1 00:00:00,000 --> 00:00:13,560 PROFESSOR: So the idea behind the encoder is increase the 2 00:00:13,560 --> 00:00:23,760 minimum distance at the cost of spectral efficiency. 3 00:00:23,760 --> 00:00:32,170 4 00:00:32,170 --> 00:00:38,070 The idea behind the decoder was the following. 5 00:00:38,070 --> 00:00:41,710 You have a received signal, y. 6 00:00:41,710 --> 00:00:49,930 And the job of the decoder is to find x hat, which belongs 7 00:00:49,930 --> 00:00:54,660 to the signal set so that you minimize your 8 00:00:54,660 --> 00:00:55,920 probability of error. 9 00:00:55,920 --> 00:01:01,840 And the probability of error x is not equal to x-hat. 10 00:01:01,840 --> 00:01:05,780 That's the probability of error for your decoder. 11 00:01:05,780 --> 00:01:09,090 And we saw a bunch of equivalence rules -- we said 12 00:01:09,090 --> 00:01:12,080 that the minimum probability of error rule, which is this 13 00:01:12,080 --> 00:01:18,900 rule here, is the same as the maximum a posteriori rule. 14 00:01:18,900 --> 00:01:22,410 And this just follows by the definition. 15 00:01:22,410 --> 00:01:28,160 This was the same as the maximum likelihood rule. 16 00:01:28,160 --> 00:01:30,350 And here we made one assumption that all the 17 00:01:30,350 --> 00:01:35,840 signals in your signal set A are equally likely. 18 00:01:35,840 --> 00:01:39,660 And this was shown to be equivalent to your minimum 19 00:01:39,660 --> 00:01:42,130 distance decision rule. 20 00:01:42,130 --> 00:01:45,580 And in order to show this equivalence, we use the fact 21 00:01:45,580 --> 00:01:48,630 that the noise was White and Gaussian. 22 00:01:48,630 --> 00:01:52,570 So this assumes equi-probable signals. 23 00:01:52,570 --> 00:01:59,290 24 00:01:59,290 --> 00:02:01,265 And this assumes AWGN channel. 25 00:02:01,265 --> 00:02:02,515 OK. 26 00:02:02,515 --> 00:02:08,259 27 00:02:08,259 --> 00:02:11,840 In fact, the minimum distance decisions has very nice 28 00:02:11,840 --> 00:02:13,090 geometrical properties. 29 00:02:13,090 --> 00:02:17,720 30 00:02:17,720 --> 00:02:20,190 We basically said that the decision regions had a 31 00:02:20,190 --> 00:02:21,250 particular shape. 32 00:02:21,250 --> 00:02:23,100 They were convex polytopes. 33 00:02:23,100 --> 00:02:27,720 And they were also known as Voronoi regions. 34 00:02:27,720 --> 00:02:30,370 However, finding the exact expression for the probability 35 00:02:30,370 --> 00:02:33,850 of error seemed quite tedious so we decided 36 00:02:33,850 --> 00:02:35,640 to settle with bounds. 37 00:02:35,640 --> 00:02:38,200 And we developed some bounds for the probability of error. 38 00:02:38,200 --> 00:02:42,530 39 00:02:42,530 --> 00:02:46,140 And the main bounds were -- we had a strict upper bound, 40 00:02:46,140 --> 00:02:46,937 which was the summation. 41 00:02:46,937 --> 00:02:48,187 OK. 42 00:02:48,187 --> 00:03:01,220 43 00:03:01,220 --> 00:03:03,230 This is a strict upper bound. 44 00:03:03,230 --> 00:03:06,890 Well, the inequality here, say it's a strict upper bound. 45 00:03:06,890 --> 00:03:10,450 We also had a union bound estimate. 46 00:03:10,450 --> 00:03:12,910 Because this is a sum of various Q functions, this is 47 00:03:12,910 --> 00:03:16,270 still a very cumbersome expression to evaluate. 48 00:03:16,270 --> 00:03:18,440 So we wanted a simpler expression. 49 00:03:18,440 --> 00:03:28,650 So we then approximated this by K_min of aj times Q of 50 00:03:28,650 --> 00:03:33,410 d_min over 2 sigma. 51 00:03:33,410 --> 00:03:37,750 OK, and now if we take the average on both sides, what we 52 00:03:37,750 --> 00:03:43,650 get is the probability of error is approximately equal 53 00:03:43,650 --> 00:03:47,980 to K_min of the constellation, which is the average number of 54 00:03:47,980 --> 00:03:55,890 nearest neighbors, times Q of d_min over 2 sigma. 55 00:03:55,890 --> 00:03:58,690 56 00:03:58,690 --> 00:04:02,600 And this expression is known as the union bound estimate. 57 00:04:02,600 --> 00:04:10,280 58 00:04:10,280 --> 00:04:13,120 We call it an estimate because it's not a bound anymore. 59 00:04:13,120 --> 00:04:16,370 We have made an approximation going from here to here. 60 00:04:16,370 --> 00:04:18,910 So this is what we came up with last time. 61 00:04:18,910 --> 00:04:23,076 Throughout the analysis today, whenever we have a probability 62 00:04:23,076 --> 00:04:25,970 of error expression, we will be estimating it by the union 63 00:04:25,970 --> 00:04:27,170 bound estimate. 64 00:04:27,170 --> 00:04:31,310 So this union bound estimate is quite important. 65 00:04:31,310 --> 00:04:34,420 Are there any questions on this? 66 00:04:34,420 --> 00:04:40,700 OK, so today we will start by introducing the notion of 67 00:04:40,700 --> 00:04:41,950 effective coding gain. 68 00:04:41,950 --> 00:04:58,240 69 00:04:58,240 --> 00:05:03,330 And we'll use the symbol gamma sub f for that. 70 00:05:03,330 --> 00:05:05,790 What do we mean by effective coding gain? 71 00:05:05,790 --> 00:05:08,630 First you have to decide what region you're operating in. 72 00:05:08,630 --> 00:05:11,300 It's defined for both bandwidth-limited regime and 73 00:05:11,300 --> 00:05:12,800 power-limited regimes. 74 00:05:12,800 --> 00:05:15,240 So let us start with the power-limited regime. 75 00:05:15,240 --> 00:05:24,610 76 00:05:24,610 --> 00:05:27,600 Remember in the power-limited regime, the trade-off we care 77 00:05:27,600 --> 00:05:35,740 about is the probability of bit error versus Eb over N_0. 78 00:05:35,740 --> 00:05:37,110 The baseline scheme is 2-PAM. 79 00:05:37,110 --> 00:05:45,100 80 00:05:45,100 --> 00:05:49,630 And for 2-PAM, we showed two lectures ago that the 81 00:05:49,630 --> 00:05:56,650 probability of bit error is given by Q of root 2 Eb N_0. 82 00:05:56,650 --> 00:06:01,160 83 00:06:01,160 --> 00:06:04,310 OK, it should be -- 84 00:06:04,310 --> 00:06:10,870 so now the idea is to plot your probability of bit error 85 00:06:10,870 --> 00:06:12,120 as a function of Eb N_0. 86 00:06:12,120 --> 00:06:16,150 87 00:06:16,150 --> 00:06:19,760 Eb N_0 is always plotted in dB on the x-axis. 88 00:06:19,760 --> 00:06:23,790 The probability of bit error we also plot on 89 00:06:23,790 --> 00:06:25,750 a logarithmic scale. 90 00:06:25,750 --> 00:06:27,590 So this is ten to the minus six. 91 00:06:27,590 --> 00:06:30,390 This is ten to the minus five, ten to the 92 00:06:30,390 --> 00:06:33,840 minus four and so on. 93 00:06:33,840 --> 00:06:37,420 The Shannon limit is at -- 94 00:06:37,420 --> 00:06:39,412 this is the Shannon limit -- 95 00:06:39,412 --> 00:06:45,830 is at minus 1.59 dB. 96 00:06:45,830 --> 00:06:46,580 OK? 97 00:06:46,580 --> 00:06:52,750 For the power-limited regime, the performance of 2-PAM, 98 00:06:52,750 --> 00:06:57,350 which is given by this expression, is here. 99 00:06:57,350 --> 00:07:01,260 And we basically go to say that when the probability of 100 00:07:01,260 --> 00:07:05,600 bit error is ten to the minus five, the EbN _0 that you 101 00:07:05,600 --> 00:07:11,310 require is 9.6 dB. 102 00:07:11,310 --> 00:07:12,560 This is a nine. 103 00:07:12,560 --> 00:07:15,510 104 00:07:15,510 --> 00:07:19,540 So what's the idea behind the effective coding gain? 105 00:07:19,540 --> 00:07:23,440 Well, suppose I give you a certain constellation, OK, a 106 00:07:23,440 --> 00:07:24,940 certain signal set. 107 00:07:24,940 --> 00:07:28,370 And you find the probability of bit error for that, 108 00:07:28,370 --> 00:07:29,070 and you plot it. 109 00:07:29,070 --> 00:07:31,970 And it comes out to be something like this. 110 00:07:31,970 --> 00:07:37,470 So this is the performance for some given signal set A. This 111 00:07:37,470 --> 00:07:38,720 is your 2-PAM. 112 00:07:38,720 --> 00:07:43,150 113 00:07:43,150 --> 00:07:46,470 And now you want to quantify how much better is the signal 114 00:07:46,470 --> 00:07:49,390 set over the baseline system. 115 00:07:49,390 --> 00:07:51,510 Now, so in other words you want to look at the gap 116 00:07:51,510 --> 00:07:53,490 between the two curves. 117 00:07:53,490 --> 00:07:55,580 Now clearly the gap is going to be a function of the 118 00:07:55,580 --> 00:07:58,100 probability of bit error, right? 119 00:07:58,100 --> 00:08:01,990 But the basic idea now is you want to fix a certain 120 00:08:01,990 --> 00:08:04,530 probability of bit error because if you're designing a 121 00:08:04,530 --> 00:08:07,070 system, the probability of bit error is something that the 122 00:08:07,070 --> 00:08:08,920 system tolerates and is going to be fixed 123 00:08:08,920 --> 00:08:10,480 throughout the analysis. 124 00:08:10,480 --> 00:08:12,320 So in this course, we'll be fixing it at ten 125 00:08:12,320 --> 00:08:13,920 to the minus five. 126 00:08:13,920 --> 00:08:16,510 You fix this probability of bit error to ten to the minus 127 00:08:16,510 --> 00:08:18,610 five, and you look at how much a gap you have 128 00:08:18,610 --> 00:08:21,520 between the two curves. 129 00:08:21,520 --> 00:08:23,630 And this gap is going to be your effective 130 00:08:23,630 --> 00:08:26,520 coding gain in dB. 131 00:08:26,520 --> 00:08:33,049 132 00:08:33,049 --> 00:08:37,010 So in other words, effective coding gain, I can write here, 133 00:08:37,010 --> 00:08:54,340 is the gap between a given signal set and concordant 134 00:08:54,340 --> 00:09:01,415 system, 2-PAM, at a fixed probability of bit error. 135 00:09:01,415 --> 00:09:05,300 136 00:09:05,300 --> 00:09:06,060 OK? 137 00:09:06,060 --> 00:09:09,470 So now let us try to quantify this effective coding gain 138 00:09:09,470 --> 00:09:12,690 using the union bound estimate that we have here. 139 00:09:12,690 --> 00:09:18,120 So if I have a certain signal set, A, then I know from the 140 00:09:18,120 --> 00:09:23,370 union bound estimate that the probability of error is given 141 00:09:23,370 --> 00:09:33,060 by K_min of A times approximately Q of 142 00:09:33,060 --> 00:09:37,160 d_min over 2 sigma. 143 00:09:37,160 --> 00:09:39,200 So since I want that expression in terms of 144 00:09:39,200 --> 00:09:43,360 probability of bit error, I will use the fact that 145 00:09:43,360 --> 00:09:46,620 probability of bit error is probability of error per 146 00:09:46,620 --> 00:09:51,130 symbols or the number of bits per symobl. 147 00:09:51,130 --> 00:09:55,730 Since I have endpoints in my signal set, I have logM bits 148 00:09:55,730 --> 00:09:56,980 per symbol. 