 

Differential equations 89 

from which we obtain B = — 12 and C = 28. Therefore, at t = 0, it is 
true that 

**(*) = M(t) - 12(5(0 + 28«(0 

x'(0 = 4d(0 - 12«(0 (4.92) 

*(*) = 4u(t) 

The w(0 terms in Eq. 4.92 gives rise to the discontinuities in the initial 
conditions at t = 0. We are given the initial conditions at t = 0— . Once 
we evaluate the A, B, C coefficients in Eq. 4.88, we can obtain the initial 
conditions at t = 0+ by referring to the coefficients of the step terms. For 
example, if 

*(<)-) = -2 

*'(0-) - -1 
*»(<)-) = 7 
Then from Eq. 4.92 we obtain 

x(0+) =-2 + 4 = 2 

x\0+) = -1 - 12 ex -13 (4.93) 

**(()+) = 7 + 28 = 35 

The total solution of Eq. 4.87 is obtained as though it were a homogeneous 
equation, since d'(t) = for t j*s 0. The only influence the doublet driver 
has is to produce discontinuities in the initial conditions at i = 0. Having 
evaluated the initial conditions at t = 0+, we can obtain the total 
solution with ease. Thus, 

z(0 = C x <r* + C*tr« (4.94) 

From Eq. 4.93 we readily obtain 

*(0 = (-9er* + ll<r-«) u(t) (4.95) 

The total solution of a differential equation with a step or impulse forcing 
function is obtained in an equally straightforward manner. For a step 
forcing function, only the highest derivative term has a discontinuity at 
t = 0. Since we do not need the initial condition of the highest derivative 
term for our solution, we proceed as if we were solving a standard non- 
homogeneous equation with a constant forcing function. For an impulse 
driver, once we determine the initial conditions at t = 0+, the equation is 
solved in the same manner as a homogeneous equation. 

90 Network analysis and synthesis 

Example 4.7 Find the step and impulse response for the equation 

2x"(f) + 4x'(t) + I0x(0 =/(f) 

where /(f) = <5(f) and/(f) = «(f), respectively. The initial conditions at f = — 

a<0-) = *'(0-) = x"(0-) = 

Solution. Let us first find the impulse response. We note that the x" term 
contains an impulse and a step ; the x' term contains a step ; the x term contains 
a ramp, and is therefore continuous at f = 0. Thus x(0+) = x(0— ) = 0. To 
obtain a;'(0+), we use Eq. 4.81 

*'(0 +) - - + *'(0 -) - \ + = \ (4.96) 

a 2 2. 2. 

Note that we actually need only x(0 +) and x'(0 +) to evaluate the constants 
for the second-order differential equation. Next, we proceed to the comple- 
mentary function x e (t). The characteristic equation is 

H(p) - 2(p* + lp + 5) - lip + 1 +jlKp + 1 -j2) (4.97) 

Since H(p) has a pair of complex conjugate roots, we use a standard form for 

x c (i) = Me~* sin (2f + # (4.98) 

Substituting the initial conditions at f = 0+, we obtain 

a<0+) =0 = Afsin^ 

x'(0+) = J = 2Mcos $ - M sin <f> 

from which we find ^ = and M = \. Thus the impulse response, which we 
denote here as x t (t), is 

**) = 1«~* sin 2f u{i) (4.100) 

Next we must solve for the step response x u {t). For convenience, let us write the 
complementary function as 

x e (t) = e-KA^ sin 2f + A t cos 2f) (4.101) 

The particular integral is evaluated by considering the forcing function a 
constant /(f) = 1 so that 

a, j t ) « _!_ = J- (4.102) 

The total solution is then 

a<f) - (A t sin 2f + A t cos 20*?-* + ^ (4.103) 

Since x'(t) and x(t) must be continuous for a step forcing function, 

a <0+)=a<0-)= 

a;'(0+)-a; , (0-) =0 

Differential equations 91 

Substituting these initial conditions into x(i) d/dt d/dt 

and *'(/), we find that A x = -0.05 A t = -0. 1 . *,(') > *»(0 >■ *j(0 

Therefore, the step response is C* C* 

*„(*)= 0.1[1 -«-*(0.5sin2/+ L_ L- 

cos 2,)]*) (4.105) *,«>— WO— >*,(/) 

FIG. 4.1. 

Note that the impulse response and the 
step response are related by the equation 

x, (0 = i. *„(<) (4.106) 


Wecan demonstrate Eq.4.106by thefollowingprocedure. Let us substitute 
x u (t) into the original equation 

2 ^ «.(*) + 4 ^ x.(0 + 10*„(0 . h(0 (4.107) 

Differentiating both sides, we have 

_| «.(*)] + 4 1[| *„(o] + io[^ *„(*)] = <H0 (4.108) 

from which Eq. 4.106 follows. 

Generalizing, we see that, if we have the step response for a differential 
equation, we can obtain the impulse response by differentiating the step 
response. We can also obtain the response to a ramp function f(t) — 
A p(t) (where A is the height of the step) by integrating the step response. 
The relationships discussed here are summarized in Fig. 4.1. 


In this section, we will consider an integrodifferential equation of the 


a„ *"(t) + a,.! *-\t) + ■ ■■ + a x(t) + a_ t f * x(r) dr = /(/) (4.109) 


where the coefficients {a„, a 1K . 1 , . . . , a_ t } are constants. In solving an 
equation of the form of Eq. 4.109 we use two very similar methods. The 
first method is to differentiate both sides of Eq. 4.109 to give 

a n x (n+1 \t) + a^ x M (t) + • • • + a x\t) + a_ t x(t) -/'(») (4.U0) 

The second method consists of a change of variables. We let y'(f) = x(t); 
Eq. 4.109 then becomes 

a, y (n+1) (t) + *„_! y (n \t) + • • • + a „ y'(t) + a_ x y(t) = f(t) (4.111) 


92 Network analysis and synthesis 

Note that from Eq. 4.110 we obtain x(t) directly. From Eq. 4.111, we 
obtain y(t), which we must then differentiate to obtain x(t). An important 
point to keep in mind is that we might have to derive some additional 
initial conditions in order to have a sufficient number to evaluate the 
unknown constants. 

Example 4.8 Solve the integrodifferential equation 

x'(t) + 3a<0 + 2 f *(t) dr = 5«(0 (4.112) 

The initial condition is x(0 — ) = 1. 

Solution. Since the characteristic equation of Eq. 4.112 is of second degree, 
we need an additional initial condition x'(0+). We obtain *'(<)+) from the 
given equation at t ■= + : 

*'(0+) + 3»<0+) + 2 J a< T ) dr = 5 (4.113) 

Since x(t) is continuous at / = 0, 

<H a<T)</T=0 (4.114) 

and «(0+)-a<0-)- 1 (4.115) 

Therefore, x'(0 +) = 5 - 3s(0 +) = 2 (4.1 16) 

-Method 1. Differentiating both sides of Eq. 4.112, we obtain 

x"(t) + 3*'(0 + 2a<0 = 5<K0 (4.117) 

The complementary function is then 

*.(0 = Car* + C*r*> (4.118) 

Using the initial conditions for a<0+) and *'(<>+), we obtain the total solution 

x(t) - 4e~' - 3e-*« (4.119) 

Method 2. Letting y'(t) = x(t), the original differential equation then 

y"(0 + MO + 2y(t) = 5«(0 (4120) 

We know that y'(P+) = «(0+) - 1 ,„,„,, 

j,'(0+)= *'(<>+)= 2 

From Eq. 4.120, at t = 0+, we obtain 

y(0 +) - i[5 - y'(0 +) - 3y'(0 +)] - (4.122) 

Without going into details, the total solution can be determined as 

j<0 - -4e- + |e-*« + \ (4- 123 > 

Differential equations 93 
Differentiating y{t), we obtain 

<i) = y'(f) = 4e~' - 3<?-*' (4.124) 


Up to this point, we have considered only differential equations with a 
single dependent variable x{t). In this section, we will discuss equations 
with more than one dependent variable. We shall limit our discussion to 
equations with two unknowns, x(t) and y(t). The methods described here, 
however, are applicable to any number of unknowns. Consider first the 
system of homogeneous equations 

«x *'(0 + a, x(0 + & y'(t) + fi, y(t) - 


yi *'(0 + Yo *(0 + <$! y'(0 + \ y(t) = o 

where a„ f} it y it d i are arbitrary constants. The complementary function is 
obtained by assuming that 

so that the characteristic equation is given by the determinant 

(a^ + oto) <fc/> + A) 
(ftP + Yo> (<V + to) 

H(p) = 


The roots of H(j>) are found by setting the determinant equal to zero, 
that is, 

(«tf> + ««)(V + *•) - OV + PoXYiP + Vo) = (4.127) 
It is seen that a nontrivial solution of H(p) = exists only if 

(«* + ««KV + <>a) * OV + &XftP + y<0 (4.128) 

Assuming that the preceding condition holds, we see that H(p) is a second- 
degree polynomial tap and can be expressed in factored form as 

H(p) = C(p- Po )(p - />0 (4.129) 

where C is a constant multiplier. The complementary functions are 

x(t) = K^e** + K*** 

y(t) - W + Kf 

94 Network anal/sis and synthesis 

and the constants K x , K t , K 9 , K t are determined from initial conditions. 
As in the case of a single unknown, if H(p) has a pair of dduble roots; i.e., 

tf Po = />i» then 

x(t) = (#, + K^e*** 

y(t) = (tf, + K t t)e*<* 
If H(p) has a pair of conjugate roots, 


= a ±jto 


x(t) = M^e"* sin (cot + <^) 
y(f) = Mte" sin (tor + <f>J 
Example 4.9. Consider the system of equations 

2x'(t) + 4x(t) + y'(t) -y(f)=0 

x\t) + 2x(0 + y'(0 + y(t) = 
with the initial conditions 

*'(0+)«2 y '(o+)=-3 
a<0+)=0 i/(0+) = 1 
Solution. The characteristic equation is 

2» + 4 » - 1 
p +2 /» + 1 

Evaluating the determinant, we find that 

#0) =/>* + 5/> + 6 = (p + 2Xp + 3) 
so that y{t) = Kjfi-" + A^- 8 ' 

a</) - KtfT* + J>-» 






With the initial conditions, z'(0+) - 2,a<0+) = 0, we obtain K, =2,K t = -2. 
From the conditions y'(0+) - -3, y(0+) = 1, we obtain K t = 0, JCg = 1. 
Thus the final solutions are 




Differential equations 95 

Next, let us determine the solutions for a set of nonhomogeneous 
differential equations. We use the method of undetermined coefficients 
here. Consider first an exponential forcing function given by the set of 

a lX ' + aye + p iy ' + p iff = Ne° t 

Yi*' + y<P + b x y' + d& = 
We first assume that x p (t) = Ae" 

y v (t) = Be 9 * 
Then Eq. 4.139 becomes 

Qtfl + xJA + (^0 + o )B = N 
(.YJ + YM + O*i0 + d )B = 
The determinant for the set of equations 4.141 is 

«i0 + «<, PiO + Po 

H(8) = A(0) = 

YiO + Y* W + \ 





where H(6) is the characteristic equation with p = 0. We now determine 
. #4he undetermined coefficients A and B from A(0) and its cofactors, namely, 

A = 

*A 1X (0) 

B = 


JVA ia (e) 

where A„ is the i/th cofactor of A(0). 

Example 4.10. Solve the set of equations 

2x' + Ax + y' - y = 3e** 



given the conditions x'(0+) = 1, a<0+) = 0, y'(0+) = 0, y(0+) = -l. 

Solution. The complementary functions * c (f) and y e (i), as well as the charac- 
teristic equation H(p), were determined in Example 4.9. Now we must find 
A and B in the equations 

*,(/) - Ae** 

y,(t) - Be** 
The characteristic equation withy* = 4 is 

2(4) +4 (4)-l 


(4) +2 (4) + l 





96 Network analysis and synthesis 

Then we obtain from Eq. 4.143 the constants 

A - A B - 

The incomplete solutions are 

*(f) = KiT* + Ktfr** + &e» 
y(t) - K&-* + K ie -* f - f e" 

Substituting for the initial conditions, we finally obtain 

y(0=K-4«" s '-3« 4 ') 

Example 4.11. Solve the system of equations 

lx' + 4x + y' + ly = 5«(/) 

x' +x +y' +3y = 56(f) 
given the initial conditions 

a<0-) = *'(<>-) = jKO-) = y'(0-) - 

Solution. First we find the characteristic equation 

2/> + 4 / + 7 

/> + 1 p + 3 







J5T(Cp> = A(/,) - 
which simplifies to give 

H(p) =(p t +2p+5)=(p + l +j2)(p + 1 -j2) 
The complementary functions x c (t) and j/ (f) are then 

x e (t) = ^e - ' cos It + A&-* sin 2/ 

y e (t) = jBiC - ' cos 2/ + B^r* sin 2f 

The particular solutions are obtained for the set of equations with t > 0, 

2x' + 4x + y' + ly = 5 


*' + a: + y' + 3y «- 

Using the method of undetermined coefficients, we assume that x v and y,, are 
constants: as, = Q; y v = C a . Since the forcing function 5 can be regarded 
as an exponential term with zero exponent, that is, 5 — 5e°', we can solve for 
C x and C a with the use of the characteristic equation H(p) with p = 0. Thus, 

#(0) = A(0) = 

4 7 
1 3 

= 5 


Differential equations 97 
n 5A U (0) 5(3) 

■ Dd c ^--5(or = — - 3 


c,« 5 - 1 

The general solution is then 

x(t) - ^e-' cos 2r + A^r* sin 2/ + 3 

y(t) = fl^* cos 2/ + B&-* sin 2r - 1 

In order to find A lt A t , B lt B* we need the values a<0+), x'(0+), y(0+), 
y'(0+). The values for a<0+) and y(0+) are first obtained by integrating the 
original differential equations between t ■» 0— and f =» 0+. Thus, 


/•0+ /VH- 

(2*' + 4* + y' + ly)dt - 5«(f)<// 
Jo- Jo- 

/•0+ f»H- 

(*' + x + y' + 3y*<ft - 5d(t)dt 


We know that only the highest derivative terms in both equations contain 
impulses at f - 0. Moreover, both x(t) and y(t) contain, at most, step dis- 
continuities at t «■ 0. Therefore, in the integration 


(4a; + ly) dt = 

(x + 3y)rff»0 

After integrating, we obtain 



Solving, we find 

a<0+) = -5 y(0+) - 10 (4.160) 

To find x'(0+) and y'(0+), we substitute the values for a<0+) and y(0+) into 
the original equations at t = 0+. Thus, 

2z'(0+) - 20 + j/'(0+) + 70 = 5 

x\0+) - 5 + y'(0+) + 30 = 

so that x'(0+) = -20 

y'(0+) = -5 

Substituting these values into Eq. 4.156, we eventually obtain the final solutions 

x(t) - ( -%e~* cos It - 14e~* sin It + 3) u(t) 

y(/) - (1 \e~* cos 2/ + 3e~* sin It - 1) «(f) 

98 Network analysis and synthesis 


4.1 Show that 

3^(0 = M x e~* cos It 

and xjj) m, M&-* sin It 

are solutions for the equation 

x'(t) + 2x'(t) + 540 = 

Show that x 1 + x t is also a solution. 

4.2 Determine only the form of the solution for the equation 

(a) **(0 + 4x'(0 + 340 = 

(6) x"(t) + 8x'(t) + 540 = 

(c) *"(/) - 540 = 

(d) x'(0 + 540 = 

(e) x"(t) + 6x'(0 + 2540 = 
(/) z'(0 + 6x'(.0 + 940 = 

4.3 Given the initial conditions 40+) = 1, *'(0+) = —1, determine the 
solutions for 

(a) x'(t) + fo'(0 + 2540 = 

(b) x\t) + &r'(0 + 16x(t) = 

(c) x'(t) + 4.8lx'(t) + 5.7640 = 

4.4 Find only the particular integrals for the equations 

(a) x"(t) + 7x'(t) + 1240 = e -8 ' 

(b) x'(t) + 3x'(t) + 240 = 2 sin 3f 

(c) *'(0 + 2x'(t) + 540 = e~* sin It 

(d) x"(t) + 2*'(0 + Sx(t) = <r"/2 

(e) x'(t) + 5.0x'(0 + 6.25x(t) = 6 

(/) **(/) + fo'(0 + 540 = 2e-* + 3e~ at 

4.5 Given the initial conditions 40+) = 1, x'(0+) =0, determine the 
solutions for 

(a) x"(t) + 4*'(0 + 340 = 5e~* sin It 

(.b) x"{t) + 6x'(t) + 254/) - 2 cos / 

(c) *"(/) + 8a;'(0 + 1640 = 2 

4.6 (a) A system is described by the differential equation 

y'(0 + 3y'(/) + 2y(t) - a(0 

The initial conditions are y(0-) = 1; y'(0-) = 2. Find y(0+) and y'(0+). 
(b) Given the differential equation 

as<*>(0 + 14*'(0 + 840 - 6<K0 

Differential equations 99 

with the intial conditions *(0-) = 12, *'(<>-) = 6, and x"(0-) = -7, find 
*<*>«)+), *'(0+), *'(0+), and a<0+). 

4.7 Given the initial conditions x'(0—) = **(0-) = 0, find the solutions 
for the equations 

(a) x'{t) + 2x'(/) + 2a</) = 3<K0 
(*) x"{t) + 7s'(0 + 12a<r) - 5«(0 

4.8 Given the initial condition x(0— ) = —2, solve the integrodifferential 

(«) x'(t) + 5a</) + 4 f a<r) </t = 2 sin / 

(b) x'(t) + 2tft) + 2 f x{r) dr = | e-« 

(c) *'(') + &<0 + 9 f *(t) oV = 2u(t) 

4.9 Given the set of equations 

*V) + *(/) + y{t) = «(/) 
*C)+y'(0 + 2y(0 = <K0 

with s(0-) - *'(0-) = y(o-) = y'(0-) = 0, find *(0+), z'(0+), y(0+), and 

4.10 Solve the set of equations 

2*'(0 + 3*(0 + y'(t) + 6y(t) = */) 
*'« + *(') + y'W + 6y(0 = «(') 
All initial conditions at f = 0— are zero. 

4.11 Solve the set of equations 

2s'(0 + 2a<0 + y'(0 - y(i) = 2<5(0 
x'(0 + a</) + y'(0 + 2y(r) = 3e~* uSj) 
The initial conditions are 

a#)-) = l y(0-)=0 

*'(0-)=-l j,'(0-)=0 

chapter 5 

Network analysis: I 


In this chapter we will apply our knowledge of differential equations to 
the analysis of linear, passive, time-invariant networks. We will assume 
that the reader is already familiar with Kirchhoff's current and voltage 
laws, and with methods for writing mesh and node equations for a-c or d-c 
circuits. 1 We will, therefore, consider only briefly the problem of writing 
mesh and node equations when the independent variable is time t. The 
problems in this chapter have the following format: Given an excitation 
signal from an energy source and the network, a specified response that 
is a current or voltage in the network is to be determined. When relating 
these problems to the mathematics in Chapter 4, we shall see that, 
physically, the forcing function corresponds to the excitation ; the network 
is described by the differential equation; and the unknown variable x(t) 
is the response. 

The problems encountered will be twofold. First, we must write the 
differential problems of the network using Kirchhoff's current and voltage 
laws. Next, we must solve these equations for a specified current or 
voltage in the network. Both problems are equally important. It is 
useless, for example, to solve a differential equation which is set up in- 
correctly, or whose initial conditions are incorrectly specified. 

The usual type of problem presented in this chapter might generally be 
described as follows. A switch is closed at t = 0, which connects an energy 
(voltage or current) source to a network (Fig. 5.1). The analog of a switch 
closing at t = is the energy source whose output is e{t) u(t). Before the 

1 For a comprehensive treatment, see H. H. Skilling, Electrical Engineering Circuits 
(2nd Edition), John Wiley and Sons, New York, 1965. 


Network analysis: I 



Energy /\ 



FIG. 5.1. Switching action. 

switch is closed, the currents and voltages in the network have known 
values. These values at t = 0- are the initial conditions. We must then 
determine the values of the currents and voltages just after the switch 
closes (at / = 0+) to solve the network equations. If the excitation is not 
an impulse function or any of its derivatives, the current and voltage 
variables are continuous at t = 0. For an impulse driver the values at 
/ = 0+ can be determined from methods given in Chapter 4. Having 
obtained the initial conditions, we then go on to solve the network 
differential equations. Unless otherwise stated, all the solutions are valid 
only for / ^ 0+. 

Since we are dealing only with linear circuits, it is essential that we bear 
in mind the all-important principle of superposition. According to the 
superposition principle, the current through any element in a linear circuit 
with n voltage and m current sources is equal to the algebraic sum of 
currents through the same element resulting from the sources taken one 
at a time, the other sources having been suppressed. Consider the linear 
network depicted in Fig. 5.2a with n voltage and m current sources. 

FIG. 5 Jo 

102 Network anal/sis and synthesis 

FIG. 5.2b 

Suppose we are interested in the current i T (f) through a given element Z, 
as shown. Let us open-circuit all the current sources and short-circuit 
n — 1 voltage sources, leaving only »/0 shown in Fig. 5.26. By i Vj (t), 
we denote the current through Z due to the voltage source »/f) alone. In 
similar fashion, by i Ck (t) we denote the current through Z due to the 
current source i k (t) alone, as depicted in Fig. 5.2c. By the superposition 

FIG. 5.2c. Superposition in linear circuits. 

Network analysis: I 


principle, the total current i T (t) due to all of the sources is equal to the 
algebraic sum 

h = \ + '.,+ •• + I., + t H + »,,+ ••• + 



In this section, we will discuss the voltage-current (v-i) relationships 
that exist for the basic network elements. Before we examine these 
relationships, it is important to assign first arbitrary reference polarities for 
the voltage across an element, and a reference direction of flow for the 
current through the element. For the purposes of our discussion, we 
assume that the positive polarity for voltage is at the tail of the current 
arrow, as shown by the resistor in Fig. 5.3. Now, let us review the voltage- 
current relationships for the resistor, the inductor, and the capacitor, 
which we first discussed in Chapter 1. 


The resistor shown in Fig. 5.3 defines a linear proportionality relation- 
ship between v(t) and i(t), namely, 

K0 - r m 

i(0 = Gt*0 
where R is given in ohms and G in mhos. 



For the capacitor shown in Fig. 5.4a the v-i relationships are 



= -\ i(r) dr 

+ »o(0-) 


' m 





FIG. 5.3. Resistor. 

Ut)\ ZZC **> i(*>\ 





FIG. 5.4. (a) Capacitor. (A) Capacitor 
with initial voltage. 

104 Network anal/sis and synthesis 

m l\ 



< ° 



(a) (b) 

FIG. 5.5. (a) Inductor. (6) Inductor with initial current. 

where C is given in farads. The initial value t> c (0— ) is the voltage across 
the capacitor just before the switching action. It can be regarded as an 
independent voltage source, as shown in Fig. 5.46. We should point out 
also that i>c(0— ) = t> c (°+) for aU excitations except impulses and 
derivatives of impulses. 


The inductor in Fig. 5.5a describes a dual relationship between voltage 
and current when compared to a capacitor. The v-i relationships are 



i f 



ir)dr+i L (0-) 

where L is given in henrys. The initial current i L (0— ) can be regarded as 
an independent current source, as shown in Fig. 5.56. As is true for the 
voltage across the capacitor, the current through the inductor is similarly 
continuous for all t, except in the case of impulse excitations. 

When the network elements are interconnected, the resulting i-v equa- 
tions are integrodifferential equations relating the excitation (voltage or 
current sources) to the response (the voltages and currents of the elements). 
There are basically two ways to write these network equations. The first 
way is to use mesh equations and, the second, node equations. 

Mesh equations are based upon Kirchhoff 's voltage law. On the mesh 
basis, we establish a fictitious set of loop currents with a given reference 
direction, and write the equations for the sum of the voltages around the 
loops. As the reader might recall from his previous studies, if the number 
of branches in the network is B, and if the number of nodes is N, the 
number of independent loop equations for the network is B — N + 1.* 

1 See Stalling, op. cit. 

Network anal/sis: I 105 

We must, in addition, choose the mesh currents such that at least one mesh 
current passes through every element in the network. 

F.™mpl« 5.1. In Fig. 5.6, a network is given with seven branches and five nodes. 
We therefore need 7 - 5 + 1 = 3 independent mesh equations. The directions 
of the mesh currents i lt i* /» are chosen as indicated. We also note that the 

I </W- 




'1 \ fCi' »» 



»R 2 vatQ 

0^9-r Qhfi-i 

FIG. 5.6 

capacitors in the circuit have associated initial voltages. These initial voltages 
are assigned reference polarities, as shown in the figure. Now we proceed to 
write the mesh equations. 

Meshi x : 

l c if* 

«i(0 - *C,(0-) - *1 h(0 + — I «» dr - — J^t) dr 

Mesh i t : 

1 C* di. 

v Cl (0-) - v c fii-) = - ^ J^r) dr+Li — 

Mesh i t : 

-vJLO + v Ct (0 -) = ^ P i/r) dr + ^r J\(t) dr+Ki tf (5.5) 

After we find the three unknowns, i lf i t , and /», we can determine the branch 
currents and the voltages across the elements. For example, if we were required 
to find the branch currents i Cl and i Ct through the capacitors, we would use the 
following relationships 

*<>*-'*-*• (5.6) 

»'c, - »i - '» 
Alternatively, if the voltage vj.t) in Fig. 5.6 is our objective, we see that 

v*t) - tff)4 (5.7) 

Network equations can also be written in terms of node equations, 

which are based upon Kirchhoff's current law. If the number of nodes in 

106 Network anal/sis and synthesis 

^ Datum 
FIG. 5.7 

the network is N, the number of independent node equations required 
for the complete solution of the network is JV — l. 3 We can therefore 
select one datum node in the network. All the node voltages will be 
positive with respect to this datum node. 

Consider the network in Fig. 5.7. Let us write a set of node equations 
for the network with the datum node shown. Since the number of nodes 
in the network is N — 3, we need JV — 1 = 2 independent node equations. 
These are written for nodes v x and v s , as given below. 

Node v^. 

i B (t) - i L (0~) = G 1 Vl (t) + j(* Vl (r) dr-jl* v t (r) dr 

L Jo- L Jo- 

Node ty. 

j i( 0-) . - - ! * v t (r) dr + 7 f ' V 2 (r) dr + C ^& + G t t> 2 (») (5.8) 
L Jo- L J o- dt 

Further examples are given in the following sections. 


In this section, we consider some methods for obtaining initial conditions 
for circuit differential equations. We also examine ways to obtain particular 
integrals for networks with constant (d-c) or sinusoidal (a-c) excitations. 
In the solution of network differential equations, the complementary 
function is called the transient solution or free response. The particular 
integral is known as the forced response. In the case of constant or periodic 
excitations, the forced response at t = oo is the steady-state or final 

* See Stalling, op. cit. 

Network analysis: I 107 

There are two ways to obtain initial conditions at / = 0+ for a network: 
(a) through the differential equations describing the network, (b) through 
knowledge of the physical behavior of the R, L, and C elements in the 

Initial conditions for a capacitor 

For a capacitor, the voltage-current relationship at t = 0+ is 



i(r)dr + vc(0-) 


If i(t) does not contain impulses or derivatives of impulses, »c(0+) = 
v d°— )• If? « the charge on the capacitor at t = 0—, the initial voltage is 

»o(0+) - vc<0-) = 


When there is no initial charge on the capacitor, »o(0+) = 0. We con- 
clude that when there is no stored energy on a capacitor, its equivalent 
circuit at t = 0+ is a short circuit. This analogy is confirmed by examining 
the physical behavior of the capacitor. As a result of the conservation of 
charge principle, an instantaneous change in voltage across a capacitor 
implies instantaneous change in charge, which in turn means infinite 
current through the capacitor. Since we never encounter infinite current 
in physical situations, the voltage across a capacitor cannot change 
instantaneously. Therefore at t = 0+, we can replace the capacitor by a 
voltage source if an initial charge exists, or by a short circuit if there is no 
initial charge. 

Example 5.2. Consider the R-C network in Fig. 5.8a. The switch is closed at 
t = 0, and we assume there is no initial charge in the capacitor. Let us find 
the initial conditions i(0 +) and i'(0 +) for the differential equation of the circuit 

Vu(.t)=Ri{t) + 





£_ I i(t) ) >B A- ' »«>+)) >R 

T I V T T 

R-C network Equivalent circuit at < - 0+ 

W m 

FIG. 5A. (a) R-C network, (b) Equivalent circuit at / — 0+. 


108 Network analysis and synthesis 

The equivalent circuit at t - 0+ is given in Fig. 5.86, from which we obtain 



To obtain /'(u+), we must refer to the differential equation 


» (0+) 


i(0+) V 

At t = 0+ we have 

We then obtain 

V8(t) -Ri'(t) + 
-Jt/'(0+) +' 

i'(0+)- -■ 


The final condition, or steady-state solution, for the current in Fig. 5.8a is 
obtained from our knowledge of d-c circuits. We know that for a d-c excitation, 
a capacitor is an open circuit for d-c current. Thus the steady-state current is 

i„(0 = »'(°°) - 

Initial conditions for an inductor 

For an inductor, the voltage-current relationship at t = 0+ is 

1 f° 

iKr)dT+i L (0-) 


If v(t) does not contain impulses, then i L (0+) — i £ (0-). If there is no 
initial current, i L (0+) = 0, which corresponds to an open circuit at 
t _ o+. This analogy can also be obtained from the fact that the current 
through an inductor cannot change instantaneously due to the conserva- 
tion of flux linkages. 

Example S3 In Fig. 5.9a, the switch closes at t = 0. Let us find the initial 
conditions i(0+) and i'(0+) for the differential equation 






-o o- 

s=: i(o+)J 

(a) W 

FIG. 5.9. (a) R-L network. (*) Equivalent circuit at / *= 0+. 

Network analysis: I 109 
From the equivalent circuit at f = 0+, shown in Fig. 5.9b, we see that 

i(0+)=0 (5.18) 

We then refer to the differential equation to obtain <"(0+). 

V-L i"(0 +) + R i(0 +) (5.19) 

V R 
Thus /'(0 +) - - - - / (0 +) 


as — 



The steady-state solution for the circuit in Fig. 5.9a is obtained through the 
knowledge that for a d-c source, an indicator is a short circuit. 

/„(') =/(«>) =•£ (5.21) 

Example 5.4. In this example we consider the two-loop network of Fig. 5.10. 
As in Examples 5.2 and 5.3, we use the equivalent circuit models at t — + 
and t = oo to obtain the initial conditions and steady-state solutions. At 
f = the switch closes. The equivalent circuits at t = 0+ and / =• <x> are 
shown in Fig. 5.1 la and b, respectively. The initial currents are 

*i(0+) = ^~ 

*i (5.22) 

ig(0+) = 
The steady-state solutions are 

«») - j^r^ = «") < 5 - 23 > 

Final conditions for sinusoidal excitations 

When the excitation is a pure sinusoid, the steady-state currents and 
voltages in the circuit are also sinusoids of the same frequency as the 
excitation. If the unknown is a voltage, for example, 1^(0. the steady-state 
solution would take the form 

»i,(0 - I KOo)l sin K< - <K<*J\ (5.24) 

where m is the frequency of the excitation, and \V(jcoJ\ and ^(<u ) 
represent the magnitude and phase of v l9 (t). A similar expression would 
hold if the unknown were a current. 

To obtain the magnitude and phase, we follow standard procedures in 
a-c circuit analysis. For example, consider the R-C circuit in Fig. 5.12. 
The current generator is 

W) - ( J o sJ n «■*) "(0 ( 5 -25) 

1 10 Network analysis and synthesis 
j VWV 

+ | h(0 ) ^=:C kM ) 




FIG. 5.10 

J VW\r 


V!= »i(0+)J 

ww* — i 

»2(0+) I O.C. 

Equivalent circuit at t = 0+ 





Equivalent circuit at t = «> 
FIG. 5.1 1, (a) Equivalent circuit at t = 0+. (6) Equivalent circuit at / = oo. 



FIG. 5.12 

Network analysis: I III 
If the steady-state voltage takes the form shown in Eq. 5.24, 




(G a + co *C*j 


and #o> ) = tan" 1 ^ (5.27) 


so that vjLO - (G>+ J ^w sin (°>o< " tan" 1 ^) (5.28) 

We refer to this problem in Section 5.5. 


As an introduction to the topic of solution of network differential 
equations, let us consider the important problem of obtaining the step and 
impulse responses for any voltage or 
current in the network. As we shall see in 
Chapter 7, the step and impulse responses 
are precise time-domain characterizations 
of the network. The problem of obtaining 
the step and impulse response is stated as 
follows : Given a network with zero initial 
energy, we are required to solve for a 
specified response (current or voltage) due F,e * 5 * ' 3 

to a given excitation function u(t) or <3(0, 

which either can be a current or a voltage source. | If the excitation is 
a step of voltage, the physical analogy is that of a switch — closing at 
time t = — which connects a 1-v battery to a circuit. The physical 
analogy of an impulse excitation is that of a very short (compared to the 
time constants of the circuit) pulse with large amplitude. 

The problems involved can best be illustrated by means of examples. 
Consider the series R-C circuit in Fig. 5.13. The differential equation of 
the circuit is 

— I ii 

KO = *«) = R 1(0 + 7 l(r) dr (5.29) 

C J»~ 

We assume Vq(0—) = 0. Since Eq. 5.29 contains an integral, we substitute 
x'(t) for j'(0 in the equation, giving us 

d(t) = Rx'(t) + ±z(t) (5.30) 

1 12 Network analysis and synthesis 
Integrating both sides between 0— and 0+ gives 


The characteristic equation is 

H(p) - Rp + £ 

and with little effort we have 


_ L ,-t/RC 



so that 


R^'RC 6 "'* "®] (5 ' 34) 

which is shown in Fig. 5.14a. We thus arrive at the current impulse 
response i(t) as the result of an impulse voltage excitation. In the process 
we have also obtained the step response x(i) in Fig. 5. 146 since, by definition, 
the derivative of the step response is the impulse response. The reader 
should check this result using a step excitation of voltage. 

In the second example consider the parallel R-C circuit in Fig. 5.12 
driven by a step current source i(f ) = 7 «(/), where /„ is a constant. 

' £aro 



FIG. 5.14. (a) Impulse response of R-C circuit. (A) Step response of R-C circuit. 

Network analysis: I 113 


FIG. 5.15. (a) Step response of parallel R-C circuit. (A) Impulse response of parallel 
Jt-C circuit. 

Assuming zero initial conditions, the differential equation is 

J o „(0 = Gt<0 + C^p 

from which we obtain the characteristic equation 

H(p) = Cp + G 
The steady-state value of KO is 

H(0) G 





FIG. 5.1*. Pulse excitation. 

1 14 Network analysis and synthesis 


FIG. 5.17. 

to pulse excitation. 

Thus the complete solution for the voltage step response is 

r(0 = (Ke- i/BC + 7oR) u(t) (5.38) 

From the initial condition »(0+) = 0, we obtain K = —I^R so that 

t>(0 = JoR(l - e-* IRC ) u(t) (5.39) 

Differentiating Eq. 5.39 gives us the voltage impulse response 

f(0 = 

l* e -tlRC 



The step and impulse responses are plotted on Figs. 5.15a and b. Suppose 
the excitation in Fig. 5.12 were a pulse 

KQ = IMt) - u(t - T)] 


shown in Fig. 5.16. Then, according to the superposition and time- 
invariance postulates of linear systems, the response would be 

K0 = V*[(l - e-" BO ) u(t) - (1 - e- (i - T)IBC )u(t - T)] (5.42) 
which is shown in Fig. 5.17. 


In this section we will apply our knowledge of differential equations 
to the analysis of linear networks. There are two important points in 
network analysis: the writing of network equations and the solution of 
these same equations. Network equations can be written on a mesh, node, 
or mixed basis. The choice between mesh and node equations depends 
largely upon the unknown quantities for which we must solve. For 

Network analysis: I 115 

instance, if the unknown quantity is a branch current, it is preferable to 
write mesh equations. On the other hand, if we wish to find a voltage 
across a certain element, then node equations are better. In many cases 
the choice is quite arbitrary. If, for example, we wish to find the voltage v 
across a resistor R, we can either find v directly by node equations or find 
the branch current through the resistor and then multiply by R. 

Example 5.5. Find the current /(f) for the 

network in Fig. 5.18, when the voltage 

source is e(t) = 2e-°-« ufjt) and »c(0-) = 0. 

Solution. The differential equation is 

*(/) = Rl(t) + ^ j i(r)dr + v^O-) 


Or, in terms of the numerical values, we 


2«~«" u{t) = i(0 + 2 »(t) dr (5.44) 

Differentiating both sides of Eq. 5.44, we obtain 


FIG. 5.18 

2(5(0 - e-° u u(t) = 

+ 2i(t) 


To obtain the initial condition i(0+), we must integrate Eq. 5.45 between 
the limits / = 0— and t = 0+ to give «(0+) = 2. From the characteristic 

H(p)=p+2*°0 (5.46) 

we obtain the complementary function as 

/<#) = JCr« (5.47) 

If we assume the particular integral to be i v (t) = Ae~° bt , then we obtain 

_1 2 


A = - 


The incomplete solution is 

»(0 = Ke~* - §<r° •" (5.49) 

From the initial condition i(0+) = 2, we obtain the final solution, 

Kt) = (f«-* - f*-°-") «(0 (5.50) 

As we already noted in Section 5.3, in the solution of network differential 
equations, the complementary function is called the free response, whereas 
the forced response is a particular integral, and in the case of constant or 
periodic excitation, the forced response at / = oo is the steady-state 
solution. Note that the free response is a function of the network elements 

1 16 Network analysis and synthesis 

alone and is independent of excitation. On the other hand, the forced 
response depends on both the network and the excitation. 

It is significant that, for networks which have only positive elements, 
the free response is made up of only damped exponential and/or sinusoids 
with constant peak amplitudes. In other words, the roots of the charac- 
teristic equation H(p) all have negative or zero real parts. For example, 
if /»! is a root of H(p) written as p x = a ±j(o, then Re (pj = a ^ 0. 
This fact is intuitively reasonable because, if a bounded excitation produces 
a response that is exponentially increasing, then conservation of energy is 
not preserved. This is one of the most important properties of a passive 
network. If a characteristic equation contains only roots whose real parts 
are zero or negative, and if theyo> axis roots are simple, then the network 
which it describes is said to be stable; otherwise, the network is unstable* 
Stability is an important property of passive networks and will be discussed 
in greater detail later. 

Example 5.6, For the R-C network in Fig. 5.19 with the excitation given by 
Eq. 5.25, find the voltage v(t) across the capacitor; it is given that v(0-) - 
p^O-) - 0. 

Solution. We have already obtained the particular integral in Eq. 5.28. 
Now let us find the complementary function. The differential equation on a 
node basis is 

C -r + Gv = I sin mtiKt) (5.51) 


from which we obtain the characteristic equation as 

H(p) -Cp+G (5.52) 

so that vJfl-KT* *® (5-53) 

and the incomplete solution is 

*/) - to-™*) + (G . + ° o)>c ^ w sin {<* - tan"* -g J «(/) (5.54) 

From the initial condition t<0 — ) — 0, we obtain 

r / a>C\ 

.(o^-KO^-Jf-^r^fe^^l^'V)- (555) 

From the argand diagram in Fig. 5.20 we see that, 

♦This is not a formal definition of stability, but it suffices for the moment. 

Network analysis: I 117 






«— o— * 



From the complementary function in Eq. 5.53, we see that the time constant of 
the circuit is T - C\G - RC. 

Next, let us examine an example of the different kinds of free responses 
of a second-order network equation that depend on relative values of the 
network elements. Suppose we are given the network in Fig. 5.21 ; let us 
find the free response v^t) for the differential equation 

' dt LJ*- 

Differentiating both sides of Eq. S.S8, we have 

VJf) - C p"(t) + G »'(0 + £ <0. 
The characteristic equation is then 

H(p) - Cp» + Gp + 1 = c(j>» + 1 P + j^) 
In factored form, H(p) is 

#(p) = G(p-J>i)0>-/>i) 






Network analysis and synthesis 

K 1 + K 2 


FIG. 5.22. Overdamped response. 




2l\c) LCI 



There are three different kinds of responses depending upon whether B is 
real, zero, or imaginary. 

case 1. Bis real, that is, 



a lc 

then the free response is 

vdt) = K ie - U - B)t + K % e- u+BU (5.63) 

which is a sum of damped exponentials. In this case, the response is said 
to be overdamped. An example of an overdamped response is shown in 
Fig. 5.22. 
case 2. 5 = 0, that is, 


\Cl LC 
then p 1 = p g = —A 

so that vc(t) = (^ + K t t)e- At (5.64) 

When B = 0, the response is critically damped, as shown in Fig. 5.23. 

o t 

FIG. 5.23. Critically damped response. 

Network analysis: I 119 

FIG. 5.24. Underdamped response. 

case 3. B is imaginary, that is, 

Letting B = jfl, we have 



»c(0 = e (*i sin fit + K t cos fit) 


In this case, the response is said to be underdamped, and is shown by the 
damped oscillatory curve in Fig. 5.24. 

Example 5.7. In this example we discuss the solution of a set of simultaneous 
network equations. As in the previous examples, we rely upon physical reasoning 
rather than formal mathematical operations to obtain the initial currents and 
voltages as well as the steady-state solutions. In the network of Fig. S.25, the 
switch S is thrown from position 1 to position 2 at t = 0. It is known that 
prior to t = 0, the circuit had been in steady state. We make the idealized 
assumption that the switch closes instantaneously at / = 0. Our task is to 
find «i(0 and /,(/) after the switch position changes. The values of the batteries 
V l and V % are V x = 2 v, V t = 3 v; and the element values are given as 

L = lh, 
C = if, 

R t = 0.5 n 

jig = 2.o n 

The mesh equations for i x {t) and i t (i) after t = are 

V t =Li\{t)+R l i 1 {i)-R 1 i£) 

-vc(0-)= -/Mito + 

l cL k 

^T)dr+{R l +R^i^t) 


120 Network analysis and synthesis 


t.o^o— nnnr^ 

- v 2 ^: hit)) 

FIG. 5.25 

Since Eq. 5.67 contains an integral, we differentiate it to give 
= -R t i\(0 + ^ i#) + (*i + **) i'#) 


Using Eqs. 5.66 and 5.68 as our system of equations, we obtain the character- 
istic equation 

Lp +'*! -Ri 



-+(R 1+ RJp 

- URx + *lp* + (| + *i*l)p + § < 569 > 

Substituting the dement values into H(p), we have 

H(p) - 2.5p* + 4p + 1.5 - 2.5(p + lXp+ 0.6) 
The free responses are then 

The steady-state solutions for the mesh currents are obtained at / » oo by 
considering the circuit from a d-c viewpoint. The inductor is then a short 
circuit and the capacitor is an open circuit; thus we have 




Now let us determine the initial currents and voltages, which, incidentally, have 
the same values at / -■ 0- and t - 0+ because the voltage sources are not 

Network analysis: I 


impulses. Before the switch is thrown at f = 0, the circuit with K x as the voltage 
source was at steady state. Consequently, 

r o (0-)-K 1 «2v 


We next find i\(0+) from Eq. 5.66 at / - 0+. 

V t ~Li\{0+) + 4*i<P+) - RiW+) 
Substituting numerical values into Eq. 5.74, we find 

«' 1 (0+) - 1 amp/sec 
From Eq. 5.68 at / <- 0+, we obtain similarly 



/'i(0+) - 0.2 amp/sec 




With these initial values of i t (t) and i/t), we can quickly arrive at the final 

/i(r) - (0.5*-* - 2.5e-° « + 6) i<0 

i/f) - ( -0.5tf-« + 0.5*-°«) n(f) 

which are plotted in Fig. 5.26. 















r i 





1 1 1 1 1 







2.5 3.0 33 40 43 M 

FIG. 5.26 

122 Network analysis and synthesis 


According to Faraday's law of induction, a current i x flowing in a coil L x 
may induce a current i t in a closed loop containing a second coil L t . The 

sufficient conditions for inducing the 
current i t are: (a) part of the flux Ox in 
the coil L x must be coupled magnetically 
to the coil L a ; (b) the flux O x must be 
changing with time. 

In this section we will analyze circuits 
containing a device made up of two 
magnetically coupled coils known as a 
transformer. In Fig. 5.27, the schematic 
FIG. 5.27. Transformer. of a transformer is given. The L t side 

of the transformer is usually referred to 

as the primary coil and the L t side as the secondary coil. The only distinction 

between primary and secondary is that the energy source is generally at 

the primary side. 

The transformer in Fig. 5.27 is described mathematically by the 


VO-L^ + M^ 
at at 

, x , , du , . di t 

at at 


where M is the mutual inductance associated with the flux linking L t to L t , 
and is related to L x and L 2 by the relationship 

M = k4I x T z (5.78) 

The constant K in Eq. 5.78 is called the coefficient of coupling. It is 
bounded by the limits <, \K\ <, 1. If \K\ = 1, then all of the flux O x in 
coil Lj is linked magnetically to L % . In this case, the transformer is a unity- 
coupled transformer. If K = 0, the coils L x and L s may be regarded as 
two separate coils having no effect upon one another. 

For circuits with transformers, we must establish reference polarities 
for the mutually induced voltages M dijdt. Usually, the references are 
given by small dots painted on the input and output leads of a transformer, 
as shown by the dots on the schematic in Fig. 5.28. The reference dots are 
placed at the time of manufacture according to the procedure outlined 
here. A voltage source v is connected to the primary Lx side of the 
transformer, as shown in Fig. 5.28. A voltmeter is attached on the second- 
ary. At the primary side, the terminal is assigned the dot reference to 

Network analysis: I 123 


FIG. 5.28. An experiment to determine dot references. 

which we connect the positive lead of the voltage source. The dot reference 
is placed on the secondary terminal at which the voltmeter indicates a 
positive voltage. In terms of the primary current i lt the positive voltage at 
the secondary dot is due to the current i t flowing into the dot on the 
primary side. Since the positive voltage at the secondary dot corresponds 
to the current /, flowing into that dot, we can think of the dot references 
in the following way. If both currents are flowing into the dots or away 
from the dots, then the sign of the mutual voltage term Mdijdt is positive. 
When one current flows into a dot and the other away from the second dot, 
the sign of Mdijdt is negative. 

If N t and N t are the number of turns of coils L^ and L % , then the flux 
linkages of Ly and L, are given by N& and iV.O,, respectively. If both i x 
and i t flow into the dots, the sum of flux linkages of the transformer is 

2 $ linkages = N&i + N^ (5.79) 

If, however, one of the currents, for example, i lt flows into a dot, and the 
other ig flows out of the other dot, then 

2 $ linkages = N& - N t Q t (5.8O) 

An important rule governing the behavior of a transformer is that the 
sum of flux linkages is continuous with time. 
The differential equations for the transformer in Fig. 5.29 are 

V u(t) = L t i\(t) + R t i&) + M i'tf) 

Q = M i\(t) + R* i 8 (0 + L, I'M (581) 

FIG. 5.29 

124 Network analysis and synthesis 

Integrating this set of equations between / = 0— and t = 0+ results in 
the determinant 

Za&(0+) - /i(o-)] MM>+) - «o-)] _ Q 

M\ii(f>+) - ix(0-)] LJMP+) - '.(0-)] " 
By evaluating this determinant, we obtain 

(I^L, - M»)[i 1 (0+) - * 1 (0-)][i,(0+) - /,(0-)] - 



If L^Lx > M*, that is, K < 1, then the currents must be continuous at 
t ■- in order for the determinant in Eq. 5.82 to be equal to zero. Thus 

fl (0+) - /,((>-), K<\ (5.84) 


»*(0+) - ig(0-), *<1 


ir««ph» 5.8. For the transformer circuit in Fig. 5.29, L t - 1A, L» - 2h, 
R t -= 3Q, R, - 80, and Jlf - 1A. The excitation is V - 6ii(0- Let us find 
f x (/) and i#), assuming that / x (0-) - //0-) - 0. 

Solution. The differential equations for the circuit are 


6«(0 - i'i(0 + 3i\(0 + /',(<) 
- iVO + 2i'&) + 8i,(0 
The characteristic equation is given by the determinant 

Wj 2^ + 8 

= 0> + 3)(2p + 8)-/>» 
= (^+14/» + 24) 
- (/> + 2)(p + 12) 
Thus the complementary functions are 

i le {t) = K x <r« + A^e-"' 
i Ie (0 = K*r« + JV" m 

To obtain the particular integral, or steady-state solution, we rely upon 
physical reasoning to arrive at 





2 amp 


Network analysis: I 125 

Since the excitation does not contain an impulse (and since I^L, — 
Af 1 jt 0, as we shall see later), we can assume that d(0+) and ^(0+) are 
also zero. Then using Eq. 5.86 we can find i\(0+) and i'J0+). 

6«'"i(0+) + f,(0+) 


o -r»<p+) + 2i',(o+) ' 

Solving, we find i' 1 (0+) - 12 and i'£0+) - 6. With these initial con- 
ditions, we obtain the final solutions of i x (t) and %(/) 

«0 - (-Ir* + *e-«0 «(0 (5 ' 91) 

Suppose, now, I^L, - M», that is, K - 1, then i,(0 and £(/) need not be 
continuous at f — 0. In fact, we will* show that the currents are discon- 
tinuous at r >■ for a unity-coupled transformer. Assuming that K = 1, 
consider the mesh equation for the secondary at i = 0+ 

*t 'i(0+) = -M i'^O-H) - L, .',(0+) (5.92) 

--?[!* i'!(0+) + M i',(0+)] (5.93) 

The mesh equation of the primary side then becomes 

V - R t i t (0+) + [L,. i\(0+) + M iV0+)) = R r h(Q+) -^R, iJ0+) 



We need an additional equation to solve for ^(0+) and /g(0+). This is 
provided by the equation 

£iPi(0+) - i t (0-)} + MM>+) - i^O-)] = (5.95) 

which we obtained from Eq. 5.82. Since i x (0-) = i^0~) = 0, we solve 
Eqs. 5.94 and 5.95 directly to give 

1^(0+) ^3 — - 

R\JL% + R^Li 


wo+) =™- 

R 1 L t + R^ 

Consider the following example. For the transformer in Fig. 5.29 
the element values are 

£i«=4h, L,= lh 

At-8 0, R 9 =*3Q 

M = 2h, K=10v 

126 Network anal/sis and synthesis 

Assuming that the circuit is at steady state before the switch is closed at 
t = 0, let us find h(t) and / 2 (0- The differential equations written on mesh 
basis are 

10-4^ + 8^ + 2^ 
dt dt 

dt dt 

The characteristic equation is 

4(p + 2) 



2p- (p + 3) 

which yields H(p) - 20p + 24 = 20(p + |) 

so that the complementary functions are 

The particular integrals that we obtain by inspection are 

t (r) = ±1 = 12 = 5 



The initial conditions are 

*i(0+) = 

i*(0+) - 


RtLi + R^ 20 


= -1.0 

We then find K x = -0.75 and K t = - 1.0 so that 

ijff) = (-O.lSe- 11 * + 1.25) u(t) 
W) = -er^u{t) 







We see that as t approaches infinity i t (t) -»■ 0, while i t (t) goes its steady- 
state value of 1.25. 

Network analysis: I 


5.1 Write the mesh equations for the network shown. 

» Cl <0-> 

Pi »2<0-) 

I — nnnp— *^\{= — i — nnnp— *~-, 



Q«Wb(« »! 


«te 1 (0-)=J;C2 


►J8 2 

PROB. 5.1 


5.2 Write a set of node equations to solve for the voltage vjt) shown in the 



(T) SJlo = = G> 

»c <0-) 

ii Ci 

*2> »2(0 

PROB. 5.2 

53 The network shown has reached steady state before the switch S is 
opened at t = 0. Determine the initial conditions for the currents 1-fy) and 
4(0 and their derivatives. 

■ o^f 

o vW- 

■^WV — i 


»2WJ ±C 

PROB. 5.3 

5.4 The network shown has reached steady state before the switch moves 
from a to b. Determine the initial conditions for i L (t) and vdt) and their first 
derivatives. Determine also the final values for i L (t) and vc(t). 


Network analysis and synthesis 


10 v 


T *t ^ 


lh_ v c (t) 
o— — 


If == 


PROB. 5.4 

5.5 The network shown has reached steady state before the switch moves 
from a to b. Determine the initial conditions for the voltages v x (,t) and v t (t) 
and their first derivatives. Determine also the final values for v^t) and vjj). 

b t«0 *i vi(t) - J:- - 


a T s 

r i 

= C R 2 * 

v 2 (t) 

PROB. 5.5 

5.6 For the network shown ii{0— ) = 0. 

(a) Find »(/); 0- < / < ». 

(A) Show that v(t) approaches an impulse as G -» 0. 

(c) Find the strength (area) of the impulse. 


PROB. 5.6 

5.7 For the network shown, »(r) - d(t) - tr*u(t) and t<0-) = 4. Find 
and sketch Jbto; 0- < r < ■«. 

Network analysis: I 129 

I v/W- 



— o— 



PROB. 5.7 

5.8 For the network shown, before the switch moves from a to b, steady- 
state conditions prevailed. Find the current /(f). 


PROB. 5.8 

5.9 For the circuit shown, switch S is opened at f = after the circuit had 
been in steady state. Find /(f); 0— <f < ». 

>Tnnp — 

/^ U\ 


PROB. 5.9 

5.10 Find the current /(f) in the network shown when the voltage source is a 
unit impulse. Discuss the three different kinds of impulse response waveforms 
possible depending upon the relative values of R, L, and C. All initial conditions 
are zero at/ —0—. 

130 Network anal/sis and synthesis 


i — nnRp 

Qm *>J 

PROB. 5.10 

5.11 An R-C differentiator circuit is shown in the figure. Find the require- 
ments for the R-C time constant, such that the output voltage v^t) is approxi- 
mately the derivative of the input voltage. 


PROB. 5. 1 1 

5.12 An R-C integrator circuit is shown in the figure. Find the requirements 
for the time constant such that the output t> (f) is approximately the integral of 
the input voltage. 


O" "«) 

ci vo(t) 

PROB. 5.12 

5.13 Find the free response for i(f) in the figure shown for (a) a current 
source; (b) a voltage source. 

Network analysis: I 131 


PROB. 5.13 

5.14 At t = 0, the switch goes from position 1 to 2. Find /(/), given that 
e(t) = e~* sin It. Assume the circuit had been in steady state for t < 0. 

PROB. 5.14 

5.15 For the circuit shown, switch S closes at t = 0. Solve for l(t) when the 
initial conditions arc zero and the voltage source is e(t) = sin (cat + 0). What 
should be in terms of R, C, and m so that the coefficient of the free response 
term is zero? 


I" ^) 

PROB. 5.15 

5.16 For the circuit shown, the switch S moves from a to b at f = 0. Find 
and sketch v^t) for - < / < «. The circuit is in a steady state at f = 0. 

132 Network analysis and synthesis 

l MA— -> ° 


t = 

1 Xs 

PROB. 5.16 

5.17 For the circuit shown, i(r) - 4«r« «(0- Find i<f); - < / < °°. 



JQ^ lf=t= ih 

PROB. 5.17 

5.18 Find the complete solution for i L (t) and vdO in Prob. 5.4. 

5.19 For the circuit shown, the switch is closed at t - 0. Find irff) ana 
ijj) for - < t < oo. Assume zero initial energy. 

10Q lh 


PROB. 5.19 

5.20 For the transformer circuit shown, the switch closes at t •= 0. Find 
ijft) and i 2 (f). Assume zero initial energy. 


PROB. 5.20 

Network analysis: I 133 



► 30 lh<3 1hg>lh 60^02(0 

PROB. 5.21 

5.21 For the transformer circuit shown, the switch S opens at t = 0. Find 
the voltage far) for - < / < °°. Assume zero initial energy. 

chapter 6 

The Laplace transform 


In Chapters 4 and 5, we discussed classical methods for solving differ- 
ential equations. The solutions were obtained directly in the time domain 
since, in the process of solving the differential equation, we deal with 
functions of time at every step. In this chapter, we will use Laplace 
transforms to transform the differential equation to the frequency domain, 
where the independent variable is complex frequency s. It will be shown 
that differentiation and integration in the time domain are transformed 
into algebraic operations. Thus, the solution is obtained by simple 
algebraic operations in the frequency domain. • 

There is a striking analogy between the use of transform methods to 
solve differential equations and the use of logarithms for arithmetic 
operations. Suppose we are given two real numbers a and b. Let us find 
the product c=axb (6.1) 

by means of logarithms. Since the logarithm of a product is the sum of the 
logarithms of the individual terms, we have 

log C = log a X b = log a + log b (6.2) 

so that C = logr 1 (log a + log b) (6.3) 

If a and b were two six-digit numbers, the use of logarithms would probably 
facilitate the calculations, because logarithms transform multiplication 
into addition. 

An analogous process is the use of transform methods to solve integro- 
differential equations. Consider the linear differential equation 

yW)) -/(») (6.4) 


The Laplace transform 135 



equation ^ 





FIG. 6.1. Philosophy of transform methods. 

where /(0 is the forcing function, x(t) is the unknown, and y(a<0) is the 
differential equation. Let us denote the transformation process by 7X-), 
and let s be the frequency variable. When we transform both sides of 
Eq. 6.4, we have 7W*(0)1 = W)] (6.5) 

Since frequency domain functions are given by capital letters, let us write 
Eq-6.5as Y(X(s),s) = F(s) (6.6) 

where X(s) = TftOL *W - T[f(t)\, and Y(X(s), s) is an algebraic 
equation in s. The essence of the transformation process is that differential 
equations in time are changed into algebraic equations in frequency. We 
can then solve Eq. 6.6 algebraically to 
obtain X(s). As a final step, we perform 
an inverse transformation to obtain 







FIG. 6.2. Linear system. 

*(0 - r- 1 [*(*)] (6.7) 

In effecting the transition between the 

time and frequency domains, a table of 

transform pairs MO, X(s)} can be very 

helpful. A diagram outlining the use of transform methods is given 

in Fig. 6.1. Figure 6.2 shows the relationship between excitation and 

response in both the time and frequency domains. 


The Laplace transform of a function of time/(/) is defined as 

CLf (01 = Hs) = {"/(&-" dt (6.8) 

where s is the complex frequency variable 

s = a +ja> (6.9) 

This definition of the Laplace transform is different from the definition 
given in most standard texts, 1 in that the lower limit of integration is 

1 M. E. Van Valkenburg, Network Analysis, 2nd Ed. Prentice-Hall, Englewood Cliffs, 
New Jersey, 1964. 

136 Network analysis and synthesis 

/ = 0— instead of t = 0+. We thus take into account the possibility 
that/(t) may be an impulse or one of its higher derivatives. It is clear that 
C[d(0] = for the 0+ definition, whereas for the 0- definition, C[($(0] = * • 
In the case when no impulses or higher derivatives of impulses are involved, 
it was shown in Chapter 4 that/(0-) =/(0+). Therefore, all of the 
"strong results" resulting from a rigorous treatment of Laplace trans- 
forms* obtained by using t = 0+ as a lower limit also apply for the 0— 

In order for a function to possess a Laplace transform, it must obey the 



\f{t)\e-"dt< oo (6.10) 

for a real, positive a. Note that, for a function to have a Fourier trans- 
form, it must obey the condition 


•/ — i 

|/(0I dt< oo (6.11) 

As a result, a ramp function or a step function will not possess a Fourier 
transform, 3 but will have a Laplace transform because of the added 
convergence factor er a *. However, the function e t% will not even have a 
Laplace transform. In transient problems, the Laplace transform is 
preferred to the Fourier transform, not only because a larger class of 
waveforms have Laplace transforms but also because the Laplace trans- 
form takes directly into account initial conditions at t = 0— because of 
the lower limit of integration in the Laplace transform. In contrast, the 
Fourier transform has limits of integration (— oo, oo), and, in order to take 
into account initial conditions due to a switch closing at t = 0, the forcing 
function must take a form as/(0 u(t) + /(0-) (3(0, where /(0-) repre- 
sents the initial condition. 

The inverse transform C- lr jFT(.0] is 

/(0 = -M FWds (6.12) 

Alt} Jn—iaa 

where a t is a real positive quantity that is greater than the a convergence 
factor in Eq. 6.10. Note that the inverse transform, as defined, involves a 

* D. V. Widder, The Laplace Transform, Princeton University Press, Princeton, 1941. 

* These functions do not possess Fourier transforms in the strict sense, but many 
possess a generalized Fourier transform containing impulses in frequency; see M. J. 
Lighthill, Fourier Analysis and Generalized Functions, Cambridge University Press, New 
York, 1959. 

The Laplace transform 137 

complex integration known as a contour integration. 1 Since it is beyond 
the intended scope of this book to cover contour integration, we will use a 
partial fraction expansion procedure to obtain the inverse transform. To 
find inverse transforms by recognition, we must remember certain basic 
transform pairs and also use a table of Laplace transforms. Two of the 
most basic transform pairs are discussed here. Consider first the transform 
of a unit step function u(t ). 

Example 6.1. /(/)-«(/). 

Next, let us find the transform of an exponential function of time. 
Example 6.2. /(/) ~ e°<«(0- 


F(j)-J e#e-«dt=- 

s —a 



With these two transform pairs, and with the use of the properties of Laplace 
transforms, which we discuss in Section 6.3, we can build up an extensive table 
of transform pairs. 


In this section we will discuss a number of important properties of 
Laplace transforms. Using these properties we will build up a table of 
transforms. To facilitate this task, each property is illustrated by con- 
sidering the transforms of important signal waveforms. First let us 
discuss the linearity property. 


The transform of a finite sum of time functions is the sum of the trans- 
forms of the individual functions, that is 

t[| /,«]= 2 OK')] (615) 

This property follows readily from the definition of the Laplace transform. 
Example 63. /(/) = sin <ot. Expanding sin cat by Euler's identity, we have 

fit) = ^.(e** - r+*) (6.16) 

4 For a lucid treatment, see S. Goldman, Transformation Calculus and Electrical 
Transients, Prentice-Hall, Englewood Cliffs, New Jersey, 1949, Chapter 7. 

1 38 Network anal/sis and synthesis 

The Laplace transform of /(f) is the sum of the transforms of the individual 

cisoidal e ±iat terms. Thus 

1/1 1 \ m 

C[sin cat] = - : — — ) = (6.17) 

2/\* — yto * +./«/ ** + «>* 

Real differentiation 

Given that £[/(f)] = *(*), then 


sF(s)-/(0-) (6.18) 

where /(0-) is the value of /(f) at f = 0-. 
Proof. By definition, 

£[A0]=J*V'7'(0<*f (6.19) 

Integrating Eq. 6.19 by parts, we have 

£[/'(')] = «-7(0 


+ s /(Oe-'df (6.20) 

Since e~" -» as f -*■ oo, and because the integral on the right-hand side is 

£[/"(')] = ^)» we have 

t[f'(t)] = sF(s)-f(0-) (6.21) 

Similarly, we can show for the nth derivative 
c |"£7(0"l = s n f(s) _ s «-i /(0 _) _ S "-V(0-) / ( - u (0-) 


where f 1 "-" is the (« — l)st derivative of /(f) at f = 0-. We thus see 
that differentiating by f in the time domain is equivalent to multiplying 
by s in the complex frequency domain. In addition, the initial conditions 
are taken into account by the terms / (i) (0— ). It is this property that 
transforms differential equations in the time domain to algebraic equations 
in the frequency domain. 

Example 6.4a. /(f) = sin cot. Let us find 

C[cos cot] = C — j sin cot) (6.23) 

By the real differentiation property, we have 

cgsin<of]=,(^) (6.24) 

The Laplace transform 139 

5 (6.25) 

sothat c[cosa ,, ]= i(_^ = _I 

In this example, note that sin <o(0—) = 0. 

Example 6.4b. /(/) = «(/). Let us find £[/'(')], which is the transform of the 
unit impulse. We know that 

Ct«(0] = - (6.26) 


Then C[*0] = *(j) = 1 (6.27) 

since «(0— ) = 0. 

Real integration 

If £[/(*)] = F(s)> then the Laplace transform of the integral of /(f) is 
flfa) divided by s, that is, 

c[£/(r)dr]=^ (6.28) 

Proof. By definition, 

c LX /(t) dr ] -.C e_rt [£- /(T) dT ] d ' (6,29) 

Integrating by parts, we obtain 

c|" P /(t) rfrl = - — f /(t) dr " + - f °° e-'/(f) dt (6.30) 
LJo- J s Jo- o- s Jo- 

Since e~" -*■ as f -»■ oo, and since 

f I 

/(t)<*V =0 (6.31) 

Jo- It-o- 

we then have 

c[£/(r)dr]=^ (6.32) 

Example 6.5. Let us find the transform of the unit ramp function, p(t) " t u(t). 

We know that j h(t) dr = p(f) (6.33) 

Since C[«(0] = - (6.34) 


£[«(')] 1 
then C[p</)] - -^ . _ (6.35) 

140 Network analysis and synthesis 

Differentiation by • 

Differentiation by s in the complex frequency domain corresponds to 
multiplication by t in the domain, that is, 

W)] = -^ (6-36) 


Proof. From the definition of the Laplace transform, we see that 

±1® = f°7(0 ± e-* dt - - f " tf(t)e-* dt - -Z[t f(t)] (6.37) 
ds Jo- ds Jo- 

Example 6.6. Given /(O = g-*\ whose transform is 

m—rrz ( 638 > 

let us find C[te~"']- By the preceding theorem, we find that 

«"-i--s(rh)-<rb •*» 

Similarly, we can show that 

"a-^-crrb* 5 (640) 

where n is a positive integer. 