149 00:09:56,980 --> 00:09:59,450 150 00:09:59,450 --> 00:10:15,980 And this is then K_min of A over logM base 2 times Q of 151 00:10:15,980 --> 00:10:17,780 d_min over 2 sigma. 152 00:10:17,780 --> 00:10:33,910 153 00:10:33,910 --> 00:10:37,100 So I'm going to rewrite my probability of bit error 154 00:10:37,100 --> 00:10:40,620 expression now in this -- 155 00:10:40,620 --> 00:10:44,080 in fact the right side in this quad. 156 00:10:44,080 --> 00:10:56,390 K_b of A times Q of root 2 gamma_c of A 157 00:10:56,390 --> 00:10:58,870 times Eb over N_0. 158 00:10:58,870 --> 00:11:00,520 So I'm just going to define the right hand 159 00:11:00,520 --> 00:11:01,880 side in this way. 160 00:11:01,880 --> 00:11:05,290 And I'm going to relate gamma_c of A and K_b of A to 161 00:11:05,290 --> 00:11:08,620 the parameters I have on the right hand side there. 162 00:11:08,620 --> 00:11:10,690 And why do I want to do it this way? 163 00:11:10,690 --> 00:11:13,620 Because I want to get an expression as close as 164 00:11:13,620 --> 00:11:17,330 possible to the performance of an uncoded 2-PAM system 165 00:11:17,330 --> 00:11:20,900 because that way things will be easier to do if I want to 166 00:11:20,900 --> 00:11:24,530 calculate the effective coding gain, OK? 167 00:11:24,530 --> 00:11:27,240 So how do the parameters relate? 168 00:11:27,240 --> 00:11:40,810 Well, K_b of A is K_min of A over logM to the base 2. 169 00:11:40,810 --> 00:11:56,430 And this is the average number of nearest neighbors per 170 00:11:56,430 --> 00:12:04,110 information bit. 171 00:12:04,110 --> 00:12:08,060 If I compare the arguments inside the Q function, what I 172 00:12:08,060 --> 00:12:20,570 have is 2 gamma_c of A times Eb N_0 is the argument d_min 173 00:12:20,570 --> 00:12:25,350 over 2 sigma squared, which I will write as d_min squared 174 00:12:25,350 --> 00:12:28,960 over 4 sigma squared. 175 00:12:28,960 --> 00:12:32,950 So remember sigma squared is N_0 over 2. 176 00:12:32,950 --> 00:12:36,397 So if I simplify this expression, what I will end up 177 00:12:36,397 --> 00:12:45,525 getting is gamma_c of A is d_min squared over 4 Eb. 178 00:12:45,525 --> 00:12:48,860 179 00:12:48,860 --> 00:12:51,500 Gamma_c of A is known as the nominal coding gain. 180 00:12:51,500 --> 00:13:03,680 181 00:13:03,680 --> 00:13:06,060 It's not the same as the effective coding gain, which 182 00:13:06,060 --> 00:13:08,880 is the distance between the two curves. 183 00:13:08,880 --> 00:13:13,730 OK, are there any questions so far? 184 00:13:13,730 --> 00:13:18,440 So far what I have done is given a constellation A, I can 185 00:13:18,440 --> 00:13:20,870 find two parameters. 186 00:13:20,870 --> 00:13:25,490 I can find the nominal coding gain, and I can find the 187 00:13:25,490 --> 00:13:30,000 average number of nearest neighbors per information bit. 188 00:13:30,000 --> 00:13:32,630 And using these two parameters, we will try to get 189 00:13:32,630 --> 00:13:35,360 a handle over the effective coding gain. 190 00:13:35,360 --> 00:13:36,990 So that's the idea. 191 00:13:36,990 --> 00:13:38,490 So how can we plot -- 192 00:13:38,490 --> 00:13:41,420 so given this constellation A, we want to plot the 193 00:13:41,420 --> 00:13:44,050 probability of bit error curve like this. 194 00:13:44,050 --> 00:13:46,430 But instead of plotting the exact curve we will be 195 00:13:46,430 --> 00:13:49,600 plotting this curve here, which is the union bound 196 00:13:49,600 --> 00:13:52,270 estimate of your probability of bit error. 197 00:13:52,270 --> 00:13:57,290 So we will always be plotting the union bound estimate. 198 00:13:57,290 --> 00:13:59,200 So how do we do that? 199 00:13:59,200 --> 00:14:02,270 So we'll again start with a curve like that. 200 00:14:02,270 --> 00:14:07,930 Probability of bit error as a function of Eb N_0. 201 00:14:07,930 --> 00:14:11,020 This is in dB. 202 00:14:11,020 --> 00:14:13,980 The y-axis is also in dB. 203 00:14:13,980 --> 00:14:15,870 Sorry, not in dB but on the log scale. 204 00:14:15,870 --> 00:14:22,995 205 00:14:22,995 --> 00:14:24,760 And so on. 206 00:14:24,760 --> 00:14:28,870 This is the Shannon limit at minus 1. 207 00:14:28,870 --> 00:14:30,120 59 dB. 208 00:14:30,120 --> 00:14:34,270 209 00:14:34,270 --> 00:14:36,000 This is the 2-PAM performance. 210 00:14:36,000 --> 00:14:39,420 211 00:14:39,420 --> 00:14:43,770 So if we want to plot this union bound estimate, you will 212 00:14:43,770 --> 00:14:45,250 be doing two steps. 213 00:14:45,250 --> 00:14:48,170 First, we'll be plotting this Q function, a curve 214 00:14:48,170 --> 00:14:49,780 corresponding to the Q function. 215 00:14:49,780 --> 00:14:52,110 And then we'll be scaling it. 216 00:14:52,110 --> 00:14:56,400 So if I just want to plot this Q function, how will it 217 00:14:56,400 --> 00:14:58,070 compare to this curve here? 218 00:14:58,070 --> 00:15:05,620 219 00:15:05,620 --> 00:15:08,660 Well, you're just scaling the argument inside the Q 220 00:15:08,660 --> 00:15:10,130 function, right? 221 00:15:10,130 --> 00:15:13,280 But now because we are plotting the x-axis on a dB 222 00:15:13,280 --> 00:15:17,770 scale, it simply means that we are translating to the left by 223 00:15:17,770 --> 00:15:20,670 an amount of gamma_c in dB. 224 00:15:20,670 --> 00:15:22,920 Is that clear? 225 00:15:22,920 --> 00:15:23,850 Well, let's see. 226 00:15:23,850 --> 00:15:28,500 What we are doing is we are going to scale the x-axis by a 227 00:15:28,500 --> 00:15:29,790 certain constant. 228 00:15:29,790 --> 00:15:33,210 So this means that we will have to scale x-axis here. 229 00:15:33,210 --> 00:15:37,480 But, our Eb N_0 is being plotted on a dB scale, right? 230 00:15:37,480 --> 00:15:41,610 So multiplying in the linear scale corresponds to an 231 00:15:41,610 --> 00:15:43,680 addition on the logarithmic scale. 232 00:15:43,680 --> 00:15:47,240 An addition on the logarithmic scale simply implies a shift 233 00:15:47,240 --> 00:15:52,170 on the x-axis so that's the constant shift on the x-axis. 234 00:15:52,170 --> 00:15:53,800 I'm going to plot it here. 235 00:15:53,800 --> 00:15:58,020 This curve is going to be parallel to the original 2-PAM 236 00:15:58,020 --> 00:16:01,310 curve to the best of my abilities. 237 00:16:01,310 --> 00:16:05,010 OK, so this is what we get as your Q function. 238 00:16:05,010 --> 00:16:08,800 Now we are going to scale the y-axis by this factor here, 239 00:16:08,800 --> 00:16:14,300 K_b of A. But remember the y-axis is also on a 240 00:16:14,300 --> 00:16:16,970 logarithmic scale, rather than the linear scale. 241 00:16:16,970 --> 00:16:21,010 So multiply [INAUDIBLE] by a factor of K_b of A implies a 242 00:16:21,010 --> 00:16:24,590 vertical shift by a certain factor. 243 00:16:24,590 --> 00:16:27,760 So I'm going to do a vertical shift here. 244 00:16:27,760 --> 00:16:34,600 245 00:16:34,600 --> 00:16:38,510 So let me make that darker now. 246 00:16:38,510 --> 00:16:40,060 So this is what I end up getting. 247 00:16:40,060 --> 00:16:43,700 248 00:16:43,700 --> 00:16:46,250 OK, now note that the distance between these two curves is 249 00:16:46,250 --> 00:16:47,960 not fixed anymore. 250 00:16:47,960 --> 00:16:51,460 If the slope is steeper, then this distance is small, 251 00:16:51,460 --> 00:16:53,790 meaning that this distance here is large. 252 00:16:53,790 --> 00:16:57,830 On the other hand, if the slope is going to be smaller 253 00:16:57,830 --> 00:17:01,330 here, then this distance is large. 254 00:17:01,330 --> 00:17:03,770 So this means that a distance between these two curves is no 255 00:17:03,770 --> 00:17:05,099 longer fixed. 256 00:17:05,099 --> 00:17:07,859 And basically what we're saying here is that if the 257 00:17:07,859 --> 00:17:11,109 probability of error is large, then the number of nearest 258 00:17:11,109 --> 00:17:13,069 neighbors matters much more. 259 00:17:13,069 --> 00:17:15,280 But if the probability of error is going to be quite 260 00:17:15,280 --> 00:17:19,400 small, then the more important factor is your exponent, how 261 00:17:19,400 --> 00:17:21,750 fast the slope decays as opposed to the number of 262 00:17:21,750 --> 00:17:26,349 nearest neighbors, OK? 263 00:17:26,349 --> 00:17:31,630 This distance here is your effective coding gain. 264 00:17:31,630 --> 00:17:37,420 And this constant horizontal distance here is your nominal 265 00:17:37,420 --> 00:17:39,364 coding gain in dB. 266 00:17:39,364 --> 00:17:45,570 267 00:17:45,570 --> 00:17:49,280 Are there any questions on this curve? 268 00:17:49,280 --> 00:17:51,842 AUDIENCE: [INAUDIBLE] 269 00:17:51,842 --> 00:17:57,800 on the right of four 2-PAM because these are the logM. 270 00:17:57,800 --> 00:18:01,340 PROFESSOR: So so what I'm really assuming is that 271 00:18:01,340 --> 00:18:03,350 gamma_c of A is greater than 1. 272 00:18:03,350 --> 00:18:05,250 So if I'm adding it, then I'm going to shift 273 00:18:05,250 --> 00:18:06,500 to the left, right? 274 00:18:06,500 --> 00:18:19,600 275 00:18:19,600 --> 00:18:21,035 OK, so a couple of remarks. 276 00:18:21,035 --> 00:18:31,980 277 00:18:31,980 --> 00:18:34,970 I'm just summarizing a few remarks I made while 278 00:18:34,970 --> 00:18:36,430 plotting the curve. 279 00:18:36,430 --> 00:18:53,430 First is the log-log scale is very convenient because any 280 00:18:53,430 --> 00:18:56,280 multiplicative factor simply involves a translation along 281 00:18:56,280 --> 00:19:00,040 the x and y directions. 282 00:19:00,040 --> 00:19:07,510 The second point is that the accuracy we have in the 283 00:19:07,510 --> 00:19:21,330 effective coding gain in this way is determined by the union 284 00:19:21,330 --> 00:19:26,020 bound estimate because that's all we are plotting. 285 00:19:26,020 --> 00:19:28,160 We are not really plotting the exact curve. 286 00:19:28,160 --> 00:19:32,210 287 00:19:32,210 --> 00:19:35,380 And the third point is that we have a rule of thumb. 288 00:19:35,380 --> 00:19:41,770 289 00:19:41,770 --> 00:19:45,190 Because the distance between the curves is not fixed, it's 290 00:19:45,190 --> 00:19:47,720 still too tedious to find the exact numerical value of 291 00:19:47,720 --> 00:19:49,610 effective coding gain. 292 00:19:49,610 --> 00:19:52,750 So what we can find is the value of the nominal coding 293 00:19:52,750 --> 00:19:57,390 gain and K_b of A exactly, given a constellation. 294 00:19:57,390 --> 00:20:00,320 So we want to have a rule of thumb for the 295 00:20:00,320 --> 00:20:02,200 effective coding gain. 296 00:20:02,200 --> 00:20:06,280 And the rule of thumb is that the effective coding gain is 297 00:20:06,280 --> 00:20:16,050 given by the nominal coding gain in dB minus 0.2 log2 of 298 00:20:16,050 --> 00:20:20,620 K_b of A. This is in dB. 299 00:20:20,620 --> 00:20:25,450 What this means is every factor of two increase is K_b 300 00:20:25,450 --> 00:20:29,590 results in a loss of 0.2 dB in the effective coding gain. 301 00:20:29,590 --> 00:20:35,520 OK, and this is our probability of bit error of 302 00:20:35,520 --> 00:20:37,600 ten to the minus five. 303 00:20:37,600 --> 00:20:42,060 So that's the region that we will be doing our analysis in. 304 00:20:42,060 --> 00:20:44,610 So this rule of thumb is quite helpful. 305 00:20:44,610 --> 00:20:46,594 Are there any questions? 306 00:20:46,594 --> 00:20:49,340 AUDIENCE: [INAUDIBLE] 307 00:20:49,340 --> 00:20:50,530 PROFESSOR: This rule of thumb? 308 00:20:50,530 --> 00:20:53,360 Well, it works quite well in practice so you 309 00:20:53,360 --> 00:20:54,860 want to use that rule. 310 00:20:54,860 --> 00:20:57,360 AUDIENCE: [INAUDIBLE] 311 00:20:57,360 --> 00:20:58,610 PROFESSOR: Right. 312 00:20:58,610 --> 00:21:06,360 313 00:21:06,360 --> 00:21:09,150 If things are clear, let us do some examples. 314 00:21:09,150 --> 00:21:17,090 315 00:21:17,090 --> 00:21:19,515 So let us start with a 2-PAM for examples. 316 00:21:19,515 --> 00:21:24,810 317 00:21:24,810 --> 00:21:26,480 The first is 2-PAM system. 318 00:21:26,480 --> 00:21:29,760 319 00:21:29,760 --> 00:21:32,950 What's the effective coding gain going to be for this 320 00:21:32,950 --> 00:21:35,790 constellation? 321 00:21:35,790 --> 00:21:38,250 It's going to be 0 dB, right? 322 00:21:38,250 --> 00:21:39,500 It just follows from the definition. 323 00:21:39,500 --> 00:21:42,718 324 00:21:42,718 --> 00:21:46,790 I mean the effective coding gain tells us how much gap we 325 00:21:46,790 --> 00:21:50,000 have between this new constellation and 2-PAM. 326 00:21:50,000 --> 00:21:52,990 If your new constellation is 2-PAM itself, we don't have 327 00:21:52,990 --> 00:21:56,100 any effective coding gain. 328 00:21:56,100 --> 00:21:57,350 What about the 4-QAM? 329 00:21:57,350 --> 00:22:01,120 330 00:22:01,120 --> 00:22:04,230 It's going to be still 0, because a 4-QAM is a Cartesian 331 00:22:04,230 --> 00:22:06,820 product of two 2-PAM constellations. 332 00:22:06,820 --> 00:22:09,290 And we argued last time that there is no coding gain if you 333 00:22:09,290 --> 00:22:11,970 use a Cartesian product. 334 00:22:11,970 --> 00:22:16,290 If you want to verify it using the framework that we have 335 00:22:16,290 --> 00:22:21,420 just developed, say we have four points. 336 00:22:21,420 --> 00:22:24,070 This point is alpha, alpha. 337 00:22:24,070 --> 00:22:30,580 This point is minus alpha, alpha, minus alpha, minus 338 00:22:30,580 --> 00:22:33,600 alpha, and alpha minus alpha. 339 00:22:33,600 --> 00:22:39,010 The nominal coding gain is given by d_min 340 00:22:39,010 --> 00:22:40,580 squared over 4 Eb. 341 00:22:40,580 --> 00:22:45,080 342 00:22:45,080 --> 00:22:49,820 The minimum distance is 3 alpha between any two points. 343 00:22:49,820 --> 00:22:52,250 So we have 4 alpha squared. 344 00:22:52,250 --> 00:22:54,040 Then what's the energy per bit? 345 00:22:54,040 --> 00:22:57,210 Well, the energy per symbol is 2 alpha squared. 346 00:22:57,210 --> 00:23:00,240 Then we have two bits per symbol, so the energy per bit 347 00:23:00,240 --> 00:23:01,520 is alpha squared. 348 00:23:01,520 --> 00:23:05,080 So we have 4 alpha squared, and so that's 1. 349 00:23:05,080 --> 00:23:08,060 So we do not have any nominal coding gain. 350 00:23:08,060 --> 00:23:10,600 K_b of A, what's that going to be? 351 00:23:10,600 --> 00:23:13,870 Well, we have two nearest neighbors per symbol, but we 352 00:23:13,870 --> 00:23:15,830 also have two bits per symbol. 353 00:23:15,830 --> 00:23:17,910 So the number of nearest neighbors per bit 354 00:23:17,910 --> 00:23:18,940 is going to be one. 355 00:23:18,940 --> 00:23:23,960 So your nominal coding gain is 1, K_b of A is going to be 1. 356 00:23:23,960 --> 00:23:28,220 And so we do not see any coding gain 357 00:23:28,220 --> 00:23:30,880 for the 4-QAM system. 358 00:23:30,880 --> 00:23:33,660 Now let me do a slightly different case. 359 00:23:33,660 --> 00:23:37,770 Let me start with a 4-QAM, but I remove two points. 360 00:23:37,770 --> 00:23:40,860 I remove this point, and I remove this point. 361 00:23:40,860 --> 00:23:43,610 And I only keep the two points here. 362 00:23:43,610 --> 00:23:48,710 So I have a constellation, say, A prime, which only has 363 00:23:48,710 --> 00:23:52,490 two points, one point here and one point here. 364 00:23:52,490 --> 00:23:55,750 So this point is alpha, alpha, and this point is minus alpha, 365 00:23:55,750 --> 00:23:58,180 minus alpha. 366 00:23:58,180 --> 00:24:00,480 Any idea what the effective coding gain 367 00:24:00,480 --> 00:24:01,730 will be for this case? 368 00:24:01,730 --> 00:24:06,000 369 00:24:06,000 --> 00:24:07,920 AUDIENCE: I think it's the same. 370 00:24:07,920 --> 00:24:08,900 PROFESSOR: It's still zero, right? 371 00:24:08,900 --> 00:24:12,130 AUDIENCE: Yeah, because it's saying that's 2-PAM. 372 00:24:12,130 --> 00:24:12,820 PROFESSOR: Exactly. 373 00:24:12,820 --> 00:24:17,600 All this is a 2-PAM constellation embedded in two 374 00:24:17,600 --> 00:24:18,910 dimensions. 375 00:24:18,910 --> 00:24:22,130 It's 2-PAM along this line, right? 376 00:24:22,130 --> 00:24:24,780 So I cannot hope to have a higher coding gain for this 377 00:24:24,780 --> 00:24:27,620 constellation over the original one. 378 00:24:27,620 --> 00:24:30,340 I mean I told you the story that we can take a subset of 379 00:24:30,340 --> 00:24:33,860 points, and we can increase the minimum distance by 380 00:24:33,860 --> 00:24:35,360 throwing away some points. 381 00:24:35,360 --> 00:24:38,700 But one has to be careful in that analysis. 382 00:24:38,700 --> 00:24:41,280 There could be examples where you're strictly improving the 383 00:24:41,280 --> 00:24:42,760 minimum distance. 384 00:24:42,760 --> 00:24:49,460 Note that the minimum distance here is going to 8 alpha 385 00:24:49,460 --> 00:24:52,230 squared now. 386 00:24:52,230 --> 00:24:54,970 OK, so the minimum distance is strictly improved. 387 00:24:54,970 --> 00:24:58,530 But what happens to be your energy per bit? 388 00:24:58,530 --> 00:25:01,780 Well, the energy per symbol is 2 alpha squared. 389 00:25:01,780 --> 00:25:05,650 But you only have one bit per symbol, so the energy per bit 390 00:25:05,650 --> 00:25:07,760 is 2 alpha squared. 391 00:25:07,760 --> 00:25:13,710 So your nominal coding gain is d_min squared over 4 Eb, which 392 00:25:13,710 --> 00:25:17,130 follows from the relation we have here. 393 00:25:17,130 --> 00:25:19,310 And d_min squared is 8 alpha squared. 394 00:25:19,310 --> 00:25:23,040 4 Eb is also 8 alpha squared, so the nominal coding gain is 395 00:25:23,040 --> 00:25:24,410 1, or 0 dB. 396 00:25:24,410 --> 00:25:27,890 397 00:25:27,890 --> 00:25:30,810 The average number of nearest neighbors, well, we only have 398 00:25:30,810 --> 00:25:34,120 one nearest neighbor, and we have one bit per symbol, so 399 00:25:34,120 --> 00:25:37,950 that's -- and this is A prime by the way -- 400 00:25:37,950 --> 00:25:39,270 is going to be 1. 401 00:25:39,270 --> 00:25:42,627 And so the effective coding gain is still 0 dB. 402 00:25:42,627 --> 00:25:46,920 403 00:25:46,920 --> 00:25:48,170 AUDIENCE: [INAUDIBLE] 404 00:25:48,170 --> 00:25:51,220 405 00:25:51,220 --> 00:25:54,170 PROFESSOR: So probability of symbol error as a function of 406 00:25:54,170 --> 00:25:56,250 alpha has definitely decreased, right. 407 00:25:56,250 --> 00:25:58,530 But the trade-off we care about is probability of bit 408 00:25:58,530 --> 00:26:00,740 error as a function of Eb N_0. 409 00:26:00,740 --> 00:26:02,620 In other words, this curve is still here. 410 00:26:02,620 --> 00:26:04,370 It's not changed. 411 00:26:04,370 --> 00:26:07,450 I just shifted right by reducing my spectral 412 00:26:07,450 --> 00:26:07,622 efficiency. 413 00:26:07,622 --> 00:26:08,872 OK? 414 00:26:08,872 --> 00:26:15,560 415 00:26:15,560 --> 00:26:19,810 So let's do one last example just to prove that it's not 416 00:26:19,810 --> 00:26:23,290 always the case that you get nothing by removing points. 417 00:26:23,290 --> 00:27:35,990 418 00:27:35,990 --> 00:27:42,770 OK, so the contellation that I have in mind is the one I 419 00:27:42,770 --> 00:27:45,420 introduced last time. 420 00:27:45,420 --> 00:27:50,170 We'll start with a 3, 4 Cartesian product of 2-PAM. 421 00:27:50,170 --> 00:27:54,980 OK, so in three dimensions it's a cube, and the 3,4 422 00:27:54,980 --> 00:27:57,840 Cartesian product will have a point on each 423 00:27:57,840 --> 00:27:59,260 vertex of the cube. 424 00:27:59,260 --> 00:28:01,150 And we only keep four points. 425 00:28:01,150 --> 00:28:03,377 So the points we keep are these four points. 426 00:28:03,377 --> 00:28:04,627 OK? 427 00:28:04,627 --> 00:28:09,800 428 00:28:09,800 --> 00:28:19,300 So I write B, the coordinates are alpha, alpha, alpha, minus 429 00:28:19,300 --> 00:28:24,540 alpha, minus alpha, alpha, minus alpha, alpha, minus 430 00:28:24,540 --> 00:28:31,770 alpha, and alpha, minus alpha, minus alpha. 431 00:28:31,770 --> 00:28:33,650 So these are the vertices. 432 00:28:33,650 --> 00:28:37,575 These are the coordinates of the four points here where we 433 00:28:37,575 --> 00:28:40,190 have the standard x, y and z-axis, which I'm not drawing 434 00:28:40,190 --> 00:28:42,360 because it will just look confusing. 435 00:28:42,360 --> 00:28:46,320 So now we'll see that for this particular signal set, we do 436 00:28:46,320 --> 00:28:49,170 have a coding gain, OK? 437 00:28:49,170 --> 00:28:50,595 Let's find the minimum distance. 438 00:28:50,595 --> 00:28:54,160 439 00:28:54,160 --> 00:28:56,740 The minimum distance here is what? 440 00:28:56,740 --> 00:29:00,010 It's the distance along a diagonal. 441 00:29:00,010 --> 00:29:04,390 And the diagonal is of length to -- each side of the cube is 442 00:29:04,390 --> 00:29:05,640 off-length to alpha. 443 00:29:05,640 --> 00:29:08,790 444 00:29:08,790 --> 00:29:13,930 So the diagonal is of length to alpha. 445 00:29:13,930 --> 00:29:16,275 So d_min squared is 8 alpha squared. 