Complex translation 

By the complex translation property, if F(s) <= £[/(/)], then 

/=X5 - a) = C[e°7(0] < 6 - 41 ) 

where a is a complex number. 
Proof. By definition, 

£1*7(0] -J" ^/(Oe"" dt = J" e" (M) 7(0 dt = F(s - a) (6.42) 
Example 6.7. Given /(t) = sin <»/, find £[«-»' sin <»/]. Since 

. qsinorfl-^^ < 643 > 

by the preceding theorem, we find that 

Similarly, we can show that 

c^coso,.]- ^;^;^ («-45) 

The Laplace transform 141 

Real translation (shifting theorem) 

Here we consider the very important concept of the transform of a 
shifted or delayed function of time. If C[/(/)] = P(s), then the transform 
of the function delayed by time a is 

£[/(' - a) u(t - a)] = e-** F(s) (6.46) 

Proof. By definition, 

C[/(r - a) u{t - a)] = f V*7(r - a) dt (6.47) 

Introducing a new dummy variable t = t — a, we have 

C[/(t) m(t)] - f °° e- lT ^f(r) dr = g— ['°f(r)e- T dr = <f~ F(s) (6.48) 

Jo- Jo- 

in Eq. 6.48/(t)u(t) is the shifted or delayed time function; therefore 
the theorem is proved. 

It is important to recognize that the term e~"' is a time-delay operator. 
If we are given the function e~" G(s), and are required to find 

C-i[r~G(,)]- ft (0, 
we can discard er*' for the moment, find the inverse transform 


and then take into account the time delay by setting 

g(t-a)u(t-a)=g 1 (t) (6.49) 

Example 6.8a. Given the square pulse /(f) in Fig. 6.3, let us first find its trans- 
form F(s). Then let us determine the inverse transform of the square of F(s), 
i.e., let us find 

Mt) = C-i[F«(5)] (6.50) 

Solution. The square pulse is given in terms of step functions as 

/(/) - «(/) -uit-a) (6.51) 

Its Laplace transform is then 


F(s)=\(l-e~«) (6.52) 

Squaring F(s), we obtain 

F\s) - -5 (1 - 2e-» + e"*") (6.53) 

To find the inverse transform of F\s), we 
need only to determine the inverse transform FIG. 4.3. Square pulse. 

142 Network analysis and synthesis 
of the term with zero delay, which is 

C-4-J -fN(<) 



Then C-»[F*(*)] = /«(/)- 2(* - a)u(t - a) +(t - 2a) «(/ - 2a) (6.55) 

so that the resulting waveform is shown in Fig. 6.4. From this example, we see 
that the square of a transformed function does not correspond to the square 
of its inverse transform. 




a 2a 

FIG. 6.4. Triangular pulse. 


K 2 


K 6 




Ti 2T! 3Ti 4Ti 5Ti 6Ti 7Ti 
FIG. 6.5. Impulse train. 

Example 6.8b. In Fig. 6.5, the output of an ideal sampler is shown. It consists 
of a train of impulses 

fit) = *o <K0 + *i <K' ~ T t ) + • " • + *» *' - « r i) ( 656 > 

The Laplace transform of this impulse train is 

C[/(/)l - *o + *!*-•*» + ^- 2,Tl + • • • + *„e-*» r i (6.57) 

In dealing with sampled signals, the substitution z = e ,r » is often used. Then 
we can represent the transform of the impulse train as 

Ai A. A« 

W)]-*. + T 1 + ^ + ■••+■? 


The transform in Eq. 6.58 is called the z-transform of /(/). This transform is 
widely used in connection with sampled-data control systems. 


The Laplace transform of a periodic waveform can be obtained in two 
ways : (a) through summation of an infinite series as illustrated in Example 
6.8d, and (*) through the formula derived below. 

£[/(*)] -} f(tyr"dt +J t f{t)e-'*dt + ■ ■ 


f(t)e-*dt + 



The Laplace transform 143 

Since /(0 is periodic, Eq. 6.59 reduces to 


= \ T f{i)e-'*dt + e- T [ 
Jo- Jo 

dt + e-' T \ f(t)e-*dt + 


+ e- nT f* f(t)e-* dt + ■ ■ ■ 

= (1 + e- T + e- aT + ■■■) \ T f{t)e-*dt 


— ~ \ T f(f)e-'*dt 

l-e- T Jo- J 

Example 6.8c. Given the periodic pulse train in Fig. 6.6 let us use Eq. 6.60 to 
determine its Laplace transform. 

1 f° 



1 1 - €~" 

s 1 - e-' T 



a T T+a 2T 2T+a 

FIG. 6.6. Periodic pulse train. 

Example 6.8d. In this example we calculate the Laplace transform of /(f) in 
Fig. 6.6 using summation of an infinite series. The periodic pulse train can be 
represented as 

/(/) = iKt) - iKt - a) + u(t - T) - u[t - (T + a)] 

+ u(t - IT) - u[t - (2T + a)] + ■ • • {pM) 
Its Laplace transform is 

F(s) = - (1 - *-" + e~ tT - e-<T+«)» + . • .) 

= - [1 - e~" + e~ ,T (l - e-») + *- 2 « T (l - «-") + • • •] 


144 Network analysis and synthesis 
which simplifies further to give 

F(s) = - (1 - «-^"Xl + e- T + e~ UT +••■) (6.64) 


We then see that F(s) can be given in the closed form 

1 1 - e-™ 

Other periodic pulse trains also can be given in closed form. The reader is 
referred to the problems at the end of this chapter. 

At the end of the chapter is a table of Laplace transforms. Most of 
the entries are obtained through simple applications of the properties just 
discussed. It is important to keep those properties in mind, because many 
transform pairs that are not given in the table can be obtained by using 
these properties. For example, let us find the inverse transform of 

F(s) = , j_ ™ , (6-66) 

(s + ay + car 

Since the s in the numerator implies differentiation in the time domain, we 

can write 

F(s)-a " . (6.67) 

(s + ay + co' 

Using the differentiation property, we obtain 

/ dt y (6.68) 

= (—«r sin oat + to cos oot)e~* 
Note that e~" sin mt at t = 0— is zero. 


Evaluation of definite integrals 

The Laplace transform is often useful in the evaluation of definite 
integrals. An obvious example occurs in the evaluation of 


c-**sin5rdr (6.69) 

If we replace tr %% by er**, the integral /then becomes the Laplace transform 
of sin 5/, which is 

Z[ S ux5t] = - r j-- (6.70) 

s + 25 

The Laplace transform 145 
Replacing J by 2, we have 

I = ¥TTs = i (6 - 71) 

Perhaps a more subtle example is the evaluation of 

/ = J + W ,W dt (6.72) 

First, we note that t % er iW is an even function; therefore 

/ = 2jV C - 2 'df (6.73) 

From the table of Laplace transforms we see that 

and the transform of 

/(0 = f tV**- (6.75) 


vm -^tv (<76) 

Later we will see that the partial-fraction expansion of Z[f(t)] in Eq. 6.76 
consists of three terms for the multiple root (s + If, as given by 

pr//A1 0.25 0.25 0.5 1 .-_ 

ELKO] = — " — ~ ^i - j^ (6.77) 

Taking the inverse transform of Eq. 6.77, we obtain 

f{t) = 0.25(1 - e-« - 2te-** - 2rtr-«) 11(f) (6.78) 

Now observe that the definite integral in Eq. 6.73 is equal to 2/(f) at 
t = 1, that is, 

t = 2/(1) = -2.5<r* + 0.5 = 0.162 (6.79) 

Solution of integrodifferential equations 

In Section 6.3 we said that the real differentiation and real integration 
properties of the Laplace transform change differential equations in time 
to algebraic equations in frequency. Let us consider some examples 
using Laplace transforms in solving differential equations. 

Example 6.9. Let us solve the differential equation 

*"(/) + 3x'(t) + Irtf) - 4e* (6.80) 

given the initial conditions a<0-) = 1, x'(Q—) = — 1. 

146 Network analysis and synthesis 

Solution. We first proceed by taking the Laplace transform of the differential 
equation, which then becomes 

[i 2 X(s) -sx(0-)- x'(0 -)] + 3[s X(s) - x(0 -)] + 2X(s) = — - (6.81) 

Substituting the initial conditions into Eq. 6.81 and simplifying, we have 

(** + 3s + : 
We then obtain X(s) explicitly as 

(s* + 3s + 2) X(s) +5+2 (6.82) 

s — 1 

™-jr=mrm+% (6 - 83) 

To find the inverse transform x(t) = f'^W], we expand X(s) into partial 

Solving for X_i, A^i, and K t algebraically, we obtain 

K_ i = § JTi = -1 K t = | 
The final solution is thf inverse transform of X(s) or 

a<0 = fe* - e-' + fe-» (6.85) 

In order to compare the Laplace transform method to the classical method of 
solving differential equations, the reader is referred to the example in Chapter 4, 
Eq. 4.64, where the differential equation in Eq. 6.80 is solved classically. 

Example 6.10. Given the set of simultaneous differential equations 

2x'(f) + 4a<r) + y'(t) + 7y(0 = 5u(t) 

x'(t) + x(t) + y'(t) + 3jK0 = 56X0 

with the initial conditions x(0 -) = y(0 -) = 0, let us find x(j) and y(t). 
Solution. Transforming the set of equations, we obtain 

2(s + 2) X(s) + (* + 7) Y(s) - - 


(s + 1) X(s) + (* + 3) Y(s) = 5 

Solving for X(s) and Y(s) simultaneously, we have 

(5/*)A u + 5^ 




(5/5)A u + 5A* 


The Laplace transform 147 

where A is the determinant of the set of equations in Eq. 6.87, and A tf is the 
zyth cofactor of A. More explicitly, X(s) is 

-5i» - 30s + 15 

™~-*FT2r+sr (689) 

Expanding X(s) in partial fractions, we have 

K x K-s + K z 

^> = 7 + ?T27T5 < 690) 

Multiplying both sides of Eq. 6.90 by s and letting * = 0, we find 

K 1 =sX(s)\,_ =3. 
K t and K s are then obtained from the equation 

3 -8s - 36 

*« "-7+5 + 5 (691) 

A further simplification occurs by completing the square of the denominator 
of X(s), that is, 

s* + 2s + 5 = (s + 1)* + 4 (6.92) 

As a result of Eq. 6.92, we can rewrite X(s) as 

3 8(i + 1) + 14(2) 

*<'>-" fr + iy+gy (693) 

so that the inverse transform is 

«(/) = (3 - 8c-' cos 2/ - 14e-' sin It) «(/) (6.94) 

In similar fashion, we obtain Y(s) as 

Wrt _ * | "* + 17 1 llfr + l)+3(2) 

y(s) = ~s + 7T27T5 - -; + (, + i)* + (2)* (6,95) 

The inverse transform is then seen to be 

y (j) = ( _ i +1 i e -* cos 2f + 3<r* sin It) u(t) (6.96) 

This example is also solved by classical methods in Chapter 4, Eq. 4.163. 
We note one sharp point of contrast. While we had to find the initial 
conditions at t = 0+ in order to solve the differential equations directly 
in the time domain, the Laplace transform method works directly with 
the initial conditions at t = 0— . In addition, we obtain both the comple- 
mentary function and the particular integral in a single operation when we 
use Laplace transforms. These are the reasons why the Laplace transform 
method is so effective in the solution of differential equations. 

148 Network analysis and synthesis 


As we have seen, the ease with which we use transform methods depends 
upon how quickly we are able to obtain the partial-fraction expansion of a 
given transform function. In this section we will elaborate on some simple 
and effective methods for partial-fraction expansions, and we discuss 
procedures for (a) simple roots, (b) complex conjugate roots, and (c) 
multiple roots. 

It should be recalled that if the degree of the numerator is greater or 
equal to the degree of the denominator, we can divide the numerator by 
the denominator such that the remainder can be expanded more easily 
into partial fractions. Consider the following example: 

= m= s » + 3s* + 3s + 2 
D(s) s' + 2s + 2 

Since the degree of N(s) is greater than the degree of D(s), we divide D(s) 
into N(s) to give 

F(s) = 5 + 1- ' (6.98) 

s +2s + 2 

Here we see the remainder term can be easily expanded into partial 
fractions. However, there is no real need at this point because the 
denominator s 2 +^2s + 2 can be written as 

s* + 2s + 2 = (5 + 1)* + 1 (6.99) 

We can then write F(s) as 

so that the inverse transform can be obtained directly from the transform 
tables, namely, 

C-W)] = <5'(0 + 3(0 + er*(sm t - cos i) (6.101) 

From this example, we see that intuition and a knowledge of the trans- 
form table can often save considerable work. Consider some further 
examples in which intuition plays a dominant role. 

Example 6.11. Find the partial-fraction expansion of 

2* + 3 

™ = (JTWTT) (6102) 

The Laplace transform 149 

If we see that F(s) can also be written as 

(s + 1) + (s + 2) 
*W- fr + Dfr+g (6103) 

then the partial-fraction expansion is trivially 

Exanpfe 6-12. Find the partial-fraction expansion of 

s +5 

j + 5 
F ( s ) = , T -, (6-105) 

We see that F(s) can be rewritten as 

m = fe±a±i = _i_ _i_^ (6 106) 

W (s + 2)» ( s + 2) T (s + 2)» 
Real roots 

Now let us discuss some formal methods for partial-fraction expansions. 
First we examine a method for simple real roots. Consider the function 

m = , tt^t: : (6-107) 

(s - s )(s — s0(s — St) 

where s<» s t , and s t are distinct, real roots, and the degree of N(a) < 3. 
Expanding F{s) we have 

F(s) = -£*- + -^- + -^- (6.108) 

S — S S — Si S — Si 

Let us first obtain the constant A,. We proceed by multiplying both sides 
of the equation by (s — j ) to give 

(s — s«)F(s) = K H 1 . (6.109) 

s — «! s — s, 

If we let 5 = * in Eq. 6.109, we obtain 

Ji-(#-J»)J^)|^ (6110) 

Similarly, we see that the other constants can be evaluated through the 
general relation 

A, = (*-*) ft*) |_., (6.111) 

1 6.13. Let us find the partial-fraction expansion for 

FV- J ' +2j - 2 .g» + J^- + J^_ (6.112) 

FW j(*+2X*-3) * + s+2+3-3 K ° Z) 


Network anal/sis and synthesis 


gEq. 6.111, we find 

Ao = sF(s) |_> 

s* + 2s -2 


~ (* + 2X* - 3) 

«-o 3 

s* +2s -2 
Kl ~ sis -3) 


»— a 5 

s 2 + 2s - 2 
*■" s(s+2) 

3 ~15 


Complex roots 

Equation 6.111 is also applicable to a function with complex roots in 
its denominator. Suppose F(s) is given by 

F(s) = 


D&Xs - * - JPK' - * + JP) 




s-x-jfi s-x+jp Djis) 
where A^/Di is the remainder term. Using Eq. 6.111, we have 


K 1 = 



where we assume that j = a ± jP are not zeros of Z>i(s). 

It can be shown that the constants K x and K % associated with conjugate 
roots are themselves conjugate. Therefore, if we denote K± as 


K 2 = A-jB = K 1 * 


If we denote the inverse transform of the complex conjugate terms as 
fx(t), we see that 


i(o = c _i r — — — + — — — l 

= e"'(K ie "» + Kft-"*) 
= 2c**(A cos Bt - B sin pi) 


The Laplace transform 151 

A more convenient way to express the inverse transform f^t) is to intro- 
duce the variables M and <X> denned by the equations 

M sin <1> = 2A 

McosO = -25 

where A and B are the real and imaginary parts of K r in Eq. 6.116. In 
terms of M and <I>, the inverse transform is 

/i(0 = Me?* sin (0t + O) (6. 120) 

To obtain M and $ from K lt we note that 

Me'* = M cos O + jM sin O = -2B + ;2,4 = 2;X X (6.121) 
When related to the original function F(s), we see from Eq. 6.115 that 

Mf = *<«+Jfl (6 . 122) 

Example 6.14. Let us find the inverse transform of 

w-vr^rkr+ii . (6123) 

For the simple root s = —2, the constant K is 

K = (*+2)F(*)|,__ 8 = i (6.124) 

For the complex conjugate roots 

**+2j+5=(* + 1 +j2Xs + 1 -y'2) (6.125) 

We see that a = — 1, = 2, thus 

«* +3 
Me** = 

The inverse transform is then 

= — c -i(tan-»j+W») (6.126) 

i— 1+*2 V5 

C-i[F(*)] = - e -** + — = <r* sin \^ ~\~ tan -1 2j 

7 2 

= - e~ u -p e~* cos (2/ — tan -1 2) 

5 V5 


Multiple roots 

Next let us consider the case in which the partial fraction involves 
repeated or multiple roots. We will examine two methods. The first 
requires differentiation; the second does not. 

152 Network analysis and synthesis 

Method A 

Suppose we are given the function 

F( S ) = ^ (6.128) 

W (s - s )" ^(s) 

with multiple roots of degree n at s = j . The partial fraction expansion 
of F(s) is 

K Kx K t ..... *«-i . *i(?) 

(s-s ) n (s-so)"" 1 (s-s )»- 8 s-s J>i(«) 


where N^jDjUs) represents the remaining terms of the expansion. The 
problem is to obtain K^K^..., K^. For the K^ term, we use the method 
cited earlier for simple roots, that is, 

Ko = (s-s rF(s)\„ t (6.130) 

However, if we were to use the same formula to obtain the factors K u 
K t ,..., #„_!, we would invariably arrive at the indeterminate 0/0 con- 
dition. Instead, let us multiply both sides of Eq. 6.129 by (s — s ) n and 

F 1 (s)s(s-s rF(s) (6.131) 


F x (s) - A, + K t (s - s ) + • • ■ + K^s - So)"" 1 + *(•*)(' - s*Y 


where R(s) indicates the remaining terms. If we differentiate Eq. 6.132 by 
s, we obtain 

4 Fi(s) - *i + 2K^s - s ) + • • • + *„_!(/! - D(s - so)""* + 

It is evident that K x = — F t (s) 


On the same basis K t = ; F^s) 

2 ds 



and in general 

K t = -tWs)! . J = 0,1, 2 n -1 (6.136) 

j\ ds' !«-«•> 

The Laplace transform 153 
Example 6.15. Consider the function 

'W-STTlJi (6 - ,37) 

which we represent-in expanded form as 

™-(7Ti? + (JTI? + 7TT + 7 <«■'»> 

The constant A for the simple root at s = is 

A=sF(s)\^=-2 (6.139) 

To obtain the constants for the multiple roots we first find F t (s). 

Fjis) = (s + l) 8 F(s) = S -^ (6.140) 


Using the general formula for the multiple root expansion, we obtain 

1 d fs -2\ I 2 I „ „,_„,. 

^i=T7T -3 =2 < 6142) 

1 ! ds \ s / !_! «■ |__! 

*-i^)L-(-5)L-; — 

sothat IW __2 gl + ^ i+ -i 1 -5 (6.144) 

Method B 

The second method for arriving at the partial-fraction expansion for 
multiple roots requires no differentiation. It involves a modified power 
series expansion.* Let us consider F(s) and F^s), defined in Eqs. 6.128 
and 6.131 in Method A. We define a new variable/; such that/> = s — Jo- 
Then we can write F^s) as 

*(p + tf-£?T^ (6145) 

Dividing N(p + s ) by D % (p + s ), with both polynomials written in 
ascending powers ofp, we obtain 

fiG* + s ) = K + K 1 p + K t p t +-- + X^j,"" 1 + n f"f 

0l(l> + *o) 


* I am indebted to the late Professor Leonard O. Goldstone of the Polytechnic 
Institute of Brooklyn for showing me this method. 

154 Network anal/sis and synthesis 

The original function F(s) is related to F x (p + s ) by the equation 

F(P + °o)- pn -,„-%»-!+ + , + Dl(p + So) 

Substituting 5 — s = p in Eq. 6.147, we obtain 

F(s) = — £*— + / g \ Bl + • • • + -^ + -£\ (6.148) 

We have thus found the partial-fraction expansion for the multiple-root 
terms. The remaining terms KjlD^s)] still must be expanded into partial 
fractions. Consider Example 6.16. 

Example 6.16 

*w- fr + ijUa (6149) 

Using the method just given, F^s) is 

F^s) - (s + If F(s) = -^ (6.150) 

Setting p = s + 1, we then have 

The expansion of F t (p - 1) into a series as given in Eq. 6.146 requires dividing 
the numerator 2 by the denominator 1 + p, with both numerator and denomi- 
nator arranged in ascending power of p. The division here is 

2 - lp + 2/ 

2 +2/> 


Since the multiplicity of the root is AT = 3, we terminate the division after 
we have three terms in the quotient. We then have 

F X { P -l)=2-2p+2p*- j£- x (6.152) 

The original function F(p — 1) is 

F x (d -1) 2 2 2 2 

F(P ~ 1) =- J 7T- - ? - ? +~ p -pTl (6153) 

The Laplace transform 155 
Substituting s + 1 = />, we have 

m - (7TT? " (7TI? + (7TT) " JT2 (6154) 

Example 6.17. As a second example, consider the function 

F(,) =^T1? (6155) 

Since we have two sets of multiple roots here (at s = 0, and s = —1) we have a 
choice of expanding F^s) about s — Ooi s = — 1. Let us arbitrarily choose to 
expand about * = 0; since /j = s here, we do not have to make any substitutions. 
F&) is then J + 2 2+s 

™=(7TT? = rT2FT7 < 6156 > 

Expanding F t (j), we obtain 

s\3s + 4) 
F^s) - 2 - 3j + ^ + iy . (6.157) 

2 3 3* + 4 
F(s) is then F(s) - ^ - j + + ^ (6.158) 

We must now repeat this process for the term 

3* +4 
(s + 1)* 
Fortunately we see that the term can be written as 

3s +4 3Qr + 1) + 1 3_ 1 

(s + 1)» = (* + iy ~ s + I + (s + 1)» (6159) 

The final answer is then 

"W-l-i+TTT + frTi? (6160) 


In this section we will discuss the many implications of a pole-zero 
description of a given rational function with real coefficients F(s). We 
define the poles of F(s) to be the roots of the denominator of F(s). The 
zeros of F(s) are defined as the roots of the numerator. In the complex s 
plane, a pole is denoted by a small cross, and a zero by a small circle. Thus, 
for the function 

_ ^-i+;D(s-i-Jl) 

(s + l)\s+j2)(s-j2) K 

the poles are at s = — 1 (double) 

s = — y*2 
s = +y*2 

156 Network analysis and synthesis 

jo | 

j 2 i « plane 

Double pole.*. ■ ,1 

.t i — i_W_ 

-<r -3 -2 -1 


i i I I 



->2 x 


FIG. 6.7. Pole-zero diagram of F(s). 
and the zeros are at s = 

s = 1 -j\ 

S = 00 

The poles and zeros of F(s) are shown in Fig. 6.7. 

Now let us consider some pole-zero diagrams corresponding to standard 
signals. For example, the unit step function is given in the complex 
frequency domain as 

C[u(0] = - ( 6162 > 


and has a pole at the origin, as shown in Fig. 6.8a. The exponential 
signal e—*, where <x > 0, has a transform 

C[e— '] — (6.163) 

S + Oq 

which has a single pole at s = -o-* as indicated in Fig. 6.86. The cosine 
function cos <a i, whose transform is 

C[coso, t] = - r - 1 - ~ t (6-164) 

has a zero at the origin and a pair of conjugate poles at s = ±7<u » 
as depicted in Fig. 6.8c. Figure 6.Sd shows the pole-zero diagram 
corresponding to a damped cosine wave, whose transform is 

= s + g ° (6.165) 

Z[e-"* cos co t ] = — r=-* i 

(s + <r r + o> 

I a 


The Laplace transform 



x — 




M, "" ******* 



FIG. 6A Poles and zeros of various functions. 

From these four pole-zero diagrams, we note that the poles corre- 
sponding to decaying exponential waves are on the — a axis and have zero 
imaginary parts. The poles and zeros corresponding to undamped 
sinusoids are on the/o> axis, and have zero real parts. Consequently, the 
poles and zeros for damped sinusoids must have real and imaginar y 
parts that are both nonzero. 

Now let us consider two exponential waves/x(0 = er'i* and/i(f) = er***, 
where a t > <r x > 0, so that/,0) decays faster than/iO), as shown in Fig. 
6.9a. The transforms of the two functions are 

*i(») = 

F&) = 


s + a x 


s + a, 

as depicted by the pole-zero diagram in Fig. 6.96. Note that the further 
the pole is from the origin on the —a axis, the more rapid the exponential 
decay. Now consider two sine waves, sin atjt and sin co t t, where <o t > 
«>! > 0. Their corresponding poles are shown in Fig. 6.10. We note here 





s plane 

*2(«K Fi(»K 

—V2 — 0"1 



FIG. *.*. Effect of pole location upon exponential decay. 

158 Network analysis and synthesis 











( -jwi 

< -j(i>2 






FIG. 6.10. Pole locations corre- 
sponding to sin to,/ and sin mj. 

FIG. 6.1 1 

that the distance from the origin on theyco axis represents frequency of 
oscUlation; the greater the distance, the higher the frequency. 

Using these rules of thumb, let us compare the time responses f x {f) and 
/ a (0 corresponding to the pole pairs {* lf s 1 *} and fo, s a *} shown in Fig. 
6.11. We see that both pairs of poles corresponds to damped sinusoids. 
The damped sinusoid/i(0 has a smaller frequency of oscillation than/ a (f) 
because the imaginary part of * x is less than the imaginary part of s t . Also, 
f z (t) decays more rapidly than f x (t) because Res 1 >Kcs 2 . The time 
responses /x(0 and/ a (0 are shown in Fig. 6.12. 

f 2 (t) - lme«»* 



= lme«»' 




_ __ __ 


— — - 

- = * - 



FIG. 6.12. Time responses for poles in Fig. 6.11. 

The Laplace transform 159 

« plane 


- m 

at a 


M (b) 

FIG. 6.13. Effect of right-half plane poles upon time response. 

Let us examine more closely the effect of the positions of the poles in 
the complex frequency plane upon transient response. We denote a 
pole pt as a complex number p ( = a t +jco i . For a given function F(s) 
with only first-order poles, consider the partial-fraction expansion 

F( s) = _*!L_ + _*!_ + 




s- Po 
The inverse transform of F(s) is 

/(*) = K e»« + K ie *« + •■■ + K n e"* 

= K^"*e im,t + K 1 e' lt e iait +■•■ + K^e*""* 

In/0), we see that if the real part of a pole is positive, that is, a t > 0, then 
the corresponding term in the partial-fraction expansion 

K i e" i *e iatt 

is an exponentially increasing sinusoid, as shown in Fig. 6.13a. We thus 
see that poles in the right half of the s plane (Fig. 6.13*) give rise to 
exponentially increasing transient responses. A system function that has 
poles in the right-half plane is, therefore, unstable. Another unstable 
situation arises if there is a pair of double poles on theyw axis, such as for 
the function 

F( "-(7f^' 


whose pole-zero diagram is shown in Fig. 6.14a. The inverse transform of 
F(s) is 

/(0 = - — sin co t 


160 Network analysis and synthesis 



f(t) m ^- sin (dot 

(a) (b) 

FIG. 6.14. Effect of double zeros on they'd) axis upon time response. 

which is shown in Fig. 6.146. It is apparent that a stable system function 
cannot also have multiple poles on theytu axis. 

Consider the system function H(s) = N(s)lD(s). If we factor the 
numerator and denominator polynomials, we obtain 

H(s) = 

Hg(s - Zq)(s - Zi) • • • (s - g„) 


(S - P )(S - Pi) • • • (S - Pm) 

It is clear that H(s) is completely specified in terms of its poles and zeros 
and an additional constant multiplier #„. From the pole-zero plot of H(s), 
we can obtain a substantial amount of information concerning the 
behavior of the system. As we have seen, we can determine whether the 
system is stable by checking the right-half plane for poles and the jm axis 
for multiple poles. We obtain information concerning its transient 
behavior from the positions of its poles and zeros; and, as we will discuss 
in Section 8.1, the pole-zero diagram also gives us significant.information 
concerning its steady-state (Jm) amplitude and phase response. We thus 
see the importance of a pole-zero description. 

Suppose we are given the poles and zeros of an excitation E(s) and the 
system function H(s). It is clear that the pole-zero diagram of the response 
function R(s) is the superposition of the pole-zero diagrams of H(s) and 
£(j). Consider the system function 


W " (i - ftK» " A) 

and the excitation 

s — p t 


Then the response function is 

The Laplace transform 161 

H( s ) = gpEofr - Zq)(s - «i) 
(s - p )(s - pj(s - p») 


It is clear that R(s) contains the poles and zeros of both H(s) and £(*), 
except in the case where a pole-zero cancellation occurs. As an example, 
let us take the system function 

H(s) = 

2(5 + 1) 

( s + 2+j4)(s + 2-j4) 


whose pole-zero diagram is shown in Fig. 6.15c. It is readily seen from the 
pole-zero diagrams of the excitation signals in Fig. 6.18a that the step 
response of the system has the pole-zero diagram indicated in Fig. 6.156. 
The response to an excitation signal 3 cos It has the pole-zero representa- 
tion of Fig. 6. 1 5c. To specify completely the given function F(s) on a pole- 
zero plot, we indicate the constant multiplier of the numerator on the 
plot itself. 

Let us examine the significance of a pole or zero at the origin. We 
know that dividing a given function H(s) by s corresponds to integrating 
the inverse transform h(t) = C- l [//(*)]. Since the division by s corresponds 
to a pole at the origin, we see that a pole at the origin implies an inte- 
gration in the time domain. Because the inverse transform of H(s) in 
Fig. 6.15a is the impulse response, placing a pole at the origin must 
correspond to the step response of the system. In similar manner, we 
deduce that a zero at the origin corresponds to a differentiation in the 
time domain. Suppose we consider the pole-zero diagram in Fig. 6.15c 
with the zero at the origin removed; then the resulting pole-zero plot is 
the response of the system to an excitation E sin It. Placing the zero at 
the origin must then give the response to an excitation E t cos It, which, of 
















-2 -1 







FIG. 6.15. (a) System function, (jb) Response to unit step excitation, (c) Response to 
excitation e(j) = 3 cos 2i. 

162 Network analysis and synthesis 

course, is true. We therefore conclude that the system function of a 
differentiator must have a zero at the origin; that of an integrator must 
have a pole at the origin. 


Poles and zeros give a powerful graphical description of the behavior 
of a system. We have seen that the poles are the complex frequencies of 
the associated time responses. What role do the zeros play? To answer 
this question, consider the partial fraction expansion 

i s — Si 

where we shall assume that F(s) has only simple poles and no poles at 
s = oo. 
The inverse transform is 

/(0 = fV (6-177) 


It is clear that the time response /(r) not only depends on the complex 
frequencies s ( but also on the constant multipliers K t . These constants K t 
are called residues when they are associated with first-order poles. We will 
show that the zeros as well as the poles play an important part in the 
determination of the residues K { . 

Earlier we discussed a number of different methods for obtaining 
the residues by partial-fraction expansion. Now we consider a graphical 
method whereby the residues are obtained directly from a pole-zero 
diagram. Suppose we are given 

f(s) = 4>( 8 ~ g oX 5 ~ ^ • • • ( s ~ 2 ») ( 6 .i78) 

(s - p )(s - Pi) • • • (s - pj 

where m > n and all the poles are simple. Let us expand F (s) as 

F(s) = -&- + -^ + • • ■ + -£=- (6-179) 

S — Po * — Pi * — Pm 

Our task is to determine the residues A,. We know that 

A (Pi — Zq)(p< - zi) • ' ' (Pi - z «) 

^ = (s-ft)^(s) 

(Pi ~ Po)-" (Pi - Pi-^iPi - ft+l) - * * (Pi - Pm) 


The Laplace transform 163 

When we interpret the Eq. 6.180 from a 
complex-plane viewpoint, we see that each one 
of the terms (p ( — z f ) represents a vector drawn 
from a zero z t to the pole in question, p t . 
Similarly, the terms (p f — />„), where i ^ k, 
represent vectors from the other poles to the 
pole p^ In other words, the residue K t of any 
pole p t is equal to the ratio of the product of 
the vectors from the zeros to p t , to the prod- 
uct of the vectors from the other poles to p { . 
To illustrate this idea, let us consider the pole- 
zero plot of 

Ms - zp)(s - Zl) 









F(s) = 

(S - p )(s - p x )(s - ft*) 

FIG. 6.16. Poles and zeros 
of F(s). 


given in Fig. 6.16. The partial-fraction expansion of F(s) is 

F(s) - 


s — Pt, s — p t 

Kl I ^1 



where the asterisk denotes complex conjugate. Let us find the residues A" 
and K x by means of the graphical method described. First we evaluate K x 
by drawing vectors from the poles and zeros to/^, as shown in Fig. 6.17a. 
The residue K x is then 

K _ XqAB 
1 CD 


where symbols in boldface represent vectors. We know that the residue 
of the conjugate pole^* is simply the conjugate of K x in Eq. 6.183. Next, 

FIG. 6.17. Determining residues by vector method. 

164 Network analysis and synthesis 



-,— r 


to evaluate K& we draw vectors from the 
poles and zeros to /> , as indicated in Fig. 
6.176. We see that 

K - A ° RL 



i With the use of a ruler and a protractor, 

we determine the lengths and the angles of 
the vectors so that the residues can be deter- 

-jl mined quickly and easily. Consider the fol- 
lowing example. The pole-zero plot of 

F(s) 55 _ 

FIG. «.I8. Pole-zero diagram of- (« + 2)(s + 1 - jl)(« + 1 + jl) 

F(s). (6.185) 

is shown in Fig. 6.18. The partial-fraction expansion of F(s) is 

F(s) = 


s + 1 - jl "*" s + 1 + jl s + 2 


First let us evaluate K x . The phasors from the poles and zeros to the pole 
at —1 +jl are shown in Fig. 6.19a. We see that K t is 

K x = 3 X -p 

^2/135° _3 

J2 1*5°_ X 2/90° 2 













r— — — 



(a) (b) 

FIG. 6.19. Evaluation of residues of F(s). 

The Laplace transform 165 
From Fig. 6.19* we find the value of the residue K t to be 

K ^ _ 3 x 2_ = _ 3 

,=! V2 /-13S° x 72 7+135° 

Therefore, the partial-fraction expansion of F(s) is 

F(s) = - + - 1— (6.187) 

s + l-jl s + l+jl s + 2 


In this section we will discuss two very useful theorems of Laplace 
transforms. The first is the initial value theorem. It relates the initial 
value of /(/) at t = 0+ to the limiting value of sF(s) as s approaches 
infinity, that is, 

lim f(t) = lim s F(s) (6.188) 

The only restriction is that /(f) must be continuous or contain, at most, a 
step discontinuity at / == 0. In terms of the transform, P(s) = t[f(t)]; 
this restriction implies that F(s) must be a. proper fraction, i.e., the degree 
of the denominator polynomial of F(s) must be greater than the degree of 
the numerator ofF(s). Now consider the proof of the initial value theorem, 
which we give in two parts. 

(a) The function /(f) is continuous at t = 0, that is,/(0— ) =/(0+). 
From the relationship 

Cf/'(0I =/ V(0«r" dt = s F(s) - /(0-) (6.189) 

we obtain hm C[/'(01 - lim s F(s) - /(0-) = (6.190) 

»-»oo •-»» 

Therefore Um s F(s) = /(0-) = /(0+) (6.191) 


(b) The function/(0 has a step discontinuity at / = 0. Let us represent 
f{t) in terms of a continuous part/ x (0 and a step discontinuity D «(/), 
as shown in Fig. 6.20. We can then write /(/) as 

/(0-/i(0 + J>«W (6^2) 

where D =/(0+) — /(0-). The derivative of /(f) is 

f'{t)=f\(f) + Do{t) (6.193) 

1 66 Network anal/sis and synthesis 





FIG. 6.20. Decomposition of a discontinuous function into a continuous function 
plus a step function. 

Since f x (t) is continuous at t = 0, we know from part (a) that 

lim s F^s) = A(0-) = /(0-) (6.194) 

Taking the Laplace transform of both sides of Eq. 6.193, we have 

s F(s) - /(0-) = s F^s) - M0-) + D (6.195) 

which simplifies to s F(s) <= s F t (s) + D (6.196) 

Now, if we take the limit ofEq. 6.196 as s-+ oo and let/(0+) -/(0-) = 

D we have 

lim s F(s) = lim s F x (s) + /(0+) - /(0-) (6.197) 

By Eq. 6.194, we then obtain 

limsF(s)=/(0+) (6.198) 

Example 6.18. Given the function 

2(s + 1) 

m = 

s* +2s + 5 


let us find/(0 -f ). Since the degree of the denominator is greater than the degree 
of the numerator of F(s), the initial value theorem applies. Thus 

lim s F(s) — lim 

2(s + \)s 

C-H^Cj)] =2c-*cos2r 

= 2 



we see that /(0+) =2. 

Example 6.19. Now let us consider a case where the initial value theorem does 
not apply. Given the function 

f(t) = (5(0 + 3r* (6.202) 

we see that/(0+) = 3. The transform of/(f) is 


F(s) = 1 + 

s + 1 


The Laplace transform 167 
so that lim sF(s) = lim si 1 + — — - = oo I (6.204) 

*-*«, .-co \ s + * / 

Next we consider the final value theorem, which states that 

lim /(0 = lim s F(s) (6.205) 

provided the poles of the denominator of F(s) have negative or zero real 
parts, i.e., the poles of F(s) must not be in the right half of the complex- 
frequency plane. The proof is quite simple. First 



f'(t)e-' dt = s F(s) - /(0-) (6.206) 

Taking the limit as s ->■ in the Eq. 6.206, we have 

/'(«) dt = lim s F(s) - /(0-) (6.207) 


Evaluating the integral, we obtain 

/(oo) - /(0-) - hm 5 F(s) - /(0-) (6 .208) 

Consequently, /(oo) = lim s F(s) (6.209) 

Example 6.20. Given the function 

/(/) = 3«(/) + 2e~* (6.210) 

which has the transform 

F(s) = l + » = ii±i (6 . 211) 

s s + 1 *(s + l) 

let us find the final value/( oo). Since the poles of F(s) are at s =0 and s = — 1 , 
we see the final value theorem applies. We find that 

]imsF(s)=3 (6.212) 

which is the final value of /(/) as seen from Eq. 6.210. 

Example 6.21. Given f(t) = le* (6.213) 

we see that lim/(f) = oo (6.214) 

But from sF(s) = ; (6.215) 

s — 1 

we have lim * F(s) = (6.216) 


We see that the final value theorem does not apply in this case, because the pole 
s = 1 is in the right half of the complex-frequency plane. 

168 Network analysis 

and synthesis 

TABLE 6.1 

Laplace Transforms 



1. fit) 


2. «i/i(0 + «./««) 


3. |/W 

»fW -/(0-) 

d n 

4- ^/<» 

a" Fis) - J f^'f^O-) 

5. f/W rfT 


6. f f fir)drda 
Jo- Jo- 


7. (-0"/W 

8. fit -a)u(t - a) 


9. **/X0 

F(s - a) 

10. 8(t) 


d n 
12. «(/) 


13. / 


14. -r 








sin cof 

( e -«« _ e -ft) 


* + a 


The Laplace transform 169 

TABLE 6.1 (cont.) 


cos cor 

5» + <0» 



sinh at 



cosh a/ 




£ "'sin or 

(* + «*) + CO 1 

*""■' cos mt 

(* + «) 


(5 + «)* + co» 




(j + a)**" 1 

t . 

— sin co/ 




0* + <*»»)» 



—/„(«/); n 

(5* + a«)W[(i» + a»)H - 5]-» 

(Bessel function of first kind, 
nth order) 





r* (A: need not be an integer) 

T(* + 1) 


6.1 Find the Laplace transforms of 
(a) /(0-sin(orf+|8) 

(6) /(/) = *-<*♦«> cos (at + P) 

(c) f(t) .- (r» + 1>-" 

(</) /(/) - JT*; A" is a real constant >1. 

(*) /« - (** + «- 0^** 

(/) f«)-td'(t) 

(g) /W - aW* - 4)] 

6 J, Find the Laplace transforms of 
(a) /(/) - cos (t - */4) u(f - w /4) 

(6) /(0-cos(/-W4)«(0 

170 Network analysis and synthesis 
6.3 Find the Laplace transforms for the waveforms shown. 


2 4 t 






3 t 

6.4 Find the Laplace transforms for the derivatives of the waveforms in 
Prob. 6.3. 

6.5 Find the inverse transforms for 


F(s) = 

2s +9 
(s + 3X* + 4) 

5s - 12 
{S) " s* + 4s + 13 

4* + 13 

Note that all the inverse transforms may be obtained without resorting to 
normal procedures for partial-fraction expansions. 

The Laplace transform 171 

6.6 Find the Laplace transform of the waveforms shown. Use (a) the infinite 
series method and (6) the Laplace transform formula for periodic waveforms. 
Both answers should be in closed form and agree. 










T 2T 3T 







2 > 

T \ 






(e ) PROB. 6.6 

Evaluate the definite integrals 


t sin It dt 

t % cos 3f dt 


6.8 Solve the following differential equations using Laplace transforms 

(a) x'{t) + fceXO + 9x(t) = cos 2t 

(b) x'(t) + 3x-(/) + 2a<0 = (5(0 

(c) x"{t) + 2^(0 + 5a<0 = u(t) 

id) x"{f) + Sx'(t) + MO = {e~* + <r**) «(0 

(e) sj-CO + 2as'(0 = «(0 + *~* «(0 

It is given that a<0— ) = as'fl)— ) = for all the equations. 

6.9 Using Laplace transforms, solve the following sets of simultaneous 

(«) 2*'(0 + 4a<0 + y'(f) + 2y(t) = <J(0 

x'(0 + 3a</) + y'(f) + 2y(r) = 
(b) 2x'{t) + «(/) + y'{t) + 4jK/) = 2e-« 

*'(0 + jr*0) + 8y(0 = (5(0 
The initial conditions are all zero at t = 0— . 

172 Network anal/sis and synthesis 

6.10 Find the inverse transforms for 




(s* + 9X' + 3) 

j + a 



s + l 
s*+2s +2 
*» + 3* + 1 

s* +s 

. s + 5 

ns) ~(s + W* + 2)» 



(* + lX*+2)* 
3j» - J* - 3s + 2 

6.11 Find the inverse transforms for 

1 +«-*' 


F(') = 
*"(*) = 

se ->* _ «r-s« 

** + 3s + 2 
s +0 

6.12 Given the pole-zero diagrams shown in (a) and (b) of the figure, write 
the rational functions F t (s) and Fjs) as quotients of polynomials. What can 
you say about the relationship between the poles in the right-half plane and the 
signs of the coefficients of the denominator polynomials? 






ju Fzb) 


A * 




+1 * 



-A * 


PROS. 6.12 

The Laplace transform 173 

6.13 For the pole-zero plots in Prob. 6.12, find the residues of the poles by 
the vector method. 

6.14 For the pole-zero plots shown in the figure, find the residues of the 

j2 : 

jl< < 

-2 -1 


f -H 


-<r-3 -2 -1 







PROB. «. 14 


6.15 Two response transforms, R^s) and R/s), have the pole-zero plots 
shown in (a) and (b) of the figures, respectively. In addition, it is known that 
»i(0 +) - 4 and /-,(/) - 4 for very large t. Find r^t) and rjf). 

-V -2 -l 




X — — 



6 i 


-1? ' 

: o 




PROB. 4.15 

174 Network analysis and synthesis 

6.16 It is given that 

_. . as* + bs + c 

F(s) = 

s* + 2j* +s + 1 

Find a, b, and c such that/(0 +) =/'(0 +) = /'(0 +) = 1 . 

6.17 Using the initial and final value theorems where they apply, find/(0 +) 
and/(oo) for 

(6) F(*) - j 

(c) FW - 

<« ^-"cr+ixr+Q 


+ 3X* 

+ 4) 


+ IX* - 1) 

s* + 3s + 2 

«* + 3* 2 + 3j + 1 
5<s + 4X* + 8) 

chapter 7 

Transform methods in 
network analysis 


In Chapter 5 we discussed the voltage-current relationships of network 
elements in the time domain. These basic relationships may also be 
represented in the complex-frequency domain. Ideal energy sources, for 
example, which were given in time domain as v(t) and i(f)> may now be 
represented by their transforms V(s) = £[t<01 and I(s) = Cfl(f)]. The 
resistor, defined by the v-i relationship 

»(0 = *«(') ( 71 ) 

is denned in the frequency domain by the transform of Eq. 7.1, or 

V(s) = RI(s) (7.2) 

For an inductor, the defining v-i relationships are 

■w- 1 ! 

»(0 = 7p t<T)dT-M(0-) 


Transforming both equations, we obtain 

F(s) = sLI(s) - Li(0-) 

1 i(O-) CM) 

/(5) = J-F( S ) + ^— ; 
sL s 


176 Network analysis and synthesis 


1 g f\HO-) 




FIG. 7.1. Inductor. 

The transformed circuit representation for an inductor is depicted in 
Fig. 7.1. For a capacitor, the defining equations are 

v(t) = - I ' i(r) dr + v(0- 
C Jo- 


*>- c 5 


The frequency domain counterparts of these equations are then 

F« -J; /« + *&=> 

sC s (7. 6 ) 

I(s) = sC V(s) - C o(0-) 
as depicted in Fig. 7.2. 

From this analysis we see that in the complex-frequency representation 
the network elements can be represented as impedances and admittances in 
series or parallel with energy sources. For example, from Eq. 7.4, we see 
that the complex-frequency impedance representation of an inductor is sL, 
and its associated admittance is XjsL. Similarly, the impedance of a 
capacitor is 1/sC and its admittance is sC. This fact is very useful in circuit 
analysis. Working from a transformed circuit diagram, we can write 
mesh and node equations on an impedance or admittance basis directly. 








=P C (l) V <» 

FIG. 7.2. Capacitor. 

Transform methods in network analysis 177 



FIG. 7.3 

The process of solving network differential equations with the use of 
transform methods has been given in Chapter 6. To analyze the circuit 
on a transform basis, the only additional step required is to represent all 
the network elements in terms of complex impedances or admittances with 
associated initial energy sources. 

Consider the example of the transformer in Fig. 7.3. If we write the 
defining equations of the transformer directly in the time domain, we have 



Transforming this set of equations, we obtain 

V&) = sLy I^s) - Ly i^O-) + sM I£s) - M ijp-) 

V&) = sMIM - M ^(0-) + sL, Us) - U i,(0-) 

This set of transform equations could also have been obtained by repre- 
senting the circuit in Fig. 7.3 by its transformed equivalent given in Fig. 7.4. 
In general, the use of transformed equivalent circuits is considered an 
easier way to solve the problem. 

v 2 W 

FIG. 7.4 

178 Network analysis and synthesis 


-nnnp — 1 




B C (0-) 

FIG. 7.5 

Example 7.1. In Fig. 7.5, the switch is thrown from position 1 to 2 at t = 0. 
Assuming there is no coupling between L^ and Z*, let us write the mesh equations 
from the transformed equivalent circuit in Fig. 7.6. The mesh equations are 

+ A'i,(0-) = -^/ 1 (5) + 

■*->■• — -l — £♦-.+.)« 




1 — 'm^Qi 



— ^ 1 vuo' — ^>— 

L1M0-) 1*1140-) 

FIG. 7.6 

Example 7.2. In Fig. 7.7, the switch is thrown from position 1 to 2 at time 
f = 0. Just before the switch is thrown, the initial conditions are /^(O— ) = 2 
amp, »o(0— ) = 2 v. Let us find the current i(t) after the switching action. 




>' 1 -—x 

FIG. 7.7 

Transform methods in network anal/sis 



3 » 

i—vw — nnnp- 

^ f /i«\ 

FIG. 7.8 

Since the switch is closed at / = 0, we can regard the S-v battery as an equiva- 
lent transformed source 5/s. The circuit is now redrawn in Fig. 7.8 as a trans- 
formed circuit. The mesh equation for the circuit in Fig. 7.8 is 

5 2 

+ 2 

s s 

Solving for I(s), we have 

(3 +s+ ?) 



2s + 3 



(j + IX* + 2) s + 1 s + 2 

i(0 = C-HAO] = e~* + «-» 

Examine 73. Consider the network in Fig. 7.9. At t = 0, the switch is opened. 
Let us find the node voltages vjCf) and v % (t) for the circuit. It is given that 

Z,=Jh C = lf 

G = 1 mho V = 1 v 



C=fco c (0-) 

FIG. 7.9 

Before we substitute element values, let us write the node equations for the 
transformed circuit in Fig. 7.10. These are 

NodeV 1 : 

(l(0-0 / 1 \ 1 

_ *_ + CPc( o-) = (sc + -)k i( o -- v M 

Node V t : (7.13) 

^-4™ ♦(£♦«)•» 


Network anal/sis and synthesis 

V 2 


1 - 

Krf,) "" s* + 25 + 2 
so that the inverse transforms are 

(s + 1)» + 1 

Vl (t) = e -« cos /, »*(/) - <r*(cos t + sin /) 


FIG. 7.10 

If we assume that prior to the switch opening the circuit had been in steady 
state, then we have v^O-) = 1 v, (t(0— ) = 1 amp. Substituting numerical 
values into the set of node equations, we have 

(*+^K 1 (i)-?K/f) 

Simplifying these equations, we obtain 

s - 1 = (s* + 2)V t (s) - 2V&) 

1 . -2V x {s) + (s + 2)V£s) 

Solving these equations simultaneously, we have 

s +1 J+l 

KlW ." s» + 2s + 2 " (s + l) 1 + 1 

s +2 * +2 





In network analysis, the objective of a problem is often to determine a 
single branch current through a given element or & single voltage across 
an element. In problems of this kind, it is generally not practicable to 
write a complete set of mesh or node equations and to solve a system of 
equations for this one current or voltage. It is then convenient to use two 
very important theorems on equivalent circuits, known as Thevenin's 
and Norton's theorems. 

Transform methods in network anal/sis 


n voltage sources 
in current sources 




FIG. 7.1 1 

Thevenin's theorem 

From the standpoint of determining the current I(s) through an element 
of impedance Z^s), shown in Fig. 7.1 1, the rest of the network N can be 
replaced by an equivalent impedance Z e (s) in series with an equivalent 
voltage source V£s), as depicted in Fig. 7.12. The equivalent impedance 
Z t (s) is the impedance "looking into" N from the terminals of Z t (s) when 
all voltage sources in JV are short circuited and all current sources are 
open circuited. The equivalent voltage source V,(s) is the voltage which 
appears between the terminals 1 and 2 in Fig. 7.1 1, when the element Z^s) 
is removed or open circuited. The only requirement for Thevenin's 
theorem is that the elements in Z x must not be magnetically coupled to any 
element in N. 

The proof follows. The network in Fig. 7.11 contains n voltage and m 
current sources. We are to find the current I(s) through an element that 
is not magnetically coupled to the rest of the circuit, and whose impedance 
is Z x (s). According to the compensation theorem, 1 we can replace Z^s) 
by a voltage source V(s), as shown in Fig. 7.13. Then by the superposition 
principle, we can think of the current /(*) as the sum of two separate parts 

Let the current I t (s) be the current due to the n voltage and m current 







FIG. 7. IX Thevenin's equivalent circuit. 

1 See H. H. Skilling, Electrical Engineering Circuits, 2nd Ed. John Wiley and Sons, 
New York, 1965. 

182 Network analysis and synthesis 

sources alone; i.e., we short circuit the source V(s), as shown in Fig. 7.14a. 
Therefore I^s) is equal to the short-circuit current 7 gc . Let IJs) be the 
current due to the voltage source V(s) alone, with the rest of the voltage 
sources short circuited and current sources open circuited (Fig. 7.146). 

With the m current and n voltage 
sources removed in Fig. 7.146, we see 
that the network N is passive so that 
7 a C0 is related to the source V(s) by the 



it voltage sources 

m current sources 


FIG. 7. 



««) = - 




where Zm(s) is the input impedance of the circuit at the terminals of the 
source V(s). We can now write /(*) in Eq. 7.18 as 

/(s) = Ms) - 



Since Eq. 7.20 must be satisfied in all cases, consider the particular case 
when we open circuit the branch containing Z x (j). Then /($) = and V(s) 
is the open-circuit voltage Kocfa). From Eq. 7.20 we have 

he(s) = 

F oc (s) 

Z ln (s) 


Transform methods in network analysis 183 

FIG. 7.14b 

so that we can rewrite Eq. 7.20 as 

Z la (s)I(s) - KocCs) = 



In order to obtain the current I(s) through Z x (s), the rest of the network 
N can be replaced by an equivalent source V e (s) = Voc(s) in series, 
with an equivalent impedance Z e (s) = Zm(s), as shown in Fig. 7.12. 

Example 7.4. Let us determine by Thevenin's theorem the current I^s) flowing 
through the capacitor in the network shown in Fig. 7.1S. First, let us obtain 



FIG. 7.15 

Z,(s) by opening all current sources and short circuiting all voltage sources. 
Then, we have in the network in Fig. 7.16, where 

Z,(s) =R+sL 


Next we find V e (s) by removing the capacitor so that the open-circuit voltage 
between the terminals 1 and 2 is V,(s), as shown in Fig. 7.17. We readily 

184 Network analysis and synthesis 




— o2 

FIG. 7.16 

determine from Fig. 7.17 that 

F,Cs) -/(*)*+ Z/(0-)- 


By Thevenin's theorem we then have 

VJto 7(j)Jt + Lt(0 -) -Pq(0-)/s 

/l(j) ~ Z.{s) + Z(s) = A+sL+ 1/sC 

FIG. 7.17 



Fvampk 7.5. For the network in Fig. 7.18, let us determine the voltage vjt) 
across the resistor by Thevenin's theorem. The switch closes at t - 0, and we 
assume that all initial conditions are zero at t = 0. 

8 Li Lj 




FIG. 7.IS 

First let us redraw the circuit in terms of its transformed representation, 
which is given in Fig. 7.19. We can almost determine by inspection that the 
Thevenin equivalent voltage source of the network to the left of AT in Fig. 7.19 is 


LiS + \jsC 


Transform methods in network analysis 

Lis La» ■*— I 

* Vo« 


FIG. 7.19 

and the input impedance to the left of AT is 


We know that 




" ZJis) + R 

(R + sL^sLi + lIsC) + LJC 

»o(0 = e-HKtCi)] 




Norton's theorem 

When it is required to find the voltage across an element whose admit- 
tance is Yifs), the rest of the network can be represented as an equivalent 
admittance Y£s) in parallel with an equivalent current source I,(s), as 
shown in Fig. 7.20. The admittance Y t (s) is the reciprocal of the Thevenin 
impedance. The current I,(s) is that current which flows through a short 
circuit across Y^s). From Fig. 7.20, 

V x {s) = 



Y t (s) + YM 

The element whose admittance is Y t must not be magnetically coupled to 
any element in the rest of the network. 

Consider the network in Fig. 7.21. Let us find the voltage across the 
capacitor by Norton's theorem. First, the short-circuit current source 
I e (s) is found by placing a short circuit across the terminals 1 and 2 of the 





FIG. 7.20 

186 Network anal/sis and synthesis 


-nppp> Pi 



' '««) 

FIG. 7.21 FIG. 7.22 

capacitor, as shown in Fig. 7.22. From Fig. 7.22 l£s) is 

e sL+R 

The admittance Y,(s) is the reciprocal of the Thevenin impedance, or 

' w sL+R 
Then the voltage across the capacitor can be given as 



V(s) = 


Y t (s) + Yds) (sL + R)sC + 1 

Example 7.6. In the network in Fig. 7.23, the switch closes at / = 0. It is given 
that v(t) "■ O.le -6 ' and all initial currents and voltages are zero. Let us find the 
current /g(/) by Norton's theorem. 

The transformed circuit is given in Fig. 7.24. To find the Norton equivalent 
current source, we short circuit points 1 and 2 in the network shown. Then 
I e (s) is the current flowing in the short circuit, or 

Us) = 




R t + sL Us + RILXs + 5) (s + S)(s + 10) 
The equivalent admittance of the circuit as viewed from points 1 and 2 is 
1 s*LC + sR x C + 1 0.5s* + 5s + 10 


Y,(s) - sC + 

J?! +sL 

R ± +sL 

* + 10 




FIG. 7.23 

Transform methods in network analysis 187 

I^s) is then 


FIG. 7.24 



( S + sy(s + 6) 

By inspection, we see that /«(«) can be written as 
(* + 6) - (s + 5) 



(s + 5)V +6) (s + 5)* (* + 6X* + 5) 
Repeating this procedure, we then obtain 


1 1 


(*+5)» j+5 j+6 
Taking the inverse transform of IJs), we finally obtain 
«0 = ('«"" - «"" + e"*')«(0 





As we discussed earlier, a linear system is one in which the excitation 
e(t) is related to the response tit) by a linear differential equation. When 
the Laplace transform is used in describing the system, the relation between 
the excitation E(s) and the response R(s) is an algebraic one. In particular, 
when we discuss initially inert systems, the excitation and response are 
related by the system function H(s) as given the relation 

R(s) = E(s)H(s) (7.42) 

We will discuss how a system function is obtained for a given network, 

and how this function can be used in determining the system response. 

As mentioned in Chapter 1, the system function may assume many forms 

and may have special names such as driving-point admittance, transfer 

188 Network analysis and synthesis 



a?-r ^ 


FIG. 7.25 

impedance, voltage or current-ratio transfer function. This is because the 
form of the system function depends on whether the excitation is a voltage 
or current source, and whether the response is a specified current or 
voltage. We now discuss some specific forms of system functions. 


When the excitation is a current source and the response is a voltage, 
then the system function is an impedance. When both excitation and 
response are measured between the same pair of terminals, we have a 
driving-point impedance. An example of a driving-point impedance is 
given in Fig. 7.25, where 

m = m = R+ «I?22L (7.43) 

HKS) I,(s) sL+l/sC 


When the excitation is a voltage source and the response is a current, 
H(s) is an admittance. In Fig. 7.26, the transfer admittance IJV, is 
obtained from the network as 

H(s)= m = — l_ — 


Voltage-ratio transfer function 

When the excitation is a voltage source and the response is also a voltage, 
then H(s) is a voltage-ratio transfer function. In Fig. 7.27, the voltage- 
ratio transfer function V (s)lV t (s) is obtained as follows. We first find the 

sL *°t fit 

i— nnnp — 1( — wv- 





FIG. 7.26 

Transform methods in network anal/sis 189 


)y t M 





FIG. 7.27 



V a (s) 

I{s)sa Z x {s)+Z t (s) 

VJs) = Z&)I(s) 
V (s) ZJs) 

V g (s) ZM + Z t (s) 


Current-ratio transfer function 

When the excitation is a current source and the response is another 
current in the network, then H(s) is called a current-ratio transfer function. 
As an example, let us find the ratio IJI a for the network given in Fig. 7.28. 
Referring to the depicted network, we know that 

I„(s) = IM + /„(s), 


Eliminating the variable I lt we find 

(7.48, 7.49) 


so that the current-ratio transfer function is 


Jofr) _ 

I„(s) R + sL+1/sC 


7 ' W Q) 4=ic 


FIG. 7.28 

190 Network anal/sis and synthesis 

FIG. 7.29 

From the preceding examples, we have seen that the system function is a 
function of the elements of the network alone, and is obtained from the 
network by a straightforward application of Kirchhoff's laws. Now let us 
obtain the response transform R(s), given the excitation and the system 
function. Consider the network in Fig. 7.29, where the excitation is the 
current source i„(t) and the response is the voltage v(t). We assume that 
the network is initially inert when the switch is closed at t = 0. Let us find 
the response V(s) for the excitations: 

1- i„(t) = (sin a>ot)u(i). 

2. i„(0 is the square pulse in Fig. 7.30. 

3. i,(0 has the waveform in Fig. 7.31. 
First, we obtain the system function as 

1 s 

H(S) sC + 1/sL + G C[s 8 + s(G/C) + 1/LC] 
1. i t (t) - (sin ou f) u(t). The transform of i„(t) is 

so that V(s) = /„(s)H(s) = 


s* + a> » C[s* + s(GIC) + 1/LC] 













FIG. 7.31 

Transform methods in network anal/sis 191 

2. For the square pulse in Fig. 7.30, i g {t) can be written as 

i„(0 = u(t) - u(t - a) (7.55) 

Its transform is j,( s ) = i (i _ e -«) (7.56) 


The response V(s) is therefore given as 

1 — p -0 * 

V(s) = — - (7 57) 

C[s* + s(G/C) + 1/LC] l ; 

Note that in obtaining the inverse transform C -1 [F(j)], the factor e-°* 
must be regarded as only a delay factor in the time domain. Suppose we 
rewrite V(s) in Eq. 7.57 as 

K(s)-^(l-0 (7.58) 


Then if we denote the inverse transform by v^t), 

*<0 = r{^] (7.59) 

we obtain the time response 

v(t) <= Vl (t) - Vl (t - a) u{t - a) (7.60) 

Observe that » x (0 in Eq. 7.59 is the response of the system to a unit step 

3. The waveform in Fig. 7.31 can be represented as 

tf = «(0 + - «(0 - ^^ u(t - a) (7.61) 

a a 

and its transform is I t (s) = - 1 1 + — — — ) (7.62) 

s \ as as / 

V(s) is then V(s) = (l + 1 _ Cl) — 1 (7 63) 

\ as as J C[s 3 + s(GIQ + 1/LC] K ' 

If we denote by v,(t) the response of the system to a ramp excitation, we 
see that 

C- 1 [F(s)] = Vl (t) + - Vi (t) - - v t (t - a) u(t - a) (7.64) 

a a 

where v t (t) is the step response in Eq. 7.59. 

192 Network analysis and synthesis 

Let us now discuss some further ramifications of the equation for the 
response R(s) = H(s) E(s). Consider the partial-fraction expansion of 

A< , v By 

R(s) = T 

i S — Si 

+ !■ 


i s — s t 

where s ( represent poles of H(s), and s, represent poles of E(s). Taking 
the inverse transform of R(s), we obtain 

rW^ZA^ + ZB^"* 


The terms Af?** are associated with the system H(s) and are called free 
response terms. The terms B^'* are due to the excitation and are known 
as forced response terms. The frequencies s t are the natural frequencies of 
the system and j, arc the forced frequencies. It is seen from our discussion 
of system stability in Chapter 5 that the natural frequencies of a passive 
network have real parts which are zero or negative. In other words, if we 
denote s t as s t = <r t + jo> { , then <x< <, 0. 

Example 7.7. For the initially inert network in Fig. 7.32, the excitation is 
v a (t) =• i cos t df). Let us find the response v (t) and determine the free and 
forced response parts of y (r). The system function is 

K (*) 1/sC 2 

VJis) R +MsC 

s + 2 

2\s* + 1/ 

Since V,{s) is 

the response is then 

W- V M™ = (s t + Ws + 2) 
We next obtain the inverse transform 

vjt) - -0.4* - " + 0.4 cos t + 0.2 sin t 

0.4 0.4* + 0.2 
J+2 + *» + l 








C-Jf=i= uoW 

FIG. 7.32 

Transform methods in network analysis 193 

It is apparent that the free response is -0.4e~*, and the forced response is 
0.4 cos r +0.2 sin/. 

As a final topic in our discussion, let us consider the basis of operation 
for the R-C differentiator and integrator shown in Figs. 7.33a and 7.336. 
We use the Fourier transform in our analysis here, so that the system 
function is given as H(ja>), where o> is the ordinary radian frequency 
variable. Consider first the system function of the differentiator in Fig. 

VJjw) _ R _ jaRC 

V,(J(o) R + 1/jcoC jmRC + 1 

Let us impose the condition that 

coRC « 1 

We have, approximately, 


=! jcoRC 

Then the response VJjm) can be expressed as 
Taking the inverse transform of V^ijco), we obtain 







Note that the derivation of Eq. 7.75 depends upon the assumption that the 
R-C time constant is much less than unity. This is a necessary condition. 
Next, for the R-C integrator in Fig. 7.33ft, the voltage-ratio transfer 
function is 

V/J*) = l/ja)C = 1 
VJijm) 1/jfoC + R = jwCR + 1 




(a) (b) 

FIG. 7.33. (a) R-C differentiator. (6) R-C integrator. 


1 94 Network analysis and synthesis 
If we assume that o>RC )J> 1, then 

VM^T^zVjUa) (7.77) 


Under these conditions, the inverse transform is 

v (t)~-±-r Va (r)dT (7.78) 

RC Jo 

so that the R-C circuit in Fig. 7.336 is approximately an integrating circuit. 


In this section we will show that the impulse response h(t) and the system 
function H(s) constitute a transform pair, so that we can obtain step and 
impulse responses directly from the system function. 

We know, first of all, that the transform of a unit impulse (5(f) is unity, 
i.e., C[<5(0] = 1. Suppose the system excitation were a unit impulse, then 
the response R(s) would be 

R(s) = E(s) H(s) = H(s) (7.79) 

We thus see that the impulse response h(t) and the system function H(s) 
constitute a transform pair, that is, 

Z[h(t)] = H( S ) 

Z-^[H(s)] = h(t) 

Since the system function is usually easy to obtain, it is apparent that we 
can find the impulse response of a system by taking the inverse transform 
of H{s). 

Example 7.8. Let us find the impulse response of the current /(/) in the R-C 
circuit in Fig. 7.34. The system function is 

~ VM ~ R + 1/*C ~ R(s + 1/RO (7 ' 81) 

Simplifying H(s) further, we have 

^^O-JTw) (7 - 82 > 

The impulse response is then 

Ht) = C-i[/f(5)] = i ^r) - -L e-wj „(,) (7.83) 

which is shown in Fig. 7.35. 

Transform methods in network analysis 195 




_ 1 



FIG. 7.34 

FIG. 7.35. Current impulse 
response of R-C network in 
Fig. 7.34. 

Since the step response is the integral of the impulse response, we can 
use the integral property of Laplace transforms to obtain the step response 



where oc(f) denotes the step response. Similarly, we obtain the unit ramp 
response from the equation 



where y(t) denotes the ramp response. From this discussion, it is clear 
that a knowledge of the system function provides sufficient information to 
obtain all the transient response data that are needed to characterize 
the system. 

Example 7.9. Let us find the current step response of the R-L circuit in Fig. 
7.36. Since I(s) is the response and V g (s) is the excitation, the system function 

Therefore H(s)/s is 

V„{s) R+sL 

h{s) = i = i n _ i \ 

s s(R+sL) R\s s + RjLJ 

The step response </) is now obtained as 


a(/)=-(l -e-W*X)ii(/) 



To check this result, let us consider the impulse response of the R-L circuit, 

196 Network analysis and synthesis 
which we found in Chapter 5. 



The step response is the integral of the impulse response, or 



»</t=-(i -*-w*>0«(i) 



It is readily seen that if we know the impulse response of an initially inert 
linear system, we can determine the response of the system due to any other 
excitation. In other words, the impulse response alone is sufficient to charac- 
terize the system from the standpoint of excitation and response. 




FIG. 7.36 

FIG. 7.37 

Example 7.10. In Fig. 7.37, the only information we possess about the system 
in the black box is: (1) it is an initially inert linear system; (2) when v&) = 6(t), 

With this information, let us determine what the excitation vjf) must be in order 
to produce a response v£f) - te~" u(f). First, we determine the system function 
to be 

H(s) = 




Vjs) 7T2 t j+3 (j + 2X« +3) 
We next find the transform of te~* «(*)• 

Vjs) - titer"] - 


<» +2)* 

The unknown excitation is then found from the equation 

JW (s + 2Ks + 3) 
VAs) ° H(s) " 20 + 2.5X* + 2)» 

Simplifying and expanding V&s) into partial fractions, we have 

(s + 3) 1 0.5 

K<W = 2(a + 2.5X« + 2) " s + 2 * + 2.5 

The system excitation is then 

vtf)-{e-* t -0.5e-* M )u(t) 






Transform methods in network analysis 197 


In this section we will explore some further ramifications of the use of 
the impulse response Mj) to determine the system response r(t). Our 
discussion is based upon the important convolution theorem of Laplace 
(or Fourier) transforms. Given two functions /,(*) and//*), which are 
zero for t < 0, the convolution theorem states that if the transform of 
/ x (f) is F t (s), and if the transform of f£i) is F£s), the transform of the 
convolution of/i(f) and/ t (r) is the product of the individual transforms, 
WTO, that is, 

t[ j*Mt - t)/,(t) dr] = F 1 (s) F&) (7.97) 

where the integral I f t (t — t)/,(t) dr 

is the convolution integral or folding integral, and is denoted operationally 

fm - t)/^t) dr - fflVMt) (798) 

Proof. Let us prove that C[/, */J = F&. We begin by writing 

C|/i(0V^0] = J V"[ J/iO ~ r)Mr) dr] dt (7.99) 

From the definition of the shifted step function 

u(t - t) = 1 r£t 

= r>t 

we have the identity 

f/i(' - t)/,(t) dr = f 7i(« " t) «(* - t)A(t) dr (7.101) 
Jo Jo 

Then Eq. 7.99 can be written as 

C[/i(0Vi(0] = f "e- f 7i(* - t) «(t - r)Mr) dr dt (7.102) 
Jo Jo 

If we let x = t ' — r so that 

e -.t = e -^»f,> (7 103) 

198 Network analysis and synthesis 
then Eq. 7.102 becomes 

£[/i(0* /*(')] = I*" f"'f 1 (x)u(x)f i (T)e-°'e->*dTdz 
Jo Jo 

= f 7i(«) u(x)e-" dx f 7»(r)e- w dr 
Jo Jo 

= F 1 (s)F 2 (s) (7.104) 

The separation of the double integral in Eq. 7.104 into a product of two 
integrals is based upon a property of integrals known as the separability 
property. 2 

Example 7.11. Let us evaluate the convolution of the functions/i(f) = e -2 ' u(t) 
and fift) = t u(t), and then compare the result with the inverse transform of 
F^s) FJs), where 


W = p 

The convolution of_/i(f) and/ 2 (f) is obtained by first substituting the dummy 
variable / — t for / in/i(/), so that 

fiff - T ) = c-* ( '- T) u(t - t) (7.106) 


f *fi(f ~ T )/*( T ) rfT = I ■r« _2 " _T) ^ T = «~ a * | T « 2r dr (7.107) 

Jo Jo Jo 

Integrating by parts, we obtain 

fi(0* fiff) = ({-\+\ *-*) «W (7-108) 

Next let us evaluate the inverse transform of F^s) F£s). From Eq. 7.105, we 

so that C-Vito *a(*)l = (^ - ^ + \ e~*) "(') ( 71 10 ) 

An important property of the convolution integral is expressed by the 

equation „ t . t 

flit - r)/ a (r) dr = MMA - r) dr (7.111) 

Jo Jo 

• See, for example, P. Franklin, A Treatise on Advanced Calculus, John Wiley and 
Sons, New York, 1940. 

Transform methods in network analysis 199 
This is readily seen from the relationships 

CC/i(0V*«] = *i(') W (7-112) 

and CWOViC)] - Wi(«) (7113) 

To give the convolution integral a more intuitive meaning, let us examine 

the convolution or folding process from a graphical standpoint. Suppose 

we take the functions - , * , . 

A(t) = m(t) 

J (7.114) 

Mr) = t «(t) 

as shown in Figs. 7.38a and 7.39a. In Fig. 7.38, the various steps for 
obtaining the integral ., 

J o A('-t)/ 2 (t)«V 

are depicted. Part (b) of the figure shows /i(— t) = «(— t). The function 

ft(t - t) - u(t - t) (7.115) 

in part (c) merely advances / x (—t) by a variable amount t. Next, the 
product A(| _ t)/ . (t) _ <f _ t)t u(t) (? n6) 

is shown in part (d). We see that the convolution integral is the area under 
the curve, as indicated by the cross-hatched area in part (</). Since the 
convolution integral has a variable upper limit, we must obtain the area 
under the curve of/i(f — T)f t (r) for all t. With t considered a variable in 
Fig. 7.38a", the area under the curve is 

AV.-/(0 = !J«(0 (7.117) 

as plotted in part (e) of the figure. 

In Fig. 7.39, we see that by folding / g (r) about a point t, we obtain the 
same result as in Fig. 7.38. The result in Eq. 7.117 can be checked by 
taking the inverse transform of F^s) Fjfa), which is seen to be 

c- 1 [F 1 (s)f»W] = £- 1 

l3j = 2 m(o (7118) 

Let us next examine the role of the convolution integral in system 
analysis. From the familiar equation 

R(s) = E(s) H(s) (7.119) 

we obtain the time response as 

r(t) = £"*[£(«) H(s)] = ('e(r) h(t - t) dr (7.120) 

where e(r) is the excitation and k(r) is the impulse response of the system. 

200 Network analysis and synthesis 





— T 





— T 

M-r) Mr) 

t T 

— T 

t T 


FIG. 7.38 












hM M-t} 


t T 



FIG. 7.39 

Using Eq. 7.120, we obtain the response of a system directly in the time 

domain. The only information we need about the system is its impulse 


EVro-pW. 7.12a. Let us find the response /(f) of the R-L network in Fig. 7.40 due 

to the excitation c(t) - 2<r* y(f) (7.121) 

The impulse response for the current is 

.AW-|*-«ra«BW (7.122) 

Therefore, for the R-L circuit under discussion 

Kt) «*-"«(*) (7-123) 

Transform methods in network analysis 201 
Using the convolution integral, we obtain the response i(i) as 

Kt) - J V ~ t) Kr) dr - 2 j V"«-rtr* </r 

- 2«-< f V"*</t - 2(«-* - *-*)«(0 





PIG. 7.40 

FIG. 7.41 




r)dr-Ke(f) (7.127) 

s 7.12b. The ideal amplifier in Fig. 7.41 has a system function H(s) — K, 
where Kis a constant. The impulse response of the ideal amplifier is then 

Kt)-Kd(t) (7.125) 

Let us show by means of the convolution integral that the response r{t) is related 
to the excitation e(t) by the equation 

K0-JT4) (7.126) 

Using the convolution integral, we have 

Since / is a variable in the expression for r(t), we see that the ideal amplifier in 
the time domain is an impulse-scanning device which scans the input «(/) from 
/ — to f — oo. Thus, the response of an ideal amplifier is a replica of the 
input e(f) multiplied by ibcgain IT of the amplifier. 


In the Section 7.5, we discussed the role that the impulse response plays 
in determining the response of a system to an arbitrary excitation. In 
this section we will study the Duhamel superposition integral, which also 
describes an input-output relationship for a system. The superposition 
integral requires the step response a(f) to characterize the system behavior. 

We plan to derive the superposition integral in two different ways. 

202 Network analysis and s/nthesis 

The simplest is examined first. We begin with the excitation-response 
relationship R(s) = E(s) H(s) (7.128) 

Multiplying and dividing by s gives 

R(s) = ^ • s E(s) (7.129) 


Taking inverse transforms of both sides gives 

C- 1 [«(5)] = C- 1 [^- ) -s£(s)] 

= £ -ip£)l* t -i [s£(s)] 


which then yields 

K0 = a(0*[e'(0 + *(0-)d(0] 

= e(0-) a(f) + J ' e'(r) a(f - t) dr 


where e'(t) is the derivative of e(t), e(0— ) is the value of e(i) at / = 0—, 
and <x.(t) is the step response of the system. Equation 7.131 is usually 
referred to as the Duhamel superposition integral. 

Example 7.13. Let us find the current i(t) in the R-C circuit in Fig. 7.42 when 
the voltage source is 

as shown in Fig. 7.43, The system function of the R-C circuit is 

Us) s 

H(s) = 

V„(s) R(s + IjRC) 

Therefore the transform of the step response is 

M£) 1 


R(s + l/RC) 
v g (t) 








Slope a A 2 

FIG. 7.42 

FIG. 7.43 

Transform methods in network analysis 
Taking the inverse transform of H(s)/s, we obtain the step response 







From the excitation in Eq. 7.132, we see that e(0 — ) =0 and 
v'^i) = A 1 d(f) + A t u{t) 


The response is then 

i(0 -jTj'V 

,(r)a(t -r)dr 
<5(t) o(/ - t) dr + 

e -Ur-r)IRC fc 

- j A x Hr)«(t - r)dr + \ A t <x(f - 

Jo- Jo- 

ic /{ Jo_ 

- [J e-"* + /<* C(l - e-** )] u(t) 



As we see from the Example 7.13, e(0— ) = 0, which is the case in many 
transient problems. Another method of deriving the Duhamel integral 
avoids the problem of discontinuities at the origin by assuming the lower 
limit of integration to be t = 0+. Consider the excitation e(t) shown by 
the dotted curve in Fig. 7.44. Let us approximate e(t) by a series of step 
functions, as indicated in the figure. We can write the staircase approxi- 
mation of e(0 as 

e(t) = e(0+) h(0 + &E 1 u(t- At) 

+ AE, i/(f - 2At) + • • • + A£ n u(t - /iAt) 








}a£ 3 





0h-AT-»| 2At 3At 4At t 

FIG. 7.44. Staircase approximation to a signal. 

204 Network analysis and synthesis 

where AE^ is the height of the step increment at t «= *At. Since we assume 
the system to be linear and time invariant, we know that if the response to a 
unit step is <x(f), the response to a step K+ u(t — r) is X< a(f — t). There- 
fore we can write the response to the step approximation in Fig. 7.44 as 

K0 = e(0+) a(0 + AE X a(/ - At) 

+ A£^a(r — 2At) + • • • + A£„a(f - nAf) 

If At is small, r(t) can be given as 

K0 - e(0+) o(0 + — * oc(* - At) At 


+ *h «(, - 2At)At + • • • + ^aO - «At)At 
At At 

which, in the limit, becomes 

n AT 

r(t) - «(0+) «(0 + lim 2 tV ~ ' At) At 

Ar-»0<-©+ AT 


= e(0+) a(0 + f e'(r) «(« - t) dr 





7.1 In the circuit shown v(t) .- 2«(f) and / £ (0-) - 2 amps. Find and 

r VW- 


I fe«| 


► 1000 


7.2 The switch is thrown from position 1 to 2 at t ■■ after having been at 1 
for a long time. The source voltage is vjif) » V 1 e~' t sin fit. 

(a) Find the transform of the output voltage vtf). 

(b) Find the initial and final values of vjt)- 

(c) Sketch one possible set of locations for the critical frequencies in the * plane 
and write tint form of the response v&). (Do not take the inverse transform.) 

Transform methods in network anal/sis 205 

MtOB. 7.2 

13 In the circuit shown, all initial currents and voltages are zero. Find 
/(f) for t > using Thevenin's theorem. 


PROS. 7.3 

7.4 In the circuit shown in (a), the excitation is the voltage source e{t) 
described in (6). Determine the response i(t) assuming zero initial conditions. 



o 'S* 4=i* 





1/2 t 


PROS. 7.4 

13 Determine the expression for vji) when /(/) — Hi), assuming zero initial 
conditions. Use transform methods. 

206 Network anal/sis and synthesis 

PROB. 7.5 

7.6 The circuit shown has zero initial energy. At t = the switch S is 
opened. Find the value of the resistor X such that the response is v(t) = 
0.5 sin Vlt n(0. The excitation is i(t) = te~ ** u(t). 





PROB. 7.6 

7.7 Use transform methods to determine the expressions for i^t) and /j(/) 
in the circuit shown. The excitation is v(f) = lOOe -10 '. Assume zero initial 



PROB. 7.7 

7.8 For the circuit shown, the switch S is opened at / = 0. Use Thevenin's 
or Norton's theorem to determine the output voltage t^f). Assume zero initial 

Transform methods in network analysis 207 

PROB. 7.8 

7.9 For the transformer shown, find i^t) and «,(/). It is given that e(t) = 
6u(0> and that prior to the switching action all initial energy was zero; also 

PROB. 7.9 

7.10 For the circuit shown, find /,(/), given that the circuit had been in steady 
state prior to the switch closing at / = 0. 

10Q Kml 400 

— WW 1 

100 vl= 


2hojlo8h «2 ) j=z=_40v 

PROB. 7.10 

7.11 Find /,(/) using Thevenin's theorem. The excitation is e(t) = 100 
cos 20 u(t). Assume zero initial energy. 

208 Network anal/sis and synthesis 

l y/W — TVTP- 

10O 2h 




PROB. 7.11 

7.12 Determine the transfer function H(s) = V^IV^s). When v^t) 
«(/), find v t {i). Assume zero initial conditions. 



Y 1 * h 


i f =j= v * G) 5V i» =r 

3f fOS v 3 

PROB. 7.12 

7.13 Using (a) standard transform methods and (b) the convolution integral, 
find v(t) when /(/) » 2e~* u(t). Assume zero initial energy. 


— o— 



lf=t ih< 

PROB. 7.13 

7.14 The impulse response of a linear system is shown in the figure. If the 
excitation were <*/) *■ 3e"*'«(0, determine the response values r(l) and HA,) 
using graphical convolution. 


PROB. 7.14 

Transform methods in network analysts 209 

7.15 The system function is given as H(s) = l/(s* + 9)*. If the excitation 
were c(/) = 3S'(t), determine the response #•(')• (Hint: Use the convolution 
integral to break up the response transform R(s).) 

7.16 Solve the following integral equations for x(t). 


«(0 + f(f - t)x(t)«/t = 1 

sinr = J a<T)e-<«- f >«/T 
«(/) +j x(jt — r) e-* dr =* t 


7.17 Using the convolution integral find the inverse transform of 




(s + aXs + b) 

(s* + 4)» 

" (s* + 1)« 

7.18 By graphical means, determine the convolution of /(/) shown in the 
figure with itself (i.e., determine/(/)*/(/)). 



7.19 Using (a) the convolution integral and (b) the Duhamel superposition 
integral, find v(t) for e(t) = 4e-**u(t). Assume zero initial conditions. 


210 Network analysis and synthesis 

7.20 Using (a) the convolution integral and (A) the Duhamel superposition 
integral, find «<0 for e(t) = 2e -s * «(/). Assume zero initial conditions. 

PRO B. 7 JO 

7.21 For the circuit in (a) f the system function H(s) = K(i)//(*) has the 
poles shown in (c). Find the element values for R and C. If the excitation »'(/) 
has the form shown in (b), use the convolution integral to find v(t). 






-«r -2 -1 


PROS. 7.21 

7.22 The excitation of a linear system is x(r), shown in (a). The system 
impulse response is Hf), shown in (ft). Sketch the system response to x(t). 
(No equations need be written. A neat, carefully dimensioned sketch will 

Transform methods in network anal/sis 21 1 



l t 

PROS. 7.22 

7.23 A unit step of voltage is applied to the network and the resulting current 
is /(/) = 0.01*^' + 0.02 amps. 

(a) Determine the admittance Y(s) for this network. 

(b) Find a network that will yield this admittance function. 

PROS. 7.23 

PROB. 7.24 

7.24 The current generator delivers a constant current of 4 amps. At 
/ = the switch 5 is opened and the resulting voltage across the terminals 1, 2 is 
c(r) =6e-«' + 12v. 

(a) Find Z(s) looking into terminals 1, 2. 

(b) Find a network realization for Z(s). 

chapter 8 

Amplitude, phase, and delay 


In this section we will study the relationship between the poles and 
zeros of a system function and its steady-state sinusoidal response. In 
other words, we will investigate the effect of pole and zero positions upon 
the behavior of H(s) along the jw axis. The steady-state response of a 
system function is given by the equation 

H(j<o) =M {my*™ (8.1) 

where Af(a>) is the amplitude or magnitude response function, and is an 
even function in o>. ^(o>) represents the phase response, and is an odd 
function of to. 

The amplitude and phase response of a system provides valuable 
information in the analysis and design of transmission circuits. Consider 
the amplitude and phase characteristics of a low-pass filter shown in 
Figs. 8.1a and 8.1 A. The cutoff frequency of the filter is indicated on the 
amplitude response curves as m c . It is generally taken to be the "half- 
power" frequency at which the system function \H(jt» a )\ is equal to 0.707 
of the maximum amplitude \H(j(Omu)\. In terms of decibels, the half- 
power point is that frequency at which 20 log \H(j(o c )\ is down 3 db from 
20 log |#(/a)max)|. The system described by the amplitude and phase 
characteristics in Fig. 8.1 shows that the system will not "pass" frequencies 
that are greater than <o a . Suppose we consider a pulse train whose 
amplitude spectrum contains significant harmonics above co . We know 
that the system will pass the harmonics below m c , but will block all 
harmonics above (o c . Therefore the output pulse train will be distorted 
when compared to the original pulse train, because many higher harmonic 
terms will be missing. It will be shown in Chapter 13 that if the phase 




Amplitude, phase, and delay 





Phase response 

FIG. 8.1. Amplitude and phase response of low-pass filter. 


response <f>{w) is linear, then minimum pulse distortion will result. We see 
from the phase response <f>(a>) in Fig. 8.1 that the phase is approxi- 
mately linear over the range —(o c <,w<, +co c . If all the significant 
harmonic terms are less than co c , then the system will produce 
minimum phase distortion. With this 
example, we see the importance of an 
amplitude-phase description of a system. 
In the remaining part of this chapter, 
we will concentrate on methods to 
obtain amplitude and phase response 
curves, both analytically and graph- 

To obtain amplitude and phase 
curves, we let s =ju> in the system 

function, and express H(ja>) in polar form. For example, for the amplitude 
and phase response of the voltage ratio VJV X of the R-C network shown 
in Fig. 8.2, the system function is 


£=r V 2 (s) 

FIG. 84 

H(s) = 

F^s) ljRC 

V^s) s + l/RC 
Letting « = jot, we see that H(Jco) is 

jm + l/RC 





Network analysis and synthesis 


FIG. 8.3. Amplitude and phase response of R-C network. 

In polar form HQco) becomes 


HO) = 

--/tan-i a>BC_ 

= M(ft>)e'* (0,) (8.4) 

(to 2 + 1/K 2 C 2 )* 

The amplitude and phase curves are plotted in Fig. 8.3. At the point 
(o = 0, the amplitude is unity and the phase is zero degrees. As a> increases, 
the amplitude and phase decrease monotonically. When co *= l/RC, the 
amplitude is 0.707 and the phase is —45°. This point is the half-power 
point of the amplitude response. Finally as <o -»■ oo, Af(o>) approaches 
zero and ^(g>) approaches —90°. 

Now let us turn to a method to obtain the amplitude and phase response 
from the pole-zero diagram of a system function. Suppose we have the 
system function 

Ms - Zq)(s - Zi) 
(s - p )(s - Pi)(« - Ps) 

ApO'cu — Zq)0'<» — gp 


H(s) = 
H(jco) can be written as 

HUco) = ^ - *,)0« >- *P (g6) 

0«> - Po)0«> - Pi)(J°> - P*) 
Each one of the factors jco — z t or jco — p } corresponds to a vector 
from the zero z i or pole/*, directed to any point jco on the imaginary axis. 
Therefore, if we express the factors in polar form, 

jco - z, = rV»", jco- p,= M,e'*' (8.7) 

then H(j(o) can be given as 

jj(jca) = j4 ° N ° Nl e i(y><>+v>i-a»-»i-ti) (g 8 ) 

u MoMiMj 

as shown in Fig. 8.4, where we note that 0! is negative. 

Amplitude, phase, and delay 215 

FIG. 8.4. Evaluation of amplitude and phase from pole-zero diagram. 

In general, we can express the amplitude response M(co) in terms of the 
following equation. 


H vector magnitudes from the zeros to the point on thejco axis 

M(.) = ^- 

XI vector magnitudes from the poles to the point on the/co axis 

Similarly, the phase response is given as 


^(eu) = J angles of the vectors from the zeros to the/co axis 


— 2 angles of the vectors from the poles to thejco axis 

It is important to note that these relationships for amplitude and phase 
are point-by-point relationships only. In other words, we must draw 
vectors from the poles and zeros to every point on thejco axis for which we 
wish to determine amplitude and phase. Consider the following example. 

F(s) = - 



5* + 2s + 2 (s + H-jl)(s+l-jl) 


Let us find the amplitude and phase for FQ2). From the poles and zeros 
of F(s), we draw vectors to the point to = 2, as shown in Fig. 8.5. From 

216 Network anal/sis and synthesis 




/\45* > 

x — i — / 



i / 


-l / 



x — 1 — . 


FIG. 8.5. Evaluation of amplitude and phase from pole-zero diagram, 
the pole-zero diagram, it is clear that 

MQ2) = 4( ,_ 2 ,- ) = 1.78 

and <f>(j2) = 90° - 45° - 71.8° = -26.8° 

With the values M(j2) and <f>Q2) and the amplitude and phase at three or 
four additional points, we have enough information for a rough estimate 


X r - 



1.0 2 


/S A5 ' 

x — 1 — 




I — J 



(a) (6) 

FIG. 8.6. Determining amplitude and phase at zero and very high frequencies. 

Amplitude, phase, and delay 217 

of the amplitude and phase response. At w = 0, we see that the vector 
magnitude from the zero at the origin to to = 0, is of course, zero. 
Consequently, M(j0) = 0. From Eq. 8.9 for F(s), F(j0) is 

lim F(ja>) = 





From this equation, we see that the zero at the origin still contributes a 
90° phase shift even though the vector magnitude is zero. From Fig. 
8.6a we see that the net phase at <w = is #0) = 90° - 45° + 45° = 90°. 
Next, at a very high frequency m h , where to h y> 1, all the vectors are 
approximately equal to co J fi i,0 °, as seen in Fig. 8.66. Then 

M(w») ~ 

4eo h _ 

ca k 



#a>») ~ 90° - 90° - 90° = -90° 

Extending this analysis for the frequencies listed in Table 8.1, we obtain 
values for amplitude and phase as given in the table. From this table we 
can sketch the amplitude and phase curves shown in Fig. 8.7. 

Next let us examine the effect of poles and zeros on the jto axis upon 
frequency response. Consider the function 

F(s) = 

s* + 1.03 _ (s + jl.015)(s - jlMS) 
s* + 1.23 ~ (s+ ;1.109Xs - jl.109) 











0123456789 10 

Frequency, w — »- 

FIG. C7. Amplitude and phase response for F(s) in Eq. 8.9. 


Network analysis 


TABLE 8.1 




Phase, degrees 

















i y'1.109 

: -y'1.109 

FIG. 8.8 

whose pole-zero diagram is shown in Fig. 8.8. At co <= 1.015, the vector 
from zero to that frequency is of zero magnitude. Therefore at a zero 
on the /o> axis, the amplitude response is zero. At co = 1.109 the vector 
from the pole to that frequency is of zero magnitude. The amplitude 
response is therefore infinite at a pole as seen from Eq. 8.11. Next, 

consider the phase response. When 
co < 1.015, it is apparent from the pole- 
zero plot that the phase is zero. When 
co > 1.015 and o>< 1.109, the vector 
from the zero at co = 1.015 is now 

pointing upward, while the vectors from 

the other poles and zeros are oriented 
in the same direction as for co < 1.015. 
We see that at a zero on the yw axis, the 
phase response has a step discontinuity 
of +180° for increasing frequency. 
Similarly, at a pole on the jco axis, the 
phase response is discontinuous by —180°. These observations are 
illustrated by the amplitude and phase plot for F(s) in Eq. 8.11 for the 
frequency range 0.9 <; co ^ 1.3, shown in Fig. 8.9. 

With a simple extension of these ideas, we see that if we have a zero at 
z — — a ± jco t , where a is very small as compared to co t , then we will have 
a dip in the amplitude characteristic and a rapid change of phase near 
co = co t , as seen in Fig. 8.10. Similarly, if there is a pole atp = — a ± jco t , 
with a very small, then the amplitude will be peaked and the phase will 
decrease rapidly near co = co t , as seen in Fig. 8.11. A contrasting situation 
occurs when we have poles and zeros far away from they'd) axis, i.e., a is 
large when compared to the frequency range of interest. Then we see that 
these poles and zeros contribute little to the shaping of the amplitude and 
phase response curves. Their only effect is to scale up or down the overall 
amplitude response. 

From stability considerations we know that there must be no poles in 
the right half of the s plane. However, transfer functions may have zeros 

Amplitude, phase, and delay 219 

1.015 1.109 
FIG. 8.9. Amplitude and phase far F(s) in Fig. 8.8. 

in the right-half plane. Consider the pole-zero diagrams in Figs. 8.12a 
and 8.126. Both pole-zero configurations have the same poles; the only 
difference is that the zeros in (a) are in the left-half plane at s = — 1 ±]1, 
while the zeros in (b) are the mirror images of the zeros in (a), and are 
located at s = +1 ±jl. Observe that the amplitude responses of the 
two configurations are the same because the lengths of the vectors corre- 
spond for both situations. We see that the absolute magnitude of the 





FIG. 8.10. Effect of zero very near FIG. 8.1 1. Effect of pole very near 
the/co axis. the/o> axis. 

220 Network analysis and synthesis 





o yi 
















1 <r 







(a) (b) 

FIG. 8.12. (a) Minimum phase function, (b) Nonminimum phase function. 

phase of (b) is greater than, the phase of (a) for all frequencies. This is 
because the zeros in the right-half plane contribute more phase shift 
(on an absolute magnitude basis) than their counterparts in the left-half 
plane. From this reasoning, we have the following definitions. A system 
function with zeros in the left-half plane, or on the jco axis only, is called a 
minimum phase function. If the function has one or more zeros in the 
right-half plane, it is a nonminimum phase ^ function. In Fig. 8.13, we see 
the phase responses of the minimum and nonminimum phase functions 
in Figs. 8.12a and 8.126. 

Let us next consider the pole-zero diagram in Fig. 8.14. Observe that 
the zeros in the right-half plane are mirror images of the poles in the 


y«- Minim 

urn phase 



— YY) 

^^/" No 






FIG. 8.13. Comparison of minimum and nonminimum phase functions. 

Amplitude, phase, and delay 221 

left-half plane. Consequently, the vector 
drawn from a pole to any point co x on 
they a> axis is identical in magnitude with 
the vector drawn from its mirror image 
to <»!. It is apparent that the amplitude 
response must be constant for all frequen- 
cies. The phase response, however, is 
anything but constant, as seen from the 
amplitude and phase response curves 
given in Fig. 8. 1 5 for the pole-zero config- 
uration in Fig. 8.14. 

A system function whose poles are only in the left-half plane and whose 
zeros are mirror images of the poles about the jm axis is called an all-pass 
function. The networks which have all-pass response characteristics are 
often used to correct for phase distortion in a transmission system. 




* n 










FIG. 8.14. All-pass function. 


In this section we will turn our attention to semilogarithmic plots of 
amplitude and phase versus frequency. These plots are commonly known 
as Bode plots. Consider the system function 


H(s) = 



+ 180 

l -I* 


1 1 1 1 1 

2 4 6 8 10 
u — *- 



FIG. 8.15. Amplitude and phase of all-pass function in Fig. 8.14. 

222 Network analysis and synthesis 

W >l 

K a 



K 2 





K positive 

K negative 




FIG. 8. 16. Magnitude and phase of constant. 
We know that the amplitude response is 

M(«) - |HC/»)I - j^} 


If we express the amplitude in terms of decibels, we have 

20 log M(w) = 20 log \N(jco)\ - 20 log | D(jco)\ (8.14) 

In factored form both N(s) and D(s) are made up of four kinds of factors: 

(a) a constant, K 

(b) a root at the origin, s 

(c) a simple real root, s + a 

(d) a complex pair of roots, s 2 + 2<w + a* + /?* 

To understand the nature of log-amplitude plots, we need only examine the 
amplitude response of the four kinds of factors just cited. If these factors 
are in the numerator, their magnitudes in decibels carry positive signs. 
If these factors belong to the denominator, their magnitudes in decibels 
carry negative signs. Let us begin with case (a). 

(a) The factor K. For the constant K, the db loss (or gain) is 

20 log K = K t (8.15) 

Amplitude, phase, and delay 223 

The constant K t is either negative if \K\ < 1, or positive if \K\ > 1. The 
phase response is either zero or 180° depending on whether K is positive or 
negative. The Bode plots showing the magnitude and phase of a constant 
are given in Fig. 8.16. 

(b) The factor s. The loss (gain) in decibels associated with a pole 
(zero) at the origin is ±20 log to. Thus the plot of magnitude in decibels 
versus frequency in semilog coordinates is a straight line with slope of 
±20 db/decade or ±6 db/octave. From the Bode plots in Fig. 8.17, we 
see that the- zero loss point (in decibels) is at to = 1, and the phase is 
constant for all co. 

(c) The factor s + a. For convenience, let us set a = 1. Then the 
magnitude is ±2Qlog ^ + ^ = ±2Qlog(o)S + , )M (g lfi) 

as shown in Fig. 8.18a. The phase is 

Arg O + l)* 1 = ±tan~ x a> (8.17) 

as shown in Fig. 8.186. 

A straight-line approximation of the actual magnitude versus frequency 
curve can be obtained from examining the asymptotic behavior of the 
factor y<o + 1. For to « 1, the low-frequency asymptote is 

20 log \j(o + 1| \ a<1 s* 20 log 1 = db (8.18) 

For to )» 1, the high-frequency asymptote is 

201og|7«o + l| 

20 log m db 



02 03 

2.0 3.0 

0.5 0.7 1.0 
Frequency, a 
FIG. 8. 17. Magnitude and phase of pole or zero at s = 0, 

5.0 7.0 10.0 

224 Network anal/sis and synthesis 





1 / 

025 0.5 1 

(ii — fc- 


FIG. «.!«. Magnitude and phase of simple real pole or zero. 

Amplitude, phase, and delay 
TABLE 8.2 










a> =- J 2 octaves below 



o) = i octave below 



o) = 1 break frequency 



m = 2 octave above 




o-4 2 octaves above 




which, as we saw in (b), has a slope of 20 db/decade or 6 db/octave. The 
low- and high-frequency asymptotes meet at m = 1, which we designate 
as the break frequency or cutoff frequency. The straight-line approximation 
is shown by the dashed lines in Fig. 8. 1 8a. Table 8.2 shows the comparison 
between the actual magnitude versus the straight-line approximation. We 
see that the maximum error is at the break frequency co = 1, or in un- 
normalized form: m = a. 

For quick estimates of magnitude response, the straight-line approxi- 
mation is an invaluable visual aid. An important example of the use of 
these straight-line approximations is in the design of linear control 

(d) Complex conjugate roots. For complex conjugate roots, it is con- 
venient to adopt standard symbols so that we can use the universal curves 
that result therefrom. We describe the conjugate pole (zero) pair in terms 
of a magnitude a> and an angle measured from the negative real axis, 
as shown in Fig. 8.19. Explicitly, the parameters that describe the pole 
(zero) positions are a>„ which we call the undamped frequency of oscillation, 
and £ = cos 0, known as the damping 
factor. If the pole (zero) pair is 
given in terms of its real and imaginary 

/>i,. - -« ±P (8-20) 

a and ft are related to £ and co by the 

a = g> cos = to«£ 




P = <u sin = o), VI — £* 

Returning to the definition of the 
damping factor, £ = cos d, we see that 



-/woVW 2 

FIG. 8.19. Pole location in terms of 
£ and <o t . 

226 Network analysis and synthesis 

the closer the angle is to 7r/2, the smaller is the damping factor. When 
the angle 6 is nearly zero degrees, the damping factor is nearly unity. 

To examine the Bode plots for the conjugate pole (zero) pair, let us set 
«> = 1 for convenience. The magnitude is then 

±20 log |1 — a) 2 +]2t,a>\ = ±20 log [(1 - a> 8 ) 8 + 4£ 8 <u 8 ]* (8.22) 

and the phase is 

<f>(w) = tan" 

-i 2gn> 
l-a> 8 


If we examine the low- and high-frequency asymptotes of the magnitude, 
we see that the low-frequency asymptote is decibels; the high-frequency 
asymptote (for <o y> 1) is ±40 log co, which is a straight line of 40 
db/decade or 12 db/octave slope. The damping factor £ plays a significant 
part in the closeness of the straight-line approximation, however. In 
Fig. 8.20 the asymptotic approximation for a pair of conjugate poles 
(a> = 1) is indicated by the dashed line. Curves showing the magnitude 
for £ = 0.1, £ = 0.6, and £ = 1.0 are given by the solid lines. We see 
that only for £ ~ 0.6 is the straight-line approximation a close one. 
Universal curves for magnitude and phase are plotted in Figs. 8.21 and 
8.22 for the frequency normalized function 

G(*) = 

(sK) 8 + 2£(s/a) ) + 1 


We see that the phase response, as viewed from a semilog scale, is an odd 
function about eo/a> = 1 . The phase at co = <w is —90° or —n\2 radians. 

+ 10 


1 1 1 








r-io-^N. ^\ 



Asymptote ^\ vft. 




, A. 






0.5 1.0 

Frequency, o> — *■ 

FIG. 8.20. Magnitude versus frequency for second-order pole. 


Amplitude, phase, and delay 227 


228 Network anal/sis and synthesis 







Amplitude, phase, and delay 229 

For a conjugate pair of zeros, we need only reverse the signs on the scales 
of the magnitude and phase curves. 

KVraMpig 8.1. Using Bode plot asymptotes, let us construct the magnitude 
versus frequency curve for 





16X10* + 10» + 1 

We see there are two first-order break frequencies at o> «- and a> — 50. In 
addition, there is a second-order break frequency at m — 400. With a quick 
calculation we find that £ — 0.2 for the second-order factor. The asymptotes 
are shown in Fig. 8.23. The magnitude and phase plots are given in Fig. 8.24 
through a microfilm plot computer program. 


FIG. 8.23. Asymptotes for G(s) in Eq. 8.25. 


We will now study a class of circuits whose system functions can be 
described by a pair of conjugate poles. These circuits are called single- 
tuned circuits because they only need two reactive elements — an inductor 
and a capacitor. The undamped frequency of oscillation of the circuit is 
then «w = {LC)- l/i . An example of a single-tuned circuit is the R-L-C 
circuit in Fig. 8.25, whose voltage-ratio transfer function is 



H(S ) = !o(i> = 
W V£s) R + sL + ljsC s* + (K/L)s + 1/LC 


230 Network anal/sis and synthesis 

FIG. 8.24a. Magnitude of G(s) in Eq. 8.2S. 
FIG. 8.246. Phase of G(s) in Eq. 8.2S. 

FIG. 8.25. Single-tuned circuit. 

Amplitude, phase, and delay 231 

FIG. 8.26 

The poles of H(s) are 

' 2L 2\LC J} J 

where we assume that (R*IL*) < (4/LC). In terms of a and /J in Eq. 8.27, 
//(5) is 


ff(s) = 

»' + ^ 

(s + a+j/f)(s + a-7/8) 


From the pole-zero diagram of H(s) shown in Fig. 8.26, we will determine 
the amplitude response \H(Jm)\. Let us denote the vectors from the poles 
to they'w axis as |M X | and |M,| as seen in Fig. 8.26. We can then write 


where K = a* + j8* and 

|MJ |M,| 

iMii - [«• + (» + m* 

|M,| - [a» + (a, - W\" 



In characterizing the amplitude response, the point eu = a>msx, at which 
\H(jco)\ is maximum, is highly significant from both the analysis and design 
aspects. Since \H(ja>)\ is always positive, the point at which \H(Jco)\* is 
maximum corresponds exactly to the point at which \H(ja>)\ is maximum. 
Since \H(jm)\* can be written as 

232 Network analysis and synthesis 

(«' + F? 

|H0«>)| 8 = 

[a* + (a, + /?)*][«* +(«,-j8) 8 ] 


(«' + /?*)' 

«> 4 + 2tt>\%* - f) + (a 8 + /?*)• 

we can find a>max by taking the derivative of | H(ja>) \ * with respect to a>* and 
setting the result equal to zero. Thus we have 

dJH(U»>)\* __ (a' + /8*)W + 2(q' - <8*)] 
<fo>« K + 2a» V " /**) + («* + /W 


From the equation ■ * |H °f )l * - (8.33) 


we determine eoSuz = jS 8 — « (8.34) 

Expressed in terms of the natural frequency of oscillation «o and the 
damping factor £, wj^ is 

^ - (o) Vl -£*)*- (C^o) 1 = <(\ ~ 2£*) (8-35) 

Since comax must always be real, the condition for a>max to exist, i.e., the 
condition for \H(jw)\ to possess a maximum, is given by the equation 

21* <, 1 (8.36) 

so that t <, 0.707. Since £ = cos 0, o)max does not exist for < 45°. 
When 8 = 45°, we have the limiting case for which ©max exists. In this 
case, £ — 0.707 and the real and imaginary parts of the poles have the same 
magnitude, i.e., a = /?, or 

£«>. = «> Vl - £* (8-37) 

We see from Eq. 8.34 that, when a = /J, then o>max = 0. This is the lowest 
frequency at which o)max may be located. For £ > 0.707, or 

£«>, > <Vl - P < 8 - 38 > 

comax is imaginary; it therefore does not exist To summarize, the key 
point in this analysis is that the imaginary part of the pole must be greater 
or equal to the real part of the pole in order for a>max to exist. Interpreted 
graphically, if we draw a circle in the s plane with the center at —a and the 
radius equal to 0, then the circle must intersect theyVo axis in order for comax 
to exist, as seen in Fig. 8.27. Moreover, the point at which the circle inter- 
sects the positive jm axis is comax- This is readily seen from the triangle 

Amplitude, phase, and delay 233 


i > 


i / 







-70 x 


FIG. 8.27. Peaking circle. 

with sides a, /S, a>max in Fig. 8.27. By the Pythagorean theorem, we find 

(oJLx - j8* - a* (8.39) 

The circle described in Fig. 8.27 is called the peaking circle. When a = /?, 
the peaking circle intersects the jot axis at o> = 0, as seen in Fig. 8.28a. 
When a > /J, the circle does not intersect the /to axis at all (Fig. 8.286); 
therefore, a>max cannot exist. When the imaginary part of the pole is 
much greater than the real part, i.e., when /J J> a, then the circle intersects 
the/a> axis at approximately a> = w , the natural frequency of oscillation 
of the circuit (Fig. 8.28c). 

A figure of merit often used in describing the "peaking" of a tuned 
circuit is the circuit Q, which is defined in pole-zero notation as 

2£ 2 cos 


From this definition, we see that poles near the ja> axis (£ small) represent 
high-g systems, as given in Fig. 8.28c, and poles far removed from the jo) 
axis represent low-Q circuits (Fig. 8.28a). Although the Q of the circuit 
given by the pole-zero plot of Fig. 8.286 is theoretically defined, it has no 
practical significance because the circuit does not possess a maximum 
point in its amplitude. 

By means of the peaking circle, we can also determine the half-power 
point, which is the frequency co c at which the amplitude response is 
\H(jw c )\ - 0.707 |#(./aw)|. 

234 Network analysis and synthesis 







J0 j- 






— «max 

FIG. 8.28. Examples of peaking circles, (a) a = P, m a 
undetermined, (c) p » a, eo max ~ p. 

■■ 0. (6) a > fi, »„ 

We will now describe a method to obtain co c by geometrical construction. 
Consider the triangle in Fig. 8.29, whose vertices are the poles {pi,pi*} 
and a point to, on they'w axis. The area of the triangle is 

Area (£piPi*aid = fa 


In terms of the vectors |MJ and |M,| from the poles to co it the area can also 
be expressed as 

\M X \ |M,| sin y > 


Area (&PiPi*<o t ) =- 


where y> is the angle at m it as seen in the figure. From Eqs. 8.41 and 8.42, 

we see that the product |Mi| |M 2 | is equal 
Ju to 

|M X | |M,| = -^ 

sin y> 
Since the amplitude response is 

\H(ja>i)\ = 

where K is a constant, then 

.Ksin y> 

FIG. 129 






Amplitude, phase, and delay 235 

For a given pole pair {puPi*} the parameters /J, a, and ^Tare prespecified. 
Therefore, we have derived \H(Jm)\ in terms of a single variable parameter, 
the angle y>. When the angle y> = tt\2 rad, then sin rp = 1, to, = Wmax, and, 

|H(/«>max)| = — 

When v = w/4 rad, then sin y> = 0.707 and 


\H(ja>J\ = 0.707 I^C/o^nu)! 

so that a>t = co c . Let us consider now a geometric construction to obtain 
o) a . Let us first draw the peaking circle as shown in Fig. 8.30. We will 
denote by A the point at which the peaking circle intersects the positive 
real axis. Now we draw a second circle with its center at A, and its radius 
equal to AB, the distance from A to either one of the poles, as seen in 
Fig. 8.30. The point where this second circle intersects the/a) axis is w c . 
The reason is that, at this point, the inscribed angle is y> x = w/4 because it 
is equal to one-half the intercepted arc, which, by construction, is w/2. 
When o)max = 0, the half-power point m c is also called the half-power 
bandwidth of the tuned circuit. In Fig. 8.31a the half-power point 's given 
when (Umax = 0. For a high-g circuit, where comw — <*><» the amplitude is 
highly peaked at m = <Um»x, as shown in Fig. 8.316. In this case, if 
\HQ0)\ < 0.707 |//(/ft>max)|, there are two half-power points m Ci and co 0i 


FIG. 8.30. Geometric construction to obtain half-power point. 

236 Network anal/sis and synthesis 

0.707 Iff, 

(a) (b) 

FIG. 8.31. (a) Low-G circuit response, (b) High-Q circuit response. 

about the point a>max, as seen in Fig. 8.316. By the construction process 
just described, we obtain the upper half-power point o> Ci . It can be shown 1 
that the point a>max is the geometric mean of m Ci and a> Ct , that is, 

As a result, the lower half-power point is 

°>Oi = ■ 




The bandwidth of the system for a high-g circuit of this type is described by 

BW = m Ct — (o 0i 


In design applications these high-g circuits are used as narrow bandpass 

Finally, there are certain aspects of the phase response of high-g circuits 
that are readily apparent. In Fig. 8.32 we see several steps in the process 
of obtaining the phase response. The phase shift at eo = is 0, as seen 
from Fig. 8.32a. At <a = oo, the phase shift is — it rad, as shown in part (c) 
of the figure. Finally, in the neighborhood of a = a> =i «>max, the phase 
shift resulting from the pole in the lower half plane is approximately 
— 0, = — tt/2 (Fig. 8.32&). The change in phase in this region is controlled 
in large by the pole p x . It is readily seen that the phase response in the 
region of a> has the greatest negative slope, as seen from a typical phase 
response of a high-g circuit shown in Fig. 8.33. 

1 See for example, F. E. Terman, Electronic and Radio Engineering, McGraw-Hill 
Book Company, New York, 1953. 

Amplitude, phase, and delay 237 

? J<* 




(*) (b) (e) 

PIG. S.32. Several steps in obtaining phase response for high-Q circuit, («) o>< ■ 

> a. 

Finally, as an example to illustrate our discussion of single-tuned 
circuits, let us find the amplitude response for the system function 

#(s) = 


s* + 6s + 34 


Now we determine the maximum and half-power points awx and <o c , 
and also the amplitudes |/f(/*comax)| and \H(jo> c )\. In factored form, H(s) is 




(s + 3+;5)(s + 3-;5) 

and the poles of H(s) are shown in Fig. 8.34. We next draw the peaking 
circle with the center at j = — 3 and the radius equal to 5. At the point 
where the circle intersects thejfa> axis, we see that eom» x — 4. To check this 
result, the equation <»*„, = /J* — a* gives 

cBm»x = (5* - 3*)* = 4 (8.52) 


FIG. 8.33. Phase response of high-Q circuit. 

238 Network analysis and synthesis 

6.78 = <o r 

The amplitude |£f(/a>max)| is then 

|H(/4)| - 

FIG. 8.34. Peaking circle construc- 
tion example. 


= — = 1.133 (8.53) 

The point A at which the peaking circle 

intersects the positive real axis is located 

at s = 2.0. With the center at A, we 

draw a circle of radius AB (equal to 

5-\/2 in this case). At the point C where 

this new circle intersects theyeo axis, we 

have o) c . By measurement, we find 

to c ~ 6.78 (8.54) 

Let us check this result. Referring to 
Fig. 8.34, we know_that the line segment 
AB is of length 5>/2 ; it follows that AC 
is also 5-v/2 units long. The line segment AO is of length 

_. AO = 5 — 3 = 2 units 

Then co c is given as 

co c = sl(AC) 1 - (AO)* = V46 = 6.782 

Finally, we obtain \H(j<o c )\ as 

|H(j'6.782)| = / . , — . = 0.802 

1 U n V(34 - 46)* + (6V46) S 

which is precisely 0.707 \H(j(Ow**l)\- 

In Section 8.3, we studied the frequency response for a pair of conjugate 
poles. Now we turn our attention to the amplitude response of two pairs 
of conjugate poles in a high-g situation. The circuit we analyze here is the 
double-tuned or stagger-tuned circuit given in Fig. 8.35. We will consider 



V 2 (s) 

FIG. 8.35. Double-tuned circuit. 

Amplitude, phase, and delay 239 

the special case when the R, L, and C elements in the primary circuit are 
equal in value to their counterparts in the secondary. Since the primary 
and secondary inductances are equal, the mutual inductance is 

M=KL (8.58) 

In this analysis we assume the coefficient of coupling K to be a variable 
parameter. Let us determine the amplitude response for the voltage-ratio 
transfer function V^V^s). From the mesh equations 

V^s) =^R + S L + J^j /.(s) - sM / 2 (s) 

= -sM h(s) + {r + sL + -M /,(*) 
we readily determine 


Vito & + WQs + 1/LC]* - s 4 K* l ' } 

Using tuned-circuit notation, we set 

2£<o = - 


H(s) can then be written as 
H(s) - ^ RM ' L \ 

(1 - K*)(s* + -^2_ , + -ȣ-)( f + -fe_ s + _22jL\ 
\ 1 + K l + K/\ l-K 1-KI 


If we set A = , * M ^ = ^2* (8.63) 

I?(l - K*) 1 - K* K ' 

then we can write 

H(s) = — (8.64) 

(s - SiXs - sfXs - s»)(s - s,*) 


where { Sl , Sl *} ^S_ ± j„J—L- £-_] * 

11 l + K J "ll + K (1 + K)*J 

1 l - k °Li - x ci — jc)*J 

Let us restrict our analysis to a high- Q circuit so that £* « 1 . Furthermore, 
let us assume that the circuit is loosely coupled so that AT« 1. Under 

240 Network analysis and synthesis 

these assumptions, we can approximate the pole locations by discarding 
the terms involving £* under the radicals in Eq. 8.65. Then the poles 
fai» *i*} can be given approximately as 

Similarly, {s„ *,*} can be given as 

{s„ s,*} ~ -£o> ± jo) ^l + —J 

The pole-zero diagram of H(s) is given in Fig. 8.36. The real part of the 
poles — £<u is greatly enlarged in comparison to the imaginary parts for 



FIG. 8.M. Poles and zeros of a dottble-tuned circuit. 

Amplitude, phase, and delay 241 

clarity purposes. Note that we have a triple zero at the origin. In terms 
of the vectors in Fig. 8.36, the amplitude response is 


]H(J(0}1 JM.I |M,| |M.| IMtl 
Since the circuit is high- 2 in the vicinity of a> = a> , we have 

|M,|-|M 4 |=i2|M |-2o, 
so that in the neighborhood of <u 






It is evident that the amplitude response 
of \H(jm)\ in the neighborhood of a> 
depends only upon the pair of vectors 
IMJ and |M,|. The double-tuned prob- 
lem has thus been reduced to a single- 
tuned problem in the neighborhood of 
o) . Consequently, we can use all the 
results on the peaking circle that were 
derived in Section 8.3. Let us draw a 
peaking circle with the center at 

s = -£«„ +/«>, (8.71) 

FIG. 8.37. Peaking circle for double- 
tuned circuit. 

and with a radius equal to tw^T/2, as 

shown in Fig. 8.37. The inscribed 

angle y> then determines the location of the maxima and half-power points 

of the response. Without going into the derivation, the amplitude response 

can be expressed as a function of y according to the equation 

|HO)l - 

A<o sin y 

sin y 


4a) K(ta> ) 2(1 - K*) 

Referring to the peaking circle in Fig. 8.37, let us consider the following 

1. a> £ > WiKfl: In this case, the peaking circle never intersects the jot 
axis; y> is always less than ir/2 (Fig. 8.38a), and the amplitude response 
never attains the theoretical maximum 

iW = *-= (8.73) 

2(1 -K*) 

as seen by the curve labeled (a) in Fig. 8.39. In this case, K < 2£, and the 
circuit is said to be undercoupled. 

242 Network anal/sis and synthesis 



2 \ 7 

i y 


-f«o A 

T / \ 

2 ly\ 


w lmax 

(a) (b) (c) 

FIG. 8.38. (a) Undercoupling. (6) Critical coupling, (c) Overcoupling. 

2. o>,£ = co^Kjl: Here the peaking circle intersects the/eo axis at a 
single point m = a> (Fig. 8.38ft). At a> , the amplitude is equal to #max 
in Eq. 8.73. In this case £ = Kj2, and we have critical coupling. 

3. <u £ < cogA/2 : The peaking circle intersects theytu axis at two points. 
o> x and to,, as seen in Fig. 8.38c. The intersecting points are given by the 

<*>!,* max = tt> ± «> ( 



Consequently, the amplitude response attains the theoretical maximum 
Hum* at two points, as shown by curve (c) in Fig. 8.39. In this situation, 
the circuit is said to be over coupled. 

Note that in Fig. 8.39 the undercoupled and critically coupled curves 
have their maximum points at m e . The overcoupled curve, however, is 

"a mix 
FIG. 8.39. (a) Undercoupled case, (A) Critically coupled case, (c) Overcoupled case. 

Amplitude, phase, and delay 243 

u Cl <<« u Ct 

FIG. 8.40. Half-power points of overcoupled circuit. 

maximum at o>! and u> t . In the case of overcoupling and critical coupling, 
we can determine the half-power points by using the geometrical con- 
struction method given in Section 8.3. Observe that there are two 
half-power points co c and co c , as shown in the overcoupled curve in 
Fig. 8.40. The bandwidth of the circuit is then 

BW = a> Ct — m Ci 


Example &2. The voltage-ratio transfer function of a double-tuned circuit is 
given as 


H ™ = (* + 2 -HylOOX* + 2 -ylOOX* + 2 +yl06X> + 2 -/106) (8/76) 

From H(s), let us determine the following: (a) the maximum points <o t ma and 
<o t „„; (b) the 3 db bandwidth BW; (c) the damping factor £; (d) the coefficient 
of coupling*; (e) the gain constant A; and (/) the maximum of the amplitude 
response //max. 

Solution, (a) The natural frequency of oscillation o> is taken to be approxi- 
mately halfway between the two poles, that is, <o = 103 radians. In the neigh- 
borhood of »„, we draw the poles s — —2 +/100 and s = —2 +/106, as 
shown in Fig. 8.41 . From the peaking circle centered at the point s = — 2 + /<»., 
shown in Fig. 8.41, we obtain 

so that 

m t max — m o = "^3* — 2* = 2.236 radians 
<°i max = °*o + 2.236 = 105.236 radians 
«»i m*r = «»o — 2.236 = 100.764 radians 



(b) Next we draw a circle centered at s = 1 +ja> with radius 3V2. Where 
this circle intersects theyVu axis, we have <o Ct so that 

">c, - 

V(3^2)» - 1 - 4.123 radians 


244 Network anal/sis and synthesis 

FIG. 8.41. Peaking circle for Example 8.2. 

The 3 db bandwidth is then 

BW - 20» Ol - o»j) - 8.246 radians (8.80) 

(c) The damping factor { is obtained from the real part of the poles £a> = 2, 
from which we obtain 

t « — 0.0194 

4 103 u '" 15 ^ 


(d) The coefficient of coupling K is obtained from the radius of the peaking 
circle, which is 


We thus have 

K = — = 0.0582 

(e) The gain constant A is equal to 

2{«.„A: 2(2X0.0582) 
* 1 - K* 1 - (0.0582)* = "•"** 

(f) Finally, the maximum amplitude H mmx is 


flmiz = 

2(1 -**) 





Amplitude, phase, and delay 245 


What is time delay? How do we relate it to frequency response? We 
will attempt to answer these questions in this section. First consider the 
transfer function of pure delay 

H(s) = <r tT (8.86) 

For a system described by Eq. 8.86, any excitation e(f) produces an 
identical response signal e(t — T), which is delayed by time T with respect 
to the excitation. This is shown by the Laplace transform relationship, 

Bis) - C[e(/ -T)] = er' T C[e(0] (8-87) 

Let us examine the amplitude and phase response of the pure delay. 
From the equation 

HO) - *-*** < 8 - 88 ) 

we obtain the amplitude response 

\H(Jw)\ - 1 (8.89) 

and the phase response <f>(a)) = —coT (8.90) 

We see that the delay T is equal to minus the derivative of the phase 
response, that is, 


T = - 



The magnitude, phase and delay characteristics of H(ja>) = er laT are 
given in Fig. 8.42a, b, and c. 

If we define delay as in Eq. 8.91, we can readily deduce that for the 
response to be nearly identical to the excitation, the system amplitude 
response should be constant, and its phase response should be linear over 
the frequency range of interest. If the phase is not linear, we have what is 
known as delay distortion. To visualize delay distortion more clearly, 




(a) (h) \ (e) 

FIG. a.42. Amplitude, phase, and delay of ideal delay function, (a) Amplitude, 
(ft) Phase, (c) Delay. 

246 Network anal/sis and synthesis 

we recall from Fourier analysis that any signal is made up of different 
frequency components. An ideal transmission system should delay each 
frequency component equally. If the frequency components are delayed 
by different amounts, the reconstruction of the output signal from its 
Fourier components would produce a signal of different shape as the input. 
For pulse applications, delay distortion is an essential design considera- 

Let us next examine how we relate delay, or envelope delay (as it is 
sometimes called) to the poles and zeros of a transfer function. For any 
transfer function 

n (»-««) 

H(s)=*tf (8.92) 


with zeros at z € = —a t ±jco t and poles at/>, = — <r y ±jco i , the phase for 
real frequencies is 

lr \ ? . -1 W ± fflj .IL . to + CO, 

<K<») = 2 tan * = * - J tan" 1 x ' (8.93) 

*-i o t J-l a, 

Envelope delay is 

dm ~ h o' + (c ± (of + A o* + (« ± co,)* (894) 

We see that the shapes of the delay versus frequency characteristic are 
the same for all poles and zeros. The zeros contribute "negative" delay; 
the poles, positive delay. However, linear physical systems do not have 
transfer functions with zeros alone. The inductor H(s) = Ls is the only 
exception. Its phase is #a>) = ir/2; thus the delay is zero. 

Now let us consider the delay due to one singularity, for example, a 
pole atp = —a +jco Q . The delay due to the one pole is 

The following points are pertinent: 

1. The maximum delay due to this pole is 

A m = - (8.96) 

and occurs at to = co . The delay versus frequency curve is symmetric 
about co = <w . 

Amplitude, phase, and delay 247 

FIG. 8.43. Graphic construction to obtain delay bandwidth. 
2. The frequency at which the delay is half the maximum, or l/2or , is 

o>x A = co ± ff 


3. The "effective delay bandwidth" is then eo — <r < m < to + cr or 
simply 2a . The upper and lower half-bandwidth points can be obtained 
graphically by drawing a circle with center at a> = co, and radius a . 
The intersections of the circle with the jco axis are the half-bandwidth 
points, as shown in Fig. 8.43. 

4. The product of the maximum delay and delay bandwidth is always 
2. Thus, if we wish to obtain large delay by placing zeros or poles near 
theyeo axis, the effective delay bandwidth is then very narrow. 

5. The delay of an all-pass function is twice the delay due to the poles 

The delay versus frequency curve for the pole is shown in Fig. 8.44. 
We see that the delay-bandwidth concept is only useful for a rough 
approximation, since the delay versus frequency characteristic only falls 
as 1/to. To calculate the delay versus frequency characteristic for a 
transfer function with a number of poles and zeros, it is convenient to 
obtain the delay curves for the individual singularities and then to add the 
separate delays. It is not as desirable to obtain the total phase response 
and then differentiate numerically. 

Finally, it should be pointed out that envelope delay only has meaning 
when the phase response goes through the origin. If it does not, there is 
a frequency-shift component in addition to the delay that is hard to 
account for analytically. 

248 Network analysis and synthesis 

wo - 4<?o ">o — 3»o wo - 2<ro wo — ^o «o «o + ffo «o + 2<ro wo + 3<ro wo + 4»o <* 

FIG. 8.44. Plot of delay versus frequency. 


8.1 Find the poles and zeros of the impedances of the following networks 
and plot on a scaled s plane. 

o — nnnr* |f- 






o M/* 1(- 

20 i M 


20 4h 


PROB. 8.1 

&2 The circuit shown in the figure is a shunt peaking circuit often used in 
video amplifiers, 
(a) Show that the admittance Y(s) is of the form 


K(s - s^s — s »> 

(s -*,) 
Express s lt s a , and s s in terms of R, L, and C. 

Amplitude, phase, and delay 249 

(6) When s t - -10 +yi0», s t - -10 -ylO», and Y(jB) - 10"* mhos, find 
the values of R, L, and C and determine the numerical value of *,. 





83 Find the amplitude and phase response for the following functions and 

(«) F(s) 
(c) F(s) . 

s +K 

(*) F(s) 

id) F(s) 

s +K 



Note that K and o> are positive quantities. 
8.4 Given the function 

<**"> C(») +71*.) 

determine the amplitude and phase of G(jm) in terms of ^, B, C, D. Show that 
the amplitude function is even and the phase function is odd. 

85 By means of the vector method, sketch the amplitude and phase response 

(a) F(s) - 

(b) F(s) 
(e) F(s) - 

j +0.5 
sis + 10) 

'■s* +2s +2 

s + 1 

' 5 -1 

(C) F(*) ■ 
(«/) F(5) 

s + 1 

s* -2s + 2 

, x « x 5* - 2j + 5 

<*> F(,) -(TTW+T) 

^ )F(j) - (,/2xV + 9) 

W F(5) = 

s* +2s +5 

(j + 2X* + 1) 

8.6 Plot on semilog paper the Bode plots of magnitude and phase for 

100(1 + 0.5*) 

F(*) = 
F(s) = 

s(s +2) 

50(1 + 0.025jX1 + O-IJ) 
(1 + 0.05*X1 + 0.01s) 

250 Network analysis and synthesis 
8.7 Plot on semilog paper the Bode plots of magnitude and phase for 




F(*) = 

(1 + O.OOZsXl + 5 • 10~ 5 * + 10~V) 

200(1 + 0.05s) 

(1 + 0.02jX1 + 4 • 10-** + 10-V) 

8.8 For the function 


s* + 2s + 5 

determine o> mu , \F(jco m *x)\, the half-power point a> , and \F(ja> )\. Sketch the 
amplitude and phase response. 

8.9 For the circuit shown, determine the current ratio IJI g and find: (a) 
the point €<>,„„, where its amplitude is maximum; (b) the half-power point <o c ; 
(c) the point to„ where |/i(a) M )// B (<o u )| = 1. Use geometric construction. 

2.5 h 

PROS. 8.9 

8.10 A network function consists of two poles at p ltt = r^*^' -9 ' = — a t ± 
jco t , as given in the figure. Show that the square of the amplitude response 
Af\to) is maximum at o> m * = /•<* |cos 20|. 

PROB. 8.10 

8.11 In connection with Prob. 8.2 plot the poles and zeros of the impedance 
function Z(s) = l/l%s). Find, approximately, the maximum point of the 
amplitude response. In addition, find the bandwidth at the half-power points 
and the circuit Q. 

Amplitude, phase, and delay 251 

8.12 The pole configuration for a system function H(s) is given in the figure. 
From the plot, calculate: 

(a) The undamped frequency of oscillation a> 

(6) The bandwidth and Q. 

f — yioo 




PROB. 8.12 

8.13 In connection with Prob. 8.10 determine the ratio Af\m aM )lM*(0). 

8.14 Determine the amplitude and phase response for the admittance Y(s) 
of the circuit shown. Is the peaking circle applicable here? What can you say 
about the shape of the amplitude response curve in a high-g situation? Deter- 
mine the bandwidth of the circuit and the circuit Q. 

20 lh 
o— — WW ^SW" 1 



PROB. 8.14 

8.15 For the overcoupled case of a double-tuned circuit, derive an expression 
for the peak-to-valley ratio that is, M(o> m »x)/M(a> ), where A#(-) denotes ampli- 
tude. Use the notation in Section 8.4. {Hint: see Prob. 8.13.) 

8.16 For the voltage ratio of a double-tuned circuit 

fri v \ ™S 

(s + 4 +y50X* + 4 -y50X* + 4 +j6Q)(.s + 4 -y60) 

252 Network analysis and synthesis 

Use the peaking circle to determine the maximum and half-power points and 
the circuit Q. Find the gain constant A and the coefficient of coupling K. 

8.17 For the double-tuned circuit shown, determine the m a x i mu m and half- 
power points and the circuit Q. Find the gain constant A and the coefficient of 
coupling K. 

10 " 9 f 

M=5xl<r 6 h 
PROB. 8.17 

8.18 Determine the delay at a = 0, 1, and 2 for 



F(s) = 
F(s) = 
F(s) = 


s +3 


(s + IX* + 2) 

s + 1 
s* + Is + 5 

* W (s+2Ks*+2s+ 2) 

chapter 9 

Network analysis II 


In electric network theory, the word port has a special meaning. A 
port may be regarded as a pair of terminals in which the current into 
one terminal equals the current out of the other. For the one-port network 
shown in Fig. 9.1, 1=1'. A one-port network is completely specified 
when the voltage-current relationship at the terminals of the port is given. 
For example, if V = 10 v and 1 = 2 amp, then we know that the input 
or driving-point impedance of the one-port is 

Zi„ = — = sa 


Whether the one-port is actually a single 5-Q resistor, two 2.5-Q resistors 
in series, or two 10-Q resistors in parallel, is of little importance because 
the primary concern is the current-voltage relationship at the port. 
Consider the example in which / = 2s + 3 and V = 1 ; then the input 
admittance of the one-port is 

Y to = ^ = 2s + 3 (9.2) 

which corresponds to a 2-f capacitor in parallel with a $-Q resistor in 
its simplest case (Fig. 9.2). 






FIG. 9.1 


PIG. 9.2 



254 Network analysis and synthesis 








v 2 



FIG. 9.3 

Two-port parameters 

A general two-port network, shown in Fig. 9.3, has two pairs of voltage- 
current relationships. The variables are V x , V 2 , J u l z . Two of these are 
dependent variables ; the other two are independent variables. The number 
of possible combinations generated by four variables taken two at a time 
is six. Thus there are six possible sets of equations describing a two-port 
network. We will discuss the four most useful descriptions here. 

The z parameters 

A particular set of equations that describe a two-port network are the 
z-parameter equations 

V x = z^ + zj t (9 3) 

V 2 = Z 21 7 l + S wJ% 

In these equations the variables V x and V t are dependent, and I v I 2 are 
independent. The individual z parameters are denned by 


z ll 












It is observed that all the z parameters have the dimensions of impedance. 
Moreover, the individual parameters are specified only when the current in 
one of the ports is zero. This corresponds to one of ports being open 
circuited, from which the z parameters also derive the name open-circuit 


z b 


FIG. 9.4 

Network anal/sis II 255 

parameters. Note that z u relates the current and voltage in the 1-1' port 
only; whereas z M gives the current-voltage relationship for the 2-2' port. 
Such parameters are called open-circuit driving-point impedances. On 
the other hand, the parameters z liS and z 2l relate the voltage in one port to 
the current in the other. These are known as (open-circuit) transfer 

As an example, let us find the open-circuit parameters for the T circuit 
in Fig. 9.4. We obtain the z parameters by inspection 



z ti 




= z« + z 6 


= z b + z e 



= Z h 


= Z 6 


Observe that z ia = z 21 . When the open-circuit transfer impedances of a 
two-port network are equal, the network is reciprocal. It will be shown 
later that most passive time-invariant networks are reciprocal. 1 

Most two-port networks, whether passive or active, can be characterized 
by a set of open-circuited parameters. Usually, the network is sufficiently 
complicated so that we cannot obtain the z parameters by inspection, as 
we did for the T circuit in Fig. 9.4. The question is now, "How do we 
obtain the z parameters for any circuit in general?" The procedure is as 
follows. We write a set of node equations with the voltages at the ports 
V x and V t , and other node voltages within the two-port V a , V 4 , . . . , V k 
as the dependent variables. The independent variables are the currents I t 
and I& which we will take to be current sources. We then proceed to 
write a set of node equations. 

h = n n V 1 + n^Vi H 

+ n lk V„ 

u r * 

+ n Sk V t 

Sk r k 

- n.1^1 + 


+ n Sk V l 

s» r t 


- "*iFi + 

+ "**n 

1 One important exception is the gyrator discussed later in this chapter. 

256 Network analysis and synthesis 

where n it represents the admittance between the ith and/th nodes, that is, 


»« = G u + sC u + 




If the circuit is made up of R-L-C elements only, then it is clear that 
n it = n ft . As a result, the ijth cofactor of the determinant of the node 
equations, A M , must be equal to they'ith cofactor, A ti , that is, A w = A H . 
This result leads directly to the reciprocity condition z 21 = z^, as we 
shall see. 