446 00:29:16,275 --> 00:29:19,450 447 00:29:19,450 --> 00:29:21,950 What is the energy per bit going to be in this case? 448 00:29:21,950 --> 00:29:33,030 449 00:29:33,030 --> 00:29:35,226 Well, what's the energy per symbol now? 450 00:29:35,226 --> 00:29:37,520 AUDIENCE: [INAUDIBLE] 451 00:29:37,520 --> 00:29:40,140 PROFESSOR: 3 alpha squared over 2 is the Eb, right? 452 00:29:40,140 --> 00:29:43,890 We have 3 alpha squared units of energy per symbol. 453 00:29:43,890 --> 00:29:48,230 We have four points, so we have two bits per symbol. 454 00:29:48,230 --> 00:29:54,900 So the energy per bit is 3 alpha squared over 2. 455 00:29:54,900 --> 00:30:01,470 So now if I look at my nominal coding gain, that's going to 456 00:30:01,470 --> 00:30:07,210 be d_min squared over 4 Eb. 457 00:30:07,210 --> 00:30:10,640 d_min squared is 8 alpha squared. 458 00:30:10,640 --> 00:30:14,170 4 dB is 6 alpha squared. 459 00:30:14,170 --> 00:30:17,080 So that's 4/3. 460 00:30:17,080 --> 00:30:21,780 On a linear scale and on a dB scale, this is going to be? 461 00:30:21,780 --> 00:30:32,680 462 00:30:32,680 --> 00:30:34,666 You guys should remember your dB tables. 463 00:30:34,666 --> 00:30:37,534 464 00:30:37,534 --> 00:30:39,446 AUDIENCE: 0.2? 465 00:30:39,446 --> 00:30:40,410 PROFESSOR: 12. 466 00:30:40,410 --> 00:30:42,510 You have to multiply by ten. 467 00:30:42,510 --> 00:30:46,150 So actually I think it's 1.3, more like 1.3 dB if you look 468 00:30:46,150 --> 00:30:47,880 at the last significant digit. 469 00:30:47,880 --> 00:30:50,730 470 00:30:50,730 --> 00:30:56,760 OK, so what's K_b of A going to be in this case? 471 00:30:56,760 --> 00:30:59,055 Or K_b of B I should say. 472 00:30:59,055 --> 00:31:01,650 I'm calling the signal set B here. 473 00:31:01,650 --> 00:31:06,140 So what is K_b if B going to be? 474 00:31:06,140 --> 00:31:13,820 475 00:31:13,820 --> 00:31:15,260 AUDIENCE: [INAUDIBLE] 476 00:31:15,260 --> 00:31:17,180 PROFESSOR: 3 by 2, right? 477 00:31:17,180 --> 00:31:19,760 You have three nearest neighbors per point, for each 478 00:31:19,760 --> 00:31:21,150 point has three nearest neighbors. 479 00:31:21,150 --> 00:31:22,840 They're all equidistant. 480 00:31:22,840 --> 00:31:23,906 And you have two bits per symbol. 481 00:31:23,906 --> 00:31:25,156 OK? 482 00:31:25,156 --> 00:31:26,680 483 00:31:26,680 --> 00:31:29,650 So if you work out your effective coding again using 484 00:31:29,650 --> 00:31:33,745 this rule of thumb here, take gamma_c of A in dB and 485 00:31:33,745 --> 00:31:37,460 subtract 0.2 log2 you will see that the effective coding gain 486 00:31:37,460 --> 00:31:39,220 is approximately 1 dB. 487 00:31:39,220 --> 00:31:55,730 488 00:31:55,730 --> 00:31:56,980 Are there any questions? 489 00:31:56,980 --> 00:32:00,970 490 00:32:00,970 --> 00:32:02,800 OK, so it might seem that this is still not a 491 00:32:02,800 --> 00:32:03,760 very impressive number. 492 00:32:03,760 --> 00:32:05,100 Yes, you had a question. 493 00:32:05,100 --> 00:32:07,810 AUDIENCE: What was it about this constellation that gave 494 00:32:07,810 --> 00:32:09,082 us coding gain? 495 00:32:09,082 --> 00:32:12,390 Was is the face that we went to higher dimensions? 496 00:32:12,390 --> 00:32:15,240 PROFESSOR: Right, so usually you do get a coding gain if 497 00:32:15,240 --> 00:32:20,400 you trade-off minimum distance with spectral efficiency. 498 00:32:20,400 --> 00:32:22,820 So that's usually the case. 499 00:32:22,820 --> 00:32:26,200 So that's why I presented you an example of that case. 500 00:32:26,200 --> 00:32:28,180 It's just that this was a very special case where you end up 501 00:32:28,180 --> 00:32:30,766 with a 2-PAM constellation to start with. 502 00:32:30,766 --> 00:32:36,010 503 00:32:36,010 --> 00:32:42,440 OK, so the comment I was going to make is that 1.2 dB still 504 00:32:42,440 --> 00:32:44,510 might not be like a very impressive number. 505 00:32:44,510 --> 00:32:47,230 I mean after all the hard work, you just get 1 dB 506 00:32:47,230 --> 00:32:48,660 effective coding gain. 507 00:32:48,660 --> 00:32:51,290 So the question to ask at this point is are there any 508 00:32:51,290 --> 00:32:54,980 constellations that have much higher coding gains? 509 00:32:54,980 --> 00:32:58,440 In particular, other constellations that come close 510 00:32:58,440 --> 00:33:01,340 to the Shannon limit at all. 511 00:33:01,340 --> 00:33:04,280 And we'll look at one interesting class of signal 512 00:33:04,280 --> 00:33:06,640 sets, which are known as orthogonal signal sets. 513 00:33:06,640 --> 00:33:18,540 514 00:33:18,540 --> 00:33:22,140 And you probably saw these examples back in 6.450, but 515 00:33:22,140 --> 00:33:24,070 we're just reviewing it again here. 516 00:33:24,070 --> 00:33:27,620 And these signal sets have a property that as you increase 517 00:33:27,620 --> 00:33:30,960 the number of dimensions, you come arbitrarily close to the 518 00:33:30,960 --> 00:33:32,770 Shannon limit. 519 00:33:32,770 --> 00:33:37,110 So the definition of the signal sets is your A -- 520 00:33:37,110 --> 00:33:43,450 it's a collection of signals aj, where j goes from 1 to M 521 00:33:43,450 --> 00:33:52,450 such that the inner product between ai and aj is E A times 522 00:33:52,450 --> 00:33:56,030 delta_i,j, meaning that if i is not equal to j, the two 523 00:33:56,030 --> 00:33:56,920 signals are orthogonal. 524 00:33:56,920 --> 00:34:01,560 And if i equals j, the energy is E(A). 525 00:34:01,560 --> 00:34:05,420 Geometrically, we have two points. 526 00:34:05,420 --> 00:34:07,220 The two points can be thought of as two 527 00:34:07,220 --> 00:34:09,870 points on the two axes. 528 00:34:09,870 --> 00:34:12,650 So this is a case when M equals 2. 529 00:34:12,650 --> 00:34:16,139 When M equals 3, the three points will lie on those three 530 00:34:16,139 --> 00:34:17,389 corresponding axes. 531 00:34:17,389 --> 00:34:19,650 532 00:34:19,650 --> 00:34:22,870 So this is the case when M equals 3 and so on. 533 00:34:22,870 --> 00:34:25,120 So generally when you have M points, your signal 534 00:34:25,120 --> 00:34:28,020 spatializes in M dimensions, and each point can be thought 535 00:34:28,020 --> 00:34:31,860 of as a point on one of the axes. 536 00:34:31,860 --> 00:34:36,760 So if you look at the distance between any two points, it's 537 00:34:36,760 --> 00:34:41,850 the norm of ai minus aj squared. 538 00:34:41,850 --> 00:34:46,210 I could simplify this simply as the norm of ai squared plus 539 00:34:46,210 --> 00:34:50,050 the norm of aj squared because the end product 540 00:34:50,050 --> 00:34:52,630 is going to be zero. 541 00:34:52,630 --> 00:34:55,760 OK, now ai squared is going to be E(A). 542 00:34:55,760 --> 00:34:57,810 aj squared is going to be E(A). 543 00:34:57,810 --> 00:35:01,280 So this is 2E(A). 544 00:35:01,280 --> 00:35:03,440 So what we're saying here is that every point is 545 00:35:03,440 --> 00:35:06,880 equidistant from every other point, and the square of the 546 00:35:06,880 --> 00:35:09,940 distance is 2 E(A). 547 00:35:09,940 --> 00:35:11,190 OK? 548 00:35:11,190 --> 00:35:13,440 549 00:35:13,440 --> 00:35:18,230 So in other words, the average number of nearest neighbors is 550 00:35:18,230 --> 00:35:20,160 always going to be a M minus 1. 551 00:35:20,160 --> 00:35:24,310 552 00:35:24,310 --> 00:35:27,700 So let us calculate the nominal coding gain for this 553 00:35:27,700 --> 00:35:28,950 constellation. 554 00:35:28,950 --> 00:35:32,190 555 00:35:32,190 --> 00:35:38,460 That's going to be d_min squared over 4 Eb. 556 00:35:38,460 --> 00:35:41,590 557 00:35:41,590 --> 00:35:49,160 d_min squared is 2E(A) over 4 Eb. 558 00:35:49,160 --> 00:35:53,090 Well, the energy per bit is the energy per symbol, which 559 00:35:53,090 --> 00:35:56,585 is E(A) over the number of bits per symbol, which is log 560 00:35:56,585 --> 00:35:59,710 M to the base 2. 561 00:35:59,710 --> 00:36:09,520 And this I'm going to write here is 1/2 log 562 00:36:09,520 --> 00:36:11,890 M to the base 2. 563 00:36:11,890 --> 00:36:15,622 So as I increase my M, my number of dimensions, my 564 00:36:15,622 --> 00:36:17,155 nominal coding gain goes to infinity. 565 00:36:17,155 --> 00:36:18,405 OK? 566 00:36:18,405 --> 00:36:21,240 567 00:36:21,240 --> 00:36:23,310 What do we expect for the effective coding gain? 568 00:36:23,310 --> 00:36:24,560 Will it also go to infinity? 569 00:36:24,560 --> 00:36:30,974 570 00:36:30,974 --> 00:36:32,408 AUDIENCE: It approaches the Shannon limit. 571 00:36:32,408 --> 00:36:34,300 PROFESSOR: Well, we don't know whether it approaches the 572 00:36:34,300 --> 00:36:36,115 Shannon limit as of yet, but we know it 573 00:36:36,115 --> 00:36:37,250 cannot go to infinity. 574 00:36:37,250 --> 00:36:38,546 It's upper bounded by the Shannon limit. 575 00:36:38,546 --> 00:36:40,220 Right? 576 00:36:40,220 --> 00:36:44,090 So the effective coding gain cannot really become 577 00:36:44,090 --> 00:36:48,220 unbounded, and what's really happening here? 578 00:36:48,220 --> 00:36:49,470 AUDIENCE: [INAUDIBLE] 579 00:36:49,470 --> 00:36:51,890 580 00:36:51,890 --> 00:36:54,280 PROFESSOR: Right, but why does the effective coding gain stay 581 00:36:54,280 --> 00:36:56,610 bounded and the nominal coding gain blow up? 582 00:36:56,610 --> 00:36:58,680 AUDIENCE: [INAUDIBLE] 583 00:36:58,680 --> 00:37:01,010 PROFESSOR: Right, the number of nearest neighbors also 584 00:37:01,010 --> 00:37:02,440 increases with M, right? 585 00:37:02,440 --> 00:37:04,359 Not just the nominal coding gain. 586 00:37:04,359 --> 00:38:32,330 587 00:38:32,330 --> 00:38:34,570 So let's write that down. 588 00:38:34,570 --> 00:38:37,460 If I look at the number of nearest neighbors per 589 00:38:37,460 --> 00:38:40,690 information bit -- 590 00:38:40,690 --> 00:38:41,510 yes? 591 00:38:41,510 --> 00:38:44,480 AUDIENCE: These don't have 0 mean, right? 592 00:38:44,480 --> 00:38:45,790 PROFESSOR: Right, they don't have 0 mean. 593 00:38:45,790 --> 00:38:47,040 That's a good point. 594 00:38:47,040 --> 00:38:51,420 595 00:38:51,420 --> 00:38:56,170 PROFESSOR: Let's see, what happens to the mean as M goes 596 00:38:56,170 --> 00:38:57,420 to infinity? 597 00:38:57,420 --> 00:39:00,140 598 00:39:00,140 --> 00:39:02,470 If we look at what happens to the mean, it will be 1 over M. 599 00:39:02,470 --> 00:39:05,410 You find the mean of this guy, it's 1 over 2 times the 600 00:39:05,410 --> 00:39:09,300 distance here times distance here is 1 over 2 E(A), E(A). 601 00:39:09,300 --> 00:39:10,260 OK? 602 00:39:10,260 --> 00:39:12,340 If you look at three dimensions, it's 1 over 3 603 00:39:12,340 --> 00:39:14,050 E(A), E(A), E(A). 604 00:39:14,050 --> 00:39:16,560 So in M dimensions, it's going to be 1 over M times all these 605 00:39:16,560 --> 00:39:17,180 coordinates. 606 00:39:17,180 --> 00:39:19,930 If we look at the mean squared, it goes to 0 as M 607 00:39:19,930 --> 00:39:21,042 goes to infinity. 608 00:39:21,042 --> 00:39:24,390 So that's a good point. 609 00:39:24,390 --> 00:39:26,276 So the mean does go to 0, and so we 610 00:39:26,276 --> 00:39:27,910 approach the Shannon limit. 611 00:39:27,910 --> 00:39:30,410 612 00:39:30,410 --> 00:39:38,260 So we have K_b of A over logM What's the 613 00:39:38,260 --> 00:39:39,610 number of nearest neighbors? 614 00:39:39,610 --> 00:39:47,990 We saw it was M minus 1 over log M. And that goes to 615 00:39:47,990 --> 00:39:51,990 infinity as M goes to infinity. 616 00:39:51,990 --> 00:39:55,980 OK, so the average number of nearest neighbors per 617 00:39:55,980 --> 00:40:00,590 information bit also go to infinity 618 00:40:00,590 --> 00:40:02,930 as M goes to infinity. 619 00:40:02,930 --> 00:40:07,060 And if we look at the mean -- just to write down the comment 620 00:40:07,060 --> 00:40:09,100 that was made. 621 00:40:09,100 --> 00:40:13,120 If we look at the mean of this constellation, that's going to 622 00:40:13,120 --> 00:40:21,960 be 1 over the number of points times root E(A) root 623 00:40:21,960 --> 00:40:26,590 E(A) and root E(A). 624 00:40:26,590 --> 00:40:28,170 So there are M coordinates. 625 00:40:28,170 --> 00:40:30,760 So that's going to be the mean if you believe this is the 626 00:40:30,760 --> 00:40:33,300 coordinate axis here. 627 00:40:33,300 --> 00:40:39,550 And if you we look at the norm of M(A) squared, then it's 628 00:40:39,550 --> 00:40:46,310 going to be E(A) over M. And as M goes to infinity, 629 00:40:46,310 --> 00:40:47,560 this goes to 0. 630 00:40:47,560 --> 00:40:54,508 631 00:40:54,508 --> 00:40:57,590 And because the mean goes to 0, it's not unreasonable to 632 00:40:57,590 --> 00:41:00,825 expect that the performance does approach to the ultimate 633 00:41:00,825 --> 00:41:03,200 Shanon limit. 634 00:41:03,200 --> 00:41:09,060 OK, so so far we have looked at K_b of A and gamma_c of A. 635 00:41:09,060 --> 00:41:10,620 If you look at the effective coding 636 00:41:10,620 --> 00:41:13,640 gain, it is still bounded. 637 00:41:13,640 --> 00:41:16,930 638 00:41:16,930 --> 00:41:18,360 It's always bounded. 639 00:41:18,360 --> 00:41:22,280 It is not clear at this point what really happens to it. 640 00:41:22,280 --> 00:41:25,750 And in fact, it's not quite straightforward to actually 641 00:41:25,750 --> 00:41:28,440 show what happens to the effective coding gain. 642 00:41:28,440 --> 00:41:32,260 If you have taken 6.450, then Professor Bob Gallager really 643 00:41:32,260 --> 00:41:35,470 goes into the details of proving how the effective 644 00:41:35,470 --> 00:41:38,490 coding gain for this signal set indeed does approach the 645 00:41:38,490 --> 00:41:40,560 Shannon limit. 646 00:41:40,560 --> 00:41:50,690 But it can be shown that the effective coding gain does 647 00:41:50,690 --> 00:41:56,790 approach 11.2 dB as M goes to infinity. 648 00:41:56,790 --> 00:41:59,560 The main trick here is that union bound is usually quite 649 00:41:59,560 --> 00:42:01,770 weak if the probability of error is small. 650 00:42:01,770 --> 00:42:04,450 So we replace the probability of error just by 1, instead of 651 00:42:04,450 --> 00:42:06,240 the union bound, at some point. 652 00:42:06,240 --> 00:42:10,300 And if you do that change, then things work out nicely. 653 00:42:10,300 --> 00:42:13,520 But in this course, we will just assert that the 654 00:42:13,520 --> 00:42:16,040 orthogonal signal sets do achieve the 655 00:42:16,040 --> 00:42:17,580 ultimate Shannon limit. 656 00:42:17,580 --> 00:42:22,180 And you can see 6.450 notes for the proof. 657 00:42:22,180 --> 00:42:30,560 658 00:42:30,560 --> 00:42:33,570 So that's the assertion that we are making right now. 659 00:42:33,570 --> 00:42:36,910 So let's step back and take a look at what's 660 00:42:36,910 --> 00:42:38,740 happening so far. 661 00:42:38,740 --> 00:42:41,990 We have a class of signal sets which are conceptually quite 662 00:42:41,990 --> 00:42:43,290 easy to describe. 663 00:42:43,290 --> 00:42:45,430 They are just orthogonal signal sets. 664 00:42:45,430 --> 00:42:47,870 They are not too hard to analyze, and they 665 00:42:47,870 --> 00:42:51,460 asymptotically approach the ultimate Shannon limit. 666 00:42:51,460 --> 00:42:52,730 So this seems quite nice. 667 00:42:52,730 --> 00:42:56,010 So why not we just implement them? 668 00:42:56,010 --> 00:42:59,180 Now it turns that these are not very common in practice. 669 00:42:59,180 --> 00:43:01,850 They are not implemented as often as you might think when 670 00:43:01,850 --> 00:43:03,740 you first look at them and because 671 00:43:03,740 --> 00:43:05,840 they have some drawbacks. 672 00:43:05,840 --> 00:43:09,610 And there are a couple of drawbacks which make them very 673 00:43:09,610 --> 00:43:12,530 less appealing than what you might otherwise think. 674 00:43:12,530 --> 00:43:15,450 675 00:43:15,450 --> 00:43:18,820 And first of all, what's the spectral efficiency for these 676 00:43:18,820 --> 00:43:20,070 signal sets? 677 00:43:20,070 --> 00:43:23,316 678 00:43:23,316 --> 00:43:24,230 AUDIENCE:0, right? 679 00:43:24,230 --> 00:43:25,150 PROFESSOR: It goes to 0. 680 00:43:25,150 --> 00:43:27,050 And how does it go to 0? 681 00:43:27,050 --> 00:43:33,370 It's 2log M to the base 2 over M. As M goes to infinity, 682 00:43:33,370 --> 00:43:34,960 this goes to 0. 683 00:43:34,960 --> 00:43:39,150 And in fact, it goes to 0 quite sharply. 684 00:43:39,150 --> 00:43:41,740 Now that's fine, but if you want to approach Shannon 685 00:43:41,740 --> 00:43:45,300 limit, our spectral efficiency should indeed go to 0. 686 00:43:45,300 --> 00:43:48,520 But if you're looking at the system design problem, what do 687 00:43:48,520 --> 00:43:50,050 you really want? 688 00:43:50,050 --> 00:43:52,330 You have a certain effective coding gain that you have in 689 00:43:52,330 --> 00:43:55,030 mind, and suppose you are presented with a list of 690 00:43:55,030 --> 00:43:57,960 several different signal sets that achieve this effective 691 00:43:57,960 --> 00:43:59,290 coding gain. 692 00:43:59,290 --> 00:44:02,700 And your job is to pick one of these possible signal sets. 693 00:44:02,700 --> 00:44:04,320 Which one will you pick? 694 00:44:04,320 --> 00:44:06,430 Well, there are two things you will look for, right? 695 00:44:06,430 --> 00:44:08,220 You will look for the spectral efficiency that these 696 00:44:08,220 --> 00:44:12,020 different signal sets require or achieve in order to get 697 00:44:12,020 --> 00:44:13,800 this effective coding gain. 698 00:44:13,800 --> 00:44:16,650 And the higher the spectral efficiency the better. 699 00:44:16,650 --> 00:44:18,660 And secondly, you will also want to look at the 700 00:44:18,660 --> 00:44:20,600 implementation complexity. 701 00:44:20,600 --> 00:44:22,760 You do not want something which really takes a lot of 702 00:44:22,760 --> 00:44:25,110 time to decode. 703 00:44:25,110 --> 00:44:25,195 OK? 704 00:44:25,195 --> 00:44:27,770 So it turns out that because the spectral efficiency goes 705 00:44:27,770 --> 00:44:31,180 to 0 so sharply, it's not very surprising that there are 706 00:44:31,180 --> 00:44:34,380 other codes that have the same effective coding gain but 707 00:44:34,380 --> 00:44:36,480 higher spectral efficiency. 708 00:44:36,480 --> 00:44:55,460 So there are binary codes that have a higher spectral 709 00:44:55,460 --> 00:45:03,050 efficiency for same effective coding gain. 710 00:45:03,050 --> 00:45:05,860 And so they would be a natural choice over the orthogonal 711 00:45:05,860 --> 00:45:06,960 signal set. 712 00:45:06,960 --> 00:45:08,130 Yes? 713 00:45:08,130 --> 00:45:09,740 AUDIENCE: [INAUDIBLE] high-spectrum efficiency to 714 00:45:09,740 --> 00:45:11,350 mean bandwidth? 715 00:45:11,350 --> 00:45:12,990 PROFESSOR: Right, because in practice, bandwidth is never 716 00:45:12,990 --> 00:45:14,320 really infinite, right? 717 00:45:14,320 --> 00:45:16,850 So if you are achieving the same -- 718 00:45:16,850 --> 00:45:18,440 the first objective is to have a certain 719 00:45:18,440 --> 00:45:19,530 effective coding gain. 720 00:45:19,530 --> 00:45:21,680 If you do have that, you want to pick something which has a 721 00:45:21,680 --> 00:45:24,090 higher spectral efficiency. 722 00:45:24,090 --> 00:45:28,120 And indeed, the subject of chapters six and eight is to 723 00:45:28,120 --> 00:45:30,850 come up with binary codes that have a much better performance 724 00:45:30,850 --> 00:45:32,970 than the orthogonal signal set for a given 725 00:45:32,970 --> 00:45:34,430 effective coding gain. 726 00:45:34,430 --> 00:45:37,240 Chapter six deals with the the block codes, whereas chapter 727 00:45:37,240 --> 00:45:39,880 eight deals with convolutional codes. 728 00:45:39,880 --> 00:45:43,220 And chapter seven basically develops all the finite field 729 00:45:43,220 --> 00:45:46,410 theory that you require to study convolutional codes. 730 00:45:46,410 --> 00:45:49,430 So the point is, in the subsequent lectures we will be 731 00:45:49,430 --> 00:45:52,510 focusing a lot on improving the performance over 732 00:45:52,510 --> 00:45:54,940 orthogonal signal sets. 733 00:45:54,940 --> 00:45:57,775 The second point I mentioned was implementation complexity. 734 00:45:57,775 --> 00:46:07,600 735 00:46:07,600 --> 00:46:10,670 Now one particular way of implementing orthogonal codes, 736 00:46:10,670 --> 00:46:15,100 which you saw in the first homework, was using this idea 737 00:46:15,100 --> 00:46:17,100 of Hadamard transforms, right? 738 00:46:17,100 --> 00:46:19,940 You have a Hardamard matrix, which is an M by M matrix. 739 00:46:19,940 --> 00:46:22,260 The rows of the Hadamard matrix are your orthogonal 740 00:46:22,260 --> 00:46:25,400 signal sets, which are elements aj. 741 00:46:25,400 --> 00:46:27,950 And at the receiver, what you would do is when you get the 742 00:46:27,950 --> 00:46:31,730 signal y, you simply multiply by the Hadamard matrix and 743 00:46:31,730 --> 00:46:34,050 look at the coordinate which is the largest. 