Returning to the set of node equations in Eq. 9.6, let us solve for V x 
and V t . We obtain 

Vl ~ A 1+ A 2 


In relating this last set of equations to the defining equations for the z 
parameters, it is clear that 

z u = 



^ A 

A 8a 

z tt = T 


Since for a passive network A^ = A w , it follows that z a = z 12 , the 
network is then reciprocal. 

As an example, let us find the z parameters of the Pi circuit in Fig. 9.5. 
First, the node equations are 

h"{T A + Y c )V 1 -Y c V l 
I a =-Y c V 1 + (Y B + Y c )V t 



v 2 



Y B 


FIG. 9S 

Network anal/sis II 257 
The determinant for this set of equations is 

Ay = Y A Y B + Y A Y C + Y B Y C (9.11) 

In terms of Ay, the open-circuit parameters for the Pi circuit are 


Now let us perform a delta-wye transformation for the circuits in Figs. 
9.4 and 9.5. In other words, let us find relationships between the im- 
mittances of the two circuits so that they both have the same z parameters. 
We readily obtain 


Y B + Y c 

^~ Ay 

Zai ~Ay 

*■ Ay 

Y A + Y C 

Zu ~ Ay 

We then find 


-z -±s- 


- z„ + z e - ^ 



r, . -T Y B + Y c 

- z a + z b - Ay 

z -■&■ 
Zo Ay 

Zc Ay 


The y parameters 

Suppose we were to write a set of mesh equations for the two port in 
Fig. 9.3. Then the voltages V x and V 2 would become independent sources, 
and the currents I x and /» would be just two of the dependent mesh 
currents. Consider the general set of mesh equations 

Vi - «uA + «iA + • • " + «i A 

V% — »»mA + "hJ* H r- m^/j 

= msi/i+ + W.A (9.15) 

= m u J l + + m kk T k 

where m it represents the sum of the impedances in the /th mesh and m ti is 
the common impedance between mesh i and meshy. We note here again 

258 Network anal/sis and synthesis 

that for an R-L-C network, m it — m H for all i and / Thus reciprocity 

Solving the set of mesh equations for / x and 7 a , we obtain the following 

h ~ A x + A * 


A * ■ A 
The equations of 9.16 define the short-circuit admittance parameters as 

h = VuVi + yit v * ( 9 17 ) 

h = y S i^i + ynVt 

where y tj = A„/A for all i andy. 

Let us find the y parameters for the bridged-T circuit given in Fig. 9.6. 
The mesh equations for the circuit are 


'. = A + (- + l)h + ~ s h 

s s \s I 


In straightforward fashion we obtain 

A = 

2(2s + 1) 

A -A ls% + 4s + 1 


A -A - - ls * + 2s + 1 


FIG. 9.6 

Network analysis II 
The short-circuit parameters are then 

2s 8 + 4s + 1 


yu = sfa = 

y«i = Vi* — - ■ 

2(2s + 1) 
2s 8 + 25+1 


2(2s + 1) 

When y lx = y n or z u = z a2 , the network is symmetrical. 2 

Returning to Eq. 9.17, which defines the y parameters, we see that the y 
parameters are expressed explicitly as 


















The reason that the y parameters are also called short-circuit admittance 
parameters is now apparent. In obtaining y u and y a , the 2-2' port must 
be short circuited, and when we find y n and y lt , the 1-1' port must be 
short circuited, as shown in Figs. 9.7a and 9.76. 

As a second example, let us obtain the y parameters of the Pi circuit in 
Fig. 9.5. To obtain y a and y M , we short circuit terminals 2-2'. We then 

Vu A ° (9.22) 

y»i = -Yc 

yn, *2i 

,Jz h>i 

m yn 




FIG. 9.7 

*A symmetrical network is easily recognized because by interchanging the 1-1' 
and 2-2' port designations, the network remains unchanged. 

260 Network analysis and synthesis 

We next short-circuit terminals 1-1' to obtain 

^22 = Y B + Y c ,g 23) 

^12 = -Yc 
The h parameters 

A set of parameters that are extremely useful in describing transistor 
circuits are the h parameters given by the equations 

Vi = *uA + hi*V* 

I» — "tl-M. "I" "22' 2 

The individual parameters are defined by the relationships 


h -h 
h - 1 * 

"21 = ~ 



h - h 

"89 = 




We see that h lt and h 91 are short-circuit type parameters, and h x% and h^ 
are open-circuit type parameters. The parameter h n can be interpreted 
as the input impedance at port 1 with port 2 short circuited. It is easily 
seen that h u is merely the reciprocal of y lt . 

hu = 



The parameter h 22 is an open-circuit admittance parameter and is related 
to z M by 

h«a = — 


Both the remaining h parameters are transfer functions; h 21 is a short- 
circuit current ratio, and h 12 is an open-circuit voltage ratio. Their 
relationships to the z and y parameters is discussed later in this chapter. 
For the Pi circuit in Fig. 9.5, the h parameters are 





+ Y C 

h 12 


Y c 

Y A 

+ Y C 

h ai 



Y c 

Y A + Y C 



. Y A Y C 


Y A +Y C 

Network analysis II 261 










FIG. 9.8. Negative impedance converter with load impedance. 

Observe that for the Pi circuit, h^ = —h lt . This is the reciprocity 
condition for the h parameters and can be derived from their relationships 
to either the z or y parameters. 

Next let us consider the h parameters of an ideal device called the 
negative impedance converter (NIC), which converts a positive load 
impedance into a negative impedance at its input port.* Consider the 
NIC with a load impedance Z L shown in Fig. 9.8. Its input impedance is 


Z.„ = -Z, 

which can be rewritten as 

¥i = ¥* 

The following voltage-current relationships hold for the NIC. 

V x - kV t 
A = kh 



If we interpret Eq. 9.31 using h parameters, we arrive at the following 

h vt = — = k 


We see that since h u j& —h tl , the NIC is nonreciprocal. 
In matrix notation, the h matrix of the NIC is 





.k . 


The NIC is a convenient device in the modeling of active circuits. It 
is not, however, a device that exists only in the imagination. Practical 

* For a lucid discussion of the properties of the NIC, see L. P. Huelsman, Circuits, 
Matrices, and Linear Vector Spaces, McGraw-Hill Company, New York, 1963, 
Chapter 4. 

262 Network anal/sis and synthesis 

realizations of NIC's have been achieved using transistors. Some of 
these are described in an article by Larky. 4 

The ABCD parameters 

Let us take as the dependent variables the voltage and current at the 
port 1, and define the following equation. 





This matrix equation defines the A, B, C, D parameters, whose matrix is 
known as the transmission matrix because it relates the voltage and current 
at the input port to their corresponding quantities at the output. The 
reason the current I t carries a negative sign is that most transmission 
engineers like to regard their output current as coming out of the output 
port instead of going into the port, as per standard usage. 
In explicit form, the ABCD parameters can be expressed as 

A = * 

C = ± 

B= - 






From these relations we see that A represents an open-circuit voltage 
transfer function; J? is a short-circuit transfer impedance; C is an open- 
circuit transfer admittance; and D is a short-circuit current ratio. Note 
that all four parameters are transfer functions so that the term trans- 
mission matrix is a very appropriate one. Let us describe the short-circuit 
transfer functions B and D in terms of y parameters, and the open-circuit 
transfer functions A and C in terms of z parameters. Using straight- 
forward algebraic operations, we obtain 


z 21 Vtl 

For the ABCD parameters, the reciprocity condition is expressed by the 


YA B~] 

= AD-BC=l (9.37) 

A = ^ 

B = 




c = i- 

D = 



C D 

* A. I. Larky, "Negative-Impedance Converters,' 
CT-4, No. 3 (September 1957), 124-131. 

Trans. IRE on Circuit Theory, 

Network analysis II 263 

Let us find, as an example, the 


ABCD parameter for the ideal trans- + 
former in Fig. 9.9, whose denning 
equations are 

V l - nV t 

/ 1 = i(_ j8 ) <*SD 




FIG. 9.9. Ideal transformer. 

If we express Eq. 9.38 in matrix form, we have 





so that the transmission matrix of the ideal transformer is 

A B" 




C D 

L «J 


Note, incidentally, that the ideal transformer does not possess an im- 
pedance or admittance matrix because the self- and mutual inductances 
are infinite. 5 

For the ideal transformer terminated in a load impedance shown in 
Fig. 9.10a, the following set of equations apply. 



Taking the ratio of V x to I l3 we find the input impedance at port 1 to be 

Zl = £ = n*Z L (9.42) 


Thus we see that an ideal transformer is an impedance transformer. If 
the load element were an inductor L (Fig. 9.106), at port 1 we would see 
an equivalent inductor of value n 2 L. Similarly, a capacitor C at the load 
would appear as a capacitor of value C\n* at port 1 (Fig. 9.10c). 
As a second example indicating the use of the transmission matrix in 

* For a detailed discussion concerning ideal transformers, see M. E. Van Valkenburg, 
Introduction ta Modem Network Synthesis, John Wiley and Sons, New York, 1960. 

264 Network anal/sis and synthesis 







v 2 


FIG. 9.10. Ideal transformer as an impedance transformer. 

network analysis, consider the ABCD parameters of the Pi circuit in 
Fig. 9.5. 


A — 

Y s +Y c 

B = 
C = 

n — 

Y c 

Y c 

Y A Y B + Y B Y C 

+ Y A Y C 

Y A +Y C 

Y c 

If we check for reciprocity from Eq. 9.43, we see that 

(Y A + Y C )(Y B + Y c ) - (Y A Y B + Y B Y C + Y A Y C ) 

AD-BC = 

Y a 




The relationships between two-port parameters are quite easily obtained 
because of the simple algebraic nature of the two-port equations. For 
example, we have seen that /t u = l/y u and h sa = l/z 2 ». To derive h lt in 
terms of open-circuit parameters, consider the z parameter equations 
when port 1 is open circuited: / x = 0. 

*\ = z i2-'a 
V t = z 22 /2 


Network analysis II 265 

Therefore we have h lt = - 1 


= 2H (9.46) 

Similarly, since h n is defined as a short-circuit type parameter, we can 
derive h 21 in terms of y parameters as 

h a = 

= ** (9.47) 

We can express all the A parameters as functions of the z parameters or y 
parameters alone. An easy way to accomplish this task is by finding out 
what the relationships are between the z and y parameters themselves. 
Certainly, by their very nature, the z and y parameters are not simply 
reciprocals of each other (as the novice might guess), since one set of 
parameters is defined for open-circuit conditions and the other for short- 

The z and y relationships can be obtained very easily by using matrix 
notation. If we define the z matrix as 

[Z] = 

and the y matrix as 


[11 = 



yu yn\ 

J/ti Vt»] 

In simplified notation we can write the two sets of equations as 

[V] = [Z][7] (9.50) 

and [/] = [Y][V] (9.51) 

Replacing [7] in Eq. 9.50 by [Y][V], we obtain 

[V] = [Z][Y)IV] (9.52) 

so that the product [Z][Y] must yield the unit matrix [U]. The matrices 
[Y] and [Z] must therefore be inverses of each other, that is, 

[ZTr 1 = [Y] and [IT-^IZ] (9.53) 

From the relationship, we can find the relations between the individual 
z and y parameters. 

Zjg Zji 

A « A * (9.54) 

A. A. 

266 Network anal/sis and synthesis 

TABLE 9.1 
Matrix Conversion Table 







A ft 



A T 


z n 

z 12 














z 22 








z ia 









A u 

A u 



















z 12 






A T 

Z 22 
















Z 22 














Z 21 







z 22 



A M 




z n 






where A„ = z u z g g — ZigZ ax ; and 
«ii = T~ 

Zia = — 


J/22 = — 

Zg!= - 



where A v = ^n2/ 22 — Vi^/n- Using these identities we can derive the h 
or ABCD parameters in terms of either the z or y parameters. Table 9.1 
provides a conversion table to facilitate the process. Note that in the 

tMe b. T = AD- BC (9.56) 


In this section we will examine how to determine driving-point and 
transfer functions of a two-port by use of two-port parameters. These 
functions fall into two broad categories. The first applies to two-ports 

Network analysis II 267 

without load and source impedances. These transfer functions can be 
described by means of z or y parameters alone. For example, let us derive 
the expressions for the open-circuit voltage ratio VJiV x by using z param- 
eters first and y parameters next. Consider the z parameter equations 
for the two-port when port 2 is open circuited. 

V t = z ai /j 


V x - z^ 

If we take the ratio of V, to V x , we obtain 

Yl = %L 

Vt *ii 

By letting / 2 of the second y parameter equation go to zero, we derive 
the open-circuit voltage ratio as 

V\ Via 

In similar manner we can derive the short-circuit current ratio of a 
two-port as 




Is _ _«M 



The open- and short-circuit transfer functions are not those we usually 
deal with in practice, since there are frequently source and load im- 
pedances to account for. The second category of two-port transfer 
functions are those including source or load impedances. These transfer 
functions are functions of the two-port parameters z, h, or y and the 
source and/or load impedance. For example, let us derive the transfer 
admittance IJV X of a two-port network that is terminated in a resistor 
of/? ohms, as given in Fig. 9.1 1. For this two-port network, the following 
equate apply. _ + 

V 2 






FIG. 9.11 


Network analysis and synthesis 







^-^ +1*12*2 1+ s~~- 

v 2 


FIG. 9.12. Two-port equivalent. 

By eliminating the variable V t , we obtain 


V 1 y u + 1/R 

Note that r sl and y tl are not the same. Y S1 is the transfer admittance 
of the two-port network terminated in a resistor R, and y n is the transfer 
admittance when port 2 is short circuited. We must be careful to make 
this distinction in other cases of a similar nature. 

In order to solve for transfer functions of two-ports terminated at 
either port by an impedance Z L , it is convenient to use the equivalent 
circuit of the two-port network given in terms of its z parameters (Fig. 
9.12) or y parameters (Fig. 9.13). The equivalent voltage sources z lg / a 
and z„I x in Fig. 9.12 are called controlled sources because they depend 
upon a current or voltage somewhere in the network. 6 Similarly, the 
current sources y lt V t and y^Yx are controlled sources. For the circuit 
in Fig. 9.12, let us find the transfer impedance Zm = VJI U with port 2 

1 1 

+ 1 


' 1 





Y 2 

v 2 

I 2' 

FIG. 9.13. Two-port equivalent. 

« For a lucid treatment of controlled sources, see E. J. Angelo, Electronic Circuits, 
2nd Ed., McGraw-Hill Book Company, New York, 1964. 

Network anal/sis II 269 

terminated in a load impedance Z L . If we write the mesh equation for 
the /* mesh, we have 

-«tA-(*ii + Z x )li (9.64) 

Since V t = —I t Z L , we readily obtain 

z « - r = ^rr < 9 - 65 > 

It also is clear that the current-ratio transfer function for the terminated 
two-port network is 

-* = "*» (9.66) 

h Z 22 + z z, 

In similar fashion, we obtain the voltage-ratio transfer function for the 
circuit represented in Fig. 9.13 as 



^ 8 + y M 

Next, suppose we are required to find the transfer function VJV„ for 
the two-port network terminated at both ends, as shown in Fig. 9.14. 
We first write the two mesh equations 

Next, we solve for I t to give 


/t = 


(«1 + 2ll)(^2 + «m) - *1**21 

From the equation V t = —R t I%, we may now arrive at the following 

Y*=_Ml = ?«£t 

V, V„ {R 1 + z^)(R t + z,,) - z nZlt 



r^aro £>* 

FIG. 9.14 

270 Network anal/sis and synthesis 

-£ — MAA" 



■WW— ^O 1- ^ 

Z22-Z12 (Z21-«12Ml 



v 2 

FIG. 9.15. Two-port equivalent circuit with one controlled-voltage source. 

Note that the equivalent circuits of the two-ports in Figs. 9.12 and 
9.13 are not unique. Two other examples are given in Figs. 9.15 and 9.16. 
Observe that the controlled sources are nonzero in these equivalent 
circuits only if the circuit is nonreciprocal. 

Finally, let us consider the hybrid equivalent circuit shown in Fig. 9.17. 
Observe the voltage-controlled source A gl F 2 at port 1 and the current- 
controlled source Ag^ at port 2. Let us find the input impedance Z ln . 
The pertinent equations are 

Vi = h lx h + hi*V* 

and V a = -Z L h = ~(h 2l h + h^V^ Z L 

Solving Eq. 9.72 for V t , we find 

V = — htiZiJi 
1 + h&Z L 

Substituting V, in Eq. 9.73 into Eq. 9.71, we have 

so that 

t^ ( i. hiah« z & \ j 

\ 1 + h^LjJ 

h ■ 1 + fc M Z £ 







.yu + yn 

y22 + yi2< Q) f Vi 


-< o 



FIG. 9.1*. Two-port equivalent circuit with one controlled-current source. 

Network anal/sis II 271 

FIG. 9.17. Hybrid equivalent circuit. 

We can easily check to see that Eq. 9.75 is dimensionally correct since A u 
has the dimensions of impedance, h 22 is an admittance, and h 12 , h tl are 
dimensionless since they represent voltage and current ratios, respectively. 


In this section we will consider various interconnections of two-ports. 
We will see that when a pair of two-ports are cascaded, the overall trans- 
mission matrix is equal to the product of the individual transmission 
matrices of the two-ports. When two two-ports are connected in series, 
their z matrices add ; when they are connected in parallel, their y matrices 
add. First let us consider the case in which we connect a pair of two- 
ports N a and N h in cascade or in tandem, as shown in Fig. 9.18. We see 


The transmission matrix equation for N a is 

k ara 



z i 









N tt 



N b 

v 2 





FIG. 9.18. Cascade connection of two ports. 

272 Network analysis and synthesis 









V 2 Vi' 

< 9 


V 2 ' 



FIG. 9.19. Gyrator in tandem with T circuit. 

Correspondingly, for N„ we have 

Substituting the second matrix into the first, we obtain 

Ma-eara <- 

We see that the transmission matrix of the overall two-port network is 
simply the product of the transmission matrices of the individual two- 

As an example, let us calculate the overall transmission matrix of a 
gyrator 7 in tandem with a T network shown in Fig. 9.19. The ideal 
gyrator is an impedance inversion device whose input impedance Z ta is 
related to its load impedance Z L by 

Zm = a*Y L 

= ■21 


The constant a in Eq. 9.80 is defined as the gyration resistance. If we 
regard the gyrator as a two-port, its denning equations are 

V x = a(-/«) 

h - - v t 


' B. D. H. Tellegen, "The Gyrator, a New Electric Network Element," Phillips 
Research Repts., 3 (April 1948), 81-101, see also Huelsman, op. eit., pp. 140-148. 

Network analysis II 
so that the transmission matrix of the gyrator is 

"0 a 



.a J 

We see that for the gyrator 

AD-BC<=-1 (9.82) 

Therefore the gyrator is a nonreciprocal device, although it is passive. 8 
The overall transmission matrix of the configuration in Fig. 9.19 is 
obtained by the product of the individual transmission matrices 

A B 

C D 



z a + z » V* + V. + V. 

z « + *» V* + Vc + V. 

az h 



If we check the configuration in Fig. 9.19 for reciprocity, we see that for 
transmission matrix in Eq. 9.83 

AD-BC=-\ (9.84) 

We thus see that any reciprocal network connected in tandem with a 
gyrator yields a configuration that is nonreciprocal. 

Next consider the situation in which a pair of two-ports N a and 
N b are connected in parallel, as shown in Fig. 9.20. Let us find the 

FIG. 9.20. Parallel connection of two ports. 

* Most gyrators are microwave devices that depend upon the Hall effect in ferrites. 


Network anal/sis and synthesis 


FIG. 9.21 

y parameters for the overall two-port network. The matrix equations 
for the individual two-ports are 





















From Fig. 9.20, we see that the following equations must hold. 

V\ ~ V\a = V\h V* = ^2o = '26 

h = Ao + Aft A = Ao + I» 

In connecting two-ports in series or in parallel, we must be careful 
that the individual character of a two-port network is not altered when 
connected in series or parallel with another two-port. For example, when 
we connect the two-ports in Fig. 9.21 in parallel, the impedances Z, and 
Zg will be short circuited. Therefore, to insure that a two-port network 
does not interfere with the internal affairs of the other, ideal transformers 
are used to provide the necessary isolation. In matrix notation, the sum 
of the individual [/J matrices of the two-ports in parallel must equal the 
[/] matrix of the overall two-port network. Thus we have 








Network anal/sis II 275 





* + 

3 e> V 2a 






N b 


C + 



FIG. 9.22. Series connection of two ports. 

so that the y parameters of the overall two-port network can be expressed 
in terms of the y parameters of the individual two-ports as 


|yii y« 1 _ [vna + Vub yita + Vin 
Ly« y»J lytia + Vtu y*ta + ytu>- 

If we connect two-ports in series, as shown in Fig. 9.22, we can express 
the z parameters of the overall two-port network in terms of the z param- 
eters of the individual two-ports as 

pil ^t] _ pUa + «U» Zl*. + 2 m "l 
LZ»i 2tJ Lzaia + Zflb z Mo + z *i J 


Z 21ft Z Mo + 222&-1 

We may summarize by the following three points. 


1. When two-ports are connected in parallel, find the y parameters 
first, and, from the y parameters, derive the other two-port parameters. 

2. When two-ports are connected in series, it is usually easiest to find 
the z parameters. 

3. When two-ports are connected in tandem, the transmission matrix 
is generally easier to obtain. 

As a final example, let us find the y parameters of the bridged-T circuit 
in Fig. 9.6. We see that the bridged-T circuit would be decomposed into 
a parallel connection of two-ports, as shown in Fig. 9.23. Our task is to 
first find the y parameters of the two-ports N a and N b . The y parameters 
of N b are obtained by inspection and are 

ym> = y*u, = — i 

Viiii = y*n = s 


N a is a T circuit so that the z parameters can be obtained by inspection. 

276 Network analysis and synthesis 

These are 

z llo — 2 Mo — 

S + l 

ZlSn — Z«]„ 1 

*12o — *Mo 

We then find the y parameters from the equations 

Az„ 2s + 1 

2 12a 

yi2a = y«io = t - 



2s + 1 

Since both JV and N b are symmetrical two-ports, we know that y lt — y at 
for the overall bridged-T circuit. The y parameters for the bridged-T 
circuit are now obtained as 

3/ii = Vila + Viu 

= s(s + 1) 1 = 2s* + 4s + 1 

2s + 1 

tflt — Vl2a + Vltb 

sf_ _ 1 

2s + 1 2 

2(2s + 1) 

2s* + 2s + 1 
2(2s + l) 



As we have seen, the system function H(s) of an R-L-C network consists 
of a ratio of polynomials whose coefficients are functions of the resistances, 
inductances, and capacitances of the network. We have considered, up 
to this point, that the inductors and capacitors are dissipationless; i.e., 

Network analysis II 277 

there are no parasitic resistances associated with the L and C. Since, at 
high frequencies, parasitic losses do play an important role in governing 
system performance, we must account for this incidental dissipation 
somehow. An effective way of accounting for parasitic resistances is to 
"load down" the pure inductances and capacitances with incidental 
dissipation by associating a resistance r, in series with every inductor L { , 
and, for every capacitor C t , we associate a resistor whose admittance is g t , 
as depicted in Fig. 9.24. Suppose we call the system function of the 
network without parasitic dissipation (Fig. 9.24a) H(s), and the system 
function of the "loaded" network H^s) (Fig. 9.246). Let us consider the 
the relationship between H(s) and H^s) when the dissipation is uniform, 
i.e., in a manner such that 

L t C t 


where the constant a is real and positive. 

When a network has uniform dissipation or is uniformly loaded, the 
sum of impedances in any mesh of the unloaded network 

becomes, after loading, 

m if <= R 4 + sL { + — 


m'a = JR, + sL t + r, + 

= R { + L&i + a) + 


C 4 (s + a) 






Hi(s) * H(s + a) 
(a) (b) 

FIG. 9.24. (a) Original network. (A) Loaded network. 

278 Network analysis and synthesis 





FIG. 9.25 


Similarly, on node basis, if the admittance between any two nodes of the 
original (unloaded) network is 

n if = G t + sC t + 1 

then the same node admittance after loading is 


»'„ = G t + sC t + g ( + 


= G t + Q(s + a) + 



L,(s + a) 

Since any system function can be obtained through mesh or node equa- 
tions, it is readily seen that the original system function H(s) becomes 
H(s + «) after the network has been uniformly loaded, that is /^(s) = 
H(s + a). 

Consider the following example. Let us first find the y parameters for 
the unloaded network in Fig. 9.25. By inspection we have 

1 , s , s 1 , 7s 

2 3 4 2 12 

Vu = 1 + — + - - 1 + — 7— 
2s 4 4s 


Via = »»i=-- 

Then, for a loading constant a = J, the loaded network is shown in Fig. 
9.26. In parallel with the capacitor C x = J f, we have an admittance 

gl = *C 1 = $mho (9.101) 

In parallel with the capacitor C t = J f, the associated admittance is 

gt — *C a = imho (9.102) 

Network anal/sis II 279 




< 1Q 




* 6Q< 


1 1 

► 10 


3 2h 

— o 

FIG. 9.26. Loaded network a = J. 
In series with the inductor L = 2 h, we have a resistor 

r, = olL = 1 Q (9.103) 

Now we determine the y parameters for the loaded network to be 
> 1 . s , 1 , s . 1 

= i + Ks + i) + K* + i) 
- 1 + A(* + t) 

y'« = i + 

= i + 

2a- +1 4 8 

(a + D* + 2 

4(s + |) 

'-- (H~*H) 


We see that the y parameters of the loaded network could have been 
obtained from the y parameters of the unloaded network by the relation- 
ship H t (s) = H(s + a). 

We will make use of the uniform loading concept to prove an important 
theorem concerning network realizability in Chapter 10. 


In this section we will consider a simple method of obtaining the 
network functions of a ladder network in a single operation.* This 
method depends only upon relationships that exist between the branch 

* F. F. Kuo and G. H. Leichner, "An Iterative Method for Determining Ladder 
Network Functions," Proc. IRE, 47, No. 10 (Oct. 1959), 1782-1783. 

280 Network analysis and synthesis 


> p 

'i * 2 I * * ^ ^ * 4 


FIG. 9.27. Ladder network. 

currents and node voltages of the ladder. Consider the network shown 
in Fig. 9.27, where all the series branches are given as impedances and 
all the parallel branches are given as admittances. If the v denote node 
voltages and i denote branch currents, then the following relationships 

v i+1 = i l+i Z i+s + v i+> 

These equations form the basis of the method we discuss here. 

To illustrate this method, consider the network in Fig. 9.28, for which 
the following node voltage and branch current relationships apply. 

/•(*) - yJA v^s) 

VJL») - '*(*) U') + v&) = [1 + Z^*) Y t (s)] v&) 

h(s) - Y 2 (s) VJLs) + Us) (9.106) 

= {Y&)[1 + ZJs) Us)] + Y,{s)} VAs) 
V 1 (s) = I 1 (s)Z 1 (s)+V a (s) 

= {zjlyai + z,yj + rj + (i+ z,n)} v&) 

Upon examining the set of equations given in Eq. 9.106, we see that 
each equation depends upon the two previous equations only. The first 
equation, V t = V % is, of course, unnecessary. But, as we shall see later, 
it is helpful as a starting point. In writing these equations, we begin at 

l h 




z 3 


Y 2 



v 2 


FIG. 9.28 

Network analysis II 281 

the 2-2' port of the ladder and work towards the 1-1' port. Each suc- 
ceeding equation takes into account one new immittance. We see, 
further, that with the exception of the first two equations, each subsequent 
equation is obtained by multiplying the equation just preceding it by the 
immittance that is next down the line, and then adding, to this product 
the equation twice preceding it. For example, we see that /,(» is obtained 
by multiplying the preceding equation V&) by the admittance Y t (s). 
The next immittance is ZJis). We obtain K B (i) by multiplying the previous 
equation IJs) by ZJs) to obtain /gZjj then we add to this product the 
equation twice preceding it, V^s), to obtain V a = I t Z, + V t . The process 
is then easily mechanized according to the following rules : (1) alternate 
writing node voltage and branch current equations; (2) the next equation 
is obtained by multiplying the present equation by the next immittance 
(as we work from one port to the other), and adding to this product the 
results of the previous equation. 

Using this set of equations, we obtain the input impedance Z {n (s) by 
dividing the equation for V^s) by the equation for I^s). We obtain the 
voltage-ratio transfer function V£s)IVi(s) by dividing the first equation 
P*C0 by the last equation V-fe). We obtain other network functions such 
as transfer immittances and current ratios in similar manner. Note that 
every equation contains V t (s) as a factor. In taking ratios of these equa- 
tions, the VJs) term is canceled. Therefore, our analysis can be simplified 
if we let K 2 (j) = 1. 

If the first equation of the set were a current variable fj(s) instead of the 
voltage Vi(s), the subsequent equations would contain the current variable 
I £s) as a factor, which we could also normalize to 7/s) = 1. An example 
in which the first equation is a current rather than a voltage equation 
may be seen by determining the y parameters of a two-port network. 

Before we embark upon some numerical examples, it is important to 
note that we must represent the series branches as impedances and the 
shunt branches as admittances. Suppose the series branch consisted of a 
resistor R = 1 £2 in parallel with a capacitor C = J f . Then the impedance 
of the branch is 

z (s ) = -^L = _i- (9 . 107 ) 

llsC + R s + 2 v ' 

and must be considered as a single entity in writing the equations for the 
ladder. Similarly, if a shunt branch consists of a resistor R 1 = 2Q and 
an inductor L^ = 1 h, then the admittance of the branch is 

y(s) = 7T~ ( 9 - 108 > 

2 + s 

282 Network analysis and synthesis 

The key point in this discussion is that we must use the total impedance 
or admittance of a branch in writing the equations for the ladder. 
Example 9.1. Let us find the voltage ratio VJV lt the current ratio Ijl u the 
input impedance Z x = VJI lt and the transfer impedance Z n = VJI-i for the 
network in Fig. 9.29. First, we must represent the series branches as impedances 
and the shunt branches as admittances, as shown in Fig. 9.30. The branch 
current and node voltage equations for the network are 





-/■fflnp-*— , 




— *- 

2f=t= V 2 


FIG. 9.29 


2«i V, 

FIG. 9.30 

V^s) = 1 

Us) = 2s 

V a (s) = 3s(2s) + 1 = 6s* + 1 

I t (s) = -(6s* + 1) + 2s = 14s + - 
s s 


{ 14g+ \) 


+ 6s 2 + l=6s* + 57+- l 


The various network functions are then obtained. 





7 — — — 


6 ^+57^+8 
14s 3 +2s 

6s» + 57s 8 + 8 

li 14s* + 2 
** ~ / x " 14s 2 + 2 


Example 9.2. Find the short-circuit admittance functions y n and j/ 2 i for the 
network in Fig. 9.31. To obtain the short-circuit functions, we must short 
circuit the 2-2' port. Then we represent the series branches as impedances and 
the shunt branches as admittances so that the resulting network is given as is 

Network analysis II 283 

l h 



»lh =j=2f 


V. I. 


FIG. 9.31 

illustrated in Fig. 9.32. The pertinent equations are 
/, = 1 V a =2 
J a = K2) + 1 = * + 1 


2$* + l/5\ 5/2 

We now obtain our short-circuit functions as 

12 2 1 

7 J + 5 + -,' 


yn ~^~~5 






° 's.aa 


«» + » 


2» 2 + l 






FIG. 9.32 

A number of other contributions on ladder networks have appeared 
in the recent literature. To name but a few, there are the works of 
Bashkow, 10 O'Meara, 11 Bubnicki," Walker," and Dutta Roy. 14 

10 T. R. Bashkow, "A Note on Ladder Network Analysis," IRE Trans, on Circuit 
Theory, CT-8, (June 1961), 168. 

11 T. R. O'Meara, "Generating Arrays for Ladder Network Transfer Functions," 
IEEE Trans, on Circuit Theory, CT-10, (June, 1963), 285. 

11 Z. Bubnicki, "Input Impedance and Transfer Function of a Ladder Network," 
IEEE Trans, on Circuit Theory, CT-10, (June, 1963), 286. 

"F. Walker, "The Topological Analysis of Non-Recurrent Ladder Networks," 
Proc. IEEE, 52, No. 7, (July, 1964), 860. 

14 S. C. Dutta Roy, "Formulas for the Terminal Impedances and Transfer Functions 
of General Multimesh Ladder Networks," Proc. IEEE, 52, No. 6, (June, 1964), 738. 

284 Network analysis and synthesis 

Because of the recursive nature of the equations involved, the digital 
computer is an ideal tool for the analysis of ladder networks. In Chapter 
15 a detailed description is given of a digital computer program based 
upon the algorithm described in this section for evaluating ladder network 


9.1 Find the z, y, h, and T parameters of the networks shown in the figures. 
Some of the parameters may not be defined for particular circuit configurations. 


— PS — 


o ' 

-OQ— > 


PROB. 9.1 


9.2 Find the z, y, h, and T parameters for the networks shown in the figures, 
lo 1 — iA/W i < \AAA/ o2 

Network analysis II 285 

93 Find the z parameters for the lattice and bridge circuits in the figures 
shown. (The results should be identical.) 

9.4 For the lattice and bridge circuits in Prob. 9.3 find the y parameters in 
terms of the admittances Y a = 1/Z„ and Y b =• MZ b . 

93 For the circuit shown, find the voltage-ratio transfer function VJV X and 
the input impedance Z x = V^l x in terms of the 2 parameters of the two-port 
network N and the load resistor R L . 



I2 2 






v 2 







286 Network anal/sis and synthesis 

9.6 For the cascade connection of two-ports depicted in the figure, show that 
the transfer impedance Zj 8 of the overall circuit is given in terms of the a param- 
eters of the individual two-ports by the equation 

In addition, show that the short-circuit admittance y ri is given by 


V-a = - 

Vu.b + Vita 






N a 

N b 

v 2 


PROB. 9.6 

PROB. 9.7 

9.7 Find the z and y parameters of the transformer (nonideal) shown in the 
figure. Determine the T- and w-equivalent circuits for the transformer in terms 
of L x , L 2 , and M. {Hint: Use the z parameters for the T-equivalent circuit, 
and the y parameters for the w-equivalent circuit.) 

9.8 The inverse hybrid parameters of a two-port network are defined by the 





Express the g parameters in terms of either the z or y parameters and give a 
physical interpretation of the meaning of these parameters; i.e., say whether a 
parameter is an open- or short-circuit parameter, and whether it is a driving 
point or transfer function. Finally, derive the conditions of reciprocity for the 
g parameters. 

9.9 Prove that for a passive reciprocal network 

AD - BC = 1 
where A, B, C, Dare the elements of the transmission matrix. 

9.10 Find the z and h parameters of the common emitter transistor repre- 
sented by its T-circuit model. 

-VA — i — Wv-^O 

r mh J 2 

. + 


V 2 

PROB. 9.10 

Network anal/sis II 287 

9.11 The circuit in part (a) of the figure is to be described by an equivalent 
input circuit shown in part (b). Determine Z eq in (b) as a function of the elements 
and voltages in (a). 


9.12 Find the T parameters of the configurations shown in parts (a) and (b) 
of the figure. 

9.13 Find the y parameters of the twin-T circuit in Prob. 9.2c by considering 
the circuit to be made up of two T circuits in parallel. 

9.14 Find the z parameters of the circuits shown. 

288 Network anal/sis and synthesis 

1 h 

o >— 



V Ideal 


1 h 








V 2 


h 2 
< o 

V 2 

Ideal 2' 


PROB. 9.14 


PROB. 9.15 

PROB. 9.16 

Network analysis II 289 

9.15 Find the z parameters of the circuit shown. 

9.16 Find the z parameters of the circuit shown. 

9.17 For the circuit in Prob. 9.26 determine the y parameters of the uniformly 
loaded circuit derived from the original circuit with the dissipation a =0.1. 
Plot the poles and zeros of both cases. 

9.18 Find the transfer impedance VJI X for the circuit in part (a) and the 
voltage ratio VJV 1 for the circuit in part (b). Plot the poles and zeros for the 
transfer functions obtained. 


o— >- 

/Tnnn^-r-^voTP . — 






10 V 2 

2 — WW- 







:ia v 2 

PROB. 9.18 

9.19 Find the short-circuit parameters for the ladder network utilizing the 
method in Section 9.6. 


HWv— i 

20 _lh 

Hwv— i-'Tnnp-^ 



PROB. 9.1* 









V 2 

PROB. 9.20 

9.20 Determine the voltage ratio VJV lt the current ratio IJI lt the transfer 
impedance VJI lt and the driving point impedance Fi// a for the network shown. 

chapter 10 

Elements <|>f realizability 



In the preceding chapters we have beep primarily concerned with the 
problem of determining the response, giver^ the excitation and the network; 
this problem lies in the domain of network analysis. In the next five 
chapters we will be dealing with the problem of synthesizing a network 
given the excitation E(s) and the response R(s). The starting point for 
any synthesis problem is the system function 



Our task is to synthesize a network from a given system function. 

The first step in a synthesis procedur^ is to determine whether H(s) 
can be realized as a physical passive network. There are two important 
considerations — causality and stability. By causality we mean that a 
voltage cannot appear between any pair of terminals in the network 
before a current is impressed, or vice ver^a. In other words, the impulse 
response of the network must be zero for t < 0, that is, 

h(t) = 
As an example, the impulse response 


h(t) = e- 

is causal, whereas 

h(t) = e- " 





Elements of realizability theory 291 


(a) (b) 

FIG. 10.1. (a) Nonrealizable impulse response, (b) Realizable impulse response. 

is not causal. In certain cases, the impulse response could be made 
realizable (causal) by delaying it appropriately. For example, the impulse 
response in Fig. 10.1a is not realizable. If we delay the response by T 
seconds, we find that the delayed response h{t — T) is realizable (Fig. 

In the frequency domain, causality is implied when the Paley-Wiener 
criterion 1 is satisfied for the amplitude function \H(jw)\. The Paley-Wiener 
criterion states that a necessary and sufficient condition for an amplitude 
function \H(ja>)\ to be realizable (causal) is that 


|log \H(ja>)\ 
1 +o> a 

dco < oo 


The following conditions must be satisfied before the Paley-Wiener 
criterion is valid: h(t) must possess a Fourier transform H(ja>); the 
square magnitude function \H(jw)\* must be integrable, that is, 


\H(jco)\ *dco < oo 


The physical implication of the Paley-Wiener criterion is that the amplitude 
\H(jco)\ of a realizable network must not be zero over a finite band of 
frequencies. Another way of looking at the Paley-Wiener criterion is that 
the amplitude function cannot fall off to zero faster than exponential 
order. For example, the ideal low-pass filter in Fig. 10.2 is not realizable 
because beyond m c the amplitude is zero. The Gaussian shaped curve 

|HO)l = e~° 
shown in Fig. 10.3 is not realizable because 

|log|tf0a>)|| = o* 



1 R. E. A. C. Paley and N. Wiener, "Fourier Transforms in the Complex Domain," 
Am. Math. Soc. Colloq. Pub., 19 (1934), 16-17. 


Network analysis and synthesis 


FIG. 10.2. Ideal filter char- 

FIG. 10.3. Gaussian filter character- 

so that the integral 

f" «> , 
i dea 

J-* 1 + eof 

is not finite. On the other hand, the amplitude function 







does represent a realizable network. Ih fact, the voltage-ratio transfer 
function of the R-C network in Fig. 10.4 has an amplitude characteristic 
given by \H(Jm)\ in Eq. 10.9. 
For the ideal filter in Fig. 10.2, the inverse transform A(f) has the form 

h{t) = 

A sin <o c t 


where A is a constant. From the sin x 
h(t) is nonzero for t less than zero. In 
it must be delayed by an infinite 
delay h(i) by a large but finite amount t d 
of h(t — t^ is less than a very small 

\Kt ~ Q\ < e 

— wwv 


fx curve in Fig. 3J4, we see that 

fact, in order to make h(t) causal, 

amount. In practice, however, if we 

such that for t < the magnitude 

quantity e, that is, 





. 10.4 

v w 

Elements of realizability theory 293 

we then can approximate h(t — t£ by a causal response h^t) which is 

zero for t < 0. (For a more detailed discussion of the Paley-Wiener 

criterion, the reader is referred to an excellent treatment by Wallman.*) 

If a network is stable, then for a bounded excitation e(t) the response 

(0 is also bounded. In other words, if 

KOKCi 0^r<oo 
then KOI < C* 0<.t<co 

where C x and C, are real, positive, finite quantities. If a linear system is 
stable, then from the convolution integral we obtain 

IKOI < CiJ"|*(t)I dr < C, (10.11) 

Equation 10.1 1 requires that the impulse response be absolutely integrable, 


""|/i(t)| dr < oo (10.12) 


One important requirement for h(t) to be absolutely integrable is that the 
impulse response approach zero as t approaches infinity, that is, 



Generally, it can be said that with the exception of isolated impulses, the 
impulse response must be bounded for all t, that is, 

|A(0I<C all* (10.13) 

where C is a real, positive, finite number. 

Observe that our definition of stability precludes such terms as sin ay 
from the impulse response because sin a> f is not absolutely integrable. 
These undamped sinusoidal terms are associated with simple poles on 
the yw axis. Since pure L-C networks have system functions with simple 
poles on the jm axis, and since we do not wish to call these networks 
unstable, we say that a system is marginally stable if its impulse response 
is bounded according to Eq. 10.13, but does not approach zero as t 
approaches infinity. 

In the frequency domain, the stability criterion requires that the system 

• G. E. Valley Jr. and H. Wallman, Vacuum Tube Amplifiers, McGraw-Hill Book 
Company, New York, 1948, Appendix A, pp. 721-727. 

294 Network anal/sis and synthesis 


function possess poles in the left-half pi 
over, the poles on theyVo axis must be sini] 
ment of simple poles on they'w axis, if 


or on theyw axis only. More- 
pie. 3 As a result of the require- 
s) is given as 

H(s) = 

a„s n + a n _ 1 s n ~ 1 + ■ ■ 

• + fl x s + a 

b m s m + b m _ lS m ~ l +!■ 

• • + bis + b 


then the order of the numerator n cannot exceed the order of the de- 
nominator m by more than unity, that is, n — m <£ 1. If n exceeded m 
by more than unity, this would imply that at s =jo> = oo, and there 
would be a multiple pole. To summarise, in order for a network to be 
stable, the following three conditions oii its system function H(s) must 
be satisfied: 

1. H(s) cannot have poles in the right-fealf plane. 

2. H(s) cannot have multiple poles in tjhey'w axis. 

3. The degree of the numerator of H(s) cannot exceed the degree of 
the denominator by more than unity. 

Finally, it should be pointed out that a rational function H(s) with poles 
in the left-half plane only has an inverse transform h(t), which is zero for 
t < 0.* In this respect, stability implies causality. Since system funct'ons 
of passive linear networks with lumped elements are rational functions 
with poles in the left-half plane or jco axis only, causality ceases to be a 
problem when we deal with system functions of this type. We are only 
concerned with the problem of causality when we have to design a filter 
for a given amplitude characteristic such as the ideal filter in Fig. 10.2. 
We know we could never hope to realize exactly a filter of this type 
because the impulse response would not^ be causal. To this extent the 
Paley- Wiener criterion is helpful in denning the limits of our capability. 


In Section 10.1 we saw that in order fot a system function to be stable, 
its poles must be restricted to the left-half plane or theyeo axis. Moreover, 
the poles on theyco axis must be simple. The denominator polynomial of 
the system function belongs to a class of] polynomials known as Hurwitz 

* In Chapter 6 it was shown that multiple poles on they'co 
i t t sin wot. 

axis gave rise to terms as 

4 G. Raisbeck, "A Definition of Passive Linear Networks in Terms of Time and 
Energy," /. Appl. Phys., 25 (Dec., 1954), 15ip-1514. The proof follows straight- 
forwardly from the properties of the Laplace transform. 

Elements of realizability theory 295 

polynomials. A polynomial P(s) is said to be Hurwitz if the following 
conditions are satisfied: 

1. P(s) is real when s is real. 

2. The roots of P(s) have real parts which are zero or negative. 

As a result of these conditions, if P(s) is a Hurwitz polynomial given by 

P(s) = a n s n + a^s"- 1 +-- + a 1 s + a (10.15) 

then all the coefficients a { must be real ; if s { = <x 4 + j@ is a root of P(s), 
then ac, must be negative. The polynomial 

P(s) = (s+ 1)(j + 1 +js/2)(s + 1 -js/2) (10.16) 

is Hurwitz because all of its roots have negative real parts. On the other 

G(s) = (s - l)(.y + 2)(j + 3) (10.17) 

is not Hurwitz because of the root s = 1, which has a positive real part. 
Hurwitz polynomials have the following properties: 

1. All the coefficients a { are nonnegative. This is readily seen by 
examining the three types of roots that a Hurwitz polynomial might 
have. These are 

s = — y< y t real and positive 

s = ±.jm t (Of real 

s — —cti±.jPi a-i real and positive 

The polynomial P(s) which contains these roots can be written as 

P(s) = (s + y< Xji + <»*)[(* + oO* + m • ■ • (10.18) 

Since P(s) is the product of terms with only positive coefficients, it follows 
that the coefficients of P(s) must be positive. A corollary is that between 
the highest order term in s and the lowest order term, none of the coeffi- 
cients may be zero unless the polynomial is even of odd. In other words, 
fln-i. a«-2> • • ■ » fl»» <*i must not be zero if the polynomial is neither even 
nor odd. This is readily seen because the absence of a term a { implies 
cancellation brought about by a root s — y< with a positive real part. 

2. Both the odd and even parts of a Hurwitz polynomial P(s) have 
roots on the ja> axis only. If we denote the odd part of P(s) as n(s) and 
the even part as m(s), so that 

P(s) = n(s) + m(j) (10.19) 

296 Network analysis and synthesis 

then m(s) and n(s) both have roots on tne jw axis only. The reader is 
referred to a proof of this property by Guillemin. 8 

3. As a result of property 2, if P(s) is either even or odd, all its roots are 
on they'co axis. 

4. The continued fraction expansion of the ratio of the odd to even 
parts or the even to odd parts of a Hurwitz polynomial yields all positive 
quotient terms. Suppose we denote the ratios as y>(s) — n(s)lm(s) or 
y(j) = m(s)ln(s), then the continued fraction expansion of y>(s) can be 
written as 

y(s) = q x s + ^ (10.20) 

««« + 

«ss + 

where the quotients q lt q%,..-,q n must be positive if the polynomial 
P(s) = «(j) + m(s) is Hurwitz. 6 To obtain the continued fraction expan- 
sion, we must perform a series of long divisions. Suppose y>(s) is 

V<s) = 2® (10.21) 


where m(s) is of one higher degree than n(s). Then if we divide n(s) into 
m(s), we obtain a single quotient and a remainder 


The degree of the term R^s) is one lower than the degree of «(*). Therefore 
if we invert the remainder term and divide, we have 

»(°L =q . + M2l (io.23) 

Inverting and dividing again, we obtain 

RM^qj + M) (10.24) 

*,(*) *.(«) 

* E. Guillemin, The Mathematics of Circuit Analysis, John Wiley and Sons, New York, 
1949. An excellent treatment of Hurwitz polynomials is given here. 

* A proof can be undertaken in connection with L-C driving-point functions; see 
M. E. Van Valkenburg, Introduction to Modern Network Synthesis, John Wiley and 
Sons, New York, 1960. 

Elements of realteability theory 297 

We see that the process of obtaining the continued fraction expansion 
of ip(s) simply involves division and inversion. At each step we obtain a 
quotient term q<s and a remainder term, R {+1 (s)/RJ[s). We then invert the 
remainder term and divide R^s) into RJs) to obtain a new quotient. 
There is a theorem in the theory of continued fraotions which states that 
the continued fraction expansion of the even to odd or odd to even parts 
of a polynomial must be finite in length. 7 Another theorem states that, if 
the continued fraction expansion of the odd to even or even to odd parts 
of a polynomial yields positive quotient terms, then the polynomial must 
be Hurwitz to within a multiplicative factor JV(s).* That is, if we write 

fTO-irO^M (10.25) 

then F(s) is Hurwitz, if W(s) and F^s) are Hurwitz. For example, let us 
test whether the polynomial 

F(s) - i* + s* + 5s* + 3s + 4 (10.26) 

is Hurwitz. The even and odd parts of F(s) are 

n(s) = s* + 3s 

We now perform a continued fraction expansion of yfa) = "»W/«C0 by 
dividing n(s) by m(s), and then inverting and dividing again, as given by 

the operation 

s* + 3s)s* + 5s* + 4(s 
s* + 3s* 

2s* + 4)j» + 3j(*/2 
j» + 2* 

s)2s* + 4(2s 


so that the continued fraction expansion of y(s) is 

*,) = 2& - 5 + ^ (10.28) 

»(«) 1 + l 


' See Van Valkenburg, be. eit. 

1 W(s) is a common factor in mfc) and n(*). 

298 Network analysis and synthesis 

Since all the quotient terms of the continued fraction expansion are 
positive, F(s) is Hurwitz. 

Example 10.1. Let us test whether the polynomial 

GO) = 5 s + 2s 9 + 3s + 6 (10.29) 

is Hurwitz. The continued fraction expansion of n(s)lm(s) is obtained from the 

2s 2 + 6)5* + 3* (s/2 
s* + 3s 

We see that the division has been terminated abruptly by a common factor 
5 s + 3s. The polynomial can then be written as 

G(s) = (s* + 3s) 1 + - (10.30) 


We know that the term 1 + 2/s is Hurwitz. Since the multiplicative factor 
s* + 3s is also Hurwitz, then G(s) is Hurwitz. The term s* + 3s is the mul- 
tiplicative factor fV(s), which we referred to earlier. 

Example 10.2. Next consider a case where W(s) is non-Hurwitz. 

F(s) = 5 7 + 2s* + 2s 5 + s* + 4s 3 + 85* + 8* + 4 (10.31) 

The continued fraction expansion of F(s) is now obtained. 
«(*) •* 1 

wiO) 2 ' 1 

fr + j^t) (1 °- 32) 


We thus see that fV(s) = s* + 4, which can be factored into 

W(s) = (s* + 2s + 2X* 8 -2^+2) (10.33) 

It is clear that F(s) is not Hurwitz. 
Example 103. Let us consider a more obvious non-Hurwitz polynomial 

F(s) = 5* + 5 s + 2t* + 35 + 2 (10.34) 

Elements of realizability theory 299 
The continued fraction expansion is 

j 3 + 3s)S* + 2s* + 2 (s 
s* + 3s* 

-j*+2)j» + 3*(-,j 




We see that F(j) is not Hurwitz because of the negative quotients. 

Example 10.4. Consider the case where F(s) is an odd or even function. It is 
impossible to perform a continued fraction expansion on the function as it 
stands. However, we can test the ratio of F(s) to its derivative, F'(s)* If the 
ratio F(s)IF'(s) gives a continued fraction expansion with all positive coefficients, 
then F(s) is Hurwitz. For example, if F(s) is given as 

F(s) = s 7 + 34 s + 2s 3 + * (10.35) 

then F'(s) is F'(s) = 7s* + 15** + 6s» + 1 (10.36) 

Without going into the details, it can be shown that the continued fraction 
expansion of F(s)IF'(s) does not yield all positive quotients. Therefore F(s) is 
not Hurwitz. 


In this section we will study the properties of a class of functions 
known as positive real functions. These functions are important because 
they represent physically realizable passive driving-point immittances. A 
function F(s) is positive real (p.r.) if the following conditions are satisfied: 

1. F(s) is real for real s; that is, F(a) is real. 

2. The real part of F(s) is greater than or equal to zero when the real 
part of s is greater than or equal to zero, that is, 

Re [F(s)] ^ for Re s ^ 

Let us consider a complex plane interpretation of a p.r. function. 
Consider the s plane and the F(s) plane in Fig. 10.5. If F(s) is p.r., then a 
point <r on the positive real axis of the s plane would correspond to, or 
map onto, a point F(p^ which must be on the positive real axis of the 
F(s) plane. In addition, a point s t in the right half of the s plane would 

* See Guillemin, he. cit. 

300 Network analysis and synthesis 


• plane 

F(t) plane 




F(ao) Ref 

FIG. 10.5. Mapping of s plane onto F(s) plane. 

map onto a point Ffo) in the right half of the F(s) plane. In other words, 
for a positive real function, the right half of the s plane maps onto the 
right half of the F(s) plane. The real axis of the s plane maps onto the 
real axis of the F(s) plane. 

A further restriction we will impose is that F(s) be rational. Consider 
the following examples of p.r. functions: 

1. F(s) = Ls (where L is a real, positive number) is p.r. by definition. 
If F(s) is an impedance function, then L is an inductance. 

2. F(s) = R (where R is real and positive) is p.r. by definition. If F(s) 
is an impedance function, R is a resistance. 

3. F(s) = K/s (K real and positive) is p.r. because, when s is real, F(s) 
is real. In addition, when the real part of s is greater than zero, Re (s) = 





o* + 




Therefore, F(s) is p.r. If F(s) is an impedance function, then the corre- 
sponding element is a capacitor of I IK farads. 

We thus see that the basic passive impedances are p.r. functions. 
Similarly, it is clear that the admittances 


are positive real if K is real and positive. We now show that all driving- 
point immittances of passive networks must be p.r. The proof depends 
upon the following assertion: for a sinusoidal input, the average power 
dissipated by a passive network is nonnegative. For the passive network 

Y(s) = 


7(5) = 


Y(s) = 






Elements of realizability theory 301 

in Fig. 10.6, the average power dissipated by the network is 

Average power = I Re [ZJtjco)] |/|» ^ (10.39) 

We then conclude that, for any passive network 

Re [Zjytt))] ^ (10.40) 

We can now prove that for Re s = a ^ 0, Re Z^o- + jm) ;> 0. Consider 
the network in Fig. 10.6, whose driving- 
point impedance is ZJ^s). Let us load 
the network with incidental dissipation 
such that if the driving-point impedance 
of the uniformly loaded network is 
ZxCO, then 

Zfc) = ZJis + a) (10.41) no. 10.6 

where a, the dissipation constant, is real and positive. Since Z^s) is the 
impedance of a passive network, 

Re Z^jco) ^ (10.42) 

so that Re ZJa + jm) ^ (10.43) 

Since a is an arbitrary real positive quantity, it can be taken to be a. 
Thus the theorem is proved. 

Next let us consider some useful properties of p.r. functions. The 
proofs of these properties are given in Appendix D. 

1. If F(s) is p.r., then l/F(s) is also p.r. This property implies that if a 
driving-point impedance is p.r., then its reciprocal, the driving-point 
admittance, is also p.r. 

2. The sum of p.r. functions is p.r. From an impedance standpoint, 
we see that if two impedances are connected in series, the sum of the 
impedances is p.r. An analogous situation holds for two admittances in 
parallel. Note that the difference of two p.r. functions is not necessarily 
p.r. ; for example, F(s) = s — 1/s is not p.r. 

3. The poles and zeros of a p.r. function cannot have positive real 
parts, i.e., they cannot be in the right half of the s plane. 

4. Only simple poles with real positive residues can exist on the yco axis. 

5. The poles and zeros of a p.r. function are real or occur in conjugate 
pairs. We know that the poles and zeros of a network function are 
functions of the elements in the network. Since the elements themselves 
are real, there cannot be complex poles or zeros without conjugates 
because this would imply imaginary elements. 

302 Network analysis and synthesis 

6. The highest powers of the numerator and denominator polynomials 
may differ at most by unity. This condition prohibits multiple poles and 
zeros at 5 = oo. 

7. The lowest powers of the denominator and numerator polynomials 
may differ by at most unity. This condition prevents the possibility of 
multiple poles or zeros at s = 0. 

8. The necessary and sufficient conditions for a rational function with 
real coefficients F(s) to be p.r. are 

(a) F(s) must have no poles in the right-half plane. 

(b) F(s) may have only simple poles on theyco axis with real and positive 

(c) Re F(jm) ^ for all a. 

Let us compare this new definition with the original one which requires 
the two conditions. 

1. F(s) is real when s is real. 

2. Re F(s) ^ 0, when Re s ^ 0. 

In order to test condition 2 of the original definition, we must test 
every single point in the right-half plane. In the alternate definition, 
condition (c) merely requires that we test the behavior of F(s) along the 
jm axis. It is apparent that testing a function for the three conditions 
given by the alternate definition represents a considerable saving of 
effort, except in simple cases as F(s) = 1/j. 

Let us examine the implications of each criterion of the second definition. 
Condition (a) requires that we test the denominator of F(s) for roots in 
the right-half plane, i.e., we must determine whether the denominator of 
F(s) is Hurwitz. This is readily accomplished through a continued fraction 
expansion of the odd to even or even to odd parts of the denominator. 
The second requirement — condition (b) — is tested by making a partial 
fraction expansion of F(s) and checking whether the residues of the poles 
on theya) axis are positive and real. Thus, if F(s) has a pair of poles at 
s = ±ja> lt a partial fraction expansion gives terms of the form shown. 

S — jCOi s + jco^ 

The residues of complex conjugate poles are themselves conjugates. If 
the residues are real — as they must be in order for F(s) to be p.r.— then 
K t = Ki* so that 

S — JO) 1 S + JCOx s* + (O* 

Elements of reaiizability theory 303 

If K x is found to be positive, then F(s) "satisfies the second of the three 

In order to test for the third condition for positive realness, we must 
first find the real part of F(ja>) from the original function F(s). To do this, 
let us consider a function F(s) given as a quotient of two polynomials 

F(s) = ^ (10.45) 


We can separate the even parts from the odd parts of P(s) and Q(s) so 
that F(s) is 

M 8 ( S ) + JV 2 (s) 

where Mj(s) is an even function and N£s) is an odd function. F(s) is now 
decomposed into its even and odd parts by multiplying both P(s) and 
Q(s) by A/g - N s so that 

r(s) _ M i + N iM*- N * 
M t + JV a M a - N t 

= M t M t - N t N t M t N x - Mj N t 

M 8 8 - N t * M* - N t * 

We see that the products M x M z and JV, N t are even functions, while 
M x N t and M s N x are odd functions. Therefore, the even part of F(s) is 

Ev[F( s) ] = M ^-^ (10.48) 

and the odd part of F(s) is 

Odd [F(s)] = ^VY' (10.49) 

If we let s = joy, we see that the even part of any polynomial is real, 
while the odd part of the polynomial is imaginary, so that if F(jw) is 
written as 

F(ja>) = Re [F(jo>)] + j Im [F(jo>)] (10.50) 

it is clear that Re [F(ja>)] = Ev [F(s)] \_ ja (10.51) 

and j Im [F(ja>)] = Odd [F(s)] \^, a (10.52) 

Therefore, to test for the third condition for positive realness, we 
determine the real part of F(ja>) by finding the even part of F(s) and then 
letting s = joy. We then check to see whether Re F{joy) ^ for all <w. 

304 Network anal/sis and synthesis 


root ' 

FIG. 10.7 FIG. 10.8 

The denominator of Re F(j(o) is always a positive quantity because 

M a (ja>)* - JVaO) 2 = M^to)* + N 2 ((o)* ^ (10.53) 

That is, there is an extra j or imaginary term in N^jco), which, when 
squared, gives —1, so that the denominator of ReF(ja>) is the sum of 
two squared numbers and is always positive. Therefore, our task resolves 
into the problem of determining whether 

A(a>) 4 M x O) MJjai) - Ntito) NJJm) £ (10.54) 

If we call the preceding function A{m), we see that A(a>) must not have 
positive, real roots of the type shown in Fig. 10.7; i.e., A(co) must never 
have single, real roots of to. However, A(a>) may have double roots 
(Fig. 10.8), because A(t») need not become negative in this case. 

As an example, consider the requirements for 

F(s) - 

s + a 


s* + bs + c 

to be p.r. First, we know that, in order for the poles and zeros to be in 
the left-half plane or on they'd) axis, the coefficients a, b, c must be greater 
or equal to zero. Second, if * = 0, then F(s) will possess poles on the 
jco axis. We can then write F(s) as 

3 • ° (10.56) 

F(s) - 


S* + C ' 5* + C 

We will show later that the coefficient a must also be zero when b = 0. 
Let us proceed with the third requirement, namely, Re F(ja>) ^ 0. From 
the equation 

M^o)) Mfim) - JVxO) NAjo>) £ (10.57) 

we have a(-(o* + c) + 1x0*^0 (10.58a) 

which simplifies to A(co) = (b — a)afi + ac ;> (10.586) 

It is evident that in order to prevent A(m) from having positive real roots 
of eu, b must be greater than or equal to a, that is, b^a. As a result, 

Elements of readability theory 305 

when b = 0, then a = 0. To summarize, the conditions that must be 
fulfilled in order for F(s) to be positive real are 

1. a, b, c ^ 0. 

2. b ^ a. 

We see that F x (s) = , s + \ , (10.59) 

sr + 3s + 2 

is p.r., while the functions 

F «< S) = TT1 (ia60) 

s + 2 

are not p.r. As a second example, let us determine the conditions for the 
biquadratic function 

F(s) = TT^TT < ia62 > 

s" + b^ + b 

to be p.r. We will assume that the coefficients a u Aq, b u b are all real, 
positive constants. Let us test whether F(s) is p.r. by testing each require- 
ment of the second definition. 

First, if the coefficients of the denominator * x and b are positive, the 
denominator must be Hurwitz. Second, if b x is positive, we have no poles 
on the yeo axis. Therefore we can ignore the second condition. 

The third condition can be checked by first finding the even part of 
F(s), which is 

(s + O ) — t»i s 


_ s* + [(a„ + ft„) - fliftjs' + a &. 
(«* + bof - ft x V 
The real part of F(j(o) is then 

Re [F o»1 - W< - ««" t ! 0) ~ °:y 8 + flA dO.64) 

( — CO + O )" + t»i to 1 

We see that the denominator of Re [F(jca)] is truly always positive 
so it remains for us to determine whether the numerator of Re [F(jco)] 
ever goes negative. Factoring the numerator, we obtain 

m , t = (a. + b ) - a 1 b 1 ± I ^ [(flo + feo) _ aAf _ 4flA (1Q65) 

306 Network analysis and synthesis 

There are two situations in which Re [F(jco)] does not have a simple real 

1 . When the quantity under the radical sign of Eq. 10.65 is zero (double, 
real root) or negative (complex roots). In other words, 

K«o + *o) - «AP - 4aA <. (10.66) 

or [(fl + b ) - aAP <, 4a<A> ( 10 - 67 ) 

If (a + K) - cA ^ (10.68) 

then (a + b ) - «A <, 2y/a^ (10.69) 

or *A ^ (V^ - yfb )* (10.70) 

If (a, + ^-aA<0 (10.71) 

then aA - (a + b ) <, 2VaA (10.72) 

but fa, + * ) - aA < < aA - (a + b ) (10.73) 

so again aA ^ (\Za — V*o) 2 (10.74) 

2. The second situation in which Re [F(jco)] does not have a simple 
real root is when eo* g in Eq. 10.65 is negative so that the roots are imagi- 
nary. This situation occurs when 

[0*o + *o) - aAl 8 - 4aA > (10.75) 

and (a + b ) - dA < (10.76) 
From Eq. 10.75 we have 

«A - 0*o + *o) > 2VoA > («o + b ) - aA (10.77) 

Thus aA > (V^ - VV) 8 (10.78) 

We thus see that Eq. 10.70 is a necessary and sufficient condition for a 
biquadratic function to be positive real. If we have 

aA = (V^o~-V6o> 00.79) 

F (s) = a T ! T ~ (10-80) 

then we will have double zeros for Re [F(jeo)] 
Consider the following example: 

s g + 2s + 25 

s a + 5s + 16 
We see that 

aA = 2 X 5 ^ (s/a - V^ ) 2 = (V25 - Vl6 )* (10.81) 
so that F(s) is p.r. 

F & = -rr7 + - (io.83) 

Elements of realizability theory 307 

The examples just given are, of course, special cases. But they do 
illustrate the procedure by which functions are tested for the p.r. property. 
Let us consider a number of other helpful points by which a function 
might be tested quickly. First, if F(s) has poles on they«w axis, a partial 
fraction expansion will show if the residues of these poles are positive 
and real. For example, 

3s 8 + 5 

F(S) = TTTT, < ia82 > 

s(s* + 1) 
has a pair of poles at s = ±j\. The partial fraction expansion of F(s), 

-2s 5 
s 2 + l s 

shows that the residue of the poles at s = ±j is negative. Therefore F(s) 
is not p.r. 

Since impedances and admittances of passive time-invariant networks 
are p.r. functions, we can make use of our knowledge of impedances 
connected in series or parallel in our testing for the p.r. property. For 
example, if Z^s) and Z^s) are passive impedances, then z\ connected in 
parallel with Z, gives an overall impedance 

Z(s) = i77^ < 1084 > 

Zi(s) + Z^s) 
Since the connecting of the two impedances in parallel has not affected 
the passivity of the network, we know that Z(s) must also be p.r. We see 
that if Fjis) and FJs) are p.r. functions, then 

m = JMM. (1085) 

must also be p.r. Consequently, the functions 

F(s) = -£*- (10.86) 

s + a 

and F(s) = — — (10.87) 

s + a 

where a and K are real and positive quantities, must be p.r. We then 
observe that functions of the type 

s + a 


s + a s + a 
must be p.r. also. 