744 00:46:34,050 --> 00:46:35,180 So this is something you did in the 745 00:46:35,180 --> 00:46:37,120 first homework exercise. 746 00:46:37,120 --> 00:46:40,680 And how many computations did you require for an orthogonal 747 00:46:40,680 --> 00:46:41,960 signal set of size M? 748 00:46:41,960 --> 00:46:46,010 749 00:46:46,010 --> 00:46:47,260 AUDIENCE: [INAUDIBLE] 750 00:46:47,260 --> 00:46:51,420 751 00:46:51,420 --> 00:46:51,506 PROFESSOR: Sorry? 752 00:46:51,506 --> 00:46:54,350 OK, so I meant M being the size of the 753 00:46:54,350 --> 00:46:55,792 orthogonal signal set. 754 00:46:55,792 --> 00:46:59,030 AUDIENCE: Yeah, [INAUDIBLE] 755 00:46:59,030 --> 00:47:01,778 PROFESSOR: Right. 756 00:47:01,778 --> 00:47:03,590 AUDIENCE: 2 to the M by 2 to the M. 757 00:47:03,590 --> 00:47:05,330 PROFESSOR: If it is 2 to the M by 2 to the M, it is M times 2 758 00:47:05,330 --> 00:47:09,470 to the M. But there are M orthogonal signal sets, then 759 00:47:09,470 --> 00:47:17,950 it will be M log M, where M is the number of signal sets. 760 00:47:17,950 --> 00:47:19,440 So that's the number of computations 761 00:47:19,440 --> 00:47:20,970 that you would require. 762 00:47:20,970 --> 00:47:26,640 Now, the question to ask is, is this fast or is this slow? 763 00:47:26,640 --> 00:47:28,670 Is this too many computations or is this too little 764 00:47:28,670 --> 00:47:30,090 computations? 765 00:47:30,090 --> 00:47:32,790 And if you're looking in the power-limited regime, what we 766 00:47:32,790 --> 00:47:36,270 really want is to see the complexity per information bit 767 00:47:36,270 --> 00:47:38,810 because we normalize things per information bit. 768 00:47:38,810 --> 00:47:41,730 Now, how many bits are sent when you have orthogonal 769 00:47:41,730 --> 00:47:43,510 signal set of size M? 770 00:47:43,510 --> 00:47:45,050 We have log M bits, right? 771 00:47:45,050 --> 00:47:47,290 So this quantity is actually exponential in the number of 772 00:47:47,290 --> 00:47:50,360 transmitted bits that we are sending, and so in fact this 773 00:47:50,360 --> 00:47:52,385 is quite slow. 774 00:47:52,385 --> 00:47:54,480 Or in other words we are saying it is we require M 775 00:47:54,480 --> 00:47:57,640 times 2 to the M computations, where M is the number of 776 00:47:57,640 --> 00:48:01,295 information bits that we are sending. 777 00:48:01,295 --> 00:48:04,220 AUDIENCE: So if M is the size of your [INAUDIBLE] 778 00:48:04,220 --> 00:48:06,127 then you have log M bits? 779 00:48:06,127 --> 00:48:07,558 PROFESSOR: Right. 780 00:48:07,558 --> 00:48:09,470 AUDIENCE: So M log M over log M is M. 781 00:48:09,470 --> 00:48:11,195 PROFESSOR: Why are you adding by M? 782 00:48:11,195 --> 00:48:13,690 AUDIENCE: You need M log M operations to decode a symbol. 783 00:48:13,690 --> 00:48:15,190 PROFESSOR: To decode a symbol. 784 00:48:15,190 --> 00:48:18,020 AUDIENCE: And each symbol gives you log M bits. 785 00:48:18,020 --> 00:48:21,650 So you have M operations per bit. 786 00:48:21,650 --> 00:48:24,080 PROFESSOR: Well, when you send, say, log M bits, right, 787 00:48:24,080 --> 00:48:26,290 you can only send one of the M possible symbols. 788 00:48:26,290 --> 00:48:29,860 789 00:48:29,860 --> 00:48:32,100 And when you want to decode the symbol, you would end up 790 00:48:32,100 --> 00:48:33,350 recovering log M bits. 791 00:48:33,350 --> 00:48:35,630 792 00:48:35,630 --> 00:48:38,928 So you should not be dividing by M here. 793 00:48:38,928 --> 00:48:42,030 AUDIENCE: You're dividing by log M because you want to do 794 00:48:42,030 --> 00:48:43,730 it per bit. 795 00:48:43,730 --> 00:48:44,710 PROFESSOR: All right. 796 00:48:44,710 --> 00:48:44,960 You're right. 797 00:48:44,960 --> 00:48:47,330 So you're dividing by log M. You have M log M bits. 798 00:48:47,330 --> 00:48:51,276 You're dividing by log M, so you still have -- 799 00:48:51,276 --> 00:48:55,240 AUDIENCE: You have M calculations per bit. 800 00:48:55,240 --> 00:48:56,490 [INAUDIBLE] 801 00:48:56,490 --> 00:49:02,870 802 00:49:02,870 --> 00:49:06,040 PROFESSOR: So we are sending -- so I need this many 803 00:49:06,040 --> 00:49:09,990 computations to get log M bits. 804 00:49:09,990 --> 00:49:12,590 805 00:49:12,590 --> 00:49:17,359 So is this expression exponential in log M or not? 806 00:49:17,359 --> 00:49:20,670 807 00:49:20,670 --> 00:49:23,310 So that's all I was saying. 808 00:49:23,310 --> 00:49:25,350 AUDIENCE: I just put it another way that you could 809 00:49:25,350 --> 00:49:27,170 divide by log M. 810 00:49:27,170 --> 00:49:28,440 PROFESSOR: M for information, right. 811 00:49:28,440 --> 00:49:32,900 812 00:49:32,900 --> 00:49:35,050 OK, are there any questions? 813 00:49:35,050 --> 00:49:37,510 I mean we are both agreeing to the fact that this is too many 814 00:49:37,510 --> 00:49:39,030 computations than you would want. 815 00:49:39,030 --> 00:49:43,730 And there exist other decoding algorithms for which the 816 00:49:43,730 --> 00:49:46,605 number of computations is much smaller than exponential. 817 00:49:46,605 --> 00:49:54,290 818 00:49:54,290 --> 00:49:58,280 OK, so let's next look at the bandwidth-limited regime. 819 00:49:58,280 --> 00:50:12,110 820 00:50:12,110 --> 00:50:15,500 Well, actually before that I wanted to mention a couple of 821 00:50:15,500 --> 00:50:18,370 other signal sets as well. 822 00:50:18,370 --> 00:50:20,780 So there are two other important class of signal sets 823 00:50:20,780 --> 00:50:24,050 which are related to the orthogonal signal sets. 824 00:50:24,050 --> 00:50:28,090 The first is this class of simplex signal sets. 825 00:50:28,090 --> 00:50:31,580 826 00:50:31,580 --> 00:50:41,065 And the other class is bi-orthogonal signal sets. 827 00:50:41,065 --> 00:50:46,960 828 00:50:46,960 --> 00:50:49,910 So the idea behind simplex signal sets goes back to the 829 00:50:49,910 --> 00:50:53,300 observation that was made that we have a non-zero mean for 830 00:50:53,300 --> 00:50:54,960 this signal set here. 831 00:50:54,960 --> 00:50:57,850 So if we do, we can indeed subtract of the mean and get a 832 00:50:57,850 --> 00:51:00,430 performance which is better than the 833 00:51:00,430 --> 00:51:02,070 orthogonal signal set. 834 00:51:02,070 --> 00:51:05,450 So in particular if I subtract of the mean here, I get a 835 00:51:05,450 --> 00:51:09,560 signal set, which is like this. 836 00:51:09,560 --> 00:51:14,090 This is the simplex signal set when M equals 2. 837 00:51:14,090 --> 00:51:18,730 When M equals 3, I'm going to subtract of the mean from this 838 00:51:18,730 --> 00:51:19,390 plane here. 839 00:51:19,390 --> 00:51:21,980 The points lie on this particular plane. 840 00:51:21,980 --> 00:51:25,570 And what I end up with is an equilateral triangle. 841 00:51:25,570 --> 00:51:29,580 So this is the case when M equals 3. 842 00:51:29,580 --> 00:51:32,960 What do you think M equals 4 would look like? 843 00:51:32,960 --> 00:51:36,020 A tetrahedron right here. 844 00:51:36,020 --> 00:51:39,930 That's a simplex signal set. 845 00:51:39,930 --> 00:51:45,410 So basically, it's also not too hard to write an algebraic 846 00:51:45,410 --> 00:51:47,820 expression for these signal sets. 847 00:51:47,820 --> 00:51:55,720 E of A prime is going to be E of A minus M of A. You 848 00:51:55,720 --> 00:52:00,590 subtract off the mean from your orthogonal signal set. 849 00:52:00,590 --> 00:52:03,460 If I want to write it in terms of inner products, the inner 850 00:52:03,460 --> 00:52:13,450 product between ai and aj is given by E(A) if i equals j. 851 00:52:13,450 --> 00:52:21,610 And it's I believe minus of 1 over M minus 1 times E(A) if i 852 00:52:21,610 --> 00:52:23,220 is not equal to j. 853 00:52:23,220 --> 00:52:26,340 So this is the inner product between ai and aj. 854 00:52:26,340 --> 00:52:28,850 And I mean there are also many properties of this simplex 855 00:52:28,850 --> 00:52:31,652 signal set, but I believe that that's going to be in your 856 00:52:31,652 --> 00:52:33,960 next homework exercise so I'm not going 857 00:52:33,960 --> 00:52:35,780 into too many details. 858 00:52:35,780 --> 00:52:38,360 What' the spectral efficiency in this case going to be? 859 00:52:38,360 --> 00:52:45,346 860 00:52:45,346 --> 00:52:46,850 AUDIENCE: I have a question. 861 00:52:46,850 --> 00:52:49,040 So what is a simplex? 862 00:52:49,040 --> 00:52:52,270 Is a simplex just a reference to anything with [INAUDIBLE]. 863 00:52:52,270 --> 00:52:54,990 PROFESSOR: Simplex signal set is derived from the orthogonal 864 00:52:54,990 --> 00:52:57,560 signal set by subtracting off its mean. 865 00:52:57,560 --> 00:52:59,935 AUDIENCE: Right, but in general simplex means anything 866 00:52:59,935 --> 00:53:01,840 [INAUDIBLE]? 867 00:53:01,840 --> 00:53:03,170 PROFESSOR: I'm not sure about that. 868 00:53:03,170 --> 00:53:08,940 869 00:53:08,940 --> 00:53:12,188 What's the spectral efficiency going to be? 870 00:53:12,188 --> 00:53:13,438 AUDIENCE: [INAUDIBLE] 871 00:53:13,438 --> 00:53:15,840 872 00:53:15,840 --> 00:53:18,536 PROFESSOR: Well, almost. 873 00:53:18,536 --> 00:53:19,030 AUDIENCE: You use one dimension. 874 00:53:19,030 --> 00:53:21,285 PROFESSOR: Right, you just use one dimension. 875 00:53:21,285 --> 00:53:27,230 876 00:53:27,230 --> 00:53:30,650 OK, and there are other properties that you'll be 877 00:53:30,650 --> 00:53:32,420 looking at it in the homework. 878 00:53:32,420 --> 00:53:34,692 Now the idea behind this -- 879 00:53:34,692 --> 00:53:35,942 AUDIENCE: [INAUDIBLE] 880 00:53:35,942 --> 00:53:39,660 881 00:53:39,660 --> 00:53:44,390 PROFESSOR: This is 2 log M over [INAUDIBLE]. 882 00:53:44,390 --> 00:53:48,510 So the idea behind the bi-orthogonal signal set is 883 00:53:48,510 --> 00:53:51,870 you start with an orthogonal signal set, and for each 884 00:53:51,870 --> 00:53:54,240 point, you also put the negative signal of it. 885 00:53:54,240 --> 00:53:56,830 So in addition to this, you will have a point here, and 886 00:53:56,830 --> 00:53:58,520 you will have a point here. 887 00:53:58,520 --> 00:54:03,770 So the case in two dimensions, M equals 2, you will have four 888 00:54:03,770 --> 00:54:09,270 points like this. 889 00:54:09,270 --> 00:54:16,510 So in this case you are increasing the number of 890 00:54:16,510 --> 00:54:25,830 signal points by a factor of two. 891 00:54:25,830 --> 00:54:30,440 892 00:54:30,440 --> 00:54:32,380 But this does come at a price, right? 