308 Network analysis and synthesis 
Finally, let us determine whether 

jr( s) = _*L_ o,K£0 (10.89) 

s + a 
is p.r. If we write F(s) as 

F(s) = ^— (10.90) 

V s/K + a/Ks 

we see that the terms s/K and ctfKs are p.r. Therefore, the sum of the 
two terms must be p.r. Since the reciprocal of a p.r. function is also p.r., 
we conclude that F(s) is p.r. 


The basic philosophy behind the synthesis of driving-point functions 
is to break up a p.r. function Z(j) into a sum of simpler p.r. functions 
Ziis), Z4s), .... ZJs), and then to synthesize these individual Z,(j) as 
elements of the overall network whose driving-point impedance is Z(j). 

Zis) = Us) + Z£s) + --- + Z n (s) (10.91) 

First, consider the "breaking-up" process of the function Z(s) into the 
sum of functions Z£s). One important restriction is that all Zj(s) must be 
p.r. Certainly, if all Z 4 (s) were given to us, we could synthesize a network 
whose driving-point impedance is Z(s) by simply connecting all the ZJs) 
in series. However, if we were to start with Z(s) alone, how would we 
decompose Z(s) to give us the individual Z/j)? Suppose Z(«) is given in 
general as 

7( * = fl - s " + ""-i* 1 "" 1 + ••• + «»' + «■ . Jfi> (1( > 92) 
W fe m s" + t^s"- 1 + • • • + hs + b a Q(s) 

Consider the case where Z(s) has a pole at s = (that is, b = 0). Let us 
divide P(s) by Q(s) to give a quotient D/s and a remainder R(s), which 
we can denote as Z 1 (^) and Z a (s). 

Z(s) = - + R(s) D £ 

= Z^s) + Z&) (10.93) 

Are Z! and Z, p.r. ? From previous discussions, we know that Z^ = D/s 
is p.r. IsZa(5)p.r.? Consider the p.r. criteria given previously. 

1. ZgCO must have no poles in the right-half plane. 

2. Poles of ZJLs) on the imaginary axis must be simple, and their 
residues must be real and positive. 

3. Re [ZJJa>)] ^ for all m. 

Elements of realizabiiity theory 309 

Let us examine these cases one by one. Criterion 1 is satisfied because 
the poles of ZJ&) are also poles of Us). Criterion 2 is satisfied by this 
same argument. A simple partial fraction expansion does not affect the 
residues of the other poles. When s —jo>, Re [Z(y«w) = D/jco] = 0. 
Therefore we have 

Re ZAjco) = Re Zjja>) ^ (10.94) 

From the foregoing discussion, it is seen that if Us) has a pole at s = 0, 
a- partial fraction expansion can be made such that one of the terms is of 
the form Kfs and the other terms combined still remain p.r. 

A similar argument shows that if U s ) has a pole at s = oo (that is, 
n — m = 1), we can divide the numerator by the denominator to give a 
quotient Ls and a remainder term R(s), again denoted as Z^s) and Zt(s). 

Us) = Ls + R(s) = Z^s) + Z^s). 


Here ZJs) is also p.r. If ZXjs) has a pair of conjugate imaginary poles on 
the imaginary axis, for example, poles at s = ±jto 1 , then Z{s) can be 
expanded into partial fractions so that 



Re (_2*M =Re (_i^L_\ =0 

so that Zj(.y) is p.r. 

Finally, if Re [Z(/<u)] is minimum at some point m^ and if the value of 
Re ZijcDf) = Kf as shown in Fig. 10.9, we can remove a constant K<, K t 
from Re [Z(ya>)] so that the remainder is still p.r. This is because Re [Z(/o>)] 
will still be greater than or equal to zero for all values of w. 

Suppose we have a p.r. function Z(s), which is a driving-point im- 
pedance function. Let Z(s) be decomposed, as before, so that 

Z(s) = 74s) + ZJs) 



310 Network anal/sis and synthesis 


Z(s) ^ 

ZzM > 

FIG. 10.10 

where both 2^ and Zj are p.r. Now let us "remove" ZjCs) from Z(,s) to 
give us a remainder Z^s). This removal process is illustrated in Fig. 
10.10 and shows that removal corresponds to synthesis of ZjO). 

Example 10.5. Consider the following p.r. function 

s* +2s +6 

Z(s) = 


s(s + 3) 

We see that Z(s) has a pole at s = 0. A partial fraction expansion of Z(s) yields 

2 j 

*(*) = - + 

* j +3 

-2&)+2fr) (10.100) 

If we remove Z t (s) from Z(s), we obtain Z 2 (s), which can be shown by a resistor 
in parallel with an inductor, as illustrated in Fig. 10.11. 

* f r— 

o 1( 



■ 1Q 


FIG. 10.11 

Example 10.6 


7s +2 


2j +4 
where ]%?) is a p.r. function. 

Let us synthesize the network by first removing min [Re Y(jco)]. The real 
part of Y(jco) can be easily obtained as 

8 + 14o>* 


We see that the minimum of Re [Y(ja>)] occurs at to = 0, and is equal to 
min [Re YQw)] =\. Let us then remove Y x =\ mho from Y{s) and denote 

Elements of realizability theory 3 1 1 

the remainder as YJs), as shown in Fig. 10.12. The remainder function Y^s) is 
p.r. because we have removed only the minimum real part of YJjm). Y^s) 

is obtained as 

1 3s 


It is readily seen that YJp) is made up of a J-ii resistor in series with a f-farad 
capacitor. Thus the final network is that shown in Fig. 10.13. 


Y(s) " 


] Y *' } , 

FIG. 10.12 

^20 " 


YM > 



FIG. 10.13 

Example 10.7. Consider the p.r. impedance 

fo 8 + 3** + 3s + 1 


6s* +3s 


The real part of the function is a constant, equal to unity. Removing a constant 
of 1 il, we obtain (Fig. 10.14) 

3^ + 1 

Z 1 (s)=Z(s)-l = 

6s* + 3s 

The reciprocal of Z t (s) is an admittance 

no) - 

fit 8 +3s 
3s* + 1 



which has a pole at s = <x>. This pole is removed by finding the partial fraction 
expansion of Yjfs); 

r 1 ( i )=25+ 5? 4n (10107) 


o — WWW — 

Z(s) „ Zi(s) 

FIG. 10.14 

312 Network analysis and synthesis 

o — WWW 


FIG. 10.15 

and then by removing the term with the pole at * = oo to give a capacitor of 
2 farads in parallel with Y£s) below (Fig. 10.15). YJs) is now obtained as 

Y&) - Tito - 25 - ^--j-j 

The reciprocal of YJs) is 

Zjfc) = 3* + • 




which is, clearly, an inductor of 3 h in series with a capacitor of 1 farad. The 
final network is shown in Fig. 10.16. 

o www- 





FIG. 10.16 

These examples are, of course, special cases of the driving-point syn- 
thesis problem. However, they do illustrate the basic techniques involved. 
In the next chapter, we will discuss the problem of synthesizing a network 
with two kinds of elements, either L-C, R-C, or R-L networks. The syn- 
thesis techniques involved, however, will be the same. 

10.1 Test the following polynomials for the Hurwitz property. 


s 8 +5* +2s + 2 


5* +5* +5 + 1 


S 1 + 5 s + 5* + 5 


5» + 45* + 55 + 2 


5 s +25» +5 


5 7 + 25* + 2/ + 5 

Elements of readability theory 3 1 3 

10.2 Determine whether the following functions are p.r. For the functions 
with the denominator already factored, perform a partial fraction expansion 

(a) F(») = ** + ! 

(&) F(s) - 

(c) F(s) = 

W F(s) = 

(«) F(s) - 

5» +4* 

2s* +2s +4 

(s + IX* 8 + 2) 

(s + 2X« + 4) 

(* + 1X5 + 3) 

.*» +4 

«* + 3«* + 3* + 1 

5** +s 

** + l 

103 Suppose Fj(5> and F t (s) are both p.r. Discuss the conditions such that 
F(s) = FjKs) - FJ.3) is also p.r. 

10.4 Show that the product of two p.r. functions need not be p.r. Also show 
that the ratio of one p.r. function to another may not be p.r. (Give one example 
of each.) 

10.5 GivenZfr)- * + *\ - 

(a) What are the restrictions on X for Z(s) to be a p.r. function? 
(A) Find A* for Re [ZQot)] to have a second-order zero at co = 0. 

(c) Choose a numerical value for X and synthesize Z(s). 

10.6 Prove that if Z t (s) and Z t (s) are both p.r., 

ZUi - WW® 
must also be positive real. 

10.7 Z(») — -j— — — — is p.r. Determine min [Re Z(/o>)] and synthesize 
Z(s) by first removing min [ReZ(/co)]. 

10.8 Perform a continued fraction expansion on the ratio 

5» + 2j* + 3j + 1 

W = S 8 + 5 s + 2* + 1 

What does the continued expansion imply if Y(s) is the driving-point admittance 
of a passive network ? Draw the network from the continued fraction. 

3 1 4 Network analysis and synthesis 

10.9 The following functions are impedance functions. Synthesize the 
impedances by successive removals of/« axis poles or by removing min [Re (/»)]. 

s 8 + 4s 

(«) 7*T2 

j + i 

(*) s(* + 2) 

2? +4 
< c > 27+1 

s* + 3s + 1 
W s» + l 

chapter 1 1 

Synthesis of one-port networks 
with two kinds of elements 

In this chapter we will study methods for synthesizing one-port net- 
works with two kinds of elements. Since we have three elements to choose 
from, the networks to be synthesized are either R-C, R-L, or L-C networks. 
We will proceed according to the following plan. First we will discuss the 
properties of a particular type of one-port network, and then we will 
synthesize it. Let us first examine some properties of L-C driving-point 


Consider the impedance Z(j) of a passive one-port network. Let us 
represent Z(s) as 

„, . M^s) + JVj(s) 

Z(S) = i^ ±^i /J J j-j 

M a (s) + N&) (U1) 

where M lt M t are even parts of the numerator and denominator, and 
N x , N s are odd parts. The average power dissipated by the one-port is 

Average power = J Re [Z(jd)] \I\* (1 \ .2) 

where J is the input current. For a pure reactive network, it is known that 
the power dissipated is zero. We therefore conclude that the real part of 
Z(/to)iszero; that is 

Re Z(» = Ev 2ija>) = (11.3) 

where Ev Z(s) = tf '('W)-^')^) n , ,, 

M t \s)-N t \s) (1L4) 


316 Network anal/sis and synthesis 
In order for Ev Kjio) = 0, that is, 

MJlJm) M 8 0'o>) ~ W») fUja>) = (11.5) 

either of the following cases must hold: 

(a) M x = - N t 

(b) M t = « N t 
In case (a), Z(i) is 


Z(s) = ^ (H.7) 

and in case (A) Z(s) - ^ (H-8) 

We see from this development the following two properties of L-C 

1. Zjj^s) or YxcC*) is tne ratio of even to odd or odd to even P 013 '" 

2. Since both Mis) and N£s) are Hurwitz, they have only imaginary 
roots, and it follows that the poles and zeros of Z^s) or Y^s) are on 
the imaginary axis. 

Consider the example of an L-C immittance function given by 

a,*' + a* 1 + a, (u 9) 

Let us examine the constraints on the coefficients a { and *,. We know, 
first of all, that in order for the impedance to be positive real, the coeffi- 
cients must be real and positive. We also know that an impedance 
function cannot have multiple poles or zeros on theyw axis. Since qo is 
defined to be on theyw axis, the highest powers of the numerator and the 
denominator polynomials can differ by, at most, unity. For example, if 
the highest order of the numerator is 2n, then the highest order of the 
denominator can either be 2n - 1 so that there is a simple pole at s = oo, 
or the order can be 2n + 1 so that there is a simple zero at s = oo. 
Similarly, the lowest orders of numerator and denominator can differ by 
at most unity, or else there would be multiple poles or zeros of Z(s) at 

s = 0. 

Another property of the numerator and denominator polynomials is 
that if the highest power of the polynomial is In, for example, the next 
highest order term must be In - 2, and the succeeding powers must 
differ by two orders all the way through. There cannot be any missing 

Synthesis of one-port networks 317 
terms, i.e., no two adjacent terms of either polynomial may differ by more 
than two powers. For example, for Z(s) given in Eq. 11.9, if ft, = 0, 
then Z(s) will have poles when 

V + *i* = (11.10) 

so that the poles will be at * = and at 

* - (^V "*" * = 0,1,2,3 dm) 

It is clearly seen that none of the poles s k are even on theyto axis, thus 
violating one of the basic properties of an L-C immittance function. 

From the properties given in Eq. 11.11, we can write a general L-C 
impedance or admittance as 

Z( S ) = Jft' + 'Ofr' + « *')•••(»' + «**)••• 

tf + coW + <»*)■ .•(,» + «,,«)... < 1112 > 

Expanding Z(j) into partial fractions, we obtain 

2(.) = S + -2p-j + -*&- + . . . + KkS (1L13) 

where the K t are the residues of the poles. Since these poles are all on 
theyco axis, the residues must be real and positive in order for Z*» to be 
positive real. Letting s =jm, we see that Z(;o>) has zero real part, and 
can thus be written as a pure reactance jX(o>). Thus we have 

\ ft) ft), — ft) 1 / 

= J x (<°) (11.14) 

Differentiating X(a>) with respect to a>, we have 

dco <o* + K <° + («,,»-cV + (1U5) 

Since all the residues K t are positive, it is seen that for an L-C function, 

dX(io) . . 

IT* < 1116 > 

A similar development shows that the derivative of Im [Y{jw)] = B(m) 
is also positive, that is, 

3 1 8 Network anal/sis and synthesis 
Consider the following example. Z(s) is given as 

™ _ **»* + <o a *) 
^'(s' + co^ + to*) 


Kco(-co a + cog 2 ) 


Letting s = jco, we obtain X{a>) as 

zo«») = ; x(co) = +j ^—^—^ <} 

Let us draw a curve of X(co) versus o>. Beginning with the zero at co = 0, 
let us examine the sequence of critical frequencies encountered as <o 
increases. Since the slope of the X(io) curve is always positive, the next 
critical frequency we encounter is when X(co) becomes infinitely large or 
the pole is at a> 2 . As we pass a> 2 , X(co) changes sign and goes from + 
to -. In general, whenever we pass through any critical frequency, 
there is always a change of sign, as seen from the wayy X(to) is written in 
the last equation. After we pass through a> 2 , with the slope of X(a>) 
always positive, it is easy to see that the next critical frequency is the zero 
at «,. Thus, if an impedance function is an L-C immittance, the poles 
and zeros of the function must alternate. The particular X(a>) under 
discussion takes the form shown in Fig. 11.1. Since the highest powers 
of the numerator and the denominator always differ by unity, and the 
lowest powers also differ by one, we observe that at s = and at s = oo, 
there is always a critical frequency, whether a zero or a pole. 

For the example just discussed, there is a zero at s = and a zero at 
s = oo. The critical frequencies at s = and s = oo are called external 
critical frequencies, whereas the remaining finite critical frequencies are 
referred to as internal. Thus, in the previous example, cog, a> 3 , and to t 
are internal critical frequencies. 

Finally, let us summarize the properties of L-C impedance or admittance 

1. Z iC (s) or Y LC &s) is the ratio of odd to even or even to odd poly- 



Synthesis of one-port networks 3 1 9 


+1 +2 +3 


2. The poles and zeros are simple and lie on theyeo axis. 

3. The poles and zeros interlace on they'eo axis. 

4. The highest powers of numerator and denominator must differ by 
unity; the lowest powers also differ by unity. 

5. There must be either a zero or a pole at the origin and infinity. 

The following functions are not L-C for the reasons listed at the left. 

Ks(s 2 + 4) 




Z(s) = 

(s 2 + l)(s a + 3) 
s 5 + 4s 3 + 5s 
3s* + 6s 2 


z( = K(s 2 + l)(s* + 9) 
(s 2 + 2)(s 2 + 10) 

On the other hand, the function Z(j) in Eq. 11.21, whose pole-zero 
diagram is shown in Fig. 11.2, is an L-C immittance. 

_ 2(s » + 1)(s » + 9) 
s(s* + 4) 



We saw in Section 11.1 that an L-C immittance is a positive real function 
with poles and zeros on the jco axis only. The partial fraction expansion 
of an L-C function is expressed in general terms as 

s s 2 + cog 

+ K x s 


The synthesis is accomplished directly from the partial fraction expansion 
by associating the individual terms in the expansion with network elements. 
If F(s) is an impedance Z(s), then the term I^/s represents a capacitor of 
l/Ko farads; the term K x s is an inductance of K„ henrys, and the term 

320 Network anal/sis and synthesis 

1 , 

2X, 1 

tf.h Wi f 

o— ^HTHf— 


2X1 . 


■— l0Q0>— ' 

2«! h 
7T5" n 

FIG. 11.3 

2^5/(5 2 + o) 4 2 ) is a parallel tank circuit that consists of a capacitor of 
\f2Kf farads in parallel with an inductance of IKjcof. Thus a partial 
fraction expansion of a general L-C impedance would yield the network 
shown in Fig. 11.3. For example, consider the following L-C function. 

_,. 2(s 8 + l)(s* + 9) 
Z(S) *- + 4) 

A partial fraction expansion of TUjs) gives 


Z(s) = 2s + * + -72- 



We then obtain the synthesized network in Fig. 11.4. 

The partial fraction expansion method is based upon the elementary 
synthesis procedure of removing poles on the jco axis. The advantage 
with L-C functions is that all the poles of the function lie on the jco axis 
so that we can remove all the poles simultaneously. Suppose F(s) in 
Eq. 11.22 is an admittance Y(s). Then the partial fraction expansion of 
Y(s) gives us a circuit consisting of parallel branches shown in Fig. 11.5. 
For example, 

s( S « + 2)(s» + 4) 
W (s 2 + l)(s* + 3) 





FIG. 11.4 

Synthesis of one-port networks 321 



FIG. 11.5 

The partial fraction expansion of Y(s) is 

r(s) = s + 



s 8 + 3 s* + 1 


from which we synthesize the network shown in Fig. 11.6. The L-C 
networks synthesized by partial fraction expansions are sometimes called 
Foster-type networks. 1 The impedance form is sometimes called a Foster 
series network and the admittance form is a Foster parallel network. 
A useful property of L-C immittances is that the numerator and the 
denominator always differ in degree by unity. Therefore, there is always 
a zero or a pole at $ = oo. Suppose we consider the case of an L-C 
impedance Z{s), whose numerator is of degree In and denominator 
is of degree 2n — I, giving Z(j) a pole at s = oo. We can remove this 
pole by removing an impedance L^ so that the remainder function Z^s) 
is still L-C: 

ZjCO = ZCO - LiS (11.27) 

The degree of the denominator of Z 2 (j) is 2n — 1, but the numerator is of 
degree 2/i — 2, because the numerator and denominator must differ in 
degree by 1. Therefore, we see that Z 8 (s) has a zero at s = oo. If we 
invert Z t (s) to give Y t (s) = l/Z^s), Y s (s) will have a pole at s = oo, 
which we can again remove to give a capacitor C*s and a remainder 
Y a (s), which is 

Y&) = Y 2 (s) - C 2 s. (11.28) 





FIG. 11.6 

1 a. M. Foster, "A Reactance Theorem," Bell System Tech. J., No. 3 (1924), 259-267. 

322 Network analysis and synthesis 

Ll £3 



FIG. 1 1.7 

FIG. 11.8 

We readily see that F 3 (s) has a zero at s = oo, which we can invert and 
remove. This process continues until the remainder is zero. Each time 
we remove a pole, we remove an inductor or a capacitor depending upon 
whether the function is an impedance or an admittance. Note that the 
final structure of the network synthesized is a ladder whose series arms 
are inductors and whose shunt arms are capacitors, as shown in Fig. 1 1 .7. 
Consider the following example. 

s + 4s 1 + 3 

We see that Z(s) has a pole at s = oo, which we can remove by first 
dividing the denominator into the numerator to give a quotient 2s and a 
remainder Z,(j), as shown in Fig. 11.8. Then we have 

Z t (s) - Z(s) - 2s = - 

4s s + 10s 

s 4 + 4s a + 3 


Observe that Z^s) has a zero at s = oo. Inverting Z^s), we again remove 
the pole at infinity. Then we realize a capacitor of J farad and a remainder 
y 3 (s), as may be seen in Fig. 11.9. 

Y 3 (s)=Y£s)-±s= *f + 3 
4 4s s + 10s 


Removing the pole at s = oo of Z^j) = llY s (s), gives a series inductor 
of | h and 

Z t (s) - Z 3 (s) -~s = FTT1 
3 f s + 3 



o — r Ginr* — i 


Y 3 (s) 

FIG. 1 1.9 

Synthesis of one-port networks 323 




Z 4> 

FIG. 11.10 

as shown in Fig. 11.10. The admittance y 4 (s) = 1/Z 4 (j) has a pole at 
s = oo, which we remove to give a capacitor of f farad and a remainder 
y 6 (j) = 3/2y, which represents an inductor of f h. Removing this in- 
ductor gives us zero remainder. Our synthesis is therefore complete and 
the final network is shown in Fig. 11.11. 

Since we always remove a pole at s = oo by inverting the remainder 
and dividing, we conclude that we can synthesize an L-C ladder network 
by a continued fraction expansion. The quotients represent the poles at 
s = oo, which we remove, and we invert the remainder successively until 
the remainder is zero. For the previous example, the continued fraction 
expansion is 

s* + 4s 2 + 3)25° + 12s 3 + 16s(2s<--> Z 
2s 5 + fo 3 + 6s 

4s 3 + I0s)s* + 4s 2 + 3(ts <-> Y 

s* + 



+ 3)4s 3 + I0s(is<->Z 

4s 3 + is 


3(|$ <-> Y 

2 s 

3)2y(fj*-> Z 

We see that the quotients of the continued fraction expansion give the 
elements of the ladder network. Because the continued fraction expansion 





-'innp— i 


FIG. 11.11 

324 Network anal/sis an synthesis 

always inverts each remainder and divides, the successive quotients alter- 
nate between Z and Y and then Z again, as shown in the preceding 
expansion. If the initial function is an impedance, the first quotient must 
necessarily be an impedance. When the first function is an admittance, 
the first quotient is an admittance. 

Since the lowest degrees of numerator and denominator of an L-C 
admittance must differ by unity, it follows that there must be a zero or a 
pole at s = 0. If we follow the same procedure we have just outlined, 
and remove successively poles at s = 0, we will have an alternate real- 
ization in a ladder structure. To do this by continued fractions, we 
arrange both numerator and denominator in ascending order and divide 
the lowest power of the denominator into the lowest power of the numer- 
ator; then we invert the remainder and divide again. For example, in 
the case of the impedance we have 

z( S ) = i * z ;r j » (H.33) 

(s* + l)(s* + 3) 
s(s* + 2) 
The continued fraction expansion to give the alternate realization is 

2s + 58)3 + As* + s\3I2s <-> Z 
3 + fs 2 

f s* + s*)2s + ^(4/5$ *-> Y 
2s + $s* 

isP^s* + 5*(25/2s <-> Z 

|j 2 


The final synthesized network is shown in Fig. 11.12. The ladder networks 
realized are called Cauer ladder networks because W. Cauer 2 discovered 
the continued fraction method for synthesis of a passive network. 

Note that for both the Foster and the 

o 1( 1 |( Cauer-form realizations, the number of 

§ f L ^ f L elements is one greater than the number of 

5 h oj 5 h oj internal critical frequencies, which we defined 

*p *p previously as being all the poles and zeros 

[ [ of the function, excluding those at s — and 

FIG. 1 1.12 s = 00. Without going into the proof of the 

•Wilhelm Cauer, "The Realization of Impedances with Prescribed Frequency 
Dependence," Arch. Electrotech., 15 (1926), 355-388. 

Synthesis of one-port networks 325 

statement, it can be said that both the Foster and the Cauer forms give 
the minimum number of elements for a specified L-C driving-point 
function. These realizations are sometimes known as canonical forms. 


The properties of R-C driving-point impedances can be derived from 
known properties of L-C functions by a process of mapping the ja> axis 
onto the — a axis. 8 We will not resort to this formalism here. Instead, 
we will assume that all driving-point functions that can be realized with 
two kinds of elements can be realized in a Foster form. Based upon this 
assumption we can derive all the pertinent properties of R-C or R-L 
driving-point functions. Let us consider first the properties of R-C 
driving-point impedance functions. 

Referring to the series Foster form for an L-C impedance given in 
Fig. 11.3, we can obtain a Foster realization of an R-C impedance by 
simply replacing all the inductances by resistances so that a general R-C 
impedance could be represented as in Fig. 11.13. The R-C impedance, 
as seen from Fig. 11.13, is 

Z(s) - ^ + K„ + -&_ + -&- + 

s s + a x s + cr g 


where C = l/A^, R a = K„, C x = \\K X , R x = Kja lf and so on. In 
order for Eq. 11.34 to represent an R-C driving-point impedance, the 
constants K\ and a t must be positive and real. From this development, 
two major properties of R-C impedances are obtained, and are listed in 
the following. 






R 2 

FIG. 11.13 

* M . E. Van Valkenburg, Introduction to Modern Network Synthesis, John Wiley and 
Sons, New York, 1960, pp. 140-145. 

326 Network anal/sis and synthesis 

1. The poles of an R-C driving-point impedance are on the negative 
real (—or) axis. It can be shown from a parallel Foster form that the 
poles of an R-C admittance function are also on the axis. We can thus 
conclude that the zeros of an R-C impedance are also on the —a axis. 

2. The residues of the poles, K it are real and positive. We shall see later 
that this property does not apply to R-C admittances. 

Since the poles and zeros of R-C impedances are on the — a axis, let 
us examine the slope of Z(<r) along the — a axis." To find the slope, 
dZ{a)lda, we first let s = a in Z(s), and then we take the derivative of 
Z(c) with respect to a. Thus we have 

Z(<r) = £ + K. + -&- + _&_ + . . • (11.35) 

a a + o-j a + a % 

and — — = ■? H 1 1 ? 1- • • • (n 36} 

da a* * (a + atf + (a + o % ? + ^^ 

It is clear that ^^^0 (11-37) 


Let us now look at the behavior of Z(s) at the two points where the 
real axis and the imaginary axis intersect, namely, at a = <a — and at 
a = (o = oo. This is readily done by examining the general R-C network 
in Fig. 11.13 at these two frequencies. At a = 0, (d-c), if the capacitor C 
is in the circuit, it is an open circuit and there is a pole of Z(s) at a = 0. 
If C is not in the circuit, then Z(0) is simply the sum of all the resistances 
in the circuit. 

Z(0) = R t + R 2 + ■ ■ ■ + R„ (11.38) 

because all of the capacitors are open circuits at a = 0. 

At a = oo, all the capacitors are short circuits. Thus, if R^ is in the 
circuit, Z(oo) = R^. If R x is missing, then Z(oo) = 0. To summarize 
these last two statements, we have 


C present 

Z(0) = 

C„ missing 

Z(oo) = 


R n missing 
R m present 

If we examine the two cases for Z(0) and Z(oo), we see that 

Z(0) £ Z(oo) (11.39) 

Synthesis of one-port networks 327 

Next, let us see whether the poles and zeros of an R-C impedance 
function alternate. We have already established that the critical frequency 
nearest the origin must be a pole and the critical frequency nearest a = oo 
must be a zero. Therefore, if Z(s) is given as 

Z( S ) = 

(5! + g a )(s + g 4 ) 
(s + ffOO + or,) 


Then, if Z(y) is R-C, the singularity nearest the origin must be a pole 
which we will assume to be at s = — g x ; the singularity furthest from 
the origin must be a zero, which we will take to be s = — g 4 . Let us plot 

z , ff ) = (g + g a )(ff + g 4 ) 
(g + a^ia + a 3 ) 


versus — g, beginning at a = and extending to a = — oo. At a = 0, 
Z(0) is equal to a positive constant 

Z(0) = ^* 


Since the slope of Z(g) is always positive as — a increases, Z(g) must 
increase until the pole s = -a x is reached (Fig. 11.14). At a = —a u 
Z(g) changes sign, and is negative until the next critical frequency is 
reached. We see that this next critical frequency must be the zero, 
* = — <V Since Z(g) increases for increasing — a, the third critical 
frequency must be the pole s = — g 8 . Because Z(g) changes sign at — g„ 
the final critical frequency must be the zero, s = — g 4 . Beyond a = — g 4 , 
the curve becomes asymptotic to Z(oo) = 1. From this analysis we see 
that the poles and zeros of an R-C impedance must alternate so that for 


— a a* 

FIG. 11.14 

328 Network anal/sis and synthesis 
the case being considered 

oo > <r 4 > <r 8 > or g > Oi ^ 0. 

In addition, we see that 


a, ex. 


which shows that Z(0) > Z(oo). 

To summarize, the three properties we need to recognize an R-C 
impedance are: 

1. Poles and zeros lie on the negative real axis, and they alternate. 

2. The singularity nearest to (or at) the origin must be a pole whereas 
the singularity nearest to (or at) a = — oo must be a zero. 

3. The residues of the poles must be real and positive. 

An example of an R-C impedance is: 

(s + l)(s + 4)(s + 8) 
s(s + 2)(s + 6) 
The following impedances are not R-C. 

(s + l)(s + 8) 
(s + 2)(s + 4) 
(s + 2)(s + 4) 

Z(s) = 


Z(s) = 
Z(s) = 
Z(s) = 


(s + D 

(s + l)(s + 2) 

s(s + 3) 

Let us reexamine the partial fraction expansion of a general R-C 

+ -^i- + • • • (11.47) 

F(s) = ^ + K a 

S + Ot 

Instead of letting F(s) represent an impedance, consider the case where 
F(s) is an admittance Y(s). If we associate the individual terms in the 
expansion to network elements, we then obtain the network shown in 

FIG. 11.15 

Synthesis of one-port networks 329 

Fig. 11.15. We sec that an R-C impedance, Z BC (s), also can be realized 
as an R-L admittance Y^is). All the properties of R-L admittances are 
the same as the properties of R-C impedances. It is therefore important 
to specify whether a function is to be realized as an R-C impedance or an 
R-L admittance. 


We postulated in Section 11.3 that the Foster form realization exists 
for an R-C impedance or an R-L admittance. Since Foster networks are 
synthesized by partial fraction expansions, the synthesis is accomplished 
with ease. An important point to remember is that we must remote the 
minimum real part of Z(/a>) in the partial fraction expansion. It can be 
shown* that min [Re Z(/co)] = Z(oo), so that we have to remove Z(oo) 
as a resistor in the partial fraction expansion. In cases where the numer- 
ator is of lower degree than the denominator, Z(oo) = 0. When the 
numerator and the denominator are of the same degree, then Z(oo) can 
be obtained by dividing the denominator into the numerator. The 
quotient is then Z(oo). Consider the following example. 

F(s) = 

3(a + 2)(s + 4) 


s(s + 3) 
The partial fraction expansion of the remainder function is obtained as 

s s + 3 

+ 3 


where F(oo) = 3. If F(s) is an impedance Z(s), it must be an R-C im- 
pedance and it is realized in the series Foster form in Fig. 11.16. On the 
other hand, if F(s) represents an admittance, we realize Y(s) as an R-L 
network in the parallel Foster form (Fig. 11.17). 

-\[ AV 

f 30 

8 f 




FIG. 11.16 

4 Van Valkenburg, loc cil. 

330 Network analysis and synthesis 






FIG. 11.17 

An alternate method of synthesis is based on the following fact. If 
we remove min Re [Z(yto)] = Z(oo) from Z(y), we create a zero at s = oo 
for the remainder ZjO). If we invert Z x (j), we then have a pole at $ = oo, 
which we can remove to give Z a (s). Since min Re [ Y^jco)] = Y t (co), if 
we remove Y 2 (co), we would have a zero at s = oo again, which we 
again invert and remove. The process of extracting Z(oo) or r(oo) and 
the removal of a pole of the reciprocal of the remainder involve dividing 
the numerator by the denominator. Consequently, we see that the whole 
synthesis process can be resolved by a continued fraction expansion. 
The quotients represent the elements of a ladder network. 

For example, the continued fraction expansion of F(s) in Eq. 11.48 is 

j 2 + 3s)3s* + l&y + 24(3 

35*+ 9s 

9s + 24>* 

+ 3s($s 


+ f* 

is)9s + 24(27 



If F(s) is an impedance Z(j), the resulting network is shown in Fig. 
11.18. If F(s) is an admittance Y(s), we have the R-L network of Fig. 11.19. 



-Wv — i 




■ j inpr^- 1 

FIG. 11.18 

FIG. 11.19 

Synthesis of one-port networks 33 1 


The immittance that represents a series Foster R-L impedance or a 
parallel Foster R-C admittance is given as 

F(s) = K K s + K + ■-&- + • • • (11.50) 

s + a t 

The significant difference between an R-C impedance and an R-L im- 
pedance is that the partial fraction expansion term for the R-C "tank" 
circuit is Kj(s + a t ); whereas, for the R-L impedance, the corresponding 
term must be multiplied by an s in order to give an R-L tank circuit 
consisting of a resistor in parallel with an inductor. 

The properties of R-L impedance or R-C admittance functions can be 
derived in much the same manner as the properties of R-C impedance 
functions. Without going into the derivation of the properties, the more 
significant ones are given in the following: 

1 . Poles and zeros of an R-L impedance or R-C admittance are located 
on the negative real axis, and they alternate. 

2. The singularity nearest to (or at) the origin is a zero. The singularity 
nearest to (or at) s = oo must be a pole. 

3. The residues of the poles must be real and negative. 

Because of the third property, a partial fraction expansion of an R-L 
impedance function would yield terms as 

Ki (11.51) 

s + o t 

This does not present any trouble, however, because the term above does 
not represent an R-L impedance at all. To obtain the Foster form of an 
R-L impedance, we will resort to the following artifice. Let us first expand 
2i_s)js into partial fractions. If Z(s) is an R-L impedance, we will state 
without proof here that the partial fraction expansion of 7^s)js yields 
positive residues. 6 Thus, we have 

Z°>=*2 + K a> + ^-+-- (11,52) 

s s s + a { 

* Actually, Z BL {s)js has the properties of an R-C impedance; see Van Valkenburg 
he. cit. 

332 Network analysis and synthesis 

°-VW — 

Z(s) T 

ifi &»={= 



FIG. 11.20 

FIG. 11.21 

where *o, K u . . . , K„ ^ 0. If we multiply both sides by s, we obtain 
Z(s) in the desired form for synthesis. Consider the following function: 


2(s + l)(s + 3) 
(s + 2)(s + 6) 


F(s) represents an R-L impedance or an R-C admittance because it 
satisfies the first two criteria cited. The partial fraction expansion of 
F{s) is i 

F(s) = 2 - 



s + 2 s + 6 


so we see that the residues are negative. The partial fraction expansion of 
F(s)js, on the other hand, is 


F(s) _ 2(5 + IX* - 3) t t 
s s(s + 2)(s + 6) s s + 2 




s + 6 

If we multiply both sides by s, we obtain 

1 is 


s + 6 



If F(s) represents an impedance Z(s), it is synthesized in series Foster 
form, giving the R-L network in Fig. 11.20. If F(s) is an admittance Y(s), 
then the resulting network is the R-C network shown in Fig. 11.21. 

To synthesize an R-L impedance in ladder form, we make use of the 
fact that min Re [Z(Jm)] = Z(0). If we remove Z(0) from Z(s), the 
remainder function Z x (s) will have a zero at s = 0. After inverting Z^s), 

FIG. 11.22 

Synthesis of one-port networks 333 


FIG. 11.23 

we can then remove the pole at s = 0. Since the value Z(0) is obtained 
by dividing the lowest power of the denominator into the lowest power 
term of the numerator, the synthesis could be carried out by a continued 
fraction expansion by arranging the numerator and denominator poly- 
nomials in ascending order and then dividing. For example, the following 
function is either an R-C impedance or an R-C admittance. 

= 2(5 + l)(s + 3) = 6 + 8s + 2s* 
(s + 2)(s + 6) 12 + 8s + s* 
The continued fraction expansion of F(s) is 

12 + 8* + s*)6 + is + 2s*(l 

6 + 4s + js* 

4s + fs 2 )12 + Ss + s\3ls 

12 + §J 

ls + s*)4s + f**(f 

4s + £r» 

iV)& + **(49/5s 




If F(s) is an impedance function, the resulting network is the R-L network 
shown in Fig. 11.22. If, on the other hand, F(s) is an R-C admittance 
Y(s), the network is synthesized as in Fig. 11.23. 


Under certain conditions, R-L-C driving-point functions may be syn- 
thesized with the use of either partial fractions or continued fractions. 
For example, the function 

s a 4- 2s + 2 

z (')~ VL it ( 1L58) 

S ' + s + 1 

334 Network analysis and synthesis 

is neither L-C, R-C, nor R-L. Nevertheless, the function can be synthesized 
by continued fractions as shown. 

s* + s+ l>* + 2s + 2(l«-Z 
s* + s+1 

s + iy + s + i(s 

S 2 + s 

1> + l(s + \*-Z 

The network derived from this expansion is given in Fig. 11.24. 

In another case, the poles and zeros of the following admittance are all 
on the negative real axis, but they do not alternate. 

Y(s) = 

(s + 2)(s + 3) 

(s + l)(s + 4) 
The partial fraction expansion for Y(s) is 


FIG. 11.24 

Y(s) = 1 + 

s+1 s +4 


Since one of the residues is negative, we cannot use this expansion for 
synthesis. An alternate method would be to expand Y(s)ls and then 
multiply the whole expansion by s. 





s s s + 1 
When we multiply by s, we obtain, 

to-*— i- 

V 2 s + 1 

s + 4 

s + 4 



Note that Y(s) also has a negative term. If we divide the denominator of 
this negative term into the numerator, we can rid ourselves of any terms 
with negative signs. 

y(j) = 3_(l__U + J! 

2 \3 s + 1/ s + - 

6 s + 1 

s + 4 




Synthesis of one-port networks 335 



FIG. 11.25 


At 4= 

The network that is realized from the expanded function is given in Fig. 

If we try to expand Y(s) by continued fractions, we see that negative 
quotients result. However, we can expand Us) = l/l%s) by continued 
fractions, although the expansion is not as simple or straightforward as 
in the case of an R-C function, because we sometimes have to reverse the 
order of division to make the quotients all positive. The continued 
fraction expansion of Z(s) is 

6 + 5 s + s*)4 + 
4 + 

5s + s%i 

¥* + !** 


| s + is*)6 + 
6 + 

5s + 

¥* + 


i** + 



+ **(¥ 

As we see, the division process giving the quotient of 1/3 involves a 
reversal of the order of the polynomials involved. The resulting ladder 
network is given in Fig. 11.26. 

o — WV 






FIG. 11.26 

336 Network anal/sis and synthesis 

In the beginning of this section, it was stated that only under special 
conditions can an R-L-C driving-point function be synthesized with the 
use of a ladder form or the Foster forms. These conditions are not 
given here because they are rather involved. Instead, when a positive 
real function is given, and it is found that the function is not synthesizable 
by using two kinds of elements only, it is suggested that a continued 
fraction expansion or a partial fraction expansion be tried first. 


11.1 (a) Which of the following functions are L-C driving point impedances? 

_,, jQ» + 4X* 8 + 16) _„ (j» + 1X«* + 8) 

A W " (** + 9X*» + 25) ' Z * S) *fc»+4) 

(6) Synthesize the realizable impedances in a Foster and a Cauer form. 

11.2 Indicate the general /orm of the two Foster and the two Cauer networks 
that could be used to synthesize the following L-C impedance. 

7( . fr* + 1X5* + 9Xi* + 25) 
w = jCi" + 4X*« + 16) 

There is no need to calculate the element values of the four networks. 
113 Synthesize the L-C driving-point impedance 

6s* + 425* + 48 

Z(s) = 

J + 18j» + 48s 

in the form shown in the figure, i.e., determine the element values of the network 
in henrys and farads. 

=FC 8 


11.4 There exists an L-C network with the same driving-point impedance as 
the network shown in the figure. This alternate network should contain only 
two elements. Find this network. 

Synthesis of one-port networks 337 

-nnnp — i 






PROB. 11.4 

11.5 The input impedance for the network shown is 

2*» +2 
Zta= "i»+2s*+25 + 2 

If Z is an L-C network: (a) Find the expression for Z . (6) Synthesize Z in a 
Foster series form. 



11.6 Indicate which of the following functions are either R-C, R-L, or L-C 
impedance functions. 


s* +2s 

z(,) " FT4?~+1 

ZW ~ s* + 4* + 3 

„ . s 2 + 45 + 3 
Z(5)= ?T6F+8 

„, . 5* + 5s + 6 
Z(4) = «■+, 


j« + Ss* + 6 

** +3 

338 Network analysis and synthesis 

11.7 An impedance function has the pole-zero pattern shown in the figure. 
If Z(— 2) = 3, synthesize the impedance in a Foster form and a Cauer form. 


-5 -3 -1 

PROB. 11.7 

11.8 From the following functions, pick out the ones which are R-C admit- 
tances and synthesize in one Foster and one Cauer form. 

2js + l)Qr + 3) 
(s + 2X* + 4) 

sjs + 4X* + 8) 
is + lXs + 6) 

Y(s) = 
Y(s) = 

4js + l)fr + 3) 
sis +2) 

(s + l)fr + 4) 
sis +2) 

11.9 Find the networks for the following functions. Both Foster and ladder 
forms are required. 

(«). Z(s) - is + 1)( * + 4) 



s(s +2) 
3is + 1)(* + 4) 

s + 3 

11.10 For the network shown, find Y when 
K 9 1 

V„ 2+Y 

sjs* +3) 
= 2s» +s* +6s + 1 

Synthesize Y as an L-C admittance. 

PROB. 11.10 

11.11 Synthesize by continued fractions the function 

5 s + 2s» + 3j + 1 


s* + s* +2s + 1 

Synthesis of one-port networks 339 

11.12 Find the networks for the following functions in one Foster and one 
Cauer form. 


(s + 2X* + 4) 
20 + 0.5)(* + 4) 

11.13 Synthesize the following functions in Cauer form. 

s* +2s* +s + 1 


11.14 Synthesize Z(s) = 

5» + J 8 +s 

s* + s*+2s + l 
V+4» + 3*«+a + l 
4^ + 3^+4^+2 
2s» +5 

(s + 2X* + 4) 

i . i\/ v - ^ mto tne f orm shown in the figure. 
(s + IX* + 5) 6 

r— *Tnnp — i i — If — 


PROB. 11.14 

11.15 Of the three pole-zero diagrams shown, pick the diagram that repre- 
sents an R-L impedance function and synthesize in a series Foster form. 

-4 -3 -2 -l 


-4 -3 -2 -1 


-4 -3 -2 -1 


10 «• 

PROB. 11.15 

340 Network analysis and synthesis 

11.16 Synthesize a driving-point impedance with the pole-zero pattern shown 
in the figure in any form you choose. (Hint: Use uniform loading concepts.) 





PROB. 11.16 

11.17 Following are four successive approximations of tanh s. 




IPs* + 105* 
s* + 455* + 105 


s* + 15s 
6s* + 15 

s* + 1055 s + 945* 
15** + 4205 s + 945 

Synthesize networks for the functions above whose input impedances approxi- 
mate tanh*. 

chapter 12 

Elements of transfer 
function synthesis 


A transfer function is a function which relates the current or voltage 
at one port to the current or voltage at another port. In Chapter ? we 
discussed various descriptions of two-port networks in terms of the open- 
circuit parameters z„ and the short-circuit parameters y u . Recall that for 
the two-port network given in Fig. 12.1, the open-circuit transfer im- 
pedances «!, and Zu were defined as 

h l/i-o (12.1) 

*1 lli-0 

In terms of the open-circuit transfer impedances, the voltage-ratio transfer 

function is given as T , 

Yl « £« (12.2) 

In terms of the short-circuit parameters, the voltage ratio is shown to be 

Yl = _ 2» (12.3) 

When the network is terminated at port two by a resistor R, as shown in 
Fig. 12.2, the transfer impedance of the overall network is 

z £ = _?S*_ (12.4) 

" h *« + * 

342 Network analysis and synthesis 







V 2 



FIG. 12.1 

FIG. 12.2 

The transfer admittance of the overall structure in Fig. 12.2 is 

Y„ = -' = 


Vi y»2 + G 

where G = l/R. When both ports are terminated in resistors, as shown in 
Fig. 12.3, the voltage-ratio transfer function V^V g is 

*a z ai°2 



(z u + RMtn. + *«) - 2gi«i8 
Other transfer functions such as current-ratio transfer functions can 
also be described in terms of the open- and short-circuit parameters. In 
Chapter 10 we discussed the various properties of driving-point imped- 
ances such as z u and z 28 . This chapter deals with the properties of the 

transfer immittances z M and y tl for a 
passive reciprocal network. First, let 
us discuss certain properties which 
apply to all transfer functions of passive 
linear networks with lumped elements. 
We denote a transfer function as T(s). 

1. T(s) is real for real s. This property 
FIG. 12.3 is satisfied when T(s) is a rational 

function with real coefficients. 

2. T(s) has no poles in the right-half plane and no multiple poles on 
they'to axis. If T(s) is given as T(s) = P(s)lQ(s), the degree of P(s) cannot 
exceed the degree of Q(s) by more than unity. In addition, Q(s) must be 
a Hurwitz polynomial. 

3. Suppose P(s) and Q(s) are given in terms of even and odd parts, 
that is, 

ns) = m = MM±IM (12 . 7) 

e( S ) m 8 ( S ) + w,(5) 

where M£s) is even and N£s) is odd. Then T(ja>) is 

MiC/oO + A^ 

MJja>) + NMa>) 
The amplitude response of T(ja>) is 

\T(j<o)\ = 

UfA/co) + N t *(ja>)] 


Elements of transfer function synthesis 343 
and is an even function in m. The phase response is 

Arg TQm) = arctan 

m—m '«— 

If arg T(jO) = 0, we see that the phase response is an odd function in <o. 
Now let us discuss some specific properties of the open-circuit and 
short-circuit parameters. 

1. The poles of z gl (,s) are also the poles of z u (j) and z it {s). However, 
not all the poles of z u (s) and z M (y) are the poles ofz n (s). Recall that in 
Chapter 9 we defined the z parameters in terms of a set of node equations 


Z^s) = ^ 

2.o — Z«l — 



If there is no cancellation between each numerator and denominator of z u , 
z lt , and z lg , then the poles are the roots of the determinant A, and all three 
functions have the same poles. Consider the two-port network described 
by the black box in Fig. 12.4a. Let z' u , z' a2 , and z' lt be the z parameters 
of the network. Let us examine the case when we attach the impedances 
Zx and Z g to ports one and two, as shown in Fig. 12.46. The z parameters 
for the two-port network in Fig. 12.46 are 

z ii = z ii + Zj 

Z 22 — Z 22 + Zj 
Z 12 = Z 12 

It is clear that the poles of z u include the poles of Z^ the poles of z M 
include the poles of Zj. However, the poles of z lt include neither the 
poles of Zi nor Z,. Consequently, we see that all the poles of z„ are also 
poles of z u and z„. The reverse is not necessarily true. 


*\l> 2*22 

l_ h 



v 2 



*'ll. *'22 

— <- 

r * £■ 

3j -o 


v 2 


FIG. 12.4 

344 Network anal/sis and synthesis 



* * r^ 

A-i " * 



y'n, y'22 

Y 2 


- 4" 

^T 7 


o > 


/ 2 



*h V* 

FIG. 1X5 

FIG. 12.6 

2. The poles of y is (s) are also the poles of y u (s) and y M (s). However, 
not all of the poles of y u (s) and y M (s) are the poles of y M (s). This property 
is readily seen when we examine the two-port network in Fig. 12.5. 
The y parameters are . , v 

y** = y'»* + y% 

yit = v'm 

Clearly, the poles of y 12 (s) do not include the poles of either Y t and Y t . 
Consider the network in Fig. 12.6. The y parameters are 


ViM = - + 3s 


y»(s) = - + 3s 


yis(s) = -3s 
Observe that y u (s) and y M (s) have poles at s = and s = 00, whereas 
y lt (s) only has a pole at s = 00. 

3. Suppose y u (s), y M (s), and y 18 (s) all have poles at s = s t . Let us 
denote by Jfc u the residue of the pole at s t of the function y u (s). The 
residue of the pole s = Sj, of y w (s) will be denoted as k m , and the residue 
of the same pole of y ia (s) will be denoted as k lt . Without going into the 
proof, 1 a general property of L-C, R-C, or R-L two-port networks is that 

k a k M - k lt * ^ 


This equation is known as the residue condition. For example, for the 
L-C network in Fig. 12.6, the residue condition applied to the pole at 
s = 00 gives 3 x 3 — 3* = 0; whereas for the pole at s = 0, we have 
2x4 — 0* = 8 > 0. Thus we see that the residue condition is fulfilled 
for both poles. 

l For a general discussion, see M. E. Van Valkenburg, Introduction to Modem 
Network Synthesis, John Wiley and Sons, New York, 1960, pp. 305-313. 

Elements of transfer function synthesis 
C / 2 


FIG. 12.7 




■< o 







FIG. 12.8 


A zero of transmission is a zero of a transfer function. At a zero of 
transmission, there is zero output for an input of the same frequency. 
For the network in Fig. 12.7, the capacitor is an open circuit at s = 0, 
so there is a zero of transmission at s = 0. For the networks in Figs. 
12.8 and 12.9, the zero of transmission occurs at s = ±//VZc. For 
the network in Fig. 12.10, the zero of transmission occurs at s = —l/RC. 

In general, all the transfer functions of a given network have the same 
zeros of transmission, except in certain special cases. For example if 
««(*) has a zero of transmission at s = s u than y«(j), V^lV^s), etc., 

v 2 

o h ■ ■ 

+ k 


FIG. 12.1* 

346 Network analysis and synthesis 



y 2 

z 3 


- Z B - 


1 ^6 

FIG. 12.11 

will also have a zero at s = s^ This fact is clearly seen when we examine 
the relationships between the transfer functions. For example, we have 

Z«i — 


yuVn. - Vi^ti 





In addition, the voltage- and current-ratio transfer functions can be 
expressed in terms of the z and y parameters as 

Yl — 5ll h — Usk (12.14) 

In Chapter 8 we saw that transfer functions that have zeros of trans- 
mission only on the jco axis or in the left-half plane are called minimum 
phase functions. If the function has one or more zeros in the right-half 
plane, then the function is nonmimimum phase. It will be shown now that 
any transfer function of a passive reciprocal ladder network must be 
minimum phase. Consider the ladder network in Fig. 12.11. The zeros of 
transmission of the ladder occur at the poles of the series branch im- 
pedances or at the zeros of the shunt branch impedances. Since these 
branch impedances are themselves positive real, the poles and zeros of 
these impedances cannot be in the right-half plane. Consequently, the 
transfer functions of ladder networks must be minimum phase. For the 
network in Fig. 12.12, a transfer function would have two zeros of 

FIG. 12.12 

Elements of transfer function synthesis 347 

FIG. 12.13 

transmission at s = oo due to the elements L x and C 8 . It would also have a 
zero of transmission at s = due to C lt a zero at s = —l/R a C t due to 
the parallel R-C branch, and a zero of transmission at s = jULtCtf* 
due to the L-C tank circuit. It is seen that none of the transmission 
zeros are in the right-half plane. We also see that a transfer function may 
possess multiple zeros on the ja> axis. 

In Section 12.4 it may be seen that lattice and bridge circuits can easily 
be nonminimum phase. It can also be demonstrated that when two ladder 
networks are connected in parallel, the resulting structure may have 
right-half-plane zeros. 2 




In this section we consider the synthesis of an L-C ladder network 
with a 1-Q resistive termination to meet a specified transfer impedance Zji 
or transfer admittance Y tl . In terms of the open- and short-circuit 
parameters of the L-C circuit, Z^C?) can be expressed as 

Zm — 


+ 1 


and Y tl (s) is 




y« + i 

as depicted in Figs. 12.13 and 12.14. 

Before we proceed with the actual details of the synthesis, it is necessary 
to discuss two important points. The first deals with the ratio of the odd 
to even or even to odd parts of a Hurwitz polynomial Q(s). Suppose 

FIG. 12.14 

« Van Valkenburg, op. cit., Chapter 11. 

348 Network analysis and synthesis 

Q(s) is given as 

Q(s) = M{s) + N(s) (12.17) 

where M(s) is the even part of Q(s), and N{s) is the odd part. We know 
that the continued fraction expansion of M(s)lN(s) or N(s)IM(s) should 
yield all positive quotients. These quotients can, in turn, be associated 
with reactances. Therefore it is clear that the ratio of the even to odd or 
the odd to even parts of a Hurwitz polynomial is an L-C driving-point 

The second point to be discussed is the fact that the open-circuit transfer 
impedance z tl or the short-circuit transfer admittance y n of an L-C circuit 
is an odd function. To show this, we must remember that in an L-C 
circuit with steady-state input, the currents are 90° out of phase with the 
voltages. Thus the phase shifts between the input currents and output 
voltages or input voltages and output currents must be 90° out of phase, 

Arctan^=±Jrad (12.18) 

and Arctan^=±?rad (12.19) 

so that Re z n (jw) = = Re j/ 21 (;a)) for an L-C network. In order for the 
real parts to be equal to zero, the functions z« and y ai of an L-C two-port 
network must be odd. 
Suppose, now, that the transfer admittance Y 21 is given as the quotient 

of two polynomials* 

Y = ^ = ^ (12.20) 

11 Q(s) M{s) + N(s) 

where P(s) is either even or odd. Now, how do we determine the short- 
circuit parameters y n and y M from the Eq. 12.20 to get it into the form 

Y = ya (12.21) 

Y " i+y» 

The answer is quite simple. We divide both the numerator P(s) and the 
denominator Q(s) by M(s) or N(s), the even or the odd part of Q{s). 
Since y tl must be odd, if P(s) is even, we divide by N(s) so that 

P(s)/N(s) (12>22) 

In 1 + [M(s)IN(s)] 

From this we obtain y n — 

Elements of transfer function synthesis 349 


y " JV(s) 

On the other hand, if P(s) is odd, we divide by M(s) so that 

y„ Z(!VMW_ (12 . 24) 

1 + [tf(s)/M(s)] 



yM M(5) 

We assume that P(s), Mis), and N(s) do not possess common roots. For 
our purposes, we will consider only the synthesis of Y n or Z tl with zeros 
of transmission either at s = or s = oo. In a ladder network, a zero of 
transmission at s = corresponds to a single capacitor in a series branch 
or a single inductor in a shunt branch. On the other hand, a zero of 
transmission at s = oo corresponds to an inductor in a series branch or 
a capacitor in a shunt branch. In terms of the transfer impedance 

Zn(s) = fW = *('" + ««-i«"- 1 + • • • + "i« + «»> (12 . 26) 

the presence of n zeros of Z, x (.y) at s = implies that the coefficients a n _ x , 

a„_, a x , a are all zero. The number of zeros of Z,^) at s = oo 

is given by the difference between the highest powers of the denominator 
and the numerator, m — n. We know that n can exceed m by at most 
unity, while m can be greater than n by more than one. For example, 
if m — n = 2, and n = 3 with «„_i, . . . , a lt a„ = 0, we know that the 
transfer function has three zeros of transmission at s = and two zeros 
of transmission at s = oo. 

We can now proceed with the matter of synthesis. Consider the 
following example. 

sr + 3s + 4s + 2 

We see that all three zeros of transmission are at j = oo. Since the 
numerator P(s) is a constant, it must be even, so we divide by the odd 

350 Network analysis and synthesis 

part of the denominator s* + 4s. We then obtain 

z as — 

s a + 4s 

3s g + 2 
i s + 4s 


We see that both z sl and z 2g have the same poles. Our task is thus simpli- 
fied to the point where we must synthesize z 22 so that the resulting network 
has the transmission zeros of z 21 . This requires that we first examine the 
possible structures of the networks which have the required zeros of 
transmission and see if we can synthesize z 22 in one of those forms. For 
the example that we are considering, a network which gives us three zeros 
of transmission at s = oo is shown in Fig. 12.15. We can synthesize z 22 
to give us this structure by the following continued fraction expansion 

of l/z 22 . 

3j« + 2)i» + 4s(ls+-Y 
s»+ |s 

^)3j* + 2(&s +-Z 



Since z 22 is synthesized from the l-£2 termination toward the input end, 
the final network takes the form shown in Fig. 12.16. Examining the 
network more closely, we see that it takes the form of a low-pass filter. 
Thus the specification of all zeros at s = oo is equivalent to the specification 
of a low-pass filter. 
As a second example, consider the transfer impedance 

*«>- , + » + 4. + 2 <12S) 

Since the numerator of Z gl (,s) is an odd function, we have to divide both 


-ffflHT 1 

<D T } 

FIG. 12.15 

FIG. 12.16 

Elements of transfer function synthesis 351 

FIG. 12.17 

numerator and denominator by the even part of the denominator so that 

-«* s* + 4s 

T (12.30) 

*n = 

*« = 

3s* + 2 "" 3s* + 2 

The network that gives three zeros of transmission at s = is a high-pass 
structure which is realized by a continued fraction expansion of « M . 
The final realization is shown in Fig. 12.17. 
Finally, consider the transfer admittance 

Y n (s) 


s* + 3s* + 4s + 2 

which has two zeros of transmission at s = 0, and one zero at s = oo. 
Since the numerator is even, we divide by s* + 4s so that 

3s' + 2 „~„„x 

y« = -rrr (12.32) 

s + 4s 

The question remains as to how we synthesize y a to give a zero of trans- 
mission at s = oo and two zeros at s = 0. First, remember that a parallel 
inductor gives us a zero of transmission at s — 0. We can remove this 
parallel inductor by removing the pole at s = of y n to give 

1 5s/2 
= i/— = ~— — 

Vi = Vu 


2s s* + 4 

If we invert y lt we see that we have a series L-C combination, which gives 
us another transmission zero at s = 0, as represented by the f-farad 
capacitor, and we have the zero of transmission at s = oo also when we 
remove the inductor of f h. The final realization is shown in Fig. 12.18. 




FIG. 12.18 

352 Network analysis and synthesis 







An important point to note in this synthesis procedure is that we must 
place the last element in series with a voltage source or in parallel with a 
current source in order for the element to have any effect upon the transfer 
function. If the last element is denoted as Z n , then the proper connection 
of Z n should be as shown in Fig. 12.19. 



In this section we will consider the synthesis of constant-resistance two- 
port networks. They derive their name from the fact that the impedance 

looking in at either port is a constant- 
resistance R when the other port is termi- 
nated in the same resistance R, as depicted 
in Fig. 12.20. Constant-resistance net- 
works are particularly useful in transfer 
function synthesis, because when two 
constant-resistance networks with the 
same R are connected in tandem, as shown 
in Fig. 12.21, neither network loads down the other. As a result, if the 
voltage-ratio transfer function of N a is V^V-l and that of N b is VJV S , the 
voltage-ratio transfer function of the total network is 

FIG. 12.20. Constant-resistance 



Equation 12.34 implies that, if a voltage-ratio transfer function is to be 
realized in terms of constant-resistance networks, the voltage ratio could 

Elements of transfer function synthesis 353 

N a 


t. *. 

V 2 «+ 



FIG. 12.21. Constant-resistance networks in tandem. 

be decomposed into a product of simpler voltage ratios, which could be 
realized as constant-resistance networks, and then connected in tandem. 
For example, suppose our objective is to realize 


s — 




s — 


X a (s 



K _ K(s - z«)(s - zi)(s - z,) 

V a (s - p )(s - Pi)(« - P«) 
in terms of constant-resistance networks. We can first synthesize the 
individual voltage ratios 

\\ _ K (s - 2 ) 


v* = 
Y* = 

V t s - p, 

as constant-resistance networks and then connect the three networks in 
tandem to realize V h jV a . 

Although there are many different types of constant-resistance net- 
works, we will restrict ourselves to networks of the bridge- and lattice-type 
structures as shown in Figs. 12.22o and 12.226. These networks are 
balanced structures; i.e., the input and output ports do not possess com- 
mon terminals. Upon a close examination, we see that the bridge and 
lattice circuits in Figs. 12.22 are identical circuits. The bridge circuit is 
merely the unfolded version of the lattice. Consider the open-circuit 
parameters of the bridge circuit in Fig. 12.23. First we determine the 
impedance z u as 

«ii = 


Next we determine the transfer impedance z n , which can be expressed as 

V * ~ V *' (12.38) 

z ai = 


354 Network analysis and synthesis 

+ '. 


+ " 

FIG. 12.22. (a) Bridge circuit, (6) Constant-resistance lattice. 

and is obtained as follows. We first obtain the current /' as 

r _ Vi _ h 
Z a + Z b 2 

Next we find that V t - V t - = (Z b - Z tt )I' 

= (Z 6 -Z a )| 



FIG. 12.23. Analysis of bridge circuit. 

Elements of transfer function synthesis 355 

so that ^ = Z> ~ Z * (12.41) 

From the lattice equivalent of the bridge circuit we note that z M = z u . 
Now let us consider the lattice circuit that is terminated in a resistance 
R, as shown in Fig. 12.226. What are the conditions on the open-circuit 
parameters such that the lattice is a constant-resistance network? In 
other words, what are the conditions upon z u and z n such that the input 
impedance of the lattice terminated in the resistor R is also equal to R7 
In Chapter 9 we found that the input impedance could be expressed as 

Zu = *ii - -SjJ-r (12.42) 

Zn + R 

Since z u = z u for a symmetrical network, we have 

Z " " z 11 + * (1243) 

In order for Zu = R, the following condition must hold. 

*u* - z«i a = * (12.44) 

For the lattice network, we then have 

iKZ, + Z,)* - (Z a - Z»)»] = ** (12.45) 

which simplifies to give Z^L b — R* (12.46) 

Therefore, in order for a lattice to be a constant-resistance network, 
Eq. 12.46 must hold. 

Next, let us examine the voltage ratio VjV g of a constant-resistance 
lattice whose source and load impedances are equal to R (Fig. 12.24). 
From Chapter 9 we can write 

V» z^R 

V 9 (zu + K)(z M + R) - ztfr* 


FIG. I2J4. Double-terminated lattice. 



356 Network anal/sis and synthesis 

which simplifies to IS- = 5*&ii 

V. (zn + Kf - **i a 

= __Zt£__ 

2Rz 11 + 2R 2 

In terms of the element values of the lattice, we have 

V, _ l(Z b - Z a )R 
V„ R(Z b + Z a ) + 2R* 

From the constant-resistance condition in Eq. 12.46, we obtain 

V, _ UZ b - (RVZ h )]R 
V t R[Z b + (R*/Z»)] + 2R a 

KZ»» - R*) 
(Z 6 a + R 8 ) + 2RZ h 

_ j(Z t a - R a ) 
(Z b + R? 

= * (Z > ~ R) (12.50) 

Z b + R 

In Eq. 12.50, the constant multiplier J comes about from the fact that 
the source resistance R acts as a voltage divider. If we let 

G(s) = ^ (12.51) 

we can express Z 6 in Eq. 12.50 in terms of G as 

HP + G(s)] 
" 1-G(5) 
In terms of Z„, the voltage ratio can be given as 

V 2 1R-Z a 

V a 2R + Z a 

In the following examples, we will usually let R be normalized to unity. 
Example 12.1. The voltage ratio is given as 

2l=l£JlI (12.54) 

V„ 2 s + 1 

Elements of transfer function synthesis 357 

which, as we recall, is an all-pass transfer function. By associating Eq. 12.54 
with Eq. 12.50, we have 

Z„ =>s, 

Since Z^Z a = 1, we then obtain 

Z a -- 


We see that Z„ is a 1-h inductor and Z a is a 1-farad capacitor. The final network 
is shown in Fig. 12.25. 


FIG. 12.25 


Example 12.2. Let us synthesize the all-pass function 

V B ~2s + 1 ' s* + 2s+2 
whose pole-zero diagram is shown in Fig. 12.26. Since the portion 

Yl _ l s ~ l 

V„ 2 s + 1 

has already been synthesized, let us concentrate on synthesizing the function 

V t j* -2s +2 
V a ~ s* +2s+2 


First, we separate the numerator and denominator function into odd and 
even parts. Thus we have 

V* (f 2 + 2) - Is 

V a = (5* + 2) + 2s 


x yi 


-<T -1 

X -fl - 

FIG. 123* 

+ 1 «■ 

358 Network analysis and synthesis 

If we divide both numerator and denominator by the odd part 2s, we obtain 

V t [(*» + 2)/2t] - 1 

We then see that 

[(5* + 2)/2*] + 1 
a* + 2 

s 1 

= 2 + I 



which consists of a $-h inductor in series with a 1 -farad capacitor. The im- 
pedance Z„ is then 

*-JT5 (1262) 

and is recognized as a J-farad capacitor in parallel with a 1-h inductor. The 
voltage ratio VjV a is thus realized as shown in Fig. 12.27. The structure that 

FIG. 12.27 

realizes the transfer function VJV g in Eq. 12.S7 is formed by connecting the 
networks in Figs. 12.25 and 12.27 in tandem, as shown in Fig. 12.28. Finally, 
it should be pointed out that constant-resistance lattices can be used to realize 
other than all-pass networks. 


Next, let us consider the constant-resistance bridged-T network in Fig. 
12.29. If the resistances in the network are all equal to R ohms, the 
network has constant-resistance if 

ZA - * (12.63) 

Elements of transfer function synthesis 





Vi .*. 





FIG. VIM. Constant-resistance bridged-T circuit. 

Under the constant-resistance assumption, the voltage-ratio transfer 
function can be given as 

V* _ * _ z > 

V x R + Z a Z t + R 
Exanple 123. Let us synthesize the voltage ratio 

V. s* + l 

j*+25 + l 



as a constant-resistance bridged-T network terminated in a 1-G resistor. First 

let us write VJV 1 as 

V 1 

1 (12.66) 

so that 


V t 1 + [2*/(*» + 1)] 

Z b ~ 

*» + l 
s* + l 



We recognize Z a as a parallel L-C tank circuit and Z» as a series L-C tank 
circuit. The final network is shown in Fig. 12.30. 