893 00:54:32,380 --> 00:54:35,280 Because if you're increasing the number of signal points, 894 00:54:35,280 --> 00:54:37,450 the number of nearest neighbors is also going to 895 00:54:37,450 --> 00:54:39,300 increase by a factor of two. 896 00:54:39,300 --> 00:54:41,410 In this case, you have two nearest neighbors now. 897 00:54:41,410 --> 00:55:07,210 898 00:55:07,210 --> 00:55:09,540 So you have both of the effects going on. 899 00:55:09,540 --> 00:55:12,680 I mean ultimately, all the signal sets, as M goes to 900 00:55:12,680 --> 00:55:15,570 infinity, are going to approach the Shannon limit. 901 00:55:15,570 --> 00:55:18,510 So again, but they do suffer from the same kind of 902 00:55:18,510 --> 00:55:21,270 drawbacks that we saw for the orthogonal signal sets. 903 00:55:21,270 --> 00:55:23,830 So these are more of theoretical interest as 904 00:55:23,830 --> 00:55:28,130 opposed to real, practical implementation points. 905 00:55:28,130 --> 00:55:29,380 Any questions? 906 00:55:29,380 --> 00:55:33,540 907 00:55:33,540 --> 00:55:34,790 AUDIENCE: Minus [INAUDIBLE] 908 00:55:34,790 --> 00:55:37,310 909 00:55:37,310 --> 00:55:38,570 PROFESSOR: It's a bit involved to show. 910 00:55:38,570 --> 00:55:42,760 I think you'll be showing it in the homework exercise or 911 00:55:42,760 --> 00:55:46,530 for an exam problem at one point. 912 00:55:46,530 --> 00:55:50,710 You basically start with the orthogonal signal sets here, 913 00:55:50,710 --> 00:55:54,590 look at the inner product here, and use the relation 914 00:55:54,590 --> 00:55:57,110 between the simplex signal sets and the orthogonal signal 915 00:55:57,110 --> 00:55:59,770 sets to do a subtraction of the mean. 916 00:55:59,770 --> 00:56:01,445 And then you can derive in a product expression. 917 00:56:01,445 --> 00:56:59,690 918 00:56:59,690 --> 00:57:02,265 OK, so let's now move on to the bandwidth-limited regime. 919 00:57:02,265 --> 00:57:18,548 920 00:57:18,548 --> 00:57:20,990 OK, so the main difference between -- 921 00:57:20,990 --> 00:57:22,946 yes? 922 00:57:22,946 --> 00:57:24,196 AUDIENCE: [INAUDIBLE] 923 00:57:24,196 --> 00:57:29,220 924 00:57:29,220 --> 00:57:31,040 PROFESSOR: You mean here? 925 00:57:31,040 --> 00:57:32,120 For each -- 926 00:57:32,120 --> 00:57:33,503 AUDIENCE: How far can you [INAUDIBLE]? 927 00:57:33,503 --> 00:57:36,270 PROFESSOR: A factor of two. 928 00:57:36,270 --> 00:57:37,550 I said a factor right? 929 00:57:37,550 --> 00:57:39,850 So if you have M points in the orthogonal signal set, you 930 00:57:39,850 --> 00:57:44,690 have 2M points in the bi-orthogonal signal set 931 00:57:44,690 --> 00:57:45,940 because for each -- 932 00:57:45,940 --> 00:57:48,460 933 00:57:48,460 --> 00:57:52,470 AUDIENCE: In the orthogonal sets you have 3. 934 00:57:52,470 --> 00:57:56,050 PROFESSOR: And in this you have 6. 935 00:57:56,050 --> 00:58:00,290 So let's look at the bandwidth-limited regime. 936 00:58:00,290 --> 00:58:03,480 The trade-off we care about is the probability of error for 937 00:58:03,480 --> 00:58:06,260 two dimensions as a function of SNR norm. 938 00:58:06,260 --> 00:58:09,410 939 00:58:09,410 --> 00:58:10,960 OK? 940 00:58:10,960 --> 00:58:15,530 The baseline scheme here is your M-PAM scheme. 941 00:58:15,530 --> 00:58:18,210 And we showed a couple of lectures ago that the 942 00:58:18,210 --> 00:58:22,460 probability of error for two dimensions is approximately 4 943 00:58:22,460 --> 00:58:24,542 Q times root 3 SNR norm. 944 00:58:24,542 --> 00:58:25,792 OK? 945 00:58:25,792 --> 00:58:30,420 946 00:58:30,420 --> 00:58:32,450 So let's plot the performance graph. 947 00:58:32,450 --> 00:58:40,576 948 00:58:40,576 --> 00:58:43,810 So on the x-axis I plot SNR norm. 949 00:58:43,810 --> 00:58:47,980 On the y-axis I plot Ps of E again on a log-log scale. 950 00:58:47,980 --> 00:58:53,310 951 00:58:53,310 --> 00:58:55,770 This is ten to the negative six, ten to the negative five, 952 00:58:55,770 --> 00:58:57,010 ten to the negative four, ten to the negative 953 00:58:57,010 --> 00:58:59,140 three and so on. 954 00:58:59,140 --> 00:59:02,900 The intercept is at 0 dB, this point here. 955 00:59:02,900 --> 00:59:05,420 So this is your Shannon limit in the 956 00:59:05,420 --> 00:59:08,320 bandwidth-limited regime. 957 00:59:08,320 --> 00:59:12,560 Now your performance curve is going to be something like 958 00:59:12,560 --> 00:59:14,010 this for the M-PAM system. 959 00:59:14,010 --> 00:59:22,710 960 00:59:22,710 --> 00:59:28,480 And we want to basically quantify now the effective 961 00:59:28,480 --> 00:59:30,650 coding gain in this regime. 962 00:59:30,650 --> 00:59:33,900 So it's the same story as in the power-limited regime. 963 00:59:33,900 --> 00:59:39,610 We will start off with the probability of error using the 964 00:59:39,610 --> 00:59:51,000 union bound estimate, which is K_min of A times Q of d_min 965 00:59:51,000 --> 00:59:52,575 over 2 sigma. 966 00:59:52,575 --> 00:59:55,620 967 00:59:55,620 --> 01:00:00,460 Now Ps of E is the probability of error for two dimensions. 968 01:00:00,460 --> 01:00:04,740 Assuming we have N dimensions here, it is two times the 969 01:00:04,740 --> 01:00:10,400 probability of error over N. So this is probability of 970 01:00:10,400 --> 01:00:12,140 error per symbol. 971 01:00:12,140 --> 01:00:15,120 We are dividing by the number of dimensions and multiplying 972 01:00:15,120 --> 01:00:18,770 by two because it is for two dimensions. 973 01:00:18,770 --> 01:00:28,980 And this is going to be 2 times K_min of A over N, times 974 01:00:28,980 --> 01:00:32,230 Q of d_min over 2 sigma. 975 01:00:32,230 --> 01:00:39,530 976 01:00:39,530 --> 01:00:41,900 So now let's do the same trick that we did in the 977 01:00:41,900 --> 01:00:43,150 power-limited regime. 978 01:00:43,150 --> 01:00:46,110 979 01:00:46,110 --> 01:00:49,860 We will write Ps of E be approximately -- 980 01:00:49,860 --> 01:00:53,980 and instead of the right hand side, I will write it as Ks of 981 01:00:53,980 --> 01:01:01,950 A times Q of root 3 SNR norm. 982 01:01:01,950 --> 01:01:06,390 983 01:01:06,390 --> 01:01:08,900 But because I have a factor of four up there, I will also 984 01:01:08,900 --> 01:01:12,790 multiply and divide by 4. 985 01:01:12,790 --> 01:01:16,024 That's not going to change anything. 986 01:01:16,024 --> 01:01:17,888 AUDIENCE: [INAUDIBLE] 987 01:01:17,888 --> 01:01:19,470 per two dimensions or by symbol. 988 01:01:19,470 --> 01:01:20,925 PROFESSOR: Per two dimensions. 989 01:01:20,925 --> 01:01:23,510 And for bandwidth-limited regime, we normalize 990 01:01:23,510 --> 01:01:24,600 everything by two dimensions. 991 01:01:24,600 --> 01:01:28,222 AUDIENCE: But for N band we used it for symbols, right? 992 01:01:28,222 --> 01:01:30,790 The calculations represent -- 993 01:01:30,790 --> 01:01:32,040 PROFESSOR: No, it was for two dimensions. 994 01:01:32,040 --> 01:01:37,720 995 01:01:37,720 --> 01:01:42,120 OK?, so now let's compare the two expressions. 996 01:01:42,120 --> 01:01:50,960 We have Ks of A, which we'll define by two times K_min of A 997 01:01:50,960 --> 01:02:05,810 over N. And this is the number of nearest neighbors per two 998 01:02:05,810 --> 01:02:08,740 dimensions. 999 01:02:08,740 --> 01:02:19,780 And I can write 3 times SNR norm is d_min squared over 4 1000 01:02:19,780 --> 01:02:27,970 sigma squared, OK? 1001 01:02:27,970 --> 01:02:30,550 Actually, I was missing this factor of gamma. 1002 01:02:30,550 --> 01:02:32,070 I knew I was missing something. 1003 01:02:32,070 --> 01:02:34,970 It's not just 3 SNR norm. 1004 01:02:34,970 --> 01:02:39,090 It's 3 times this nominal coding gain times SNR norm. 1005 01:02:39,090 --> 01:02:42,080 1006 01:02:42,080 --> 01:02:47,550 So 3 times SNR norm times gamma_c of A here. 1007 01:02:47,550 --> 01:02:50,020 So this is d_min squared over 2N_0. 1008 01:02:50,020 --> 01:02:54,850 1009 01:02:54,850 --> 01:03:02,470 So what do I have for the gamma_c c o A is d_min squared 1010 01:03:02,470 --> 01:03:06,070 over 6 N_0 times SNR norm. 1011 01:03:06,070 --> 01:03:08,740 1012 01:03:08,740 --> 01:03:12,480 Remember SNR norm is SNR over 2 to the rho minus 1. 1013 01:03:12,480 --> 01:03:15,730 It's normalized signal-to-noise ratio. 1014 01:03:15,730 --> 01:03:20,890 So this is the d_min squared times 2 to the rho minus 1 1015 01:03:20,890 --> 01:03:24,300 over 6 times N_0 times SNR. 1016 01:03:24,300 --> 01:03:27,390 But N_0 times SNR is just Es. 1017 01:03:27,390 --> 01:03:30,180 So this is 6 times Es. 1018 01:03:30,180 --> 01:03:31,740 So this is the expression you have for the 1019 01:03:31,740 --> 01:03:32,990 nominal coding gain. 1020 01:03:32,990 --> 01:03:47,760 1021 01:03:47,760 --> 01:03:52,140 So now again, given a signal set A, I can find those two 1022 01:03:52,140 --> 01:03:59,370 parameters, the nominal coding gain and the number of nearest 1023 01:03:59,370 --> 01:04:01,440 neighbors per two dimensions. 1024 01:04:01,440 --> 01:04:05,686 And I can use those to plot on the Ps of E 1025 01:04:05,686 --> 01:04:07,990 versus SNR norm curve. 1026 01:04:07,990 --> 01:04:09,405 Again, the exact same story. 1027 01:04:09,405 --> 01:04:12,940 1028 01:04:12,940 --> 01:04:17,920 If I have a certain nominal coding gain, I will simply 1029 01:04:17,920 --> 01:04:23,460 shift my curve around the x-axis by that factor as a 1030 01:04:23,460 --> 01:04:25,550 pure translation. 1031 01:04:25,550 --> 01:04:32,120 And then because I have a factor of Ks A over 4, I'm 1032 01:04:32,120 --> 01:04:33,396 going to plot -- 1033 01:04:33,396 --> 01:04:35,340 how did that expression go? 1034 01:04:35,340 --> 01:04:37,590 This expression on the top. 1035 01:04:37,590 --> 01:04:41,440 Remember this curve here is 4 times root 3 SNR norm, right? 1036 01:04:41,440 --> 01:04:44,740 So the multiplicative factor I have is Ks(A) over 4. 1037 01:04:44,740 --> 01:04:46,880 So I will shift my curve up by that factor. 1038 01:04:46,880 --> 01:04:50,180 1039 01:04:50,180 --> 01:04:52,360 And I will get something which is like this. 1040 01:04:52,360 --> 01:04:57,110 1041 01:04:57,110 --> 01:05:00,430 And that's my union bound estimate for this new 1042 01:05:00,430 --> 01:05:04,470 constellation A. 1043 01:05:04,470 --> 01:05:06,340 The distance here is gamma effective. 