-a/v\ — i-vw- 

10 k 10 



FIG. I2J0 

360 Network analysis and synthesis 
Example 12.4. Let us synthesize the voltage ratio 

Vs (s + 2X* + 4) 
F x " (5 + 3X3* + 4) 


in terms of two constant-resistance bridged-T circuits connected in tandem. 
At first, we break up the voltage ratio in Eq. 12.69 into two separate voltage 

Va S+1 (12.70) 







v a ~ 

5 +4 


For the 




we have 
s + 2 


= z bl + 1 

so that Z n =» s 

+ 2 and 

z al 



For the 



vjv a 

we have 




1 +z ot 





FIG. 12.31 


Elements of transfer function synthesis 361 

s +4 

Zha — ' 



The final synthesized network is shown in Fig. 12.31. 


12.1 Give an example of a network where: (a) a transfer function has 
multiple zeros on they'to axis; (b) the residue of a pole of a transfer function on 
thsjo> axis is negative. 

12.2 Show that the residue condition holds for the networks shown in the 

I — 'TflHP — I 

















PROB. 12.2 

123 For the network shown, find by inspection the zeros of transmission 
and plot on a complex plane. 

i — nnnp— i 

lf^= ^=2f S10 

ff in. 




PROB. 12.3 

12.4 For the networks in the figure show that the driving point impedances 
Z ln are equal to R when Z„Z t - R*. 

362 Network anal/sis and synthesis 




Vi Zfci-* 

z b 


*£ * 



Z ta -ii 

*^ v 2 

PROS. 12.4 

123 For the networks in Prob. 12.4, find the voltage-ratio transfer functions 


12.6 For the network shown in the figure (a) show that 

Yl. . 1 

(b) Synthesize Ywhen 

V 2+Y 

l\ O.S(*» + 2) 

K ™ s* + 2s + 2 

10 < V 2 

PROB. 12.4 

12.7 (a) For the constant-resistance bridged-T circuit, show that if Z tt Z h — 1, 

Vt L_ 

n i +z a 

Elements of transfer function synthesis 363 

(b) Synthesize Z a and Z h if 

_ s* +3s + 2 

V l ~ s* +45* +5j + 2 


12.8 Synthesize the following voltage ratios in one of the forms of the 
networks in Prob. 12.4 

V, s+2 



V l * + 3 

Vt 2(s*+3) 
V x " 2** + 2* + 6 

Vt 3(jr + 0.5) 
V x " 4j + 1.5 

12£ Synthesize iV with termination resistors /t t =4fi,i? 1 ~into give 

Yl 12j* 

V, " 15i» + 7* + 2 


12.10 For the network in Prob. 12.9, realize network N a to give 

Yl X _L_ 
F„ 2 2s + 3 

(a) Synthesize AT. as a constant-resistance lattice. (R =- 1 n.) 

(6) Synthesize AT. as a constant-resistance ladder as in Prob. 12.4. (Jt - 1 Q.) 

(c) Synthesize A', as a constant-resistance bridged-T circuit. (R - 1 Q.) 

12.11 Synthesize the following functions into the form shown in the figure 

Z - 1 





s* +3s* +3s +2 


*» + 3s* + 3s + 2 


s* + 3s* + 3s + 2 


s* + 3s* + 3s +2 

' (* + 2)« 

364 Network anal/sis and synthesis 


12.12 Synthesize as a constant-resistance lattice terminated in a 1-Q resistor. 

V, s* - s + 1 


12.13 Synthesize the functions in Prob. 12.8 as constant-resistance bridged-T 


s* + s + 1 


s* - 3s + 2 

v t 

~ s* + 3s +2 


s* - 20s* + 5s - 20 

V x (s) 

^ s* + 20s* + 5s + 20 

chapter 13 

Topics in filter design 


In the preceding chapters we examined different methods for synthesizing 
a driving point or transfer function H(s). Most problems have as their 
initial specification an amplitude or phase characteristic, or an impulse 
response characteristic instead of the system function H(s). Our problem 
is to obtain a realizable system function from the given amplitude or 
phase characteristic. For example, a typical design problem might be to 
synthesize a network to meet a given low-pass filter characteristic. The 
specifications might consist of the cutoff frequency m c , the maximum 
allowed deviation from a prescribed amplitude within the pass band, and 
the rate of fall off in the stop band. We must then construct the system 
function from the amplitude specification. After we obtain H(s), we 
proceed with the actual synthesis as described in the Chapter 12. Another 
problem might consist of designing a low-pass filter with a linear phase 
characteristic within the pass band. Here, both amplitude and phase are 
specified. We must construct H(s) to meet both specifications. Problems 
of this nature fall within the domain of approximation theory. In this 
chapter we will consider selected topics in approximation theory and then 
present examples of filter design where both the approximation and the 
synthesis problems must be solved. 


The essence of the problem is the approximation of a given func- 
tion/(rr) by another function/^t; «!,..., a„) in an interval x x ^ x ^ x t . 
The parameters a x , . . . , <x„ in the approximating function are fixed 


366 Network analysis and s/nthesis 

by the particular error criterion chosen. When we let e = f{x) — 
/at*; <*i» • • • » *n)» the following error criteria are most common: 

1. Least squares. The value of /(a lf . . . , aj is minimized where 

and w(x) is a weighting function which stresses the error in certain sub- 

2. Maximally flat. The first n — \ derivatives of e are made to vanish 
at x = *o. 

3. Chebyshev. The value of p is minimized in the interval *i<,x <,x t 

where /* = |«ln«x- 

4. Interpolation. The value of e is made to vanish at a set of n points 
in the interval x^£x <, x t . 

After an error criterion is chosen, we must determine the particular 
form of the approximating function. This depends upon whether we 
choose to approximate in the time or frequency domain. Suppose /(a;) 
represents a magnitude function in the frequency domain and the approxi- 
mating function is to be rational in co*; then 

/„(*; «!,..•,*»)= „ . — (13.1) 

where * = to*. In addition, the values of x k must be restricted to insure 
that/ (x; a lf . . . , O ^ 0. In the time domain, /(&) might represent an 
impulse response of a system to be synthesized. In the case of an R-C 
transfer function, we have 

f a (x; 04, ... , oc„) = «**' + oe, e"«* + • • • (13.2) 

where x = t. Since an R-C transfer function must have its poles on the 
negative real axis, the values of <x t , k even, are restricted to negative real 

The keystone of any approximation problem lies in the choice of a 
suitable error criterion subject to realizability restrictions. The problem 
can be simplified when some of the oc's are assigned before applying the 
error criteria. All the error criteria cited, except the Chebyshev, can then 
be reduced to a set of linear algebraic equations for the unknowns 
«ii • • • > *»• 

Time-domain approximation 

The principal problem of time-domain approximation consists of 
approximating an impulse response h(0 by an approximating function 

Topics in filter design 367 
h*(t) such that the squared error 

e = ["[fc(0-**(')f * 

is minimum. 

A generally effective procedure in time-domain approximation utilizes 
orthonormal functions ^(r). 1 The approximating function h*{t) takes 
the form 


so that the error 



is minimized when 

Jo L *-i 

«*&(<)] dt 


h(t)<f> k (t)dt k = l,2,...,n 


as we saw in Chapter 3. If the orthonormal set is made up of a sum 
of exponentials e**', then the approximate impulse response 




has a transform 


*=1 s — s k 


Realizability is insured if in the orthonormal set {a t e***}, Re ^ ^ 0; 
k = 1, 2, . . . , n. Synthesis then proceeds from the system function H*(s) 
obtained in Eq. 13.S. 

Frequency-domain approximation 

In frequency-domain approximation 
the principal problem is to find a 
rational function H(s) whose magnitude 
|/f(/Yu)| approximates the ideal low-pass 
characteristic in Fig. 13.1 according to 
a predetermined error criterion. In the FIG . I3X Ideal , fihcrchar _ 

next few sections we examine several acteristic. 



1 W. H. Kautz, "Transient Synthesis in the Time Domain," IRE Trans, on Circuit 
Theory, CT-1, No. 3, September 1954, 29-39. 

368 Network analysis and synthesis 

different ways to approximate the ideal low-pass: the maximally flat or 
Butterworth approximation, the equal-ripple or Chebyshev approximation, 
and the optimal or Legendre approximation. Another major problem is 
that of obtaining a transfer function H^s), whose phase is approximately 
linear or whose delay is approximately flat over a given range of frequencies. 
Here again, there are two different methods : the maximally flat or the 
equal-ripple methods. Our discussion will center around the maximally 
flat method. The joint problem of approximating both magnitude and 
phase over a given frequency range is possible, but will not be discussed 


In Chapter 10 we saw that the ideal low-pass filter in Fig. 13.1 is not 
realizable because its associated impulse response is not zero for t < 0. 
However, if we use a rational function approximation to this low-pass 
filter characteristic, the Paley- Wiener criterion will be automatically 
satisfied. We will therefore restrict ourselves to rational function approxi- 

In low-p.ass filter design, if we assume that all the zeros of the system 
function are at infinity, the magnitude function takes the general form 

where K is the d-c gain constant and/(<u 2 ) is the polynomial to be selected 
to give the desired amplitude response. For example, if 

/(«) a ) = a) 2 " (13.7) 

then the amplitude function can be written as 

(1 + o> 2n ) 

M(») = ,. \ n ^ < 13 - 8 ) 

We see that M(0) = K,,, and that M(a>) is monotonically decreasing with 
eo. In addition, the 0.707 or 3-decibel point is at to = 1 for all n, that is, 

M(l) = -^ alln (13.9) 

The cutoff frequency is thus seen to be <w = 1. The parameter n con- 
trols the closeness of approximation in both the pass band and the stop 
band. Curves of M(m) for different n are shown in Fig. 13.2. Observe 

Topics In filter design 369 
















Radian frequency, a 
0.3 0.4 0.6 0.8 




n = 3 

n = 5 

n = 7 


I* \ 



^ \ 

\\ \ 


\ \ 

\ \ 

\ \ 

FIG. 13.2. Amplitude response of Butterworth low-pass filters. 

that the higher n is, the better the approximation. The amplitude approxi- 
mation of the type in Eq. 13.8 is called a Butterworth or maximally flat 
response. The reason for the term "maximally fiat" is that when we 
expand M(to) in a power series about <o = 0, we have 

Jlf(eo) - Ao(l - l<o tn + |co«» - A<o«" + i^eo"" + • • •) (13.10) 

We see that the first 2n — 1 derivatives of M{o>) are equal to zero at 
(0 = 0. For «> » 1, the amplitude response of a Butterworth function 
can be written as (with Kg normalized to be unity) 

M(w) =! — 



We observe that asymptotically, Af(to) falls off as wr n for a Butterworth 
response. In terms of decibels, the asymptotic slope is obtained as 

20 log M(co) = — 20/i log <w 


370 Network analysis and synthesis 

Consequently, the amplitude response falls asymptotically at a rate of 
6m db/octave or 20n db/decade. 

One question remains. How do we obtain a transfer function H(s) 
from only the amplitude characteristics M(co)l The procedure is as 
follows. We first note that the amplitude response M{m) and the complex 
system function H(jco) are related by 

M\w) = H(j<o) H(-jco) (13.13) 

If we define a new function A(s 2 ) such that 

A(s 2 ) - H(s) H(-s) (13.14) 

we see that M 2 (<o) = ^(-ot) 2 ) (13.15) 

From A(— eo 2 ) all we need do is to substitute s* = —to 2 to give «(s 2 ). 
Then we factor A(s 2 ) into the product H(s) H(—s). Since the poles and 
zeros of H(s) are the mirror images of the poles and zeros of H(—s), we 
simply choose the Hurwitz factors of A(j s ) as H(s). An example will serve 
to clarify this discussion. Consider the third-order (n = 3) Butterworth 
response given by 

l (13.16) 


JW \OJ) — 

1 + 

to 6 

«(s a ) = 


"We see that AC? 2 ) is 
Factoring «(**), we 

1 - 


1 - 

-(s 8 ) 8 

«(**) = 



1 + 2s + 2s 2 

+ s' 

» 1 - 2s + 2s 2 - s* 

We then have 

= H(s)H(-s) 

if(s) = 


s 8 + 2s s + 2s + 1 



(» + l)(s + i + A/3/2X* + i - A/3/2) (13-20) 

The poles of #(y) and fT(-*) are shown in Fig. 13.3. Observe that the 
poles of H(—s) are mirror images of the poles of H(s), as given by the 
theorem on Hurwitz polynomials in the Chapter 2. 

Topics in filter design 371 

HM^.- -^ «f-»> 

FIG. 13.3. Poles of H(s) H(-s) for an n = 3 Butterworth filter. 

For a Butterworth response, the poles of H(s) H(—s) are the roots of 

(-1)V B = -1 

= «*«*-"» k = 0, 1, 2 In - 1 

The poles s k are then given by 

s k = e*<»-u/ft* „ even 

= e i(i/n) ' n odd 

or simply by s k = e «<»+'-i>/an] I fc = o, 1, 2, 






Expressing s t as s k = a k + ja> k , the real and imaginary parts are given 

ow = cos 

2fc + n - 

ct)v = sin ■ 


+ n- 1 . /2fc - 1\tt 

ir = sin I 1 - 

2n \ n /2 

+ n- l /2fc- 1\tt 

IT = cos I 1 — 

In \ n 12 


It is seen from Eqs. 13.22 and 13.23 that all the poles of H(s) H(—s) are 
located on the unit circle in the s plane, and are symmetrical about both 
the a and the jco axes. To satisfy readability conditions, we associate the 
poles in the right-half plane with H(—s), and the poles in the left-half 
plane with H(s). 

As an example, consider the construction of an H(s) that gives an n = 4 
Butterworth response. From Eq. 13.23, it is seen that the poles are 
given by 

= ^(«fc+*)/8]» (13.25) 

s t = e" 

372 Network anal/sis and synthesis 
H(s) is then given as 

H($) = ( S + ««»«.)(, + «*««»«Xa + e i{ ™')(s + e> (, *>*) (13 " 26) 
If we express s k in complex form and expand, we obtain 

H( s \ — _ (12 27) 

(s 2 + 0.76536s + l)(s* + 1.84776s + 1) 

To simplify the use of Butterworth functions, H(s) is given in Tables 
13.1 and 13.2 for n = 1 to n = 8, in factored form as in Eq. 13.27, or 
multiplied out as 

H(s) = -^ (13.28) 

a n s n + fln-xs"- 1 + • • • + a t s + 1 

TABLE 13.1 
Butterworth Polynomials (Factored Form) 



s + 1 


5 s + V2s + 1 

(**+* + 1)(* + 1) 

(s 2 + 0.76536* + IX** + 1.84776* + 1) 

(* + 1)(* 2 + 0.6180* + IX* 8 + 1.6180* + 1) 

(** + 0.5176* + IX* 2 + ^2* + IX* 2 + 1.9318* + 1) 

(* + IX* 2 + 0.4450* + IX* 2 + 1.2456* + IX* 2 + 1.8022* + 1) 

(s 2 + 0.3986s + IX* 2 + 1.1110* + IX* 2 + 1.6630* + IX* 2 + 1.9622* + 1) 

TABLE 13.2 
Butterworth Polynomials * 






















































a = !• 

Topics in filter design 373 


In Section 13.3, we examined the maximally flat approximation to a 
low-pass filter characteristic. We will consider other low-pass filter 
approximants in this section. 

The Chebyshev or equal-ripple approximation 

We have seen that the maximally flat approximation to the ideal low- 
pass filter is best at co = 0, whereas, as we approach the cutoff frequency 
<w = l, the approximation becomes progressively poorer. We now 
consider an approximation which "ripples" about unity in the pass band 
and falls off rapidly beyond the cutoff co = 1. The approximation is 
equally good at co = and co = 1, and, as a result is called an equal- 
ripple approximation. The equal-ripple property is brought about by the 
use of Chebyshev cosine polynomials defined as 

C„(a>) = cos (n cos -1 co) |co| <, 1 

= cosh (n cosh -1 co) |co| > 1 

For n = we see that C^co) = 1 (13.30) 

and for n = 1, we have Ci(<*>) = <° (13.31) 

Higher order Chebyshev polynomials are obtained through the recursive 
formula CJfu) = ^ ^^ _ ^ ^ (B 32) 

Thus for n = 2, we obtain Cg(co) as 


Cg(co) = 2<o((o) — 1 
= 2eo 8 - 1 


In Table 13.3, Chebyshev polynomials of orders up to n = 10 are given. 

TABLE 13.3 


Chebyshev polynomials C n (m) = cos (n cos -1 <o) 





2co* -1 


4e» 3 — 3o> 


8a>* - 8a>* + 1 


16a> 6 - 20a> 8 + 5o> 


32o>* - 48<o« + 18a>* - 1 


64o> 7 - 112a>« + 56eo 8 - lea 


128a> 8 - 256w« + 160w 4 - 32o>* + 1 


256o>» - 576o» 7 + 432o)« - 120a. 8 + 9a> 


512a» 10 - 1280w 8 + 1120a>* - 400eo* + 50«* - 1 

374 Network analysis and synthesis 

FIG. 13.4. C,(co) and C,(u>) Chebyshev polynomials. 

The pertinent properties of Chebyshev polynomials used in the low-pass 
filter approximation are: 

1. The zeros of the polynomials are located in the interval |eo| <, 1, 
as seen by the plots of C 3 (o)) and CJio)) in Fig. 13.4. 

2. Within the interval \w\ <, 1, the absolute value of C B (co) never 
exceeds unity; that is, |C„(eo)| <, 1 for \co\ <, 1. 

3. Beyond the interval \a>\ < 1, |C„(w)| increases rapidly for increasing 
values of \a>\. 

Now, how do we apply the Chebyshev polynomials to the low-pass 
filter approximation? Consider the function e 2 C„%w), where e is real and 
small compared to 1. It is clear that e 2 C n 2 ((o) will vary between and e 2 
in the interval \u>\ ^ 1. Now we add 1 to this function making it 1 + 
e 2 C„ 2 (g>). This new function varies between 1 and 1 + e 2 , a quantity 
slightly greater than unity, for |<w| <, 1. Inverting this function, we obtain 
the function which we will associate with \H(jco)\ 2 ; thus 

IHO"))!* = 

1 + *C n \a>) 


Within the interval |eo| < 1, \H(ja>)\ 2 oscillates about unity such that the 
maximum value is 1 and the minimum is 1/(1 + e 2 ). Outside this interval, 
C n 2 (w) becomes very large so that, as to increases, a point will be reached 
where e 2 C B 2 (e>) » 1 and \H(jcu)\ 2 approaches zero very rapidly with 
further increase in <o. Thus, we see that \H(jw)\ 2 in Eq. 13.34 is indeed a 
suitable approximant for the ideal low-pass filter characteristic. 

Figure 13.5 shows a Chebyshev approximation to the ideal low-pass 
filter. We see that within the pass band ^ <u <, 1, \H{jw)\ ripples 
between the value 1 and (1 + e 2 )- 1 -*. The ripple height or distance between 

Topics in filter design 375 

(l + e 2 ) - * -^— ^-— 

l « 

FIG. 1 3 .5. Chebyshev approximation to low-pass filter. 

maximum and minimum in the pass band is given as 


Ripple = 1 — 



At <o = 1, |H(;«>)| is \H(jl)\ = + ^a 

because C n \\) = 1. 

In the stop band, that is, for \<o\ ^ 1, as a> increases, we reach a point 
io k , where e* C n *{(o) » 1 so that 


O) > 0) k 

eC n (a>) 
The loss in decibels is given as 

Loss = -20 log 10 \ H (j m )\ 

^ 20 log e + 20 log C B (o) 



But for large <w, C B (co) can be approximated by its leading term 2" _1 <w", 
so that 

Loss = 20 log e + 20 log 2*- 1 g>" 

6 5 (13.39) 

= 20 log e + 6(n - 1) + 20/t log a> 

cosh 0k 

376 Network anal/sis and synthesis 

We see that the Chebyshev response also falls off at the rate of 20/i 
db/decade after an initial drop of 20 log e + 6(n — 1) decibels. However, 
in most applications, e is a very small number so that the 20 log e term is 

actually negative. It is necessary, therefore, 
to compensate for this decrease in loss in 
the stop band by choosing a sufficiently 
large n. 

From the preceding discussion, we see 
that a Chebyshev approximation depends 
upon two variables, e and n, which can be 
determined from the specifications directly. 
The maximum permissible ripple puts a 
bound on e. Once e is determined, any 
desired value of attenuation in the stop 
band fixes n. 

The derivation of the system function 
H(s) from a Chebyshev amplitude approxi- 
mation \H(jco)\ is somewhat involved and 
will not be given here. 2 Instead, we will 
simply give the results of such a derivation. 
First we introduce a design parameter. 

sinh 0* 

FIG. 13.6. Locus of poles of 
Chebshev filter. 

fi h — - sinh * - 
n e 


where n is the degree of the Chebyshev polynomial and e is the factor 
controlling ripple width. The poles, s k = a k + jco k , of the equal-ripple 
approximant H(s) are located on an ellipse in the s plane, given by 


co k 

sinh a /S fc cosh* /?* 

= 1 


The major semiaxis of the ellipse is on the jco axis and has a value co = 
±cosh p k . The minor semiaxis has a value a = ±sirih(i k , and the foci 
are at co = ±1 (Fig. 13.6). The half-power point of the equal-ripple 
amplitude response occurs at the point where the ellipse intersects the^eo 
axis, i.e., at co = cosh fi k . Recall that for the Butterworth response, the 
half-power point occurs at co = 1. Let us normalize the Chebyshev poles 
s k such that the half-power point also falls at co = 1 instead of at co = 
cosh fi k ; i.e., let us choose a normalizing factor, cosh P k , such that the 

* Interested parties are referred to M. E. Van Valkenburg, Introduction to Modern 
Network Synthesis, John Wiley and Sons, New York, 1960, Chapter 13. 

Topics in filter design 377 
normalized pole locations s' t are given by 

cosh (t h 

cosh /?* cosh t 

= <**+./«>* 

The normalized pole locations can be derived as 

u „ • /2fc - lW 
<r' t = tanh & sin ^ ^j 

a>t = cos ^__j_ 


Comparing the normalized Chebyshev pole locations with the Butterworth 
pole locations in Eq. 13.24, we see that the imaginary parts are the same, 
while the real part a' k of the Chebyshev pole location is equal to the real 
part of the Butterworth poles times the factor tanh /3». For example, with 
n = 3 and tanh /?,. = 0.444, the Butterworth poles are 

*i- -1 +J0 
j 23 = _0.5 ±/0.866 
so that the normalized Chebyshev poles are given by 
4 --1(0.444)+ JO 
= -0.444 +;0 
s zs = -0.5(0.444) ±;0.866 
= -0.222 ±y0.866 

Finally, to obtain the denormalized Chebyshev poles, we simply multiply 
s' k by cosh /3 t , that is, 

Sk = (a\ + /»',) cosh & (1 3.44) 

There is an easier geometrical method to obtain the Chebyshev poles, 
given only the semiaxis information and the degree n. First we draw two 
circles, the smaller of radius sinh /?* and the larger of radius cosh /?„ as 

378 Network anal/sis and synthesis 

FIG. 13.7. » = 3 Chebyshev filter poles. 

pole locus 


pole locus 

shown in Fig. 13.7. Next, we draw radial lines according to the angles of 
the Butterworth poles (Eq. 13.22) as shown. Finally, we draw vertical 
dashed lines from the intersections of the smaller circle and the radial lines, 
and horizontal dashed lines from the intersections of the large circle and 
the radial lines. The Chebyshev poles are located at the intersection of 
the vertical and horizontal dashed lines, as shown in Fig. 13.7. 

Consider the following example. We would like to obtain a system 
function H(s) that exhibits a Chebyshev characteristic with not more than 
1 -decibel ripple in the pass band and is down at least 20 decibels at w = 2. 

When we design for 1-decibel ripple, we know that at to = 1, \H(j\)\ is 
down 1 decibel so that 

20 log \H(jl)\ = 20 log —-t-zu = -1 ( 134 5) 

We then obtain 


(1 + * a ) 1A 


= 0.891 

e = 0.509 


Our next task is to find n from the 20 decibels at m = 2 specification. 
From Eq. 13.39 the loss can be given as approximately 

20 ^ 20 log 0.509 + 6(n - 1) + 20« log 2 


Solving for «, we obtain « = 2.65. Since n must be an integer, we let n = 3. 
With the specification of n and e, the pole locations are completely speci- 
fied. Our next task is to determine these pole locations. First we must 

Topics in filter design 379 
find P k . From Eq. 13.40 we have 

n e (13.49) 

= I sinh" 1 1.965 = 0.476 

In order to find the normalized Chebyshev poles from the Butterworth 
poles, we must first determine tanh fi k . Here we have 

tanh p k = tanh 0.476 = 0.443 (13.50) 

From Table 13.1, the n = 3 Butterworth poles are 

*! = -1.0, s 23 = -0.5 ±y0.866 (13.51) 

Multiplying the real parts of these poles by 0.443, we obtain the normalized 
Chebyshev poles. 

s\ = -0.443, s' 2S = -0.222 ±/0.866 

Finally, the denormalized Chebyshev poles are obtained by multiplying 
the normalized ones by cosh (i k = 1.1155 so that the denormalized poles 
are s x = -0.494 and $ 2S = -0.249 ±/0.972. 
H(s) is then 

H(s) = °^02 

(s + 0.494)(s + 0.249 - ;0.972)(s + 0.249 + /0.972) 

= 0.502 v ' 

s* + 0.992s* + 1.255s + 0.502 

In Fig. 13.8, the amplitude responses of the Chebyshev and an n = 3 
Butterworth filter are shown. 

Monotonic filters with optimum cutoff 

In comparing Butterworth filters with Chebyshev niters, the following 
can be said. The Butterworth response is a maximally flat, monotonic 
response, whereas the Chebyshev response is equal ripple in the pass band. 
In the stop band, the Chebyshev response falls off more rapidly than the 
Butterworth (except when e is very, very small). In this respect, the Cheby- 
shev filter is a better filter than the Butterworth. However, as we shall see 
in Section 13.5, the transient response of the Chebyshev filter is very poor. 
If we require sharp cutoff characteristics for a given degree n, however, 
the Butterworth filter is quite unsatisfactory. In 1958, Papoulis 8 proposed 

* A. Papoulis, "Optimum Filters with Monotonic Response," Proc. IRE, 46, No. 3, 
March 1958, pp. 606-409. 

380 Network analysis and synthesis 


Radian frequency, o> 
0.1 02 0.3 0.4 0.6 0.8 

1 i 

I 3 












Amplitude response of n = 3 

Chebyshev filter with 

1.0-db ripple in pass band 





— But 



>e (n 

= 3) 






— 11 

I \ 

— 12 

\ \ 

— 13 


FIG. 13.8. Amplitude response of n = 3 Chebyshev filter with 1.0-decibel ripple in pass 
band and Butterworth response (n = 3). 

a class of filters called Optimum or "L" filters, which have the following 

1. The amplitude response is monotonic. 

2. The fall-off rate at m cutoff is the greatest possible, if monotonicity is 


3. The zeros of the system function of the L filter are all at infinity. 

Recall that the magnitude response of a low-pass filter with all zeros at 
infinity can be expressed as 

*• (13.53) 

M(o>) = 

[1 +J V)l* 

Let us denote the polynomial generating the L filter by 

/(>*) = L„(ft> 8 ) 


Topics in filter design 381 
The polynomial L n (m x ) has the following properties: 

(a) LJO) = 

(b) L„(l) = 1 

(0 ^^0 


«0 dL > 2 > 


= M (M maximum) 

Properties a, b and c are the same as for the Butterworth generating poly- 
nomial/^*) = to 2 ". Property c insures that the response M(a>) is mono- 
tonic and property d requires that the slope of L n (caP) at tu = 1 be the 
steepest to insure sharpest cutoff. 

Papoulis originally derived the generating equation for the polynomials 
L n (for n odd) to be 

Ln((ot)= L Llo a<p<(a;) J dx (13,55) 

where n = 2k + 1 and the PJx) are the Legendre polynomials of the 
first kind 4 

/>,>(*) = 1 

P^x) = x 

P 2 (x) = K3* 2 - 1) 

P 9 (x) = i(5x* - 3x) (1356) 

and the constants a, are given by 

a« = — = — = • • ■ = = — 1= (13.57) 

3 5 2k + 1 V2(fc + 1) 

Later Papoulis 6 and, independently, Fukada,' showed that the even- 
ordered L n polynomials can be given by 

rim*-! r * -| 8 

W (a>8) -J (* + 1)[2>« *>«(*)] dx (13.58) 

n = 2k + 2 

• E. Jahnke and F. Emde, Tables of Functions, Dover Publications, New York, 1945. 

• A. Papoulis, "On Monotonic Response Filters," Proc. IRE, 47, February 1959, 

• M. Fukada, "Optimum Filters of Even Orders with Monotonic Response," 
Trans. IRE, CT-6, No. 3, September 1959, 277-281. 

382 Network anal/sis and synthesis 

where the constants a t are given by: 
Case 1 (k even) : 

a - 7 - 

«i = «» = 
Case 2 (k odd): 



2fc + 1 V(fc + l)(fc + 2) 
= a M = 






2fc + l V(fc + l)(k + 2 ) 
flo = «2 =•=«.* = ° 
Fukada tabulated the L B (w 2 ) polynomials up to n = 7 together with 
dLjai^ldco evaluated at to = 1 to give an indication of the steepness of 
the cutoff. This is shown in Table 13.4. 

To obtain the system function H(s) for the L filter, we must factor the 
equation for his 2 ) and choose the Hurwitz factors as H(s). 

For example, for n = 3, the magnitude response squared is 



M\a>) = 

1 + L^co*) 

1 + a>* - 3«> 4 + 3eo 6 
Substituting — <w 2 = s 2 , we obtain 

fc(s 2 ) = H(s) H(-s) = 


1 - s* - 3s 4 - 35* 



TABLE 13.4 
L„(a> 2 ) Polynomials 

L n (a> 2 ) 

dL n (l) 


3o> 8 - 3o> 4 + <u 2 

6eo 8 - 6o>* + 3to 4 

20o 10 - 40a> 8 + 28a> 8 - 8a> 4 + to 2 

50to 12 - 120a> 10 + 105e» 8 - 40w* + 

175*o M - 525a>» + 615«> 10 - 355o> 8 

6o> 4 

+ 105a>« 

-15eo 4 


a) 2 


Topics in filter design 383 




Frequency, rad/sec 
0.4 0.6 0.8 



Amplitude i 

— OpB 


mum vers 
srworth fit 





Frequency i \ 
0.4 0.5 0.6 0.8 1.0 1.2 \ \ 

— — ■ 







\ \ 
\ \ 

















FIG. 13.9. Amplitude response of Optimum versus Butterworth filters 
After we factor A(s*), we obtain 


if(s) = 


s* + 1.31s 2 + 1.359s + 0.577 
where the numerator factor, 0.577, is chosen to let the d-c gain be unity. 
The poles of H(s) are s x = -0.62; s 2>3 = -0.345 ±;0.901. The ampli- 
tude response of third-order Optimum (L) and Butterworth niters are 
compared in Fig. 13.9 Note that the amplitude response of the Optimum 
filter is not maximally flat, although still monotonic. However, the cutoff 
characteristic of the Optimum filter is sharper than the cutoff of the 
Butterworth filter. 

Linear phase filters 

Suppose a system function is given by 

H(s) = Ke~' T (13.65) 

where K is a positive real constant. Then the frequency response of the 

system can be expressed as 

H{jm) = Ke-"" T 


384 Network analysis and synthesis 

so that the amplitude response M(<co) is a constant K, and the phase re- 

#w) = -mT (13.67) 

is linear in co. The response of such a system to an excitation denoted by 
the transform pair {e(t), E(s)} is 

R(s) = K E(s)e-' T (13.68) 

so that the inverse transform r(t) can be written as 

KO-c-W)] (1369) 

= Ke(t - T)u(t - T) 

We see that the response r(t) is simply the excitation delayed by a time T, 
and multiplied by a constant. Thus no signal distortion results from 
transmission through a system described by H(s) in Eq. 13.65. We note 
further that the delay T can be obtained by differentiating the phase 
response <f>((o), by <u; that is, 

Delay _ - f^2> - r (13.70) 


Consequently, in a system with linear phase, the delay of the system is 
obtained by differentiating the phase response #co). 

A system with linear phase and constant amplitude is obviously desirable 
from a pulse transmission viewpoint. However, the system function H(s) 
in Eq. 13.65 is only realizable in terms of a lossless transmission line called 
a delay line. If we require that the transmission network be made up of 
lumped elements, then we must approximate H(s) = Ke~ ,T by a rational 
function in s. The approximation method we shall describe here is 
due to Thomson. 7 We can write H(s) as 


sinh sT + cosh sT 

where K is chosen such that H(0) =1. Let the delay T be normalized to 
unity and let us divide both numerator and denominator of H(s) by sinh s 
to obtain 

Kjsmhs (13?2) 

coth s + 1 

' W. E. Thomson, "Network with Maximally Flat Delay," Wireless Engrg., 29, Oct. 
1952, 256-263. 

Topics in filter design 385 

If sinh s and cosh * are expanded in power series, we have 

s* s* s* 
coshs = l+- + - + - + --- 

» .' ,' (1373) 

s s 6 s 
sinh5- S + - + - + - + '-- 

From these series expansions, we then obtain a continued fraction expan- 
sion of coth s as 

coth s = - + 

S 3 + 

s 5 1 (13.74) 

~* z+... 


If the continued fraction is terminated in n terms, then H(s) can be 
written as 

where B n (s) are f?es.«>/ polynomials defined by the formulas 
B t = s+1 (13.76) 

2?„ - (2/i - 1)3^ + s*B n _ t 
From these formulas, we obtain 

if, = s* + 3s + 3 

2?, = j» + 6s* + 15s + 15 


Higher order Bessel polynomials are given in Table 13.5, and the roots of 
Bessel polynomials are given in Table 13.6. Note that the roots are all in 
the left-half plane. A more extensive table of roots of Bessel polynomi- 
als is given by Orchard. 8 

The amplitude and phase response of a system function employing an 
unnormalized third-order Bessel polynomial 

Ms) = ^ (13.78) 

W s* + 6s* + 15s + 15 

• H. J. Orchard, "The Roots of Maximally Flat Delay Polynomials" IEEE Trans, 
on Circuit Theory, CT-12 No. 3, September 1965, 452-454. 

386 Network analysis and synthesis 

TABLE 13.5 
Coefficients of Bessel Polynomials 






b t 


h b 7 










































28 1 

are given by the solid lines in Figs. 13.10 and 13.11. These are compared 
with the amplitude and phase of an unnormalized third-order Butterworth 
function given by the dotted lines. Note that the phase response of the 
constant-delay function is more linear than the phase of the Butterworth 
function. Also, the amplitude cutoff of the constant-delay curve is more 
gradual than that of the Butterworth. 

TABLE 13.6 
Roots of Bessel Polynomials 
Roots of Bessel Polynomials 





-1.0 +y0 

-1.5 ±y0.866025 
[-2.32219 +y0 
i — 1.83891 ±yl.75438 
/ -2.89621 ±y0.86723 
\ -2.10379 ±y2.65742 
C -3.64674 +j0 

-3.35196 ±yl.74266 
[-2.32467 ±/3.57102 

C -4.24836 ±y0.86751 
-3.73571 ±/2.62627 

[ -2.51593 ±y4.49267 

r -4.97179 +j0 

-4.75829 ±yl .73929 

i -4.07014 ±y3.51717 

1-2.68568 ±y5.42069 

Topics in filter design 387 

Frequency, rad/sec 
0.1 0.2 0.4 0.6 0.8 1 


h 2 

! -14 

3 4 5 6 7 


V >^ 












rth — A 














FIG. 13.10. Amplitude response of n = 3 Bessel and Butterworth filters. 

0.2 0.4 0.6 

Frequency, rad/sec 
0.8 1.0 1.2 1.4 1.6 1.8 2.0 

FIG. 13.1 1. Phase responses of low-pass filters. 

388 Network analysis and synthesis 


In this section we will compare the transient response of the filters 
discussed in Section 13.4. In particular, we will compare the step response 
of the niters according to the following figures of merit: 

1. Rise time t R . The rise time of the step response is defined here as 
the time required for the step response to rise from 10% to 90% of its 
final value as depicted in Fig. 13.12. 

2. Ringing. Ringing is an oscillatory transient occurring in the response 
of a filter as a result of a sudden change in input (such as a step). A 
quantitative measure of the ringing in a step response is given by its 
settling time. 

3. Settling time. The settling time is that time t, beyond which the 
step response does not differ from the final value by more than ±2%, 
as depicted in Fig. 13.12. 

4. Delay time, t D . Delay time is the time which the step response 
requires to reach 50% of its final value as shown in Fig. 13.12. 

5. Overshoot. The overshoot of the step response is defined as the 
difference between the peak value and the final value of the step response 
(see Fig. 13.12) expressed as a percentage of the final value. 

8 10 12 

Time t, sec 

FIC. 13.12. Figures of merit for step response. t a = rise time; /. = setting time; 
t B = delay time. 

Topics in filter design 389 

Most of the foregoing figures of merit are related to frequency response, 
particularly bandwidth and phase linearity. Some of the quantities, such 
as rise time and delay time are intimately related to each other but have 
rather tenuous ties with overshoot. Let us examine qualitatively the 
relationships between the transient response criteria just cited and 
frequency response. 

Rise time and bandwidth have an inverse relationship in a filter. The 
wider the bandwidth, the smaller the rise time; the narrower the band- 
width, the longer the rise time. Physically, the inverse relationship could 
be explained by noting that the limited performance of the filter at high 
frequencies slows down the abrupt rise in voltage of the step and prolongs 
the rise time. Thus we have 

T B X BW = Constant (13.79) 

Rise time is a particularly important criterion in pulse transmission. In 
an article on data transmission,* it was shown that in transmitting a pulse 
of width 7\ through a system with adjustable bandwidths, the following 
results were obtained: 


Rise Time 

(/. - UTJ 












The table shows a definite inverse relationship between rise time and 

A definition of time delay is given by Elmore as the first moment or 
centroid of the impulse response 



t h{t) dt (13.80) 

provided the step response has little or no overshoot. Elmore's definitio 
of rise time is given as the second moment 

T B = \2nj\t - T D f h(t) a] (13.81) 

•R. T. James, "Data Transmission— The Art of Moving Information," IEEE 
Spectrum, January 1965, 65-83. 

390 Network anal/sis and synthesis 

These definitions are useful because we 
can obtain rise time and delay time directly 
from the coefficients of the system function 
H(s). Without going into the proof, which 
is in Elmore and Sands, 10 if H(s) is given as 

FIG. 13.13. R-C network. 

His) = 1 + ° lS + a * s2 + ' 

1 + b x s + 6js 2 + 

T D = b l — a x 

the time delay T D i% 
and the rise time is 

T R = {2tt[V - af + 2(a 2 - b 2 )]}* 
For the R-C network in Fig. 13.13, H(s) is 

V(s) R 

H(s) = 

■ + a*s n 
+ b n s m 



so that 


I(s) 1 + sRC 
T D = RC_ 
T B = yJlnRC 

It should be emphasized that Elmore's definitions are restricted to step 
responses without overshoot because of the moment definition. The more 
general definition of rise time is the 10-90% one cited earlier, which has 
no formal mathematical definition. 

Overshoot is generally caused by "excess" gain at high frequencies. By 
excess gain we normally mean a magnitude characteristic with a peak 
such as the shunt peaked response shown by the dashed curve in Fig. 
13.14. A magnitude characteristic with no overshoot is the magnitude 
characteristic of an R-C interstage shown by the solid curve in Fig. 13.14. 

Log frequency 
FIG. 13.14. Comparison of shunt-peaked and simple R-C magnitudes. 

10 W. C. Elmore and M. Sands, Electronics, National Nuclear Energy Series, Div. 
V, 1, McGraw-Hill Book Company, New York, 1949, pp. 137-138. 

Topics in filter design 391 



=5 0.7 

5 0.6 

I 0.5 

f 6.4 

■I 0.3 



In J 3 ~A> = 7/il=lo" 

2 4 6 8 10 12 14 16 18 20 

Time t, sec 

FIG. 13.15. Step response of normalized Butterworth low-pass filters. 

The step responses of the n = 3, n = 7, and n = 10 Butterworth filters 
are shown in Fig. 13.15. Note that as n increases, the overshoot increases. 
This is because the higher order Butterworth niters have flatter magnitude 
characteristics (i.e., there is more gain at frequencies just below the cutoff). 

Ringing is due to sharp cutoff in the filter magnitude response, and is 
accentuated by a rising gain characteristic preceding the discontinuity. 
The step response of an n =■ 3 Bessel (linear phase) filter is compared 
to the response of an n = 3 Chebyshev filter with 1-decibel ripple in 
Fig. 13.16. We cannot compare their rise times since the bandwidths of 
the two filters have not been adjusted to be equal. However, we can 
compare their ringing and settling times. The Chebyshev filter has a 
sharper cutoff, and therefore has more ringing and longer settling time 
than the Bessel filter. Note also the negligible overshoot of the Bessel 
filter that is characteristic of the entire class of Bessel filters. 

The decision as to which filter is best depends upon the particular 
situation. In certain applications, such as for transmission of music, 
phase is not important. In these cases, the sharpness of cutoff may be the 
dominant factor so that the Chebyshev or the Optimum filter is better 
than the others. Suppose we were dealing with a pulse transmission 
system with the requirement that the output sequence have approximately 
the same shape as the input sequence, except for a time delay of T = 
T t — T x , as shown in Fig. 13.17a. It is clear that a filter with a long rise 
time is not suitable, because the pulses would "smear" over each other 

392 Network analysis and synthesis 


i — 




/ ' 













-Step response of n ■ 3 Bessel fitter 
-Step response of n « 3 Chebyshev 
filter with 1-db ripple 

•S 0.6 
a nr. 

■a 0.5 









2 4 6 8 10 12 14 16 18 

Time, sec 
FIG. 13.16. Comparison of filter transient responses. 



Input pulse train 


with short 

rise and 





T 2 

Output pulse train 


Input pulse train 


with long 

rise and 




Output pulse train 

FIG. 13.17. Smearing of pulses in systems with long rise and setting times. 

as seen in Fig. 13.176. The same can be said for long settling times. 
Since a pulse transmission system must have linear phase to insure un- 
distorted harmonic reconstruction at the receiver, the best filter for the 
system is a linear phase filter with small rise and settling times. 


We present here a method to reduce the overshoot and ringing of a 
filter step response. The step response of a tenth-order Butterworth filter 
is shown in Fig. 13.18. It is seen that the overshoot is about 18%. We 

Topics in filter design 393 




I 0.6 
I 0.4 




tep re 








Second derivative 






8 10 12 14 16 18 20 
Time t, sec 

FIG. 13.18 

note that after the first peak, the ringing of the step response has an 
approximate sinusoidal waveshape. Let us now consider the second 
derivative of the step response shown by the dashed curve in Fig. 13.18. 
Beyond the first peak of the step response, the second derivative is also 
(approximately) sinusoidal, and is negative when the step response is 
greater than unity, and positive when the step response is less than unity. 
If we add the second derivative to the step response, we reduce the over- 
shoot and ringing. 11 

Suppose a(0 is the step response and H(s) is the system function of the 
filter. The corrected step response can be written as 

a 1 (0 = «(0 + «^ L> 


where K is a real, positive constant. Taking the Laplace transform of 
Eq. 13.87, we have 

tW*)] - 

Xs L— J 




11 F. F. Kuo, "A Method to Reduce Overshoot in Filters," Trans. IRE on Circuit 
Theory, CI-9, No. 4, December 1962, 413-414. 

394 Network analysis and synthesis 














— - — Tenth order Butterworth 

Zeros at a) = i 1.0 

Zeros at a>= * 1.5 





8 10 12 

Time t, sec 

FIG. 13.19 





From Eq. 13.88 we see that by adding a pair of zeros on the jm axis at 
s = ±//v K, the overshoot and ringing are reduced. 

For low-pass niters, the factor l/\K must in general be greater than 
the bandwidth of the system. For normalized Butterworth filters the 
bandwidth is co = 1 so that K <, 1. The factor AT also controls the amount 
of overshoot reduction. If K is too small, adding the zeros on they'co axis 



.o -12 


c -14 
<3 _i 6 



— « Tenth order Butterworth 

" Zeros at co = t 1.0 

.— — Zeros at w = * 1.5 

— — Zeros at co = ± 1.8: with 

original bandwidth also 

equal to 1.8 


0.2 0.3 0.4 0.6 0.8 1.0 
Radian frequency, u 

FIG. 13.20 

1.5 2.0 3.0 

Topics in filter design 395 

will have negligible effect. Therefore the zeros should be added some- 
where near the band edge. Figure 13.19 shows the effects of adding 
zeros at <w = ±1 (i.e., right at the band edge), and at m = 1.5. We see 
that the further away the zeros are placed from the band edge, the less 
effect they will have. The addition of the zeros will decrease the 3-decibel 
bandwidth of the filter, however, as seen in Fig. 13.20. Therefore, a 
compromise must be reached between reduction in bandwidth and 
reduction in overshoot. 

An effective way to overcome this difficulty is to scale up the bandwidth 
by a factor of, for example, 1.8. Then the zeros are placed at co = ±1.8, 
which will reduce the 3-decibel bandwidth to approximately its original 
figure <w = 1.0, as shown in Fig. 13.20. The overshoot, however, will be 
reduced as though the zeros were at the band edge. 


In this section we will examine an interesting result, which is due to 
Budak. 1 * The result deals with linear phase approximation with con- 
trollable magnitude. In Section 13.4 we discussed approximation of a 
flat delay using Bessel polynomials. The resulting rational approximant 
was an all-pole function (all transmission zeros at s = oo), whose de- 
nominator was a Bessel polynomial. There was no control of the magni- 
tude using the all-pole approximant. In Budak' s method the magnitude 
is controllable, while the phase is as linear as the standard Bessel approxi- 

Budak's approximation is obtained by introducing the parameter k to 
split e~~* into two parts such that 

*- = pSr. < k <. 1 (13.89) 

and then approximate independently c - ** and e~ lk ~ lU with all-pole Bessel 
polynomial approximations. Thus the resulting approximation for e~* 
will have Bessel polynomials for both numerator and denominator. The 
poles of the e~ ( * -1) ' approximant will be the zeros in the final approximant, 
while the poles of the e~** approximant remain as poles in the final 
approximant. For realizability, the degree of the c- <t-1) * approximant 
should be less than the degree of the e - ** approximant. 

11 A. Budak, "A Maximally Flat Phase and Controllable Magnitude Approximation," 
Trans, of IEEE on Circuit Theory, CT-12, No. 2, June 1965, 279. 

396 Network anal/sis and synthesis 




■ 0.7^ 


2 3 4 


ft -1.0 




= 0,8- 






FIG. 13.21 

As an example, consider the approximation with three zeros and four 



(ks)* + 10(fcs) 8 + 45(fcs)* + 105(jfcs) + 105 


[(fe - l)sf + 6[(fc - l)s] a + 15[(ik - l)s] + 15 
We then perform the operations as indicated by Eq. 13.89 to obtain 

-._ 7{[(fc - 1)5]' + 6[(k - l)s]« + 15[(fe - l)s] + 15} 
6 — (ks)* + 10(fcs) 3 + 45(fcs)» + 105(Jb) + 105 

In Fig. 13.21a the magnitude characteristic of Eq. 13.92 is plotted with 
£ as a parameter. The phase characteristic is given in terms of deviation 
of phase from linearity A<f> = eo — <£(co) and is shown in Fig. 13.216. 
The improvement in phase linearity over the all-pole Bessel approximation 
k = 1 is shown by these curves. Note as the bandwidth is increased 
(A: decreasing), phase linearity is improved. Figure 13.22 shows the step 
response of Eq. 13.92 also with A: as a parameter. Since the effect of 
decreasing k increases bandwidth, the corresponding effect in the time 
domain is to decrease rise time. 

Budak also observes that as k decreases from unity, the poles and zeros 
migrate to keep the phase linear. The zeros move inward from infinity 
along radial lines, while the poles move outward along radial lines. 

Topics in filter design 397 






■ 0.6- 







02 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 

FIG. 13.22 


Given the system function of the low-pass filter as derived by the 
methods described in Section 13.4, we can proceed with the synthesis of 
the filter network. If we consider the class of filters terminated in a 1-Q 
load, and if we let the system function be a transfer impedance, 

Z«(s) = -^a- 



or a transfer admittance Y n (s) = 

1 + y» 

we can synthesize the low-pass filter according to methods given in 
Chapter 12. For example, consider the n = 3 Optimum (L) filter function 

398 Network analysis and synthesis 
given as a transfer impedance 

Z21W = 



s 8 + 1.31s a + 1.359s + 0.577 

We see that the zeros of transmission are all at infinity. Since the numer- 
ator of Z 21 is even, we divide both numerator and denominator by the 
odd part of the denominator s 3 + 1.359*. Thus we have 


Za s 8 + 1.359s 
1.31s 8 + 0.577 


Z S2 — 

s s + 1.359s 

The structure of the low-pass filter with three zeros of transmission at 
infinity is given in Chapter 12. We must synthesize z M to give the it 
reactance structure. This we accomplish through the following continued 
fraction expansion of l/« 22 : 

1.31s 2 + 0.577 )s s + 1.359s (0.763s 
s 8 + 0.440s 

0.919s ) 1.31s 8 + 0.577 ( 1.415s 

1.31s 8 

0.577) 0.919s ( 1 .593s 

The optimum filter is shown in Fig. 13.23. For the n = 3 Butterworth filter 
given by the transfer impedance 

Z2l(s) ~s s + 2s 8 + 2s-rl 

we have 

z n (s) = 

s s + 2s 

,. 2s 8 + 1 

-t» - 7T7s 


We then synthesize « M (s) by a continued fraction expansion to give the 
filter shown in Fig. 13.23. 

(a) Optimum filter 

(b) Butterworth filter 

FIG. 13.23 

Topics in filter design 399 

c 7 :i >ia v 2 

FIG. 13.24. Canonical form for niters described in Tables 13.7, 13.8, and 13.9. 

In Tables 13.7, 13.8, 13.9 are listed element values (up to n = 7) for 
single-terminated Butterworth, Chebyshev (1-decibel ripple), and Bessel 
filters, respectively. 13 These apply to the canonical realization for a 
transfer impedance Z 21 (s) shown in Fig. 13.24. If a Y 21 (s) realization is 
desired, we simply replace all shunt capacitors by series inductors and 
vice versa. The element values all carry over. 

In Chapter 14, we will consider some examples of synthesis of double 
terminated filters. To stimulate the curiosity of the reader, note that the 
voltage-ratio transfer function V 2 jV of the network in Fig. 13.25 is 
precisely the n = 3 Butterworth function. 

Recall that in Chapter 12, when we cascaded two constant-resistance 
networks, the overall system function H (s) was the product of the in- 
dividual system functions H t (s) H 2 (s). We can apply this property to 
networks which are not constant-resistance if we place an isolation ampli- 
fier between the networks, as shown in Fig. 13.26. Since pentodes provide 
the necessary isolation, our task is simplified to the design of the individual 
structures H^s), H 2 (s), . . . , H n (s), which we call interstage networks. 

Some common interstage structures are shown in Fig. 13.27. In Fig. 
1 3.27a a structure known as the shunt-peaked network is shown. The trans- 
fer impedance of the shunt-peaked network is 

Za(s) = - 

1 s + RjL 

Cs 2 + sR/L+(llLC) 


FIG. 13.25. n = 3 double-terminated Butterworth filter. 

u More extensive tables are given in L. Weinberg's excellent book, Network Analysis 
and Synthesis, Chapter 13, McGraw-Hill Book Company, 1962. 

400 Network analysis and synthesis 

TABLE 13.7 

Normalized Element Values for a Single Terminated 
Butterworth Filter 


C x 

L t 




L t 

C 7 




































TABLE 13.8 

Normalized Element Values for a Single Terminated 
Chebyshev Filter with l-db Ripple 

C x L, C, L 4 C s 4 C 




































TABLE 13.9 
Normalized Element Values for a Single Terminated 

Bessel Filter 












































Topics in filter design 401 





FIG. 13.26. Pentodes used as isolation amplifiers. 

We sec that Z tl (s) has a real zero and a pair of poles which may be com- 
plex depending upon the values of R, L, and C. In Fig. 13.27A, a simple 
R-C interstage is shown, whose transfer impedance is 

Z n (s) - 



C s + 1/RC 

Observe that all the filter transfer functions considered up to this point 
are made up of pairs of conjugate poles and simple poles on the —a axis. 
It is clear that if we cascade shunt-peaked stages and R-C stages, we can 
adjust the R, L, and C elements to give the desired response characteristic. 
The only problem is to cancel the finite zero of the shunt-peaked stage. 
For example, if we wish to design an amplifier with an n = 3 low-pass 
Butterworth characteristic, we first break up the system function into 
complex pole pairs and real pole terms, as given by 

Z«(s) = 


(s* + s + l)(s + 1) 
s+ 1 1 • 1 


s* + s+ls + ls + l 

We then associate the individual factors with shunt-peaked or simple 
R-C stages and solve for the element values. The n *= 3 Butterworth 
amplifier is given in Fig. 13.28. 



ci ^* 

(o> (b) 

FIG. 13.27. (a) Shunt-peaked interstage. (f>) R-C interstage. 

402 Network analysis and synthesis 

FIG. 13.28. Butterworth amplifier. 


In Section 13.8, we discussed the synthesis of low-pass niters with a 
cutoff frequency of 1 rad/sec and a load impedance of 1 Q. Filters designed 
with these restrictions are considered to be normalized in both cutoff 
frequency and impedance level. We will now discuss methods whereby 
the normalized filters can be converted into filters which meet arbitrary 
cutoff frequency and impedance level specifications. Let us denote by a 
subscript n the normalized frequency variable s n and the normalized 
element values L„, jR„, and C„. The normalized frequency variable s n is 
related to the actual frequency s by the relation 

s„ = 


where <u , the normalizing constant, is dimensionless and is often taken to 
be the actual cutoff frequency. 

Since the impedance of an element remains invariant under frequency 
normalization, we obtain the actual element values from the normalized 
values by setting the impedances in the two cases equal to each other. For 
example, for an inductor, we have 

s„L„ = sL = a>o5 n L 


From this equation we then obtain the denormalized value of inductance 



Similarly, from the impedance l/s n C„ of a frequency normalized capacitor 
C„, we obtain the denormalized value of capacitance through the equation 

1 _ 1 
s n C n sC 


Topics in filter design 403 
so that the actual value of the capacitance is 

C = — " (13.106) 


Since resistances, ideally, are independent of frequency, they are unaf- 
fected by frequency normalization. 

Consider, next, impedance denormalization. Suppose the actual 
impedance level should be Rq ohms instead of 1 Q. Then a denormalized 
impedance Z is related to a normalized impedance Z n by 

Z = 2?oZ„ (13.107) 

where R is taken to be dimensionless here. Thus, for a normalized resistor 
R n , the denormalized (actual) resistance is 

R = RoR„ (13.108) 

For an inductance, the corresponding relationship is 

sL = R^sL n ) (13.109) 

so that the actual inductance value is 

L = R L n (13.110) 

Similarly, for a capacitor we have 

_L = Jk 
sC sC„ 
so that the actual capacitance is 


C = ^ (13.112) 


For combined frequency and magnitude denormalization, we simply 
combine the two sets of equations to give 

R = RoR n 

C = R^o (13.113) 

L = 


co a 

Let us consider an actual example in design. In Section 13.8, we 
synthesized a transfer impedance Zn with an n = 3 Butterworth amplitude 
characteristic with a cutoff frequency of 1 rad/sec and a load impedance of 
1 ii. Let us redesign this filter for a cutoff frequency of 10* rad/sec to work 

404 Network anal/sis and synthesis 


66.7 mh 

^=0.3 nf 

=4=0.1 nf 

500 a 

V 2 

FIG. 13.29. Denormalized low-pass filter. 

into a load of 500 Q. From the original network in Fig. 13.23, we take the 
element values and denormalize with the normalizing factors, <o = 10* 
and Rt = 500. 
Then the denormalized element values are 

R = 500* £ - 500 


50000 4 ) 

= 0.1 /*f 

L = 1(500) = 0.0067 h 


C, = 


ST = 0.3^f 

500(10 4 ) 
The final design is shown in Fig. 13.29. 


Up to this point, we have discussed only the design of low-pass filters, 
while neglecting the equally important designs of high-pass, band-pass, 
and band-elimination filters. We will remedy this situation here, not by 
introducing new design procedures but through a technique known as a 
frequency transformation, whereby, beginning from a normalized low-pass 
filter, we can generate any other form of filter. Using frequency trans- 
formations, the elements of the normalized low-pass filter are changed 
into elements of a high-pass, band-pass, or band-elimination filter. 

Analytically, a frequency transformation simply changes one L-C 
driving-point function into another I^C driving-point function. Therefore, 
the transformation equations must be LrC functions themselves. Also, 
since we proceed from normalized low-pass filters, the transformation 
equations include built-in frequency denormalization factors so that the 
resulting networks need only be scaled for impedance level. Consider the 

Topics in filter design 405 
simplest transformation equation, that of low-pass to high-pass, which is 

s = 



where s n represents the normalized low-pass frequency variable, s is the 
regular frequency variable, and to is the cutoff frequency of the high-pass 
filter. In terms of real and imaginary parts, we have 

a + jm = 


ff» +./'«>« 


Since we are interested principally in how the j<o n axis maps into theycu 
axis, we let <r„ = so that 




which is the equation that transforms normalized low-pass filters to 
denormalized high-pass filters. From Eq. 13.117 we see that the point 
w n = ±1 corresponds to the point m = ±a> . It is also clear that the 
transformation maps the segment \a>„\ ^ 1 on to the segments denned by 
m o ^ M ^ °°> as shown in Fig. 13.30. 

Now let us see how the frequency transformations change the network 
elements. For convenience, let us denote the normalized low-pass network 
elements with a subscript n, the high-pass elements with a subscript h, the 
band-pass elements with a subscript b, and the band-elimination elements 
with a subscript e. For the low-pass to high-pass case, let us first consider 
the changes for the capacitor C„. The transformation is given by the 


+j ■„ plane 




/<*> * plane 


FIG. 13.30. Low-pass to high-pass transformation. 

406 Network anal/sis and synthesis 


C„s n 

fo C, 

= L h S 

For the inductor L n , we have 

s C h s 



We observe that a capacitor changes into an inductor and an inductor 
changes into a capacitor in a low-pass to high-pass transformation (Fig. 
13.3 1). The element values of the high-pass filter are given in terms of the 
normalized low-pass filter elements as 


L h = 

<o C n 



C h = 

(o L n 


Consider the following example. From the normalized third-order 
Butterworth filter given in Fig. 13.23, let us design a corresponding high- 
pass filter with its cutoff frequency cu = 10« rad/sec and the impedance 
level of 500 Q. From the low-pass filter, we can draw by inspection the 






o 1{— 

Ca = 



C d = 


o — npnr- 




<">0 2L » 


L h - 

woC n 





r„. BW 

Cb2= -gyp 

FIG. 13.31. Element changes resulting from frequency transformations. 

Topics in filter design 407 

C h = 1.5 x 10~ 9 f 


FIG. 13.32. Transformation of low-pass filter in Fig. 13.23 into high-pass filter, 
high-pass-filter circuit shown in Fig. 13.32. Its element values are: 

^ 10 6 (i) 

1 . (13.122) 

10 4 (f) 

Next, let us examine the low-pass to band-pass transformation (also 
an L-C function): 

S - = S(- + -°) < 13123 > 

BW \a> s I 

where, if m ct and wp, denote the upper and lower cutoff frequencies of the 
band-pass filter, BWis the bandwidth 

BW^cocz-cOd (13.124) 

and <o is the geometric mean of <o C2 and co cl 

m = Vo)c2«>ci (13.125) 

The low-pass to band-pass transformation maps the segment \a> n \ <, 1 to 
the segments m^ ^ |a»| ^ eu^, shown in Fig. 13.33. The normalized 
low-pass elements are then modified according to the following equations : 

**. - bs + °^ 



408 Network anal/sis and synthesis 


«» Plane 





s plane 


FIG. 13.33. Low-pass to band-pass transformation. 

We note that the inductor L„ is transformed into a series-tuned tank, 
shown in Fig. 13.31, whose elements are given as 

L M = 

C M = 



«>0 L n 

The capacitor C„ is transformed into a parallel-tuned tank (Fig. 13.31), 
whose elements are 


■LfiS — 



Let us transform the third-order Butterworth low-pass filter in Fig. 13.23 
into a band-pass filter with a 1-ii impedance level, whose bandwidth is 
BW =6x10* rad/sec, and its band-pass is "centered" at w = 4 x 10* 
rad/sec. We draw the band-pass filter shown in Fig. 13.34 by the rules 

Li Ct 

nroip — 1(- 

(T\ isg d=c 3 yCi 

d=c 3 "=r-Ci Sjij i* 

v 2 

FIG. 13.34. Band-pass filter transformed from low-pass filter in Fig. 13.23. 

Topics in filter design 409 

given above. The element values of the band-pass filter are given in the 
following equations: 

U « 6x 10 * = 0.75 x KT'h 

** (4 x io*m) 


— X 10-*f 

6 X 10* 12 

-^— = ^xl(T 4 h 
6 x 10* 9 

6 X 10* 

(4 X 10»At) 32 

- x 10-*f 

= 6X10 4 = o.25 x l(T*h 
^ (4 X lO^f) 


Q = 

= 0.25 X KT*f 

6x 10* 

Finally, the band-elimination filter is obtained through the transfor- 


s. = 

Is eu \ 

\O) S I 


where BW and a> are defined in a manner similar to that for the band-pass 
filter. The transformation maps the segment of the _/«>,, axis in Fig. 13.35a 
onto the segments shown on the ja> axis in Fig. 13.356. For the low-pass 







(a) (b) 

FIG. 13.35. Low-pass to band-elimination transformation. 

410 Network analysis and synthesis 

to band-elimination transformation we, therefore, have the following 
element changes: 

Ls 1 

n " (s/L n BW) + ML n BWs) 

A 1 


C tl s + (1/L a s) 


C n s n C n BW\a> 

= L si s -\ 

° C <gS 

Observe that the normalized low-pass inductor goes into a parallel- 
tuned circuit and the capacitor C„ goes into a series-tuned circuit, as 
shown in Fig. 13.31. In Table 13.10, we have a composite summary of the 
various transformations. 

TABLE 13.10 

Table of Various Frequency Transformations 

Low-Pass to Equation 



s n =■ 

■fe + ?) 


13.1 Find the transfer impedance Z n = VJI X for the filter shown in the 
figure. What should L be in order for \Z 21 (ja>)\ to be maximally flat? 

PROB. 13.1 

Topics in filter design 41 1 

13.2 Find the poles of system functions with n = 3, » = 4, and n = 5 
Butterworth characteristics. (Do not use the tables.) 

13.3 Show that the half-power point of a Chebyshev low-pass amplitude 
response is at m = cosh /?* for e « 1. 

13.4 Determine the system function for the following filter specifications: 

(a) Ripple of \ db in band |o>| <> 1 ; 

(b) at co = 3, amplitude is down 30 db. 

13.5 Compare the slopes at to = 1 of the following polynomials (for n — 3): 

(a) /(a>*) = a>*» 

(6) /(a,*) - iC„(2a»* - 1) + J = Q«(«.) 

(c) /(«>») = Ln(«» 8 ). 

13.6 Determine the polynomials L 4 (co 2 ) and !»(«>*). 

13.7 Expand cosh j and sinh s into power series and find the first four terms 
of the continued fraction expansion of cosh * /sinh s. Truncate the expansion at 
« = 4 and show that H(s) = KjB^s). 

13.8 Synthesize the n = 3 linear phase filter as a transfer impedance termin- 
ated in a l-O load. 

13.9 Synthesize the low-pass filter, which, when terminated in a 1-il resistor, 
will have a transfer admittance whose poles are shown in the figure. 

PROB. 13.9. 

13.10 Determine the asymptotic rate of falloff in the stop band of: (a) 
optimum filters; (b) linear phase filters. 

13.11 Synthesize the n = 3 and n = 4 Butterworth responses as transfer 
impedances terminated in a load of 600 SI with a cutoff frequency of 10* rad/sec. 

13.12 Synthesize a Chebyshev low-pass filter to meet the following speci- 

(a) load resistor, R L = 600 Q 

(b) &-db ripple within pass band 

(c) cutoff frequency = 5 x 10 5 rad/sec 

(d) at 1.5 x 10* rad/sec, the magnitude must be down 30 db. 

412 Network analysis and synthesis 

13.13 Synthesize n = 3 Optimum and linear phase filters to meet the following 

(a) load resistor = 10* Q 

(b) cutoff frequency - 10* rad/sec. 

13.14 Design an n = 4 Butterworth amplifier with the following specifi- 

(a) impedance level = 500 ft 

(ft) cutoff frequency = 10* rad/sec. 

13.15 Synthesize a high-pass filter for a given transfer admittance terminated 
in a 10*-O load, whose amplitude characteristic is Optimum (L), with a cutoff 
frequency of a> =■ 10* rad/sec. 

13.16 Synthesize: (a) a band-pass filter; (b) a band-elimination filter, with 
maximally flat (n = 4) amplitude response with ce w - 8 x 10* and o>ci = 
2 x 10*. 

chapter 14 

The scattering matrix 


In this chapter, we will devote our attention to certain power relation- 
ships in one- and two-port networks. The characterization of a network in 
terms of power instead of the conventional voltage-current description is a 
helpful analytical tool used by transmission engineers. It is especially 
important in microwave transmission problems where circuits can no 
longer be given in terms of lumped R, L, and C elements. In the power- 
flow description, we are concerned with the power into the network, which 
we call the incident power and the power reflected back from the load, 
which is the reflected power. A convenient description of the network in 
terms of incident and reflected power is given by the scattering matrix, 
which is the main topic of discussion in this chapter. 

It is convenient to think of incident and reflected power when dealing 
with transmission lines. Therefore, we will briefly review some concepts in 
transmission line theory. For a more comprehensive treatment of trans- 
mission lines, the reader is referred to any standard text on wave propa- 
gation. 1 Consider the transmission line shown in Fig. 14.1. The voltage 
at any point down the line is a function of x, the distance from the source. 
The parameters which describe the transmission line are given in the 

R = resistance per unit length 
G = conductance per unit length 
L = inductance per unit length 
C = capacitance per unit length 

1 See, for example, E. C. Jordan, Electromagnetic Waves and Radiating Systems, 
Prentice-Hall, EngUwood Cliffs, New Jersey, 1950. 



Network anal/sis and synthesis 

i h: l- 

FIG. 14.1. Transmission line. 

Given these parameters, we can now define the impedance per unit 

length as 

Z = R+joL (14.1) 

and the admittance per unit length as 

Y=G+jcoC (14.2) 

The characteristic impedance Z of the line is given in terms of Z and Fas 

z = Vz/r (14.3) 

and the propagation constant is 

y = sfZY (14.4) 

With these definitions in mind, let us turn to the general equations for 
the current and voltage at any point x down the line 

V{x) . Vfi->* + Vs yx 
I(x) = !£*" - iy 


Z Z 

The terms with the subscript / refer to the incident wave at point * and the 
terms with subscript r refer to the reflected wave at x. Solving Eqs. 14.5 
simultaneously, we obtain explicit expressions for the incident and reflected 

^■«*[K(*) + Z,/(*)] (146) 

V r e* x = h[V(x)-Z l(x)] 

Consider the case when a transmission line of length L is terminated in 
its characteristic impedance, that is, 

V(L) _ 



The scattering matrix 415 
Then we see that the reflected wave is zero. 

f>» L = (14.8) 

Since e yL cannot be zero, we see that the coefficient V T is identically zero 
for this case. As a result, the reflected wave at any point x is zero. Also, 
the impedance at any point x down the line is equal to Z as seen from 
Eq. 14.5 with V r = 0. With these brief thoughts of transmission lines in 
mind, let us turn our attention to the main topic of this chapter, namely, 
the scattering parameters. 


For the one-port network shown in Fig. 14.2a, consider the following 
definitions. The incident parameter a is defined as 



and the reflected parameter b, is defined as 

where R^ is an arbitrary, positive, dimensionless constant called the 
reference impedance factor. For the transmission line described in 
Section 14.1, if the characteristic impedance Z = /{<,, then we can describe 
the incident parameter in terms of the incident voltage as 

a - ^=- (14.11) 

Similarly, b can be expressed in terms of the reflected wave as 

b = ^=r (14.12) 



• network 



a * 

, b 

(a) m 

FIG. 142. Scattering parameters of a one-port network. 

416 Network analysis and synthesis 

Thus we see that the parameters a and b do indeed describe an incident- 
reflected wave relationship in a one-port network, as depicted in Fig. 14.26. 
To give further meaning to a and b, consider the power dissipated by the 
one-port network 

P = iRe VI* (14.13) 

where /* denotes the complex conjugate of /. From Eqs. 14.9 and 14.10 
we solve for V and / in terms of the incident and reflected parameters to 
give _ 

K=(a + b)V« 

a - b (1414) 

Then the power dissipated by the one-port network is 

P = Haa* - bb*) 
-KM* -1*1*) 


where, again, the asterisk denotes complex conjugate. The term Jaa* 
can be interpreted as the power incident, while $bb* can be regarded as the 
power reflected. The difference yields the power dissipated by the one-port 

The incident parameter a and the reflected parameter b are related by the 

b = Sa (14.16) 

where S is called the scattering element or, more commonly, the reflection 
coefficient. From the definitions of a and b we can make the following 

Solving for S, we obtain S = Z ~ **° (14.18) 

Z + R 

where Z is the impedance of the one-port network 

Z = | (14.19) 

A further useful result is that when the impedance R<, is set equal to the 
impedance Z, the reflected parameter b — 0. 
For the one-port network excited by a voltage source V 9 with a source 

The scattering matrix 417 

resistor R,, as depicted in Fig. 14.3, we 
will show that when we choose the 
reference impedance Rq to be equal to + 

-& (,4 - 20> 

The proof follows. By definition, the 
incident parameter a is given as 


From Fig. 14.3, we have 

V, -IR B =V 

1 = 

Ra + Z 







FIG. 14.3 



Substituting these equations for V and / into the expression for a in Eq. 
14.21, we obtain 

_ l 17., V B R, \ , RqV,_~\ 

_ 2V^ r r,+z } ^+ zJ 

2VV R ' + zJ 


Consider once more the expression for the power dissipated in a one- 
port network: 
V P = l(aa* - bb*) (14.25) 

Factoring the term aa* = \a\* from within the parentheses, P becomes 

2 \ |a|V 

= l^- a (i-isi») 


When we choose the reference impedance to be equal to the source resist- 
ance, i.e., when R* = R„, then S = and 

P n 2 _ nil! 

e A. ~ ~Z~ ~ on 



418 Network analysis and synthesis 

** A where P A represents the available gain or 

~»vV ° I available power of a voltage source V a with 

a source resistance R„. For the case of a 
A v > one-port network, the available gain is 

^*^ *" ^ defined as the power dissipated in the one- 

port network when the impedance of the 
network Z is equal to the resistance of the 

FIG. 14.4 source/?,. As a result of this definition, we 

see that for the one-port network shown in 
Fig. 14.4, the power dissipated in Z with Z = R t is 

Pa = ^- (14.28) 

The available gain thus represents the maximum available power at the 
terminals of the voltage source. 

From this discussion, it is apparent that the value of the reference im- 
pedance Rq should be chosen equal to the source impedance R„. A 
standard procedure is to assume a 1-Q source impedance and denormalize 
when necessary, that is, let 

S - 2^-i (14.29) 

z + 1 

where z = — (14.30) 

Next, let us briefly consider some of the important properties of the 
scattering parameter S for a one-port network. 

1. The magnitude of S along theyco axis is always less or equal to unity 
for a passive network, that is, 

|S(/«>)| <. 1 (14.31) 

This property follows from the fact that the power dissipated in a passive 
network is always greater or equal to zero. Since the power can be 
expressed as 

P-^fl-ISft^O (14.32) 

we see that \S(ja>)\* ^ 1 (14.33) 

2. For a reactive network |S(jo>)| = 1. This property follows from the 
fact that the power dissipated in a purely reactive network is zero. 

The scattering matrix 419 

3. For an open circuit 5=1, and for a short circuit S = — 1. This is 
shown to be true from the equation 

S = ^| (14.34) 

z + 1 

For an open circuit z = oo, so that 5=1. For a short circuit z = 0; 
therefore 5= —1. 

Before we proceed to the next property, let us consider the following 

DEFINITION. A bounded real function P(s) is defined by the con- 

(«) |^)| ^ AT for Rea^O 

(6) F(s) is real when s is real. 

In (a), K denotes any positive real constant. 

4. If z — Z/Rq is a positive real function, then 5 is a bounded real 

The proof follows from the equation, 

S = (z- l)/(z + 1). 

From the positive real condition (a) Re z(s) ^ 0, when Re s ^ 0, we see 


s j Im z(s)-[l- Re z(s)] 
jlmz(s)+ [I + Rez(s)] 

so that |S(s)| = ( Im'^ + U-Re^ 

llm , z(s) + [l + Re2( S )]«/ * y ' 3 ° f 

when Re j ;> 0. (b) Where s is real, z(s) is real. Then 

S = (z- l)/( 2 + 1) 

must be real. Thus the scattering parameter for a passive network is a 
bounded real function. 


In this section we will extend the concepts developed for one-port 
networks discussed in Section 14.2 to two-port networks. In the two-port 
network shown in Fig. 14.5, we are concerned with two sets of incident and 
reflected parameters {a t , * x } at the 1-1' port, and {a,, b t } at the 2-2' port. 
These parameters are defined in similar manner as for the one-port 

420 Network anal/sis and synthesis 

-o lo— >- 




-o l'o- 


-*— o2 o- 


v 2 


D2' o- 


FIG. 14.5. Scattering parameters for a two-port network, 
network, that is, 

"• = l(vfe +vr "'') 

where R 01 and i? w are the reference impedances at the input and output 
ports respectively. 

The scattering parameters S tj for the two-port network are given by the 

b t = iSjitfi + S i2 a 2 
In matrix form the set of equations of Eqs. 14.38 becomes 

where the matrix 

r 6i i = \ Si1 ^im 




is called the scattering matrix of the two-port network. From Eqs. 14.38, 
we see that the scattering parameters of the two-port network can be 
expressed in terms of the incident and reflected parameters as 








S M 






The scattering matrix 421 

I I 

In Eqs. 14.41, the parameter S u is called the input reflection coefficient; 
5 M is the forward transmission coefficient; S lt is the reverse transmission 
coefficient; and S M is the output reflection coefficient. Observe that all 
four scattering parameters are expressed as ratios of reflected to incident 

Now let us examine the physical meaning of these scattering parameters. 
First, consider the implications of setting the incident parameters a t and a t 
to zero in the defining relations in Eqs. 14.41. Let us see what the con- 
dition a 4 = implies in the definition for the forward reflection coefficient 



Figure 14.6 shows the terminating section of the two-port network of Fig. 
14.5 with the parameters a, and b a of the 2-2' port shown. If we treat 
the load resistor R t as a one-port network with scattering parameter 

S 8 = 

R» + Ro» 


where Rgx is the reference impedance of port 2, then o i and b % are related 

a a = SA (14.43) 

When the reference impedance R^ is set equal to the load impedance Rt, 
then S t becomes 


Rot — Ri 

Rat + Rt 

*8 = 



so that a t «■ under this condition. Similarly, we can show that, when 
a x — 0, the reference impedance of port 1 is equal to the terminating 

* From the viewpoint of the load resistor /?„ the incident parameter is b, and the 
reflected parameter is a x . 

422 Network anal/sis and synthesis 

impedance (i.e., R 01 = /?i). We see as a result of this discussion that the 
conditions a x = and a a = merely imply that the reference impedances 
R 01 and /? M are chosen to be equal to the terminating resistors R t and R t , 

Next, let us consider the relationship between the driving-point imped- 
ances at ports 1 and 2 and the reflection coefficients 5 U and S u . Let us 
denote the driving-point impedances at ports 1 and 2 as 

From the equation 

We can write 

S„ = 

h h 

S -M 

a l lai-0 

which reduces easily to S n = 

Similarly, we have 

*K W*oi) + 

2q — Rqi 
Z x + Rqx 

z» — Rq» 








Z t + R oi h,_h„ 

These expressions tell us if we choose the reference impedance at a given 
port to equal the driving-point impedance at that port, the reflection 
coefficient of that port will be zero, provided the other port is terminated 
in its reference impedance. 

Next, let us derive some physically meaningful expressions for the 
forward and reverse transmission coefficients S n and S lt . Consider the 
definition for <S 81 

S -*l 

•321 — — 



As we have just seen, the condition a, = implies that the reference im- 
pedance Roi is set equal to the load impedance R t , as seen in Fig. 14.7. 
If we connect a voltage source V gl with source impedance Rn = R lt 


FIG. 14.7 

The scattering matrix 423 
then we can express a x as 

""vfe < 14 - 5I > 

Since a, = 0, we have the equation 


from which we obtain —^ = —>/*»'» (14.53) 

Consequently, b t = "(-/= - V*» z « ) 




Finally, we can express the forward transmission coefficient as 

s _ vjj**_ 

Krl^ ^* Ri-fi.i,Ri-Boi 

In similar fashion, we find that when port 1 is terminated in Rqi = R t and 
when a voltage source V, t with source impedance R t is connected to port 2, 

S " - ?MF (1456) 

We see that both S lt and S M have the dimensions of a voltage-ratio transfer 
function. Indeed, if R^ = R^, then S lt and S tl are simple voltage ratios. 
It is seen that for a passive reciprocal network, S n = S lt . 

Now let us consider as an example the scattering matrix of the n: 1 ratio 
ideal transformer in Fig. 14.8a. Recall that for an ideal transformer 

'Vi-jtV* A---/, (14.57) 


Assuming first that Jin'— Rn ■» 1, let us find 5 U by terminating the 
2-2' port in a l-Cl resistor, as shown in Fig. 14.86. Then 

-^- - 1 (14.58) 


Network anal/sis and synthesis 

1 h _Jf_ 





— o 

V 2 

Ideal transformer 

FIG. 14.8. Determination of scattering parameters for an ideal transformer, 
so that Z 1 = ^ = n* (14.59) 

From Eq. 14.48, we have S n = 



n* + l 

Next we terminate the 1-1' port in a 1-G resistor (Fi£. 14.8c). We obtain, 
as we did for S u , 

(1/w*) - 1 _ 1 - n* 

(!/«*) + 1 1 + n 1 

S M = 


The scattering matrix 425 

We obtain S n by connecting a voltage source V gl with a source impedance 
jRo! = 1 Q at the 1-1' port and terminating the 2-2' port with a resistance 
jRo, = 1 ii, as seen in Fig. 14.8rf. Since V^h - «*» the equivalent circuit 
of the ideal transformer as seen from the voltage source as a 1-Ii imped- 
ance in series with an «*-ohm resistance (Fig. 14.8c). Then V x can be 
expressed in terms of V tl , as 


Since V a = Vjn, we have 
Since R^ = R^ = 1 Q, S M is 

n 8 + l 

n* + l 

2F, 2it 

r »i 

n* + l 




We can show in similar fashion that 


n + 1 
Therefore the scattering matrix for the ideal transformer is given as 




n* + l 

n* + l 



M* + l fl* + l 


As a second example, let us find the scattering matrix for a lossless 
transmission line of length L terminated in its characteristic impedance, as 
shown in Fig. 14.9. If we assume that R^ = !*« = Z* then the reflection 
coefficients are 

c = ?oZlZo _ Q 

z, + z 





FIG, 14.9. Lossless transmission line. 


426 Network anal/sis and synthesis 

This result is not implausible, because a transmission line terminated in 
its characteristic impedance has zero reflected energy. To determine S u , 
we terminate the line in Z at both ends and connect at the 1-1' port a 
voltage source V tl , as depicted in Fig. 14.9. Since the transmission line 
has zero reflected energy, that is, 


In similar fashion, we find that 

Sit - 2e- rL (14.71) 

Therefore the scattering matrix for the lossless transmission line is 

Z> x = a 2 = 


v, = *>- ri 

From the equation 

Stl ~ 2 \*J v gl 

we obtain 

S n - 2e-* 

r o 2«-" z, i 

~~ VW- J 



Having defined the scattering matrix of a two-port network in 
Section 14.3, let us consider some important properties of the scattering 
matrix. From the general restriction for a passive network that the net 
power delivered to all ports must be positive, we obtain the condition 

P = H«i«i* + a*fh* - W - W) £ (14.73) 

Equation 14.73 follows from the fact that the power delivered to the 1-1' 
port is 

Pi - KW - bib,*) (14.74) 

and the power delivered to the 2-2' port is 

P, - i(^»* - W) (14.75) 

The total power delivered to the network is then 

P = P 1 + P t (14.76) 

which is exactly the expression in Eq. 14.73. In matrix notation, the power 
delivered to the network is 

P - Hla'fW ~ lb*flb]} £ (14.77) 

The scattering matrix 427 
where T denotes the transpose operation, and 




Since [b] = [S][a], then 

[b*] T = [a*] T [S*] T (14.79) 

Equation 14.77 can now be rewritten as 

IP = {[a*] T [a] - [a*] T [S*] T [S][a]} 
- la*] T llu] - [S*] T [Smal > 


This then implies that the determinant of the matrix [[«] — [5*] T [S]] 
must be greater or equal to zero, that is, 

Dct [[«] - [S*] T [S]] £ (14.81) 

Consider the special but, nevertheless, important case of a lossless 
network. In this case P = 0, so that 

[S*] T [S] = [u] (14.82) 

A matrix satisfying the condition in Eq. (14.82) is unitary. For a lossless 
two-port network 

From this equation, we have the following conditions for the scattering 

Su*S 11 + S tl *S tl = 1 (14.84) 

Sj»*Su + Sn'Su = (14.85) 

S n *Si, +■ S n *S„ = (14.86) 

Su*Su + S„*S„ - 1 (14.87) 

Note that Eqs. 14.85 and 14.86 are conjugates of each other. If the 
network is reciprocal, then S tl = S lt and 

|Sii(/«>)|» + |S„(/«>)|» = 1 

IVHI 1 + |5„(/«))|« - 1 

428 Network analysis and synthesis 


A <*2 

FIG. 14.10 

from which it follows that for a lossless reciprocal network |S u (ya>)| = 
\SJ(ja>)\ <. 1 and \S u (ja>)\ ^ 1. Also it is clear that when \Sjj<o)\ = 
(i.e., when there is a zero of transmission), then \S u (ja>)\ = 1. This 
condition states that all the power that has been delivered to the network 
from port 1-1' is reflected back to port 1-1'. 

At this point, it might be profitable to discuss why we use scattering 
matrices. What are the advantages of the scattering description over 
conventional descriptions? Let us discuss three major reasons for the 
scattering formalism. 

1. Many networks do not possess an impedance or admittance matrix. 
For example, an ideal transformer has no Z or Y matrix because its 
elements are not finite. However, as we have seen, the ideal transformer 
can be described by a scattering matrix. Carlin states 8 that all passive 
networks possess scattering matrices. 

2. At high frequencies, incident and reflected parameters play dominant 
roles in problems of transmission, while voltage-current descriptions are 
relegated to the background. Then the scattering matrix is necessarily the 
more powerful description of the system. Note that the voltage standing- 
wave ratio (VSWR) is given in terms of a reflection coefficient S as 

VSWR = i±M (14.89) 

3. In networks where power flow is a prime consideration (e.g., filters), 
the scattering matrix is very useful. For example, in the network given in 
Fig. 14.10, ifP A represents the available power from the generator and P t 
is the power dissipated in the load R t , then we that the magnitude- 
squared forward transmission coefficient is 

is«o)i*-£* < 14 - 90 > 



We will discuss this point in more detail in Section 14.5. 

* H. J. Carlin, "The Scattering Matrix in Network Theory," Trans. IRE, CT-3, 
No. 2, June 1956, 88-96; see his extensive bibliography. 

The scattering matrix 429 


In Section 14.4, we described the forward and reverse transmission 
coefficients in terms of voltage ratios. Perhaps a more appropriate 
description of a transmission coefficient is in terms of a power ratio rather 
than a voltage ratio. In this section we will show that |S M (/«o)|* and ]S U 
(J(o)\* can be expressed in terms of power ratios. We will then introduce 
the very important concept of insertion loss and finally relate |£ti(/<»)l* to 
the insertion power ratio. 

Consider the equation for S tl in the two-port network shown in Fig. 

From the equation l-S^OOl* = S n (/w)S a *(/a>) (14.92) 

we obtain |M-)|- = l^>» 




where P t is the power dissipated by the load R t , and P Al is the available 
gain of the generator V gl . Similarly, we have 

|Si.O)l* - £- (14.94) 

We see that both \S u (j(o)\* and ISutyco)!* are power transfer ratios which 
relate the power dissipated at a given port to the available power in the 
other port. 

Now let us examine the idea of insertion loss. Consider the network 
shown in Fig. 14.1 la. Between the terminals 1-1' and 2-2' we will insert a 
two-port network, as shown in Fig. 14.116. Let us denote by V M the 
voltage across the load resistor R, before inserting the two-port network, 
and by V t the voltage across R t after inserting the two-port network. A 
measure of the effect of inserting the two-port network is given by the 
insertion voltage ratio IVR, which is denned as 

IVR 4 ^ (14.95) 

V t 

Another method of gaging the effect of inserting the two-port network 
is to measure the power dissipated at the load before and after inserting the 

430 Network analysis and synthesis 

-J t 













>V 2 



FIG. 14.11 

two-port network. If P M is the power dissipated at the load before the 
two-port network is inserted, and if P t is the power dissipated after inser- 
tion, then the insertion power ratio of the two-port network is denned as 

e u _ Ew 


If we take the logarithm of both sides, we obtain 

a =10 log ^ 

where a is the insertion loss of the two-port network. In terms of the 
circuit given in Fig. 14.11, we can calculate P x from the relation 


V„ = 


Then P m is 

P» = 

|K„| 8 



2R t 


2(R t + RJ* 
The power dissipated by the load after inserting the two-port network is 


given by 



2R t 


The scattering matrix 
The insertion power ratio can then be expressed as 


2a _ £_80 _ 

w„r v 

P.-VXOk + V <14101) 

In the special case when the source and load impedances are equal, that 
R x = R 2 = R n = R^ (14.102) 

the reciprocal of the squared magnitude of the forward transmission 
coefficient in Eq. 14.93 is equal to the insertion power ratio 


When R t j& R t , then 


|S*iO)| a 
4R 1 R 2 



(*i + R*) 2 |s*iO)l a 



In any event, we see that the magnitude-squared transmission coefficients 
i\S a (jai)\ t and {S^jai)]* can be regarded physically as equivalent insertion 
power ratios. In Section 14.6, we will use this relationship in the synthesis 
of double-terminated filter networks. 


In this section, we will consider a filter synthesis procedure first proposed 
by Darlington in a classic paper in 1939. 4 We will use scattering matrix 
notation to describe the essence of Darlington's original work. Our 
coverage will be restricted to the class of low-pass filters which are termi- 
nated in equal source and load impedances, Rn = R oi = R , as shown in 
Fig. 14.12. For normalizing purposes we will let /^ be equal to 1 12. 

FIG. 14.12 

4 S. Darlington, "Synthesis of Reactance 4-Poles which Produce Prescribed Insertion 
Loss Characteristics," /. Math. Phys., 18, 1939, 257-353. 

432 Network analysis and synthesis 

Recall that when the source and load 
impedances are equal, then the insertion 
power ratio is equal to the reciprocal of 
|S M (yG>)| 2 , that is, 



P* |S 21 (>>)| a 


Expressed as a loss function, the insertion 
power ratio is 

A =10 log %*■ 

= -101og|S 21 (»| a db (14.106) 

In circuit design, the specification of an 
insertion loss A (Fig. 14.13a) is equivalent 
to the specification of the amplitude- 
squared transmission coefficient shown in Fig. 14.13ft. One of the most 
ingenious techniques given in Darlington's synthesis procedure is the re- 
duction of insertion loss synthesis to an equivalent L-C driving-point 
synthesis problem. This technique can be developed in terms of scattering 
parameters. Our initial specification is in terms of |S u (/<u)|. For an L-C 

two-port network 

\Sn(jcoW = 1 - |S*iO)l 8 

Next, S n (s) is obtained from the magnitude-squared function 
S u (s)Sii(- l s) = 1 - \S ia (Jo>)\ 2 \ im ^, 

Then from the equation S u = — — — — - 

^i + «o 

we obtain the driving-point impedance 

1 + Su(s) 

Z^s) = R 

1 - S u (s) 




shown in Fig. 14.12. We then synthesize the network from Z^s). 

We will restrict our discussion here to low-pass filters given by the 
lossless ladder structure terminated at both ports by 1-Q resistors in 
Fig. 14.14. These low-pass filters can take the form of a Butterworth or 
Chebyshev specification for |S sl (/a))| 2 , that is, 

|S 21 0)f = 

2 _ 

1 + CO 



10 Li 


| + 

T- Zi(s) 

c 2 z 

The scattering matrix 433 

~*np T-nnnp — 

C-1-— - 


v 2 

FIG. 14.14. Canonical form for double-terminated low-pass niters. 

1 + c C B 8 (<u) 
where C n (co) represents an nth-order Chebyshev polynomial. 
Example 14.1. Let us synthesize a low-pass filter for the specification 

._. 1 




which represents a third-order Butterworth amplitude characteristic. The load 
and source impedances are R^ = R^ = 1 Q. First we find |5 u (y'o>)| s as 

IWI* = !-—-* 

1 +0) 6 

Lettingy'eo = s in |S u (/a>)| a , we obtain 

<Sii(»>S_(-*) = - 


1 -s* 
A-* 8 ) 



which factors into 

S n (s) Sil (- S ) - g +2> + ^ + " ^ 1^ +2j2 _ ^ (14.116) 


sothat5 u (5)is s ll(f ) = _______ (14 . 117) 

Next, Z x (j) is obtained from the equation 

Zfy) = 

1 + S u (s) 
1 - SuM 

2s* +2s* +25 + 1 

2s? + 2s + 1 


434 Network analysis and synthesis 

FIG. 14.15 

We next perform a Cauer ladder expansion for Z^s). 

2s* + 2* + l)^ + 2s» + 2s + l(s 


s + 1)2** +2s + l{2s 
2s* + Zs 

1)5 + 1(5 

The low-pass filter is thus synthesized in the structure shown in Fig. 14.15. 

An equivalent realization for the double-terminated filter is obtained if 
we use the equation 

S u (5) = 

Then, assuming G = 1 mho 

Y 1 (s) = 

YM - G 
Y t (s) + G 

1 + S n (s) 
1 - S u (s) 



The canonical realization for Yx(s) is shown in Fig. 14.16. Tables 14.1, 
14.2, and 14.3 list element values (up to n = 7) for double-terminated 
Butterworth, Chebyshev (1-db ripple) and Bessel filters, respectively. 
These apply to the canonical realization for Y t (s) given in Fig. 14.16. If 
a ZjCs) realization in Fig. 14.14 is desired, we simply replace all shunt 
capacitors by series inductors and vice versa. 

FIG. 14.16. Canonical form for filters in Tables 14.1, 14.2, and 14.3. 

The scattering matrix 435 

TABLE 14.1 
Normalized Element Values for a Double-Terminated 
Butterworth Filter (Equal Terminations) 
n Ci Lf C3 L t Cj Lg C? 




































TABLE 14.2 

Normalized Element Values for a Double-Terminated 

Cheb/shev Filter with I -decibel Ripple (Equal Terminations) 




C 8 


c 5 


c 7 





















TABLE 14.3 
Normalized Element Values for a Double-Terminated 
Bessel Filter (Equal Terminations) 
Cj L a C 3 L t C 3 L e Cj 




































Note that the even orders for the double-terminated Chebyshev niters 
are not given. This is because the even-ordered Chebyshev filters do not 
meet realizability conditions for minimum insertion loss at s = 0. 5 We 
have only given tables for equal source and load terminations. For other 
possible realizations, the reader should consult L. Weinberg's excellent 
book. 8 

£ L. Weinberg, Network Analysis and Synthesis, McGraw-Hill Book Company, 
New York, 1962, p. 589. 
'Ibid., Chapter 13. 

436 Network anal/sis and synthesis 


14.1 Determine the reflection coefficient S for the one-port networks shown 
in the figure. 

°-VW — r— vW — I 




14.2 For the one-port_network in Fig. 14.3, let iJ, . R B . If the incident 
parameter is a = VJl^R^ find the reflected parameter A. 

14.3 For the network in Prob. 14.1, determine \S(jio)\. Show that the 
scattering elements S for the networks in Prob. 14.1 are bounded real functions. 

14.4 For each of the networks shown, find the scattering matrix for ik, = 








— -*- 





V L 



^ 2 W* 

PROB. 14.4 




r 1. 




v 2 



— o 


The scattering matrix 437 

14J5 Find the insertion voltage ratio and insertion power ratios for each of 
the networks shown. These networks are to be inserted between a source 
impedance R„ = 2 ft and a load impedance R L = 1 ft. From the insertion 
power ratio, find |S»i(j«>)|*. 

2h_ 2_h 

i o-'TRRTH-^WnP- 02 


io — . — nptpP 1 - 







PROB. 14.5 

14.6 Synthesize low-pass niters for the specifications 
(a) |S a Q'«>)l* = 


l-W)! 2 

1 +0)* 


1 +afl 

14.7 Synthesize an equal-ripple low-pass filter such that 20 log |5 n (/a>)| has 
at most i-db ripple in the pass band and an asymptotic falloff of 12 db/octave in 
the stop band. 

chapter 15 

Computer techniques in 
circuit analysis 


The advent of the high-speed computer has made routine many of the 
formerly tedious and difficult computational aspects of circuit theory. 
Digital computers have become widely used in circuit analysis, time and 
frequency-domain analysis, circuit (filter) design, and optimization or 
iterative design. We will discuss these aspects in general in this section. 
In succeeding sections, we will discuss some specific circuit-analysis 
computer programs. 

Circuit analysis 

The primary objective of a linear circuit-analysis program is to obtain 
responses to prescribed excitation signals. These programs are based on 
many different methods: nodal analysis, mesh analysis, topological 
formulas, and state variables. Most of them can handle active elements 
such as transistors and diodes by means of equivalent circuit models. 

The state-variable programs based upon Bashkow's A matrix formula- 
tion 1 perform their calculations directly in the time domain via numerical 
integration and matrix inversion. The outputs of these programs provide 
impulse and step response in tabular form. If the excitation signal were 
given in data form, the state-variable programs would calculate the 
response directly in the time domain. 

1 T. R. Bashkow, "The A Matrix— New Network Description," IRE Trans, on Circuit 
Theory, CT-4, No. 3, September 1957, 117-119. 


Computer techniques in circuit anal/sis 439 

The majority of circuit-analysis programs, however, perform their 
calculations in the frequency domain. The program user is only required 
to specify the topology of the network, the element values, and what 
transfer functions he wishes to obtain. The computer does the rest. It 
calculates the specified transfer functions in polynomial form, calculates 
the poles and zeros of these functions, and can also provide transient 
response and steady-state response, if desired. With versatile input- 
output equipment, the output can also provide a schematic of the original 
network, as well as plots of time- and frequency-response characteristics. 

Time- and frequency-domain analysis 

The time- and frequency-domain analysis programs can be used in 
conjunction with the circuit-analysis programs or independently. The 
time-domain programs depend upon solving the convolution integral 

FIG. 15.1. Frequency response of fifth-order Butterworth filter. 
FIG. 15.2. Phase response of fifth-order Butterworth filter. 

440 Network analysis and synthesis 

FIG. 15.3. Impulse response of fifth-order Butterworth filter evaluation by Laplace 

numerically. This approach obviates the necessity of finding roots of 
high-order polynomials. It has the advantage that the excitation signal 
need not be specified analytically, but merely in numerical form. 

The frequency-domain programs usually consist of finding transient 
and steady-state responses, given the transfer function in factored or 
unfactored form. The program user must specify the numerator and 
denominator polynomials of the transfer functions, the types of transient 
response he wishes (i.e., impulse or step response), and the types of 
steady-state responses he wishes (amplitude, amplitude in decibels, phase, 
delay, etc.). In addition, he must specify the frequency and time data 
points at which the calculations are to be performed. This may be done 
in two ways. If he requires evenly spaced data, he need only specify the 
minimum point, the increment, and the number of points. If he wishes 
to obtain data at certain points, he must supply the list of data points at 
the input. 

Examples of outputs of a steady-state and transient analysis computer 
program are shown in Figs. 15.1, 15.2, and 15.3. In Fig. 15.1, the magni- 
tude of a fifth-order Butterworth filter is plotted via a microfilm plotting 
subroutine. Figure 15.2 shows the phase of the filter, while Fig. 15.3 
shows its impulse response. 

Computer techniques in circuit analysis 441 

In Section 15.2 we will examine further details of a typical steady-state 
analysis program. 

Circuit (filter) design 

The filter design programs are probably the most convincing argument 
for the use of computers in circuit design. Designing insertion loss filters" 
to meet certain amplitude requirements requires considerable numerical 
calculation even in the simplest cases. The use of digital computers in 
insertion loss filter design is clearly a logical alternative. The amount of 
programming time for a general filter synthesis program is considerable. 
However, the ends certainly justify the means when large numbers of 
filters must be designed to meet different specifications. 

An outstanding example of a digital computer program for filter design 
is the one written by Dr. George Szentirmai and his associates.* The 
program is complete in that it handles the approximation as well as the 
synthesis problem. It is capable of dealing with low-, high-, and band-pass 
filters with prescribed zeros of transmission (also called attenuation poles 
or loss peaks). There are provisions for either equal-ripple or maximally 
flat-type pass-band behavior, for arbitrary ratios of load to source im- 
pedances, and for predistortion and incidental dissipation. 

In addition, Dr. Szentirmai has a modified program that synthesizes 
low- and band-pass filters with maximally flat or equal-ripple-type delay 
in their pass band, and monotonic or equal-ripple-type loss in the stop 

In the specifications, the designer could specify both the zeros of 
transmission (loss peaks) and the network configuration desired. 4 If 
his specifications include neither, the computer is free to pick both con- 
figuration and zeros of transmission. The computer's choice is one in 
which the inductance values are kept at a minimum. The program was 
written so that the same network could be synthesized from both ends. 

Finally, the computer prints out the network configuration, its dual, 
the normalized element values, and the denormalized ones. It also 
provides information such as amplitude and phase response, as well as 
plots of these responses obtained from a microfilm printer. 

Figures 15.4, 15.5, 15.6, 15.7, 15.8, and 15.9 show the results of a 
band-pass filter synthesis using Dr. Szentirmai's program. Figure 15.4 

» R. Saal and E. Ulbrkh, "On the Design of Filters by Synthesis," Trans. IRE on 
Circuit Theory, CT-5, December 1958, 287-327. 

• G. Szentirmai, "Theoretical Basis of a Digital Computer Program Package for 
Filter Synthesis," Proceedings of the First Allerton Conference on Circuit and System 
Theory, November 1963, University of Illinois. 

4 Saal and Ulbrich, op. cit. 

442 Network analysis and synthesis 










6.2000000E+03 4.621 2071E-01 





FIG. 15.4 




0.09000 DB. 
1.0000000E+04 CPS. 
1.3416408E*04 CPS. 

i.80oooooe*04 cps. 


6.0000000E«02 OHMS 
0. OHMS 



gives the specification of the problem. The pass-band magnitude is to be 
equal ripple with ripple magnitude of 0.05 db, and the degree of the filter 
is to be 13. As we indicated in Chapter 14, odd-degree, equal-ripple 
filters are nonrealizable. In this example the designer utilized an ingenious 
device — an extra pole — to accomplish the synthesis. The program logic 
then provided two extra zeros: one to cancel the extra pole and the other 
to provide the odd degree. The extra pole is called a special pole in Fig. 
15.4, and is located at s = —1.0. 

Further specifications call for the lower band-edge frequency to be 10* 
cycles; the upper, 1.8 X 10* cycles; and the midband frequency to be 
Vl.8 X 10* = 1.3416408 X 10* cycles. In the stop band, there are to be 
zeros of transmission at/= (three), /= oo (two), and four finite zeros 
of transmission: one below the pass band and three above the pass band. 
The positions of these finite zeros of transmission are chosen by the 
designer as indicated by the notation "arbitrary" stop band requested. 


Computer techniques in circuit analysis 443 








■ 0.0500 DB 

- i.ooooooos»o* cps 

• i.aooooooE»04 cps 

• 1.3416408E*04 CPS 


•03 901 *01 100 002 *00 

200 100 300 















9. 089744 OE-08 











L C 













2. 310000 0E»04 









3.2382 800E-08 








FIG. 15.5 


The terminations are: input = 60 Q, output = 0Q which means the 
filter terminates in a short circuit. In this example, the filter configuration 
is chosen by the computer, and is a minimum inductance configuration. 8 
In Fig. 15.4 there are, in addition, listings of the finite zeros of trans- 
mission (loss peaks), which the designer specified. 

Figure 15.5 is a printout of the configuration of the filter as shown by 
the dotted lines flanked by the associated element values, both normalized 
(left column) and unnormalized (right column). Since there are four 
finite zeros of transmission, there must be associated four L-C tank circuits. 

S W. Saraga, "Minimum Inductance or Capacitance Filters," Wireless Engineer, 
30, July 1953, 163-175. 

444 Network anal/sis and synthesis 





■NAG 2 IN 




TIO IN 08 


























































65. 165 



























































































































































FIG. 15.6 




















0.000 0.400 0.800 1.200 
Add 1.00000£ + 03 to abscissa 

1.600 2.000 2.400 2.800 
Frequency, cycles x 10* 
FIG. 15.7 

3.200 3.600 4.000 

Computer techniques in circuit analysis 445 




7 4 


J 3 


►; 2 






0.000 0.400 0.800 1.200 1.600 2.000 2.400 2.800 3.200 3.600 4.000 
Add 1.00000S + 03 to abscissa Fl6queflcy ^ x „, 

FIG. 15.8 




r s 0.200 

J 0.000 

(*" -0.200 



| -0.400 















0.000 0.400 0.800 1.200 1.600 2.000 2.400 2.800 3.200 3.600 4.000 
Add 1.00000E + 03 to abscissa }nmm ^ x 10 4 

FIG. 15.9 

446 Network anal/sis and synthesis 

The peak frequencies are listed under the column entitled "termination," 
and the associated L-C tank circuits are on the same line two columns to 
the left. 

Figure 15.6 is a listing of a small portion of the frequency response, 
which was calculated after the filter had been synthesized. Figure 15.7 
is a plot of voltage ratio in decibels versus frequency. Note that 1000 
cycles must be added to every frequency value in the abscissa. This 
merely reflects an idiosyncrasy of the plot routine. 

Figures 15.8 and 15.9 show the real and imaginary parts of the input 
admittance of the filter. Note particularly the shapes of these character- 
istics. If a number of these band-pass filters were connected in parallel 
and if the pass bands of the filters were adjacent but nonoverlapping, 
then the input conductance of the filter system would be essentially 
constant over the entire pass band, while the input susceptance would 
cancel out, as shown in Figs. 15.10a and b. This then means that the 
input admittance of the filter system would be real and could then be 
driven by any arbitrary source impedance without fear of reflections. 

Filters obtained using computer programs of the type we have just 
described are rapidly supplanting conventional hand- and handbook- 
designed filters. The filter synthesis programs provide filters that are 
orders of magnitude more sophisticated than niters designed by the 



FIG. 15.10. Effects of paralleling several band-pass filters with adjacent pass bands. 

Computer techniques in circuit anal/sis 447 

conventional manner. Moreover, the designs are completed in minutes 
rather than days and at a typical cost of twenty dollars rather than two 
thousand (not including initial programming costs, of course). 


Network design cannot always be accomplished by way of analytical 
means. Quite often networks are designed through a trial-and-error 
process. The network designer begins with a set of specifications. He 
then selects a network configuration, and makes an initial guess about 
the element values. Next he calculates or measures the desired responses 
and compares them with the specifications. If the measured responses 
differ by a wide margin from the specified responses, the designer changes 
the values of the elements and compares again. He does this a number 
of times until (hopefully) the measured responses agree with the specified 
responses to within a preset tolerance. 

This process of cut and try can be made to converge, sometimes quite 
rapidly, if one uses a method of steepest descent.* To get a rough idea 
of the steepest descent method, suppose there are n parameters in the 
network. Let us regard each parameter x { as a dimension in an n- 
dimensional euclidean space. We define a function f(x u x a , x s , . . . ,x n ) 
that assigns a functional value to each point of the euclidean space. We 
then ask, at point p t , what direction of motion decreases the value of 
f(x lt x it *»,...,*„) most rapidly? The function f(x t ) may be defined as 
a least squared error, or as an absolute error between calculated response 
and specified response. The direction of steepest descent is the direction 
denned by the gradient 

mdf=T-*i + T-*» + --- + ^*n (15.D 

OXi OX 2 OX n 

Therefore, incremental value of change for each parameter is 

6x t =-C— (15.2) 


where C is a constant. After all parameters have been changed by the 

incremental value in Eq. 15.2, a new gradient and new incremental values 

are obtained. The process continues until the optimum is obtained. 

An excellent paper by C. L. Semmelman 7 describes a steepest descent 

* Charles B. Tompkins, "Methods of Steep Descent," in Modern Mathematics for the 
Engineer (Edwin F. Beckenbach, ed.), Chapter 18, McGraw-Hill Book Company, New 
York, 1956. 

7 C. L. Semmelman, "Experience with a Steepest Descent Computer Program for 
Designing Delay Networks," IRE International Convention Hec, Part 2, 1962, 206-210. 

448 Network analysis and synthesis 

Fortran program used for designing delay networks. The specifications 
are the delay values R t given at the frequency data points/,. The program 
successively changes the parameters x i so that the squared error 

<*,)=i[K,-n*<,/,)] a (i5.3) 


is minimized. In Eq. 15.3, T{x it f t ) represents the delay, at frequency f f , 
of the network with parameters x t . 

In order to use the program, the network designer must first select the 
initial values for the parameters x ( . He must also provide the specified 
delays R, and the frequency data points/^. The program provides for 128 
match points and 64 parameter values. It is capable of meeting require- 
ments simultaneously in the time and frequency domains. The designer 
is not restricted to equal-ripple approximations or infinite Q requirements. 
He is free to impose requirements such as nonuniform dissipation and 
range of available element values on the design. For related methods of 
optimization, the reader should refer to a tutorial paper by M. R. Aaron. 8 

Machine-aided design 

The concept of real-time interaction between man and computer holds 
much promise in the field of network design. Multiple-access computing 
systems, such as the project MAC of Massachusetts Institute of Tech- 
nology, make cut and try design procedures practicable. The initials 
MAC could describe either the term multiple-access computer or the term 
machine-aided cognition. In an article describing the MAC computer 
system, 9 Professor R. M. Fano states "The notion of machine-aided cog- 
nition implies an intimate collaboration between a human user and a 
computer in a real-time dialogue on the solution of a problem, in which 
the two parties contribute their best capabilities." 

A simple multiple-access computer system is shown in Fig. 15.11. 
There are n data links and n terminals connected to the central processor. 
Located at each terminal is input-output equipment, such as teletype- 
writers, teletypes, and oscillographic displays. The sequence in which 
the central processor accepts programs from the terminals is controlled by 
a built-in queueing logic. 

The main reason a multiple-access computer is effective is that the 
central processor computes thousands of times faster than the user's 
reaction time. When a user feeds in a program, it seems only a "moment" 

8 M. R. Aaron, "The Use of Least Squares in System Design," IRE Trans, on Circuit 
Theory, CT-3, No. 4, December 1956, 224-231. 

•R. M. Fano, "The MAC System: The Computer Utility Approach," IEEE 
Spectrum, 2, No. 1, January 1965, 56-64. 

Computer techniques in circuit analysis 449 









link . 
















FIG. 15.1 1. Block diagram showing typical multiple-access computing system. 

before he gets the results through the teletypewriter or oscilloscope 
display. He examines the results, changes some parameters, and feeds the 
program into the central processor again. The queueing time and proc- 
essing time on the multiple-access computer may amount to only a 
minute or two, which, for the user, is probably not significantly long. 
Dr. H. C. So has written a paper on a hybrid description of a linear 
ii-port network 10 in which he has shown that the hybrid matrix is ideally 
suited for such problems such as multielement variation studies and 
iterative design. Suppose there are n variable elements in the network. 
These can be "extracted" and the rest — the bulk of the network — can be 
described by the n-port hybrid matrix. 

10 H. C. So, "On the Hybrid Description of a Linear i»-Port Resulting from the 
Extraction of Arbitrarily Specified Elements," Trans. IRE on Circuit Theory, CT-12, 
No. 3, September 1965, pp. 381-387. 

450 Network analysis and synthesis 



i ■ 








— *j 




' ' 





FIG. 15.12. Iterative design of network using machine-aided cognition. 

Dr. So has written a computer program to formulate the hybrid matrix 
for the H-port network automatically. 11 The inputs to this program are 
(1) the node connections specifying the network topology; (2) the im- 
pedance functions of the elements; and (3) the specifications of the 
special elements to be extracted. This program was written with man- 
machine interaction in mind. The process is described in Fig. 15.12. 
First the computer reads in the n-port program with initial specifications. 
It performs the calculations and feeds information, such as transient or 
steady-state responses, back to the engineer via a visual display console. 
The engineer assesses the data and then changes the parameter values of 
the extracted elements, and, in some instances, the frequency range over 
which the calculations are made. The process is repeated a number of 
times until the engineer has obtained the desired results. Such a program 
could also be controlled by a steepest descent steering program if the 
engineer wished to obtain his results with less "eyeballing." 

Now let us examine some details of specific network analysis programs. 



Our purpose is to compute the amplitude and phase of a rational 

" (s) - ?Hr? (154) 

1 H. C. So, unpublished memorandum. 

Computer techniques in circuit anal/sis 451 

over a set of frequencies (o^ In Eq. 15.4, C„ and C d are real constants, 
and E(s) and F(s) are polynomials in s = a +ja>. The program described 
here computes the amplitudes M(co,) and phase #a> t ) over a set of 
frequencies co k , where 

«W = 


\CnE(jlO h )\ 

\C a FOco k )\ 

<f>((o k ) = arctan 


Fj(ja> k ) 




where in Eq. 15.6 the r and i subscripts 
indicate real and imaginary parts, 
respectively. Note that in Eq. 15.6, 
<Ko>k) * s limited to the range — it <> 
<f> <,it radians. In order to circumvent 
this restriction on the phase, we use the 
following method to compute amplitude 
and phase. 


FIG. 15.13. Calculation of magni- 
tude and phase at a> ( due to zero z. 

Method used 

Let H(s) be factored into poles and zeros such that 

H(s) = 

C n (s - z t )(s - z t ) 
C d (s ~ Pi)(s - P») 



Consider the amplitude and phase due to any pole or zero, for example, 
« = — * +jP, shown in Fig. 15.13. The magnitude is 

M s {a>,) = [«« + (<»< - ft*}* 
and the phase is 4>.{a>d = arctan | ft, '~ " l 

The amplitude and phase of the overall function is 


M(wJ = 




<K°>i) = 2 tJfOi) - j £,,(«>,) 


452 Network analysis and synthesis 

where the subscripts p and z indicate the contribution due to a pole and 
zero, respectively. 


1. The numerator E(s) and denominator F(s) can be read in either as 
polynomials (high degree to low) or in terms of their roots. If numerator 
and/or denominator are read in as polynomials, their roots will be printed. 
If they are given in terms of their roots, the program could generate 
polynomials from these roots. 

2. The data points can be read in if unequally spaced; if equally spaced, 
only the minimum point and the spacing need be read in. 

Read Nl, ND, NN 
KK, LI, L2, L3 

KK = 

Read W (I) 

L1 = 

Read A (I), B (I) 

L2 = 

Read C (I), D (I) 

Read CN, CD 

Test KK 

Test LI 




Read WMIN, DW 

Ll = l 

Read E(l) 



Read F(l) 

To main program 
FIG. 15.14. Flow chart of input instructions for magnitude and phase program. 

Computer techniques in circuit analysis 453 

A flow chart listing the input instructions is given in Fig. 15.14. Def- 
initions of the symbols in the flow chart follow. 

Nl = number of frequency points 

ND = degree of denominator 

NN = degree of numerator 

KK = indicates equally spaced data points. We then read in only 

Wjqn, Aa>, and number of points. 
= 1 indicates unequally spaced data points. We then read in all 

LI =0 Read in zeros (if complex conjugate, both zeros must be 

read in). 
= 1 Read in numerator polynomial, high degree to low. 
L2 = Read in poles (if complex conjugate, both poles must be 

read in). 
= 1 Read in denominator polynomial, high degree to low. 
CN = numerator multiplier 
CD = denominator multiplier 


The main output of the program is the frequency response consisting 
of the following columns: frequency in radians W(I); amplitude, 
M(W(I)); magnitude in decibels, 20 log 10 Af; and phase in degrees, 
^(W(I)). A typical printout of an amplitude and phase program is given 
in Fig. 1S.6, which shows the magnitude and phase of the band-pass filter 
designed using Szentirmai's program. 


In this section we will discuss the methods and organization for a 
Fortran computer program for the analysis of linear ladder networks. 
The analysis proceeds by computing the voltage and current at the input 
of successive L sections, beginning with the terminating section. 

The program calculates the branch impedances of the ladder by com- 
bining R, L, C series impedance arms according to the instructions of the 
individual programmer. 

The poles and zeros and the frequency responses of the input im- 
pedance and voltage transfer ratio at the input of each L section are 
found. In addition, the program provides for the analysis of short- and 
open-circuited networks as well as for those with normal terminations. 
Separate problems may be run consecutively if so desired. 

454 Network analysis and synthesis 

v m 



Z n 



l Vl 

Zi ■ 


1 ° 






Y m 





o — 



FIG. 15.15 

The network is initially decomposed into separate L sections as in 
Fig. 15.15. These L sections are then added successively to the terminating 
L section to form the complete network. With the addition of an L 
section, the voltage and current at the input of the resulting network are 
computed by the equations 

^ n = ^n^n + V n -1 


which were originally discussed in Chapter 9 of this book. Thus the 
program proceeds toward the front of the ladder requiring only the branch 
immittances of the present L section and the voltage and current of the 
previous L section to make its calculations. 

Suitable assumptions for the initial voltage and current V and / allow 
for analysis of short- and open-circuited networks as well as for those 
with normal terminations. These initial values of voltage and current are 
determined by control instructions. 

For the calculation of the voltage and current at an L section, the 
branch impedances of the section are required. To simplify the prepara- 
tion of the input data specifying these impedances, the following procedure 
is used. Each branch impedance is formed by the addition in series or 
parallel of basic R, L, C impedance arms. A basic R, L, C impedance 
arm is a series combination of a resistance, an inductance, and a capac- 
itor — any or all of which may be absent. The impedance arms are speci- 
fied by the element values R, L, C and an instruction indicating how the 
arm is to be combined. 

The elements of the impedance arms are frequency and impedance 
normalized to aid computation, and then the arms are combined as 
specified to form the appropriate branch impedance. 

The roots of the voltage and .current polynomials for each L section 
are computed by a root-finding routine. 

Computer techniques in circuit analysis 455 

For the normal load and open circuit termination, the frequency 
responses of the input impedance, and voltage-transfer ratio are computed ; 
for short-circuit termination, the input impedance and current gain are 
computed. Provision is made for either a linear or logarithmic increment 
in frequency. The frequency boundaries and increments are read by the 
program along with the input data. 

According to instructions given by the programmer, the various cal- 
culations are printed for each L section or only for the network as a whole. 

Program operation 

The ladder network-analysis program requires two sets of input data. 
The first set consists of the control instructions. These tell the computer 
which of the various options, such as information concerning polynomial 
roots or frequency responses, are to be exercised. 

In addition, the control instructions provide the computer with necessary 
parameters such as normalizing factors, and the number of L sections in 
the ladder. The second set of data determines the branch impedances of 
the network by specifying the R, L, C impedance arms and their combining 

Impedance data 

The data specifying the branch impedances consist of an ordered series 
of instruction cards, each determining an impedance arm and/or a com- 
bining instruction. The first arm of any branch impedance requires no 
combining instruction. The branch impedance is set equal to the im- 
pedance of this arm. Subsequent impedance arms are added in parallel 
or series with the existing branch impedance, according to the combining 
instructions of the arms. A blank card tells the computer that a complete 
branch impedance has been formed and that the next arm begins a new 
branch impedance. If blank cards are not properly inserted, the computer 
will calculate a single giant branch impedance. 

The order of the data determining the branch impedances isZ x , ljY lt 

^a» 1 / Jg, • • • • 

The values of R, L, C and the combining instruction XIN are written 
on one card in the order XIN, R, L, C. 

The values of XIN and the associated operation are shown in the 
table at the bottom of page 456. 

Although the instructions XIN = 1.0, 2.0 will suffice for the con- 
struction of many branch impedances, they are inadequate for other 

For example, suppose the series combination of two tank circuits (Fig. 
15.16) is desired as a branch impedance, this impedance may only be 

456 Network anal/sis and synthesis 


Tank circuit A 


.05 h 
■-^OOOy- 1 

Tank circuit B 



5.00000£ - 11 


2.00000E - 03 








FIG. 15.16. Construction of branch impedances using instructions for XIN. 

constructed with use of three more instructions (XIN = 3.0, 4.0, and 5.0) 
and an additional set of computer storage locations. 

Tank circuit A is formed in the usual manner, but must then be tem- 
porarily stored during the calculation of tank circuit B. The two tank 
circuits are then added to form the final branch impedance. Examples of 
impedances that cannot be formed are shown in Fig. 15.17. Observe that 
a series branch impedance of zero ohms often simplifies the formation of 
branch impedances. Two blank cards will indicate a short-circuited 



Blank card 




A complete branch impedance has been calculated, and 
the next arm begins a new branch impedance. 

This arm is added in series with the existing branch 

This arm is paralleled with the existing branch impedance. 

This arm begins a new internal branch impedance. 
Subsequent XIN = 1.0 and 2.0 refer to this new in- 
ternal branch. 

Add the two internal branches in series to form the final 
branch impedance. 

Parallel the internal branches. 

Computer techniques in circuit analysis 457 





FIG. 15.17. Branch impedances that cannot be formed by program. 


The coefficients of the voltage and current polynomials are printed in 
frequency normalized form. To calculate the denormalized coefficients, 
divide by the normalizing frequency raised to the power of the corre- 
sponding exponent. 

The poles and zeros of the voltage and current polynomials are in 
frequency normalized form to facilitate zero-pole plots. 

Frequency responses appear in denormalized form. The frequency 
variable is in radians per second. The impedance level is in decibels, 
the phase in degrees. The gain and phase of the transfer ratios are 
expressed similarly. 


Two double precision Fortran programs have been written that help 
in the synthesis of double-terminated filters using the Darlington pro- 
cedure. The mathematics of the programs is described here. 

Given a forward transmission coefficient 

S 8 i(s) - 



458 Network analysis and synthesis 

we generate an input reflection coefficient S u (s) from the equation 

Su(s) Sn(-s) = 1 - S»(s) S gl (-5) 

D(s) D(-s) - K* N(s) N(-s) 
D(s) D(-s) 


The denominator of S u (j) is D(s) because all the poles of S u (s) must be 
in the left-half plane. The zeros of S n (s) are not so restricted. If S n (s) 
Su(—s) has zeros &t—a±jb and a ± jb, the zeros of S n (s) can be chosen 
as either — a ±jb or a ±jb. The only difference is that certain choices 
lead to niters with unity-coupled coils and others without. 12 

Once the zeros of S n (s) are chosen, the input impedance of the filter 
is then 

1 ~ Sii(*) 

Zi(s) = 

1 + S u (s) 




Read CD 



A (I), B(I),N(I) 

Read LD 

H(I),G(I), M(l) 

Numerator multiplier 

Denominator multiplier 

Number of zeros 
(complex conjugate zeros 
count as one zero) 

Read real and imaginary 
parts of zeros and 
multiplicity, N (I) 

Number of poles 
(conjugate poles 
count as one pole) 

Read real and imaginary 
parts of poles and 
multiplicity, P(l) 

To program 
FIG. 15.18. Input to Sn(5) 5 u (—j) program. 

11 T. Fujisawa, "Realizability Theorem for Mid-Series or Mid-Shunt Low Pass 
Ladders without Mutual Induction," Trans. IRE on Circuit Theory, CT-2, December 
1955, 320-325. 

Computer techniques in circuit anal/sis 459 

Program descriptions 

We next proceed with brief descriptions of the two programs. The 
first program finds the roots of S u (s) S^—s) given the roots of N(s), 
N(—s), D(s), and D(— s). A flow chart for the input of the program is 
given in Fig. 15.18. Note that we must read in both the zeros of N(s) and 
N(—s), although only half of a complex conjugate pair need be read in. 
The same applies for the roots of Z>($) and /)(—$). The constant CN is 
K* in Eq. 15.14. 

The second program performs the computations in Eq. 15.15 for 
different combinations of zeros of S xl (s), given the fact that the zeros of 
Su(s) need not be in the left-hand plane. The previous program finds all 
the zeros of S n (s) S u (— s). We must choose only half the number of 
zero pairs for 5 u (s). Certain combinations of zeros of S n (s) S u (—s) 
lead to filters with coupled coils; others do not. If one wishes to try a 
number of combinations of zeros for 5 u (f ), the program has the facility 
to enable him to do so. He need only supply those zeros of S n (s) S u (— s) 
in the right-hand plane, C(I) +y'B(I), and a list of constants S(I), which 
are either 1.0 or —1.0 so that the zeros of S xl (s) may be represented as 
S(I) • C(I) +y'B(I) — A(I) +y'B(I). The routine forms numerator and 
denominator polynomials of S u (s) 


S u (s) = 




Number of pole pairs of Su 
(real poles count as one pair) 


Real and imaginary parts of poles 
of Sn and multiplicity, M (I) 



C(l), G(l), N(l) 



Read SO) 

Number of right-hand plane 
zeros of Sn (a) Sn(-s) 

Real and imaginary parts of right-hand 
plane zeros and multiplicity, N 0) 

Multiplicative constants (i 1.0) (each 
combination of zeros has LN cards) 

To program 
FIG. 15.19. Input to Z ta (j) program. 

460 Network analysis and synthesis 

and then computes the numerator and denominator of Z^s) as 

ZM= i^ = m^m (15 . 17) 

The input instructions are summarized on the flow chart in Fig. 15.19. 


15.1 Organize a flow-chart for computing magnitude and phase given only 
the unfactored numerator and denominator polynomials. Do not use a root- 
finding subroutine. 

15.2 Write a program to calculate the delay of 

_,. 3(5 + 1) 

at the points o> = 0, 1, 2, . . . , 10. 

15.3 Repeat Prob. 15.2 for a general transfer function given in terms of its 
unfactored numerator and denominator polynomials. The frequency points co t 
are to be read in by the computer. 

15.4 Suppose you have calculated the phase of a transfer function at points 
m = 0, 1, 2, .... 50. Devise an algorithm to test for phase linearity. The 
deviation from phase linearity is to be called phase runoff. 

15.5 Write a program to analyze a two-mesh network made up of only 
R, L, and C elements. 

15.6 Repeat Prob. 15.5 for nodal analysis. 

15.7 Write a program to calculate the step and impulse response of a linear 
system whose system function contains only simple, real poles and zeros. 

15.8 Repeat Prob. 15.7 if simple complex poles and zeros are allowed. 

15.9 Write a program to calculate the residues of a transfer function with 
multiple as well as simple poles. The poles and zeros are to be real. 

appendix A 

Introduction to matrix algebra 


Matrix notation is merely a shorthand method of algebraic symbolism 
that enables one to carry out the algebraic operations more quickly. 
The theory of matrices originated primarily from the need (1) to solve 
simultaneous linear equations and (2) to have a compact notation for 
linear transformations from one set of variables to another. 

As an example of (2), consider the set of simultaneous linear equations 

«u*i + «i**a + • • • + a ln x n = y x 
^l^X + a M" ; 3 + • • • + a* " — ** 

»*» = y* 


ami^i + a mi x a + • • • + a mn x n = y m 

These may express a general linear transformation from the x t to the y t . 
In general, mj&n. An example where the numbers of variables in the 
two sets are unequal is that of representing a three-dimensional object in 
two dimensions (in a perspective drawing). Here, m = 2 and n = 3. 


A matrix is an ordered rectangular array of numbers, generally the 
coefficients of a linear transformation. The matrix is denned by giving 
all its elements, and the location of each. 

The matrix of the equations in Eq. A.1 written as 

Oil «12 

fl gl a tt 


is of order m x n. (The first number here is the number of rows; the 


462 Network analysis and synthesis 

second is the number of columns.) The matrix may be denoted by a 
single capital letter A or by [a w ]. 

A matrix is a single complete entity, like a position in chess. Two 
matrices A and B are equal only if all corresponding elements are the 
same: a tj = b tj for all i andy. A matrix may consist of a single row or 
single column. The complete matrix notation applied to Eq. A.1 is 

<hi o lt 
On. o n 


o in 





x a 



i Xn . 

. Vm . 


which says, if put crudely, that A operates on x t to yield y,. This 
emphasizes the similarity between a matrix and a transformation. The 
x- and y-matrices are column matrices. 

Row and column matrices are called vectors (specifically, row vectors 
and column vectors) and their similarity to the more usual type of vector 
is discussed later. Here, vectors will be written as small letters, such as 
x or y. Note that the elements of vectors need only one subscript, while 
elements of matrices need two. 


Square matrix 

A square matrix has the same number of rows as columns (i.e., a 
matrix of order rat). The Y matrix is an example of a square matrix 


Diagonal matrix 

A diagonal matrix is a square matrix whose elements off the main 
diagonal are zero (i.e., one in which <x w = for 1 ?*/)• The following 
matrix is diagonal. 

'1 01 


_0 3 
Unit matrix 

A unit matrix is a diagonal matrix for which a u 
denoted as U. For example, U is 

U = 







1 for 1 =j, and is 


Appendix A 463 


Two matrices are equivalent if they have the same number of rows and 
columns and if the elements of corresponding orientation are equal 
Suppose y x = 2, y t = —3, and y a = —6. If we write this set of equations 
in matrix form, we have 



The transpose of a matrix A denoted as A r , is the matrix formed by 
interchanging the rows and columns of A. Thus, if we have 


" 2" 






A = 









Determinant of a matrix 

The determinant of a matrix is defined only for square matrices and is 

formed by taking the determinant of the elements of the matrix. For 

example, we have 

1 2 

f 12 l 
L-5 4j 

■5 4 

= 14 


Note that the determinant of a matrix has a particular value, whereas the 
matrix itself is merely an array of quantities. 


The cofactor A i} of a square matrix is the determinant formed by 
deleting the fth row andy'th column, and multiplying by (—l) i+i . For 
example, the cofactor A tl of the matrix 



A n = (-iy*+i x 6= -6 


Adjoint matrix 

The adjoint matrix of a square matrix A is formed by replacing each 
element of A by its cofactor and transposing. For example, for the 

464 Network analysis and synthesis 
matrix in Eq. A.10, we have 

adjA = 

l: h 
-l; i 


Singular and nonsingular matrices 

A singular matrix is a square matrix A for which det A = 0. A non- 
singular matrix is one for which det A ?* 0. 



Two matrices may be added if both matrices are of the same order. 
Each element of the first matrix is added to the element of the second 
matrix, whose row and column orientation is the same. An example of 
matrix addition is shown in Eq. A. 13. 


Thus if matrices A, B, C K are all of the same order, then 

A + B + • • • + K = [a u + b it + • • • + k tj ] 

The associative and commutative laws apply. 

Associative: A + (B + C) = (A + B) + C 
Commutative: A + B = B + A 

Multiplication by a scalar 

We define multiplication by a scalar as 

XA = A[a w ] = [AaJ 





Thus to multiply a matrix by a scalar, multiply each of its elements by 
the scalar. 

Example Al 

" 2 0" 

" 6 0" 

-1 1 


-3 3 

-3 2 

-9 6 


Appendix A 465 

Linear combination of matrices 

If the rules of addition and multiplication by a scalar are combined, 
for two matrices of the same order we have 

oA + /SB = [*a it + pb it ] (A.18) 

where « and /9 are scalars. 


In order for matrix multiplication AB to be possible, the number of 
columns of the first matrix A must equal the number of rows of the second 
matrix B. The product C will have the number of rows of the first and 
the number of columns of the second matrix. In other words, if A has 
m rows and n columns, and B has n rows and/> columns, then the product 
C will have m rows and p columns. The individual elements of C are 
given by 

Example A.2 


Example A3. The system of equations 

«u/i + Ws = V x 
ZtJi + z«/s = V t 
can be written in matrix notation as 






We see that systems of equations can be very conveniently written in 
matrix notation. 
Matrix multiplication is not generally commutative, that is, 

A mn B B „ * B^A,™ (A.23) 

Observe that the product BA is not defined unless p = m. Even a product 
of square matrices is generally not commutative, as may be seen in the 
following example. 

r i onr-i <n _ r-i oi 
L-i 2JL o 2j~L i f] 


466 Network anal/sis and synthesis 

If we interchange the order of multiplication, we obtain 







Because of the noncommutative nature of matrix multiplication, we 
must distinguish between premultiplication and postmultiplication. In 
BA, A is premultiplied by B; B is postmultiplied by A. For matrix 
multiplication, the associative and distributive laws apply. 


A(BC) = (AB)C = ABC 
A(B + C) = AB + AC 


Transpose of a product 

The transpose of a product AB is equal to the product, in reverse order, 
of the transpose of the individual matrices A and B, that is, 


(AB) - B r A T 
(ABC) r = C T (AB) T = C^B^A 2 " 


The product x T x, if x is a column vector, is a scalar number equal to 
the sum of the squares of the elements of x. Thus we have 

x T x = [ Xl x t x„] 


= *i 2 + *,* + ••• + x* 
The product xx r is a square matrix C such that C = C T . 

Common expressions in simple notation 

(a) The sum of products aA + ajb t + • • • + a n b„, which are the 
typical element of a product matrix, may be written as a^b or b r a where 
a and b are column vectors. 

(b) The sum of squares x^ + x t * + • • • + x n * is thus x T x (x, column 
vector). It is also xx r if x is a row vector. 

(c) The expression a u xf + a ti x t a + • • • + a nn x n * is x r Ax (x, column 
vector), where A is a diagonal matrix with elements a it . Similarly, the 
expression a a xjf x + a^e^y t H h a nn x„y n is x T Ay or y T Ax. 

(d) An expression such as 

2 Z a u x i x t> 

= a, 

Appendix A 467 
is a quadratic form. This is 

«ii*i* + 1a x% x x x t + + 2a ln x 1 x„ 

+ ««*!* + 2a u x t x a H h 2a, B a; 2 x B (A.30) 

+ • • • + <J»»*« 2 = x T Ax 


Division is not defined in matrix algebra. The analogous operation is 
that of obtaining the inverse of a square matrix. The inverse A -1 of a 
matrix A is defined by the relation 

A-*A = AA- 1 = U (A.31) 

To obtain A -1 , we first obtain the adjoint of A, adj A. Then we obtain 
the determinant of A. The inverse A -1 is equal to adj A divided by |A|, 
that is, 




A" 1 = — adj A 


Example A.4. Let A be 

given as 

Its determinant is 

|A| =3 

and the cofactors are 

An = 1 A u = 1 

The adjoint matrix is 

Afn = —1 A n = 2 

adJA = [-l 2j 

-C I 

so that A -1 is 

-»C 1 



46.8 Network anal/sis and synthesis 
As a check we see that 

p -r i 

" 2 r 

} K 

-l i 

' 2 V 

P _i l 

-1 1 

* § 


If the determinant of the matrix is zero, then the inverse is not defined. In 
other words, only nonsingular square matrices have inverses. 


Consider a set of linear algebraic equations, to be solved simultaneously. 

011*1 + <*iss*» H r- a ln x n = h x 

asi^i + a^x t + ■•• + a Sn x„ = h s 


«m*i + a n **i + ••• + a nn x n = h n 
The h t are constants. It is desired to solve for the x f . In matrix notation 
we have 

Premultiplying by A -1 gives 
In expanded form, this is 



Ax = h 
x = A x h 

■™ia -"m 










h t 



h a 

a S2 


and similarly for x t and * 3 . This is the familiar Cramer's rule for solving 
such equations. 

Example A.5. Solve for x, y, z. 

x — y +2 =2 

2x +y = 1 (A.43) 

-* + 3y + « = -1 

Appendix A 469 
In matrix form, these equations are written as Ax = h where 

■ 1 -1 r 

2 1 

-1 3 1 

Also, |A| =» 10. Thus, we have x = A _1 h 

*1 r i 4 -r 

y -& -2 2 2 
z 7-2 3 


■ r 


X = 



, h = 



r r 

■ r 


= i X o 






x = 0.7, y = -0.4, z = 0.9 


The following is a short list of books on matrices that the reader might wish to 
A. C. Aitken, Determinants and Matrices, 9th Ed., Interscience Publishers, New York, 

R. Bellman, Introduction to Matrix Analysis, McGraw-Hill Book Company, New 

York, 1960. 
R. L. Eisenman, Matrix Vector Analysis, McGraw-Hill Book Company, New York, 

D. K. Faddeev and V. N. Faddeeva, Computational Methods of Linear Algebra, W. H. 
Freeman and Company, San Francisco, 1963. 

F. R. Gantmacher, Applications of the Theory of Matrices, Interscience Publishers, 

New York, 1959. 
F. E. Hohn, Elementary Matrix Algebra, The Macmillan Company, New York, 1964. 
L. P. Huelsman, Circuits, Matrices, and Linear Vector Spaces, McGraw-Hill Book 

Company, New York, 1963. 
P. LeCorbeiller, Matrix Analysis of Electric Networks, John Wiley and Sons, New York, 

M. Marcus and H. Mine, Survey of Matrix Theory and Matrix Inequalities, Allyn and 

Bacon, Boston, 1964. 

E. D. Nering, Linear Algebra and Matrix Theory, John Wiley and Sons, New York, 

S. Perlis, Theory of Matrices, Addison-Westey, Reading, Massachusetts, 1952. 

L. A. Pipes, Matrix Methods for Engineering, Prentice-Hall, Englewood Cliffs, NJ., 

A. M. Trapper, Matrix Theory for Electrical Engineering Students, Harrop, London, 

A. von Weiss, Matrix Analysis for Electrical Engineers, D. Van Nostrand, Princeton, 

NJ., 1964. 

appendix B 

Generalized functions and 
the unit impulse 


The unit impulse, or delta function, is a mathematical anomaly. P. A. 
M. Dirac, the physicist, first used it in his writings on quantum mechanics. 1 
He denned the delta function d(x) by the equations 


d(x) dx = 1 


d(x) = for x jt 
Its most important property is 


V — i 

f(x)6(z)dx=f(0) (B.2) 

where f(x) is continuous at x — 0. Dirac called the delta function an 
improper function, because there existed no rigorous mathematical justi- 
fication for it at the time. In 1950 Laurent Schwartz* published a treatise 
entitled The Theory oj Distributions, which provided, among other things, 
a fully rigorous and satisfactory basis for the delta function. Distribution 
theory, however, proved too abstract for applied mathematics and 

1 P. A. M. Dirac, The Principles of Quantum Mechanics, Oxford University Press, 

1 L. Schwartz, Theorie des Distributions, Vols. I and II, Hermann et Cie, Paris, 1950 
and 1951. 


Appendix B 471 

physicists. It was not until 1953, when George Temple produced a more 
elementary (although no less rigorous) theory through the use of general- 
ized functions,* that this new branch of analysis received the attention it 
deserved. Our treatment of generalized functions will be limited to the 
definition of the generalized step function and its derivative, the unit 
impulse. The treatment of these functions follows closely the work of 
Temple 4 and Lighthill. 8 

To get an idea of what a generalized function is, it is convenient to use 
as an analogy the notion of an irrational number a beng a sequence {oc„} 
of rational numbers <x n such that 

a = lim a„ 


where the limit indicates that the points <x„ on the real line converge to 
the point representing a. All arithmetic operations performed on the 
irrational number a are actually performed on the sequence {a„} defining 
a.' We can also think of a generalized function as being a sequence of 
functions, which when multiplied by a test function and integrated over 
(—00, oo) yields a finite limit. Before we formally define a generalized 
function, it is important to consider the definition of (1) a testing function 
and (2) a regular sequence. 

DEFINITION B.I A function #f) of class C[#f) G C] is one that (1) 
is differentiable everywhere, any number of times and that (2) when it or 
any of its derivatives are multiplied by t raised to any power, the limit is 

lim [t m <f> M (t)]-*0 for all m&fe^O (B.3) 


Any testing function is a function of class C. 

Example B.l. The Gaussian function e-*' ln% is a function of class C. It is 
obvious that if a function is of class C then all of its derivatives belong to class C. 

*G. Temple, "Theories and Applications of Generalized Functions," J. London 
Math. Soc., 28, 1953, 134-148. 

4 G. Temple, "The Theory of Generalized Functions," Proc. Royal Society, A, 228, 
1955, 175-190. 

6 M. J. Lighthill, Fourier Analysis and Generalized Functions, Cambridge University 
Press, 1935. Lighthill dedicated his excellent book to "Paul Dirac, who saw it must be 
true, Laurent Schwartz, who proved it, and George Temple, who showed how simple 
it could be made." 

* This defines an irrational number according to the Cantor definition. For a more 
detailed account see any text on real variables such as E. W. Hobson, The Theory of 
Functions of a Real Variable, Vol. I, third edition, Chapter 1, Cambridge University 
Press, Cambridge, England, 1927. 

472 Network anal/sis and synthesis 

DEFINITION B.2 A sequence {/„(*)} of functions of class C is said 
to be regular if for any function <f>(t) belonging to C, the limit 

lim(/„,#=lim ( W f n (t)<Kt)dt (B.4) 

n-*ao n-*oo J— oo 

exists. Note that it is not necessary that the sequence converge pointwise. 
For example, the sequence {e~ nt \nlir)^} approaches infinity as n -»■ oo at 
the point t = 0. However, the limit lim (/"„, <f>) exists. 


DEFINITION B.3 Two regular sequences {/„} and {g„} are equiv- 
alent if for all 6 £ C 

lim(/ B ,#=lim(g B ,# (B.5) 

n-*oo n-*ao 

Example B.2. The regular sequences {e- n **(«/f) w } and {<r' ,/aB *(l/V2^i)} are 

DEFINITION B.4 A generalized Junction g is denned as a total, or 
complete, class of equivalent regular sequences. The term total implies 
here that there exists no other equivalent regular sequence not belonging 
to this class. Any member of the class, for example, {g n }, is sufficient to 
represent both g and the total class of equivalent regular sequences 
denning g. We denote this symbolically by the form g ~ {g n }. 

Example B.3. All of the equivalent, regular sequences 

[{*-'*/»•}, Or* 1 '" 4 }, {e-"'-} {«--«»/»•*}] 

represent the same generalized function g ~ {e~** lnt }. 

DEFINITION B.5 The inner product (g, <f>) of a generalized function, 
g and a function <f>(t) € C is defined as 

(g,#=lim f"g n (t)<Kt)dt (B.6) 

n-*oo J— co 

The inner product is often given the following symbolic representation. 

g(t)<Kt)dt (B.7) 

(g, +) = f ° 

Note that the integral here is used symbolically and does not imply 
actual integration. 

DEFINITION B.6 If g and h are two generalized functions rep- 
resented by the sequences g ~ {g n } and h ~ {h n }, the sum g + h is 
defined by the representation g + h ~ {g „ + A»}. 

Note that the set of sequences {g n + h n } represents a total class of 
equivalent regular sequences made up of the sum of sequences defining g 
and h; therefore g + h is a properly defined generalized function. 

Appendix B 473 

DEFINITION B.7 The product a.g of a generalized function; ~ {#„} 
and a constant a is defined by the representation ag <~ {«g n }. 

DEFINITION B.8 The derivative g' of a generalized function g~ 
{#„} is defined by the representation g' ~ {g' n }. 

Example B.4. For the generalized function g x ~ {*-* 1/b *} the derivative is 
represented by 

and (g\, *) = lim f " ( - ^ e"**'" 1 J *(/) «// (B.8) 

In Definitions B.6, B.7, and B.8 we have defined the operations of addition, 
multiplication by a scalar and differentiation. It must be pointed out that 
the operation of multiplication between two generalized functions is not 
defined in general. 

We next consider an important theorem, whose proof is given in 
Lighthill, 7 which will enable us to represent any ordinary function, such 
as a step function by a generalized function equivalent. 

Theorem B.l. Given any ordinary function /(f) satisfying the condition 



■ r dt < co (B.9) 

for some N ^ 0, there exists a generalized function 8 / ~ {/,(/)} such that 

for all ^ € C. In other words, an ordinary function satisfying Eq. B.9 is equiva- 
lent in terms of inner products to a generalized function. Symbolically, we 
write / =/. If, in addition, / is continuous in an interval, then lim f n =/ 
pointwise in that interval. ""*" 

Furthermore, it can be shown that all the operations of addition, scalar 
multiplication, and differentiation performed on both /and /yield equivalent 
results, that is, 

(«a + m' = («/i + 6ti' <M D 

when differentiation is permitted on the ordinary function. 

' Lighthill, op. cit., Section 2.3. 

* Note that when we represent an ordinary function by generalized function equivalent, 
we use a bold face italic letter to denote the generalized function. 

474 Network analysis and synthesis 

DEFINITION B.9 The generalized step function u is defined as the 
total class of equivalent regular sequences {«„(*)} such that 

(«,# = lim f°uJit)4(t)dt 

n-*oo J— ao 


J— 00 


where u(i) is the unit step defined in Chapter 2. That {«„(/)} exists is 
guaranteed by the previous theorem allowing representations of ordinary 
functions by generalized functions. Hence, we write u = u. 

Example B.5. The sequence 




which is plotted in Fig. B.l, is one member of the class of equivalent regular 
sequences which represents the generalized step function. 

2 3 4 5 6 7 8 t 

FIG. B.l. The generalized step sequence, «,(/)• 

DEFINITION B.I0 The unit impulse, or Dirac delta function <$(r), 
is denned as the derivative of the generalized step function d(t) ~ {«'„(*)}. 

Appendix B 475 

It should be stressed that <J(f) is merely the symbolic representation for a 
total class of equivalent regular sequences represented by {u' n (t)}. Thus 
when we write the integral 



we actually mean 

f " <5(0 <Kt) dt = (d, <j>) = lim f " u' n (0 #t) dt 

J— oo n-»oo J— oo 

Example B.6. The sequence 

^-(5-2r)«p[-i(J + |.)] ,>o 

= t £0 

in Fig. B.2 is one member of the class of equivalent regular sequences which 
represents the unit impulse. Other members of the class are the sequences 
{e-»«*(«MK} and {^1*^(1/ VJ^n)}. 










FIG. B.2. The generalized sequence, «'„(/). 

476 Network anal/sis and synthesis 



The most important property of the unit impulse is the sifting property 
represented symbolically by 


W(0*-/(0); (M,lj8|<«>) 


where /is any function differentiable over [a, /J]. The left hand side of 
Eq. B.16 is defined formally by 

\ f> \t)f(t)dt slim \"*u'JLt)f{t)dt (B.17) 

Js<0 n-»ao Ja<0 

The proof of the sifting property is obtained by simply integrating by 
parts, as follows. 

lim r >O u'„(0/(0 dt = lim u n (t)f(t)\- lim f'ii,(0/'(0 dt 

n—<x> J«<0 n-»oo l« n-»oo J* 

-/tf)-f'limu.(0/'(OA _,„ 

= /C8)-[/(/5)-/(0)]=/(0) 

Pictorially we represent <$(0 by a spike as shown in Fig. B.3. If the 
impulse is centered at t = a, then the sifting property is given symbolically 


d(.t - fl )/(f) dt - /(a) (|a|, HI < ao) 



FIG. B.3. The unit impulse. 

Appendix B 477 

where/'(0 most exist over [a, /?]. Note that when the limits of integration 
are infinite, we actually mean 

(°°S(t)f(t)dts lira \\t)f{t)dt (B.20) 

J— oo a-*— oo Ja 

In the sifting property, if both a, (I > or a, /? < 0, then 


<K0/(0 dt = (B.21) 

The proof of this property is similar to the original proof of the sifting 
property, and will be left as an exercise for the reader. 


The defining equations of the delta function according to Dirac are 



d(x) dx = 1 
«<o (B.22) 

d(x) = 0; x i* 

These are actually properties of the delta function as viewed from the 
generalized function standpoint. The proof can be obtained directly from 
the sifting property. Suppose we have the integral 


f> \t)f(t)dt=f(0) (B.23) 

and we let f{t) = 1. Then we have 


d(t)dt=f(p) = l (B.24) 


If both a, /? are greater than zero or both are less than zero, then 


d(t) dt = (B.25) 

This property is stated symbolically by the conditions d(t) = for / ft 0. 

Differentiation across a discontinuity 

Consider the function /(0 in Fig. B.4. We see that /(f) has a discon- 
tinuity of A at t = T. If we let/i(0 =f(f) for t < T, and/i(f) =■/(*) - A 
for t ^ T, then we have 

At)=MQ + Au(t-T) (B.26) 

478 Network analysis and synthesis 



FIG. B.4. Function with discontinuity. 

Since /(0,/i(0» and u(t) satisfy the condition 



« (1 + O 


dt < co 

for some N; we can represent these ordinary functions by generalized 

(ft) =Mt) + n(0 

Taking derivatives on both sides of Eq. B.27 yields 

fV)=f\(.t) + Au'(.t) 
which symbolically can be written as 

no =A(o + A m 


We thus see that whenever we differentiate across a discontinuity, we 
obtain a delta function times the height of the discontinuity. 

Example B.7. The step response of an R-C network is given as 

Hf) = Ae-" T u(t) (B.30) 

shown in Fig. B.Sa. The impulse response is 

h'(t)=Ad(t)-j,e- l l T u(t) 


and is shown in Fig. B.Sb. 

Appendix B 479 

FIC. B.5. Differentiation across a discontinuity. 


The derivative of a delta function, which we call a doublet, is defined 
symbolically as d\t) ~ {u" n (t)}. It has the following property, where 
f"(i) exists over [a, fit]. 


J ' 


<J'(0/(0 dt = -/'(0); (|«|, \fi\ < oo) 

The proof is obtained through successive integration by parts. 
T" V(0/(0 dt = lim f >0 «" B (0/(0 dt 

Ja<0 n-»oo Ja<0 

- Km « '.(0/(0 

fi>0 Ct>9 

-lim u' n (t)f'(t)dt 

«<0 n-»oo J«<0 



We see that since lim u'Jfi) — lim u' n (a) = 0, 

limH' B (0/0)' >0 =0 

n-*oo «<p 



Network analysis and synthesis 


FIG. B.6. The doublet <$'(')■ 
We then integrate by parts again so that 

-lim f >0 u' B (0/'(0^=-lim« n /'| +lim ['u n (t)f\t)dt 


-/'(/?)+ u{t)f{t)dt 


= -fW +fW -/'(0) = -/'(0) 
In general, the derivative-sifting property can be stated symbolically as 

J * 


«-(*-«)/(») A -(-DT"'(«) 


where/ <B+1) (0 exists over [a, /3].» 

The generalized function d'(t) is sometimes called a doublet. The 
pictorial representation of a doublet is given in Fig. B.6. 

Other properties of the unit impulse 

Dirac and others have obtained a host of identities concerning the unit 
impulse. We will merely give these here without proof. 

<5(-o - ,5(0 

d'(-t) - -<5'(0 

t <3(0 = o 
td'(t) = -d{t) 





d(t* -a*) = i \a\- 1 {d(t - a) + d(t + a)} 

The proofs of these properties are obtained through the inner product 
with a testing function <f>(t) € C. 

• The condition on/'" +1 ' is sufficient, but not necessary. 

appendix C 

Elements of complex variables 


A complex variable z is a pair of real variables (x, y) written as 

z = x+jy (CI) 

where j can be thought of as v — 1. 

The variable x is called the real part of z, and y is the imaginary part of z. 
Written in simpler notation, we have 

z = Re(z), y = Im(z) (C.2) 

The variable z can be plotted on a pair of rectangular coordinates. The 
abscissa represents the x or real axis, and the ordinate represents the y or 
imaginary axis. The plane upon which x and y are plotted is called the 
complex plane. Any point on the complex plane, such as z = 3 +j2, can 
be represented in terms of its real and imaginary parts, as shown in Fig. C. 1 . 
From the origin of the complex plane, let us draw a vector to any point z. 
The distance from the origin to z is given by 

|x| = (x* + y»)* (C.3) 

and is known as the modulus of z. The 
angle which the vector subtends is known 
as the argument of z or 

-i V 

arg z = tan * - 


Letting 6 = arg z and r = \z\, we can 
represent z in polar coordinates as 


3 * 

z = re» 




482 Network anal/sis and synthesis 

Expanding this last equation by Euler's formula, we obtain 

z == r cos 6 +jr sin 6, (C.6) 

so that x = r cos 8 

y = r sin 

The rule for addition for two complex numbers is given as 

(a +jb) + (c +jd) = (a + c) +j(b + d) (C.8) 

When two complex numbers are multiplied, we have 

(a + jb)(c +jd) = (ac - bd) + j(ad + be) (C.9) 

where j* = — 1. If we express the complex numbers in polar form, we 

(a + jb) = ri e m (CIO) 

and (c+jd) = r^ m (C.ll) 

When we multiply the two numbers in polar form, then 

r^rtf"* = jv^**- 1 -*' (C.12) 

If we divide these two numbers in polar form, then 

7^ = 7^'^ (C13) 

In rectangular coordinates, the operation of division can be expressed as 

a+jb _ (a+ jb)(c - jd) 

c+jd (c+jd)(c-jd) 

ac + bd .be — ad 


+ d* J c* + d* 

In connection with the modulus of a complex number, it is useful to note 
the following rules: 

l«i«2l = Uil • tal 

1*1*1*1 = |«il • l*i*l = l*il 2 (CIS) 

z-z* = |z|» 

where z* is the complex conjugate of z and is defined as 

z* = x +jy = x —jy (C.16) 

Appendix C 483 
The following rules deal with operations involving the conjugate definition : 

Zi + z t = z x * + z,* 

^ = *!*•*.* (C.17) 

Finally, if z has a modulus of unity, then 

z - 1 (C.18) 


The operations of raising a complex number to the nth power, or taking 
the nth root of a complex number, can be dealt with most readily by using 
the polar form of the number. Thus, we have z B = (re 1 ")" = r n e SnB , 

and z 1/n = jMn e sn»+tk,)M k0l n _ l (C 19) 


If to each z = x +jy, we assign a complex number w = u +jv, then w 
is a function of z or 

w =/(*) (C.20) 

The following are examples of complex functions, i.e., functions of a 
complex variable: 

w = 2z 

w = log, z 

w = 1/z (C.21) 

w = z a + 4 

w = |z| 

We see that w may be complex, pure real, or pure imaginary, depending 
upon the particular relationship with z. In general, the real and imaginary 
parts of w are both functions of x and y. That is, if we let w = j + jv, 

u-A*,V) and v=f(x,y) (C.22) 

As an example, let us find u and v for the function w = z* + 4. 

w = z * + 4 = (x +jy)* + 4 (C.23) 

Simplifying, we obtain 

w = (*« - y* +/'2ry) + 4 (C^4) 

484 Network analysis and synthesis 



z + Az 



« + A? 




FIG. C.2 

Path 2 

FIG. C.3 

so that u = a:* — y l + 4 and i> = 2«y 

The derivative of a complex function /(z) is defined as 



If one restricts the direction or path along which Az approaches zero, then 
we have what is known as a directional derivative. However, if a complex 
function is to possess a derivative at all, the derivative must be the same 
at any point regardless of the direction in which Az approaches zero. In 
other words, in order for/(z) to be differentiable at z = z , we must have 

= constant 



for all directions of approach of Az. 

Consider the two directions in which Az approaches zero in Figs. C.2 
and C.3. For path 1, we have 

,/(z + Az)-/(z) 

If we substitute 

/'(z) =lim lim' 

A*->0 Av->0 Az 

into Eq. C.28, we obtain 

f'(z) = lim lim 

Aac-»0 Aff-*0 

Az = Ax + j Ay 
f[x + A* +j(y + Ay)] -f(x+jy) 


As + j Ay 
/(z) = u +jv 
f(z + Az) = u + Aw +j(v + A») 




Appendix C 485 

we finally arrive at 

Am +jAi> 

/'(«) = lim lim 

a»-.o Av->o Aa; + j'Ay 


For path 2, we have 

,. Au + /Ao 9u , . 9» 

= hm -* — = — + j — 

a*-»o Ax dx dx 

Au + / Ad 
/'(z)=lim Km " T J A 
a»-»oa«s-oAx + J Ay 

==lim A iL±M£ (C.34) 

a»-o /'Ay 

dt> . du 
dy dy 

Since we assume that the function /(z) is differentiable, the derivatives 
must be independent of path. Thus, we have 

— _•— = — 4- / — (C 35) 

dy dy dx 9x 

From this last equation, we obtain the Cauchy-Riemann equations, which 

dv _ du 
dy dx 

dy dx 

We have just seen that in order for a function to have a derivative, the 
Cauchy-Riemann equations must hold. A function which is single valued 
and possesses a unique derivative is called an analytic function. A set of 
sufficient conditions for analyticity is that the Cauchy-Riemann equations 
are obeyed. For example, consider the function 

/(z) is analytic because 

/(z) = z* + 4 (C.37) 

^ = 2x = ^ 
dy dx 

9 (C.38) 

— = — 2 = — — 
dy dx 

486 Network anal/sis and synthesis 

On the other hand,/(z) = z* is not analytic because 

u = x and v = —y 

du j., a *> 1 (C39) 

— = +1 and — = -1 

ox ay 


If/(z) is analytic within a region or domain in the complex plane except 
at a point z , then/(z) has an isolated singularity at z . Suppose /(z) has a 
singularity at z , then we can expand /(z) about z in a Laurent series 

(z - z ) B z - z 

+ Om(2 - So)"* + • * * 


In the expansion, if m is finite, then z is called a pole of order m. 1 The 
term a_i is called the residue of the singularity. 

Example C.l. Consider the Laurent series for the function /(z) = e*\z about 
the pole at the origin. We can expand e* in a power series to give 

— *= I/i +* + !«• + -*»+ ■■) 

z z\ 2! 3! / 


= l+l + Lz+l.z* + -- 
z 2! 3! 

According to the definition, the residue of the pole at z = is equal to 1. 

Example C.2. Expand the function /(z) = l/z(z — l) 2 about the pole at z = 1, 
and find the residue of the pole at z = 1 . 

1 1 1 

z(z-l)« (« - i)M + (z - 1) 

(z - 1)» 

1 1 

[1 - (z - 1) + (z - 1)» - (z - I)* + • • •] (C.42) 

, „, , + 1 - (z - 1) + (z - 1)» + • • • 

(z — 1)* z — 1 

for < |z — 1| < 1. Here, the residue of the pole at z = 1 is equal to —1. 

1 Note that if we have an infinite number of nonzero terms with negative exponents, 
then *o is an essential singularity. 

Appendix C 487 

Example C3. Find the residues of the poles at s = and s =■ 

s +2 

To find the residues, we simply perform a partial fraction expansion 
«x 2 3 1 3 

-1 of the 


Thus the residue of the pole at s =» is —3, and the residue of the pole at 
s = -lis +3. 


In complex integration the integral is taken over a piecewise smooth path 
C and is defined as the limit of an infinite summation 


f(z) dz = lim 2 f( z i) A*/ 

C n-»oo j— 1 


where z t lies on C. Unlike the process of differentiation, the path along 
which we take the integral makes a difference as to the ultimate value of 
the integral. Thus the integral 




in general, has different values depending upon whether we choose to inte- 
grate along path Q or path C 8 , as shown in Fig. C.4. If we integrate along 
a closed path, say from a to b and then to a again, we are integrating along 
a closed contour. The path shown in Fig. C.5 is an example of a closed 
contour. The following theorem, known as Couch/ s residue theorem gives 
a method for rapid evaluation of integrals on closed paths. 








FIG. C.4 

FIG. C.5 

Theorem C.l. If C is a simple closed curve in a domain D, within which /(z) is 
analytic except for isolated singularities at z^, z,, ... , z„, then the integral along 

488 Network anal/sis and synthesis 
the closed path C is 

f(z)dz = 2n/(JKi +K. + ---+KJ 


where K t represents the residue of the singularity z ( . 

Example C.4. Consider the integral 



sH.s + 1? 




along the circle \s\ = 2, as given in Fig. C.6. Since there are two singularities 
within the circle, at s = and at s = —1, whose residues are respectively —3 
and +3, then the integral along the circle is 

S + 2 ■ ds - 2*j(-3 + 3) = (C.49) 


s\s + iy 


Example C.5. Find the integral of f(s) along the closed contour shown in Fig. 
C.7. The function /(«) is given as 

35+5 (C.50) 

■'*"-(* + IX* + 2) 





— * 

-2 -1 




. C7 

Appendix C 489 
A partial fraction expansion of f(s) shows that 

m — ri + Txi (C51) 

s + 1 s + 2 

so that the residues are 1 and 2. The value of the integral along the closed 
path within which both the singularities lie is then 

j> /(s) ds = iTTJil + 2) = 6tt/ (C.52) 

If f(z) is analytic in a domain with no singularities, then the integral 
along any closed path is zero, that is, 


f(z)dz = (C.53) 

This result is known as Cauchy's integral theorem. 

appendix D 

Proofs of some theorems on 
positive real functions 

Theorem D.l. If Z(s) and W(s) are both positive real, then Z(fV(s)) is also 
positive real. 

Proof. When Re s £ 0, both Re Z(s) and Re W(s) ^ 0, then Re Z( W(s)) £ 0, 
also. When s is real, both Z(s) and ff%s) are real, hence Z(W(s)) is real. Since 
Z(W(s)) satisfies both conditions of positive realness, it is positive real. 

Theorem D.2. If Z{s) is positive real, then Z(\/s) is positive real. 
Proof. W(s) = 1/* is positive real, hence Z(tV(s)} = Z(l/s) is positive real. 

Theorem D.3. If lV(s) is positive real, then l/fV(s) is also positive real. 

Proof. Z(s) = l/s is positive real, hence Z(W(s)) = l/fV(s) is positive real by 
Theorem D.l. 

Theorem D.4. The sum of positive real functions is positive real. 
Proof. Suppose Z x (s) and Z 2 (s) are both positive real. When Re s ^ 0, then 

ReZ x k and ReZ 2 ^ 

so that ReZj + ReZ a = ReZ ^ 0. 

Also, when s is real, both Z x and Z 2 are real. The sum of two real numbers 
is a real number. Therefore, Z x + Z t is positive real. 

Theorem D.5. The poles and zeros of Z(s) cannot have positive real parts 
(i.e., lie in the right half of the s plane). 

Proof Suppose there is a pole s m the right-half plane. Let us make a 
Laurent series expansion about s so that 

(* - 5o) n (S - * )" * 



Appendix D 491 


where n is real and finite. In the neighborhood of the pole S& Z(s) can be 
approximated by 

Z(s) »,-^V- 
(s - s ) n 

We can represent Z(s) in polar form by substituting each term by its polar form; 
i.e., let (j - «o)+" = rV»» and k_ n = Kt** so that 

z <*) = ^ **~**. Re Z{s) = * cos (* - nfl) 

which is represented in Fig. D. 1 . When 9 varies from to lit, the sign of Re Z(j) 
will change 2n times. Since ReZ(s) t. when Re s 2. 0, it is seen that any 
change of sign of Re Z(s) in the right-half plane will show that the function is not 
positive real. Therefore, we cannot have a pole in the right-half plane. Since the 
function 1/Z(s) is positive real if Z(s) is positive real, it is obvious that there 
cannot be any zeros in the right-half plane also. 

Theorem D.6. Only simple poles with positive real residues can exist on the 
ya> axis. 

Proof. As a consequence of the derivation of Theorem D.5, it is seen that 
poles may exist on they'd) axis if n = 1, and ^ = 0. The condition n — 1 implies 
that the pole is simple and the condition <j> = implies that the residue is positive 
and real. It is readily seen that zeros on they' to axis must also be simple. 

Theorem D.7. The poles and zeros of Z(s) are real or occur in conjugate pairs. 
Proof. If a complex pole or zero exists without its conjugate, Z(s) cannot be 
real when s is real. As a result of this theorem and Theorem D.S, it is seen that 
both the numerator and denominator polynomials of Z(s) must be Hurwitz. 

Theorem D.8. The highest powers of the numerator polynomial and the 
denominator polynomial of Z(s) may differ by at most unity. 
Proof. Let Z(s) be written as 


««■?" + fln-iJ" -1 + • • • + OyS + a. 

*>m* m + bn.lS" 

+ V + b 


492 Network anal/sis and synthesis 

FIG. D.2 

If m — n ^ 2, when s = », Z(«) will have a zero of order 2 or greater at s = oo, 
which is on the/w axis. Similarly, if n - m £ 2, then, at s = », Z(s) will have a 
pole of order 2 or more at * = oo. Since Z(s) cannot have multiple poles or 
zeros on the/a> axis, these situations cannot exist; therefore \n — m\ <.\. 

Theorem D.9. The lowest powers of P(s) and Q(s) may differ by at most unity, 
Proof. The proof is obtained as in Theorem D.8 by simply substituting 1/s 
for s and proceeding as described. 

Theorem D.10. A rational function F(s) with real coefficients is positive real if: 

(a) F(s) is analytic in the right-half plane. 

(b) If F(s) has poles on they'a> axis, they must be simple and have real, positive 

(c) Re F(ja>) ^ for all o>. 

Proof. We need only show that these three conditions fulfill the same require- 
ments as Re Z(s) ^ for Re s ^ 0. We will make use of the minimum modulus 
theorem which states that if a function is analytic within a given region, the 
min imum value of the real part of the function lies on the boundary of that 
region. The region with which we are concerned is the right-half plane which is 
bounded by a semicircle of infinite radius and the imaginary axis with small 
indentations for theyVu axis poles. If the minimum value on theya) axis is greater 
than zero, then Re Z(s) must be positive over the entire right-half plane (Fig. D.2). 

appendix E 

An aid to the improvement 
of filter approximation 


The introduction of an additional pole and zero in the second quadrant 
of the complex frequency plane, and at their conjugate locations, can give 
amplitude or phase corrections to a filter approximant over some desired 
band of frequencies without significantly changing the approximant at 
other frequencies. However, a cut-and-try procedure for finding the best 
positions for such a pole-zero pair can be tedious. A visual aid is presented 
herein which reduces the amount of labor required to make modest 
corrections of this type. 

Constant phase and constant logarithmic gain contours for the correc- 
tion by a pole-zero pair 1 are plotted on transparent overlays. One of these 
may be placed over a suitably scaled sheet of graph paper representing the 
complex frequency plane. Then the pair-shaped phase and gain correc- 
tions along the jm axis are indicated by the intersections of the overlay 
contours with this axis. Corrections which best reduce the errors in the 
original approximant are then sought by variation of the overlay position 
and orientation. Either phase or amplitude may be corrected. However, 
it is not always possible to simultaneously improve both the phase and the 
amplitude characteristics of an approximant by a single pair-shaped 

F. F. Kuo and M. Karnaugh, reprinted from the IRE Transactions on Circuit Theory, 
CT-9, No. 4, December 1962, pp. 400-404. 

1 This will be called, hereafter, a pair-shaped correction. 


494 Network analysis and synthesis 


Suppose we begin with a transfer function G(s) with certain deficiencies 
in its amplitude or phase response. Let us consider the transfer function 
of a corrective network G^s) such that the product 

GAs)-GM<K») CE.1) 

will have better gain or phase characteristics. For the purposes of this 
paper, we will restrict G^s) to have the form 

O^-C <'-«)<'-«> (E.2) 

(s - p)(s - p) 

where C is a constant, and q and p are a zero and a pole, respectively, in 
the second quadrant of the complex frequency plane. If the correction 
is to be applied at sufficiently high frequencies such that s — qo*s — p 

GMcsZtzA (E.3) 


Let us consider the effect when the pole-zero pair in Eq. E.3 is used to 
augment any given rational function in the complex frequency plane. 
The added gain, in decibels, due to this pole-zero pair is 

D = 20 log 10 


s -q 
s- p 

where we have neglected the effect of the constant C in Eq. E.3. If we let 

fc = (lO)*' 20 (E.5) 

then k = 



For fixed k, this is the equation of a circle with inverse points* at q and 
p. Its radius is 

kjp^q\ (E7) 

P II - *"l 
and its center is at 

* E. C. Titchmarsh, The Theory of Functions, Oxford University Press, Oxford, 
England, second edition 1939, pp. 191-192. 



Appendix E 495 


P — 

II - k*\ 

— £5© 



Here, s is the midpoint between the pole and zero, and their separation 
is e. It is easy to see from Eq. E.l 1 that the center of each circle of constant 
gain is externally collinear with the pole and zero. For the purpose of 
drawing the family of constant gain contours, we may let s be the origin 
of coordinates and the scale factor e may be set equal to unity. 

Furthermore, only half the pattern need be drawn because the function 
D has negative symmetry with respect to a reflection about the perpen- 
dicular bisector of the pole-zero pair. 


Figure represents a pole at/>, a zero at q and an arbitrary point 
s in the complex frequency plane. When the pole and zero are used to 
correct a given phase characteristic, the added phase at s is 

4> - a - 9, (E.12) 

where B Q and 6 9 are measured with respect to a single arbitrary reference. 

In Fig. E.l A, a circle has been drawn through p, q, and s. Angle ^ is 

FIG. E.l. Derivation of constant phase contours. 

496 Network analysis and synthesis 

equal to one half the subtended arc ps'q. Therefore, the arc qsp is a 
constant phase contour. The angle at the center of the circle between cp 
and the perpendicular bisector of chord pq is also equal to <f>. All of the 
circular contours of constant phase have their centers on this perpendicular 

Note that minor arc ps'q is also a contour of constant phase, but the 
phase angle 

« = <f> - * (E.13) 

is negative, as indicated by the clockwise rotation from s'p to s'q. The 
convention for positive rotation is herein taken to be counterclockwise. 


Sets of constant phase and constant logarithmic gain contours are 
drawn in Figs. E.2 and E.3. The curves are symmetric about the zero- 
decibel line, except that the gain curves are of opposite sign. Therefore, 
only one half of each figure has been drawn. 




















1.71 1 

5| ]l 

4 |] 


1 °- 




























FIG. E.2. Constant amplitude and phase contours. 

Appendix E 497 

FIG. E.3. Constant amplitude and phase contours. 

The zero-decibel line is the perpendicular bisector of the line segment 
joining the pole and zero. It is a gain contour of infinite radius. 

The line through the pole and zero is a zero phase contour. Together, 
these perpendicular axes form a useful reference system. The signs of the 
phase and gain in the four quadrants formed by these axes are shown in 
Fig. E.4. 

The corrections in Figs. E.2 and E.3 do not carry algebraic signs because 

498 Network analysis and synthesis 





Zero gain 



-x^-*-Zero phase 


FIG. E.4. Signs of the gain G and phase <f> corrections. 

only half of the symmetrical pattern has been drawn. The signs may be 
obtained from Fig. E.4, and are important in selecting the orientation of 
the pole-zero pair. 


In correcting a given approximant, it is first necessary to plot the desired 
magnitude and/or phase corrections versus frequency. When the contours 
of Fig. E.2 or E.3 are overlaid on a second sheet representing the complex 
frequency plane, the contour intersections with the ja> axis indicate the 
corrections that will actually be realized. 

There are several variables at the designer's disposal. The first is the 
distance between the correcting pole and zero. Since the scale of the 
contours in Figs. E.2 and E.3 is not specified, the frequency scale of 
the underlying complex plane determines the distance between the pole 
and zero. The second design variable is the location of the center of the 
pole-zero pair. The distance of the center from a given band on theyeo 
axis determines the magnitude of the correction. 

The third variable is the orientation of the pole-zero pair. Figure E.4 
shows how the orientation affects the gain and phase corrections. As a 
simple example, suppose one wishes to have zero phase correction at 
co = 1.0, negative phase correction above and positive correction below 
that frequency. The attack would be to point the zero phase axis at 
co = 1.0 with the pole nearest to they'to axis. If it is desired to have equal 
phase correction above and below co = 1.0, the zero phase axis should be 
oriented parallel to the real axis of the complex frequency plane. If one 
wishes to have more phase correction above co = 1.0 and less below, the 

Appendix E 499 

zero phase axis should be rotated clockwise with respect to the a (real) 

We thus see that by varying the frequency scale of the complex frequency 
plane, the position of the center of the pole-zero pair, and the orientation 
of the pole-zero pair, the pair-shaped correction can be made to approxi- 
mate the desired correction. 

It must be emphasized that the method suggested herein is an aid to 
cut-and-try correction. As such, it is easier to use the method than to 
precisely set down rules for applying it. However, a few rather general 
statements may be helpful. 

Unless the pole and zero both lie on the real axis, one must remember 
that another pole and zero are located at conjugate positions. The con- 
tributions from both pole-zero pairs may be added algebraically. In most 
practical cases, the desired correction will have a band-pass character. 
Therefore, only one pole-zero pair will normally contribute significantly 
at any frequency. 

The shape and magnitude of the desired correction will dictate the way 
in which the/co axis must intersect the correction contours. The broadness 
of the desired correction will dictate the proper scaling of the jco axis. 
Usually, only a few trials are needed to fix the pole and zero locations for 
the best fit. 

It will be found that a worthwhile correction can be made in either the 
phase or the gain characteristic. Only fortuitously can they be improved 
simultaneously by a single pair-shaped correction. 

Example E.1. The amplitude response of a third-order Butterworth filter is 
given by the solid curve in Fig. E.5. It is desired to steepen the gain roll-off 
near the cutoff frequency «o c = 1 . This is done by increasing the gain just below 
o) = m e and decreasing it above that frequency. Figure E.6 illustrates the type 
of correction desired. Figure E.7 shows a pole-zero pair that achieves this type 
of correction. The gain at to = 1 remains unchanged by this particular choice. 
Other pole-zero pairs that "aim" the zero-decibel line at to = 1 but give asym- 
metrical corrections about that point might also be used. The dashed curve in 
Fig. E.S shows the corrected gain. 

A different pole-zero pair, also shown in Fig. E.7, has been chosen to minimize 
the deviation of the slope of the phase response from its slope at to = 0. This 
pair-shaped correction decreases the phase for m < 0.7 and increases it for 
co > 0.7. 

Figure E.8 shows the deviation of the phase responses from linear phase. 

It is clear from Figs. E.5 and E.8 that the gain corrected approximant has a 
poorer phase response than the original, while the phase corrected approximant 
has a gentler gain roll-off than the original. 

500 Network anal/sis and synthesis 

0.2 0.3 0.5 0.7 1.0 1.5 2.0 

Frequency, u 

FIG. E.5. Amplitude response of corrected and uncorrected Butterworth filters. 

This will not surprise the experienced filter designer. It is possible, however, 
to achieve moderate corrections in both gain and phase by using two pair- 
shaped corrections. A useful approach to this objective lies in localizing the 
gain correction further out of band, and the phase correction further in-band 
than in the separate corrections just discussed. This can be done by shifting 
the pole-zero centers to higher or lower frequencies and also by experimenting 
with nonsymmetrical corrections. 

FIG. E.6. Amplitude correction to steepen fall-off of third-degree Butterworth filter. 

Appendix E 501 

Loss — 1 

Butterworth _ 

X o 




-1.0 -0.8 -0.6 -0.4 -0.2 
\ x o 
















FIG. E.7. Poles and zeros of the original filter and correction equalizers. 



s -10 















/ / 
/ / 





er zL 

■ / 

\ 1 








th amp 




0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 
Frequency, rad/sec 

FIG. E.8. Phase deviation from slope at zero frequency. 

502 Network anal/sis and synthesis 


A few words are in order concerning the synthesis of the equalizer from 
the pole-zero pair obtained by the method described. 

In order to provide optimum power transfer and to facilitate cascading 
the correction network with the original filter, the correction network 
should be of the constant-resistance type. We will restrict our discussion 
to the bridged-T network in Fig. E.9, whose voltage transfer function is 
given as 



provided the network is of the constant resistance type as given by the 

Z 1 (j)Z 2 (i) = J R » (E.15) 

For normalization purposes, we will let R = 1 Q. 
Let the pole-zero correction be written in general as 

G(s) = 

s 2 + a t s + a 


fc(s* + b lS + b ) 

Since the correction has minimum phase, we know that {a ( , b t ^ 0}. In 
addition, since the d-c gain cannot exceed unity, kb ^ a . 
We can express the impedance Z^s) in terms of G(s) in Eq. E.14 as 

Zl(s) = _L_ 1= (± 


l)s* + (kb, - ajs + (fcb„ - a„) 

T~, ~, (Ai') 

s + a t s + a 

Since Z^s) must be positive real, the coefficients must all be nonnegative 
so that 

k- 1 ^0 

kb i — <*i ^ 
kb — a ^ 









v 2 

FIG. E.9. Bridged-T network. 

Appendix E 503 

FIG. E.I0. Poles and zeros of Z t (s). 

Moreover, in order for a biquadratic driving-point immittance to be 
positive real, the following condition must apply. 8 

Details concerning the synthesis of Z^j) are also given in Seshu's paper. 
Let us plot the pole-zero pair of Z t {s) as in Fig. E.10. We can represent 
the locations of the poles and zeros in terms of the polar angles <£„ and 0, 
and their distances from the origin, n p and n t . Rack* has shown that in 
order for in-band loss to be finite 



In addition he has shown that for an unbalanced bridged-T circuit, if 
a = max [0 M , <f> 9 ] then the larger the angle a, the larger the in-band loss. 
In particular, « should be less than 70° to restrict the in-band loss to 
reasonable proportions. If one considers other network configurations 

*S. Seshu, Minimal Realizations of the Biquadratic Minimum Function, IRE 
Trans, on Circuit Theory, CT-6, December 1959, 345-350. 
4 A. J. Rack, private communication. 

504 Network analysis and synthesis 

such as lattice networks, it is conceivable that Rack's restrictions might 
be relaxed. For lattice network design, Weinberg's method is applicable, 5 
although here again, the in-band loss restrictions are severe. 


We have presented here a simple visual aid to the correction of the 
amplitude or phase response of filters. The method has the advantage of 
facilitating the commonly used cut-and-try approach to this problem. It 
has the disadvantage, in many cases, of failing to provide simultaneous 
amplitude and phase corrections in a single step. 

1 L. Weinberg, "RLC lattice networks," Proc. IRE, 41, September 1953, 1139-1144. 



Goldman, S., Frequency Analysis, Modulation and Noise, McGraw-Hill, 

New York, 1948. . ' 

Javid, M. and E. Brenner, ^w/y^w, Transmission and Filtering of Signals, 

McGraw-Hill, New York, 1963. 
Lathi, B. P., Signals, Systems and Communications, John Wiley and 5>ons, 

New York, 1965. 
Lighthill, M. J., Introduction to Fourier Analysis and Generalized Functions, 

Cambridge University Press, New York, 1958. 
Mason, S. J. and H. J. Zimmerman, Electronic Circuits, Signals and 

Systems, John Wiley and Sons, New York, 1960. 
Papoulis, A., The Fourier Integral and Its Applications, McGraw-Hill, New 

York, 1962. 
Rowe, H. E., Signals and Noise in Communications Systems, D. van 

Nostrand, Princeton, N.J. 1965. 
Schwartz, M., Information Transmission, Modulation and Noise, McGraw- 
Hill, New York, 1959. 


Balabanian, N., Fundamentals of Circuit Theory, Allyn and Bacon, Boston, 

Bohn, E. V., The Transform Analysis of Linear Systems, Addison-Wesley, 

Reading, Massachusetts, 1963. 
Brenner, E. and M. Javid, Analysis of Electric Circuits, McGraw-Hill, New 

York, 1959. 


506 Network analysis and synthesis 

Brown, R. G. and J. W. Nilsson, Introduction to Linear Systems Analysis, 
John Wiley and Sons, New York, 1962. 

Brown, W. M., Analysis of Linear Time-Invariant Systems, McGraw-Hill, 
New York, 1963. 

Carlin, H. J. and A. Giordano, Network Theory, Prentice-Hall, Englewood 
Cliffs, N.J., 1964. 

Cassell, W. L., Linear Electric Circuits, John Wiley and Sons, New York, 

Chen, W. H., The Analysis of Linear Systems, McGraw-Hill, New York, 

dePian, L., Linear Active Network Theory, Prentice-Hall, Englewood 
Cliffs, N.J. 1962. 

Friedland, B., O. Wing, and R. B. Ash, Principles of Linear Networks, 
McGraw-Hill, New York, 1961. 

Gardner, M. and J. L. Barnes, Transients in Linear Systems, Vol. 1, John 
Wiley and Sons, New York, 1942. 

Guillemin, E. A., Introductory Circuit Theory, John Wiley and Sons, New 
York, 1953. 

Guillemin, E. A., Theory of Linear Physical Systems, John Wiley and Sons, 
New York, 1963. 

Harman, W. W. and D. W. Lytle, Electrical and Mechanical Networks, 
McGraw-Hill, New York, 1962. 

Hayt, W. H., Jr. and J. E. Kemmerly, Engineering Circuit Analysis, 
McGraw-Hill, New York, 1962. 

Huelsman, L. P., Circuits, Matrices and Linear Vector Spaces, McGraw- 
Hill, New York, 1963. 

Kim, W, H. and R. T. Chien, Topological Analysis and Synthesis of Com- 
munication Networks, Columbia University Press, New York, 1962. 

Ku, Y. H., Transient Circuit Analysis, D. Van Nostrand, Princeton, N.J., 

Legros, R. and A. V. J. Martin, Transform Calculus for Electrical Engineers, 
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LePage, W. R. and S. Seely, General Network Analysis, McGraw-Hill, 
New York, 1952. 

Ley, B. J., S.G. Lutz, and C. F. Rehberg, Linear Circuit Analysis, McGraw- 
Hill, New York, 1959. 

Lynch, W. A. and J. G. Truxal, Introductory System Analysis, McGraw- 
Hill, New York, 1961. 

Paskusz, G. F. and B. Bussell, Linear Circuit Analysis, Prenctice-Hall, Inc. 
Englewood Cliffs, N.J., 1964. 

Pearson, S. I. and G. J. Maler, Introductory Circuit Analysis, John Wiley 
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Bibliography 507 

Pfeiffer, P. E., Linear System Analysis, McGraw-Hill, New York, 1961. 
Reza, F. M. and S. Seely, Modem Network Analysis, McGraw-Hill, New 

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Sanford, R. S., Physical Networks, Prentice-Hall, Englewood Cliffs, N.J., 

Schwarz, R. J. and B. Friedland, Linear Systems, McGraw-Hill, New York, 

Scott, R. E., Elements of Linear Circuits, Addison-Wesley, Reading, 

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Seely, S., Dynamic Systems Analysis, Reinhold, New York, 1964. 
Seshu, S. and N. Balabanian, Linear Network Analysis, John Wiley and 

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Skilling, H. H., Electrical Engineering Circuits, Second Edition, John 

Wiley and Sons, New York, 1965. 
Van Valkenburg, M. E., Network Analysis, Second Edition, Prentice-Hall, 

Englewood Cliffs, N.J., 1964. 
Weber, E., Linear Transient Analysis, John Wiley and Sons, New York, 

Zadeh, L. A. and C. A. Desoer, Linear Systems Theory, McGraw-Hill, New 

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Balabanian, N., Network Synthesis, Prentice-Hall, Englewood Cliffs, N.J., 

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Nostrand, Princeton, N.J., 1945. 
Calahan, D. A., Modern Network Synthesis, Hayden, New York, 1964. 
Chen, W. H., Linear Network Design and Synthesis, McGraw-Hill, New 

York, 1964. 
Geffe, P. R., Simplified Modern Filter Design, John F. Rider, New York, 

Guillemin, E. A., Synthesis of Passive Networks, John Wiley and Sons, 

New York, 1957. 
Guillemin, E. A., The Mathematics of Circuit Analysis, John Wiley and 

Sons, New York, 1949. 
Hazony, D., Elements of Network Synthesis, Reinhold, New York, 1963. 
Kuh, E. S. and D. O. Pederson, Principles of Circuit Synthesis, McGraw- 
Hill, New York, 1959. 
Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters, 

Impedance-Matching Networks and Coupling Structures, McGraw-Hill, 

New York, 1964. 

508 Network analysis and synthesis 

Saal, R., Der Entwurf von Filtern mit Hilfe des Kataloges Normierter 

Tiefpdsse, Telefunken GMBH, 1961. 
Skwirzynski, J. K., Design Theory and Data for Electrical Filters, D. Van 

Nostrand, Princeton, 1965. 
Storer, J. E., Passive Network Synthesis, McGraw-Hill, New York, 1957. 
Truxal, J. G., Control System Synthesis, McGraw-Hill, New York, 1955. 
Tuttle, D. F., Network Synthesis, Vol. 1, John Wiley and Sons, New York, 

Van Valkenburg, M. E., Introduction to Modern Network Syntheses, 

John Wiley and Sons, New York, 1960. 
Weinberg, L., Network Analysis and Synthesis, McGraw-Hill, New York, 

Yengst, W. C, Procedures of Modern Network Synthesis, Macmillan, New 

York, 1964. 

Name Index 

Aaron, M. R., 448 
Aitken, A. C, 469 
Angelo, E. J., 268 
Ash, R. B., S06 

Balabanian, N., 505, 507 
Barnes, J. L., 506 
Bashkow, T. R., 283, 438 
Bellman, R., 469 
Bode, H. W., 221, 507 
Bohn, E. V., 505 
Brenner, E., 505 
Brown, R. G., 506 
Brown, W. M., 506 
Bubnicki, Z., 283 
Budak, A., 395 
Bussell, B., 506 

Calahan, D. A., 507 
Carlin, H. I., 428 
Cassell, W. L., 506 
Cauer, W., 324 
Chen, W. H., 506 
Chien, R. T., 506 

Darlington, S., 431, 457 

Davis, H. F., 51 

de Pian, L., 506 

Desoer, C. A., 507 

Dirac, P. A. M., 33, 470, 471 

Dutta Roy, S. C, 283 

Eisenman, R. L., 469 
Elmore, W. C, 390 
Ende, F., 381 

Faddeeva, V. N., 469 
Faddev, D. K., 469 
Fano, R. M., 448 
Feshbach, H., 86 
Foster, R. M., 321 
Franklin, P., 198 
Friedland, B. O., 506 
Fujisawa, T., 458 
Fukada, M., 381 

Gantmacher, F. R., 469 

Hayt, W. H., Jr., 506 

Geffe, P. R., 507 

Giordano, A., 506 

Goldman, S., 137, 505 

Goldstone, L. O., 153 

Guillemin, E. A., 296, 299, 506, 507. 

Harman, W. W., 506 

Hayt, W. H., Jr., 506 

Hazony, D., 507 

Hobson, E. W., 471 

Hohn, F. E., 469 

Huelsman, L. P., 261, 469, 506 

Jahnke, E., 381 
James, R. T., 389 
Javid, M., 505 


510 Name Index 

Jones, E. M. T., 507 
Jordan, E. C, 413 
Justice, G., 27 

Karnaugh, M., 493 

Kautz, W. H., 47, 367 

Kemmerly, J. E., 506 

Kim, W. H., 506 

Ku, Y. H., 506 

Kuh, E. S., 507 

Kuo, F. F., 279, 393, 493 

Larky, A. I., 262 

Lathi, B. P., 505 

Le Corbeiller, P., 469 

Legros, R., 506 

Leichner, G. H., 279 

Le Page, W. R., 506 

Ley, B. J., 506 

Lighthill, M. ., 65, 136, 471, 473, 505 

Lutz, S. G., 506 

Lynch, W. A., 506 

Lytle, D. W., 506 

Maler, G. J., 506 
Marcus, M., 469 
Martin, A. V. J., 506 
Mason, S. J., 505 
Matthaei, G. L., 507 
Mine, H., 469 
Morse, P. M., 86 

Nering, E. D., 469 
Nilsson, J. W., 506 

O'Meara, T. R., 283 
Orchard, H. J., 385 

Paley, R. E. A. C, 291 
Papoulis, A., 379, 381, 505 
Paskusz, G. F., 506 
Pearson, S. I. 506 
Pederson, D. O., 507 
Perlis, S., 469 
Pfeiffer, P. E., 507 
Pipes, L. A., 469 

Rack, A. J., 503 
Raisbeck, G., 8, 294 
Rehberg, C. F., 506 

Reza, F. M., 507 
Rowe, H. E., 505 

Saal, R., 441 

Sands, M., 390 

Sanford, R. S., 507 

Saraga, W., 443 

Schwartz, L., 470, 471 

Schwartz, M., 505 

Schwarz, R. J., 507 

Scott, R. E., 507 

Seely, S., 506, 507 

Semmelman, C. L., 447 

Seshu, S., 503, 507 

Skilling, H. H., 100, 104, 106, 181, 507 

Skwirzynski, J. K., 508 

So, H. C, 449, 450 

Storer, J. E., 508 

Szentirmai, G., 441, 453 

Tellegen, B. D. H., 272 
Temple, G., 34, 471 
Terman, F. E., 236 
Thomson, W. E., 384 
Titchmarsh, E. C, 494 
Tompkins, C. B., 447 
Tropper, A. M., 469 
Truxal, J. G., 506, 508 
Tuttle, D. F., 508 

Ulbrich, E., 441 

Valley, G. E., Jr., 293 

Van Valkenburg, M. E., 135, 263, 296, 

297, 325, 329, 331, 344, 347, 376, 

507, 508 
Von Weiss, H., 469 

Walker, F., 283 
Wallman, H., 293 
Weber, E., 507 
Weinberg, L., 400, 435, 504 
Widder, D. V., 136 
Wiener, N., 291 

Yengst, W. C, 508 
Young, L., 507 

Zadeh, L. A., 507 
Zimmerman, H., 505 

Subject Index 

ABCD parameters, 262 

Admittance, driving-point, 15, 187 

All-pass network, 221, 357 

Amplifier, 11 

impulse response of, 201 

Amplitude response, 212, 342 
computer program for, 450 
evaluation by vector method, 215 

Amplitude spectrum, 3 

Analytic function, 485 

Approximation problem, 17, 365 

Available gain, 418 

Available power, 418 

Band-elimination filter transformation, 

Band-pass filter transformation, 407 
Bandwidth, half-power, 235 

spectral, 67, 389 
Bessel filter, 383, 435 

phase response of, 387 

tables of, 400, 435 

transient response of, 392 
Bessel polynomial, 386, 395 
Biquadratic immittance, 305, 503 
Black box, 15 
Bode plots, 221 
Bounded real function, 419 
Break frequency, 225 
Bridge circuit, 285, 354 
Butterworth filter, 368, 434, 500 

amplitude response of, 369, 387, 394, 

pole locus of, 371 

step response of, 391, 394 

tables of, 372, 400, 435 

Canonical form, 325, 400, 433 
Capacitor, 13, 103, 176 
Cauchy integral theorem, 489 
Cauchy residue theorem, 487 

Cauchy-Riemann equations, 485 
Cauer ladder expansion, 324 
Causality, 9, 290 
Characteristic equation, 77 
Characteristic impedance, 414 
Characteristic value, 78 
Cbebyshev filter, 373 
approximation, 366 
locus of poles, 376, 378 
tables of, 373, 400, 435 
transient response, 392 
Chebyshev polynomial, 373 
Circuit, bridged-T 258, 275, 502 
double-tuned, 238 
reciprocal, 8, 255 
single-tuned, 229 
symmetrical, 259 
Coefficient of coupling, 122 
Compensation theorem, 181 
Complementary function, 82 
Complex variables, 481 
analysis, 483 
differentiation, 484 
integration, 487 
Computer programs, 18, 438, 439, 

441, 447 448, 450 453, 457 
Constant-resistance network, 352, 502 
Contour integration, 487 
Controlled source, 268 
Convergence in the mean, 49, 51 
Convolution integral, 197 
Crest factoT, 27 
Critical coupling, 242 
Current source, 12 
Cutoff frequency, 17, 225 

Damping factor, 225 

d-c value, 25 

Delay, 32, 212, 245, 384 

distortion, 245 

time, 388 


512 Subject Index 

Delta function, see Unit impulse 
Delta-wye transformation, 257 
Differential equation, 75, 145 

forcing function of, 75 

homogeneous, 75, 76 

integrodifferential equation, 91, 145 

linear, 76 

nonhomogeneous, 75 

ordinary, 76 

simultaneous, 93, 146 
Differentiator, 11, 18, 193 
Digital computer, 18, 438 
Distributions, theory of, 470 
Dot reference for transformer, 123 
Doublet, 40, 479 

Duhamel superposition integral, 201 
Duty cycle, 26 

Energy density, 71 

Energy spectrum, 71 

Equal ripple approximation, see Cheb- 

yshev approximation 
Essential singularity, 486 
Excitation, 1 

Faraday's law of induction, 122 
Filter approximation, 368, 373, 379, 

383, 493 
Filter design, 17, 365, 397, 433, 457 

computer programs for, 441, 457 
Final conditions, 106, 109 
Final value theorem, 165 
Flux linkage, 122 
Forced response, 106 
Foster network, 321, 325 
Fourier series, 1, 46, 50 

amplitude spectrum, 57 

complex form, 55 

cosine series, 53 

evaluation of coefficients, 52, 58 

orthogonality conditions, 49 

phase spectrum, 57 

symmetry conditions, 53 
Fourier transform, 1, 63 

amplitude spectrum, 64, 65 

discrete, 56, 63 

inverse, 64 

of unit impulse, 65 

of unity, 68 

Fourier transform, phase spectrum, 64, 

properties of, 67 

symmetry conditions, 68 
Free response, 106 
Frequency, angular, 2 

complex, 4 

domain, 14, 134, 367 

forced, 192 

natural, 192 
Frequency normalization, 18, 402 
Frequency transformation, 18, 404 

Gain contours, logarithmic, 494, 496 

Gaussian filter, 291 

Generalized functions, 34, 64, 470 

g parameters, 286 

Green's function, 86 

Gyrator, 272, 287 

Hall effect, 273 

High-pass filter transformation, 405 
Hurwitz polynomial, 294, 316 347 
Hybrid (h) parameters, 260 
Hybrid matrix, 261, 449 

Ideal low-pass filter, 292, 367 
Ideal transformer, 263, 273, 275, 424 
Immittance, 15, 315 
Impedance, 15, 176 

driving-point, 15, 253, 315 

matrix, 254 

transfer, 188 

transformer, 263 
Impulse function, see Unit impulse 
Impulse response, 44, 85, 111, 194 
Incident power, 416 
Incidental dissipation, 276 
Inductor, 13, 104, 176 
Initial conditions, 13, 77, 86 107, 176 
Initial value theorem, 165 
Insertion loss, 429 

filter synthesis, 431 
Insertion power ratio, 430 

voltage ratio, 429 
Integrator, 11, 18, 193 
Integrodifferential equation, 91, 145 

Kirchhoffs laws, 100, 104 

Subject Index 513 

Ladder network, 279, 346, 347, 453 
Laguerre polynomial, 48 
Laplace transform, 1, 134 

definition of, 135 

inverse, 136 

properties of, 137 

table of, 168 

uses of, 144 
Lattice network, 285, 354, 503 
Laurent series, 486 
L-C immittance, 315 

properties of, 315 

synthesis of, 319 
Least squares, principle of, 49, 366 
Legendre polynomial, 381 
Linear phase filter, see Bessel filter 
Linear system, 8 

derivative property, 9 

ideal elements, 12 

ideal models, 10 

MAC, Project, 448 
Machine-aided design, 448 
Magnitude normalization, 18, 402 
Matrix algebra, 461 

definitions, 462 

operations, 464 

references, 469 
Maximally flat response, see Butter- 
worth filter 
Mean squared error, 48 
Minimum inductance transformation, 

Minimum modulus theorem, 492 
Modulation, amplitude, 70 
Monotonic filter, see Optimum filter 
Multiple-access computer, 448 
Mutual inductance, 122 

Negative impedance converter (NIC), 

Network, analysis, 7, 100, 175, 253 

linear, 8, 100 

n-port, 449 

passive, 8, 100 

reciprocal, 8, 100, 255 

symmetrical, 259 

synthesis, 290, 315, 341, 397, 431 

time-invariant, 9, 100 

Node equations, 106, 255 

Norm, 47 

Normalization, frequency, 402 

magnitude, 402 
Norton's theorem, 180, 185 

Open-circuit impedance parameters, 

254, 343 
Optimization techniques, 447 
Optimum (L) filter, 379 

amplitude response of, 380 

polynomials, 382 
Orthogonal set, 47 
Orthonormal set, 47 
Overcoupling, 242 
Overshoot, 388, 392 

Paley-Wiener criterion, 291 
Parseval's equality, 50 

theorem, 71 
Partial fraction expansion, 148 

conjugate poles, 150 

multiple poles, 151 

real poles, 149 
Particular integral, 82 
Passive network, 8 
Peaking circle, 233, 241 
Peak-to-valley ratio, 251 
Phase contours, logarithmic, 495 
Phase response, 212, 343 

computer program for, 450 

evaluation by vector method, 215 

linearity of, 213, 383 
Phase shift, 1 

distortion, 213 

minimum, 220, 346 

spectrum, 3, 57 
Phasor, 5 

Plancheral's theorem, 71 
Pole, definition of, 155 
Pole-zero diagram, 156 
Port, 12, 253 
Positive real (p.r.) function, 16, 299, 

Power, average, 301, 315 

available, 418 
Propagation constant, 414 
Proportionality, principle of, 8 
Pulse transmission, 389, 392 

514 Subject Index 

Q, circuit, 233, 236 

Ramp function, 31 

R-C admittance, properties of, 331 

R-C impedance, properties of, 325 

synthesis of, 329 
Realizability conditions, 16, 290, 301, 

342, 490 

for driving-point functions, 301 

for transfer functions, 342 
Reciprocal network, 8, 255 
Rect function, 65 
Reference impedance factor, 415 
Reflected parameter, 415 
Reflected power, 416 
Reflection coefficient, 416, 421, 458 
Remainder function, 308 
Residue, definition of, 162 

evaluation by vector method, 162 
Residue condition for two-ports, 344 
Resistor, 13, 103, 175 
Response, critically damped, 118 

overdamped, 118 

underdamped, 119 
Ringing, 388 
Rise time, 388 
R-L admittance, properties of, 325 

synthesis of, 329 
R-L impedance, properties of, 331 

synthesis of, 331 

Sampled-data system, 142 

Sampler, 18, 142 

Scattering parameters, 415, 419 

element, 416 

matrix, 420 
Settling time, 388 
Short-circuit admittance parameters, 

257, 344 
Shunt peaked circuit, 248, 400 
Signal, continuous, 23 

decomposition into even and odd 
parts, 21 

deterministic, 20 

periodic, 20, 46 

symmetrical, 21, 54 
Signum (sgn) function, 30 
Sine function, 65 
Singularity, 486 

Spectra, continuous, 3, 63 
line (discrete), 3, 57 

Stability, 116, 290 
marginal, 293 

Steady-state solution, 106, 109 

Steepest descent method, 447 

Step function, see Unit step 

Step response, 44, 45, 85, 111, 194 

Synthesis procedures elementary, 308 
filter synthesis, 397 
L-C functions, 315, 319 
R-C functions, 325, 329, 331 
R-L functions, 325, 329, 331 
R-L-C functions, 333 
transfer functions, 347, 352 

System function, 14, 187, 194 

Terminals, 12 
Thevenin's theorem, 180 
Time constant, 24 
Time delay, 32, 212, 245, 384, 388 
Time delayer, 11 
Time domain, 14, 134, 366 
Time-invariant system, 8 
Transfer function, 14, 16, 188, 266, 
341, 352 

admittance, 188, 347 

impedance, 16, 188, 347 
Transformed circuit, 175 
Transformer. 122. 177 

ideal, 263 

unity coupled, 125 
Transient response of low-pass filters, 

Transient solution, 106 
Transistor, common emitter, 286 
Transmission coefficient, 421, 457 
Transmission line, 413, 425 
Transmission matrix, 262 
Transmittance, 16 
Two-port, 12, 16, 254, 264 

cascade connection, 271 

equivalent circuits of, 268, 271 

matrix representation, 264 

parallel connection of, 273 

series connection of, 275 

Undercoupling, 241 

Subject Index 515 

Undetermined coefficients, method of, Voltage ratio, 16, 188 

g2 Voltage source, 12 

Uniform loading, 278 Voltage standing wave ratio, 428 

Unit coupled transformer, 121 

Unit impulse (delta) functions, 33, 43, y-parameters, 257, 344 

47( j see also Short-circuit parameters 

derivative of, 40 

properties of, 34, 36, 476, 480 z -parameters, 254 343 
Unit ramp function, 31 see also Open-circuit parameters 

Unit step function, 28, 474 z-transform, 142 

derivative of, 34, 37, 475 Zero, definition of , 155 

Zero of transmission, 345 

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