1044 01:05:06,340 --> 01:05:08,990 1045 01:05:08,990 --> 01:05:19,400 This distance here is gamma_c of A. Are there any questions? 1046 01:05:19,400 --> 01:05:21,870 So now we also have a similar rule of thumb as in the 1047 01:05:21,870 --> 01:05:24,450 power-limited regime. 1048 01:05:24,450 --> 01:05:27,140 I will write that rule of thumb here. 1049 01:05:27,140 --> 01:05:34,430 We have that gamma effective is going to be gamma_c in dB 1050 01:05:34,430 --> 01:05:43,020 minus 0.2 times log2 times Ks of A over 4 in dB. 1051 01:05:43,020 --> 01:05:46,670 1052 01:05:46,670 --> 01:05:51,400 OK, so again, a factor of 2NKs will decrease your nominal 1053 01:05:51,400 --> 01:05:54,830 coding gain by a factor of 0.2 in dB. 1054 01:05:54,830 --> 01:05:58,620 1055 01:05:58,620 --> 01:05:59,870 Are there any questions? 1056 01:05:59,870 --> 01:06:03,260 1057 01:06:03,260 --> 01:06:06,260 Well, so I mean this is the end of chapter five. 1058 01:06:06,260 --> 01:06:08,830 We still have like ten minutes remaining, so I thought I 1059 01:06:08,830 --> 01:06:11,480 would give you a preview of the next chapter. 1060 01:06:11,480 --> 01:06:13,510 Professor Forney will be coming next class, and he will 1061 01:06:13,510 --> 01:06:15,970 be starting chapter six all over I believe. 1062 01:06:15,970 --> 01:06:19,030 1063 01:06:19,030 --> 01:06:21,303 This chalk is much nicer to erase than that other one. 1064 01:06:21,303 --> 01:06:43,000 1065 01:06:43,000 --> 01:06:48,650 So in chapter six, we'll be looking at the 1066 01:06:48,650 --> 01:06:49,900 binary block codes. 1067 01:06:49,900 --> 01:07:00,130 1068 01:07:00,130 --> 01:07:04,540 OK, and what is the main architecture of these codes? 1069 01:07:04,540 --> 01:07:09,700 Well, for an encoder, remember we have K bits coming in. 1070 01:07:09,700 --> 01:07:12,315 And we pass it now through a binary block code. 1071 01:07:12,315 --> 01:07:16,070 1072 01:07:16,070 --> 01:07:23,220 And what we get out is a binary sequence 1073 01:07:23,220 --> 01:07:25,425 of length N. OK? 1074 01:07:25,425 --> 01:07:31,750 So x belongs to c, where c is a subset of binary sequences 1075 01:07:31,750 --> 01:07:33,550 of length N. 1076 01:07:33,550 --> 01:07:37,240 So we start with K bits, and we convert them to N bits, 1077 01:07:37,240 --> 01:07:42,610 where N is going to be greater than or equal to K. Once we 1078 01:07:42,610 --> 01:07:46,590 have this sequence of N bits, what we end up doing is we 1079 01:07:46,590 --> 01:07:49,820 pass it through a binary signal constellation. 1080 01:07:49,820 --> 01:07:54,710 1081 01:07:54,710 --> 01:07:56,690 So I'm writing in as binary signaling. 1082 01:07:56,690 --> 01:08:02,200 And what we get out is S of x, which is a 1083 01:08:02,200 --> 01:08:04,940 sequence of N symbols. 1084 01:08:04,940 --> 01:08:07,110 Each symbol can either take value minus alpha or alpha. 1085 01:08:07,110 --> 01:08:12,130 So this binary signalling is actually quite a trivial step. 1086 01:08:12,130 --> 01:08:16,560 Basically, the relation is S of 0 is going to be alpha, S 1087 01:08:16,560 --> 01:08:18,500 of 1 is going to be minus alpha. 1088 01:08:18,500 --> 01:08:21,520 If I have a 0 coming in, I will produce an alpha. 1089 01:08:21,520 --> 01:08:24,520 if I have a 1 coming in, I will produce a minus alpha. 1090 01:08:24,520 --> 01:08:26,689 So this part here is trivial. 1091 01:08:26,689 --> 01:08:29,253 All our energy will be focused on will be 1092 01:08:29,253 --> 01:08:31,340 to find a good code. 1093 01:08:31,340 --> 01:08:33,970 Given a sequence of input bits, we want to find a good 1094 01:08:33,970 --> 01:08:39,080 binary code in order to optimize the minimum 1095 01:08:39,080 --> 01:08:40,880 distance and so on. 1096 01:08:40,880 --> 01:08:43,540 So that's the architecture that we will be using for the 1097 01:08:43,540 --> 01:08:45,210 binary linear codes. 1098 01:08:45,210 --> 01:08:46,800 Both chapters six and eight will be only 1099 01:08:46,800 --> 01:08:48,250 focusing on this part. 1100 01:08:48,250 --> 01:08:50,550 And this binary signalling scheme, for the most part, 1101 01:08:50,550 --> 01:08:52,265 will be implicit in our architecture. 1102 01:08:52,265 --> 01:08:55,069 1103 01:08:55,069 --> 01:08:58,399 So at this point, one might ask is there anything to lose 1104 01:08:58,399 --> 01:09:01,359 in having imposed this type of architecture? 1105 01:09:01,359 --> 01:09:03,899 Remember, the nice thing about this architecture is we have 1106 01:09:03,899 --> 01:09:06,910 coding, which is going on here, and we have modulation 1107 01:09:06,910 --> 01:09:08,460 that is happening here. 1108 01:09:08,460 --> 01:09:10,960 And the modulation step is quite simple. 1109 01:09:10,960 --> 01:09:23,156 So we have a separation between coding and modulation. 1110 01:09:23,156 --> 01:09:24,406 OK? 1111 01:09:24,406 --> 01:09:29,689 1112 01:09:29,689 --> 01:09:32,600 And the question is, is this the right thing to do? 1113 01:09:32,600 --> 01:09:34,520 Now, it turns out that if we are going to live in this 1114 01:09:34,520 --> 01:09:36,984 binary world where we are only constrained ourselves to 1115 01:09:36,984 --> 01:09:40,134 sending binary signals over the channel, this is in fact 1116 01:09:40,134 --> 01:09:41,800 an economical structure. 1117 01:09:41,800 --> 01:09:44,310 There isn't much improvement we can get, and that is quite 1118 01:09:44,310 --> 01:09:45,424 obvious, right? 1119 01:09:45,424 --> 01:09:48,410 But it turns out that if you are going to go for non-binary 1120 01:09:48,410 --> 01:09:51,870 signaling, particularly in the bandwidth-limited regime, 1121 01:09:51,870 --> 01:09:55,460 there is a cost to be paid for this kind of separation. 1122 01:09:55,460 --> 01:09:57,950 In fact, in the 1980's there was this whole idea of 1123 01:09:57,950 --> 01:10:01,060 turbo-coded modulation or trellis-coded modulation. 1124 01:10:01,060 --> 01:10:03,680 And the idea there was to combine coding and modulation 1125 01:10:03,680 --> 01:10:05,110 in one step. 1126 01:10:05,110 --> 01:10:19,440 So this is not optimal for non-binary signalling, but's 1127 01:10:19,440 --> 01:10:25,270 it's OK for binary. 1128 01:10:25,270 --> 01:10:30,570 1129 01:10:30,570 --> 01:10:33,820 So we will be focusing on this type of an architecture. 1130 01:10:33,820 --> 01:10:37,130 The other issue is OK, nobody told us that you can only send 1131 01:10:37,130 --> 01:10:39,080 binary signals over the channel. 1132 01:10:39,080 --> 01:10:42,270 If you want to achieve the Shannon limit, when we derived 1133 01:10:42,270 --> 01:10:44,580 the capacity expression -- 1134 01:10:44,580 --> 01:10:46,970 or we just stated it but you can derive it. 1135 01:10:46,970 --> 01:10:49,660 We said that the Shannon limit is always less 1136 01:10:49,660 --> 01:10:52,830 than log2 1 plus SNR. 1137 01:10:52,830 --> 01:10:55,980 And we never said that we can only send binary signals over 1138 01:10:55,980 --> 01:10:56,840 the channel. 1139 01:10:56,840 --> 01:10:59,070 So it could be that this is some inherence of optimality 1140 01:10:59,070 --> 01:11:00,770 in this type of architecture. 1141 01:11:00,770 --> 01:11:03,730 Why send binary signals in the first place? 1142 01:11:03,730 --> 01:11:06,940 Again, it turns out that the regime that we are interested 1143 01:11:06,940 --> 01:11:10,130 in here is the power-limited regime because the spectral 1144 01:11:10,130 --> 01:11:12,220 efficiency can never be more than two bits per two 1145 01:11:12,220 --> 01:11:13,810 dimensions, right? 1146 01:11:13,810 --> 01:11:16,340 Remember, rho is 2K over N here, and K is going to be 1147 01:11:16,340 --> 01:11:17,290 less than that. 1148 01:11:17,290 --> 01:11:22,130 So we are inherently in the power-limited regime. 1149 01:11:22,130 --> 01:11:24,770 And so a natural thing to do in order to understand how 1150 01:11:24,770 --> 01:11:28,120 much fundamental loss we have in imposing this type of 1151 01:11:28,120 --> 01:11:32,360 binary constraint over the signals is to not look at this 1152 01:11:32,360 --> 01:11:36,470 expression, but to find the best possible performance we 1153 01:11:36,470 --> 01:11:39,390 can have if we constrain ourselves to binary signals 1154 01:11:39,390 --> 01:11:41,090 over the channel. 1155 01:11:41,090 --> 01:11:43,910 If we impose a certain signaling constraint, we can 1156 01:11:43,910 --> 01:11:49,330 find another fundamental limit on the spectral efficiency. 1157 01:11:49,330 --> 01:11:53,070 So the point being, the best possible binary code can only 1158 01:11:53,070 --> 01:11:55,450 achieve this upper bound here. 1159 01:11:55,450 --> 01:11:57,050 Now, this upper bound has to be less than 1160 01:11:57,050 --> 01:11:58,300 log2 1 plus SNR, right? 1161 01:11:58,300 --> 01:12:03,170 1162 01:12:03,170 --> 01:12:06,070 Because the Shannon code assumes the binary signaling 1163 01:12:06,070 --> 01:12:07,640 is a special case. 1164 01:12:07,640 --> 01:12:10,700 So it turns out that this expression is actually quite 1165 01:12:10,700 --> 01:12:12,310 tedious to write out. 1166 01:12:12,310 --> 01:12:15,090 I'm not even sure if the closed-form expression exists, 1167 01:12:15,090 --> 01:12:16,820 but you can calculate this numerically. 1168 01:12:16,820 --> 01:12:19,760 It's just some kind of a convex optimization problem. 1169 01:12:19,760 --> 01:12:22,350 And now, to understand how much loss we have, we can 1170 01:12:22,350 --> 01:12:26,090 compare the two expressions in the power-limited regime. 1171 01:12:26,090 --> 01:12:30,690 So I'm going to plot the curves for the two cases. 1172 01:12:30,690 --> 01:12:35,350 I'm going to plot the spectral efficiency on the y-axis as a 1173 01:12:35,350 --> 01:12:36,500 function of Eb N_0. 1174 01:12:36,500 --> 01:12:39,000 You can easily convert from SNR to Eb N_0 using the 1175 01:12:39,000 --> 01:12:42,210 relations we discussed in chapter four. 1176 01:12:42,210 --> 01:12:45,030 Now we first plot the Shannon limit. 1177 01:12:45,030 --> 01:12:48,920 The ultimate Shannon limit is minus 1.59 dB. 1178 01:12:48,920 --> 01:12:52,000 This is going to be 0 dB here. 1179 01:12:52,000 --> 01:12:54,100 And if I'm plot the Shannon limit, it will be 1180 01:12:54,100 --> 01:12:56,260 some code like this. 1181 01:12:56,260 --> 01:12:58,080 It increases with Eb N_0. 1182 01:12:58,080 --> 01:13:01,210 And minus 1.59 dB is 0. 1183 01:13:01,210 --> 01:13:03,130 This point here is 1. 1184 01:13:03,130 --> 01:13:05,070 Some point here is 2. 1185 01:13:05,070 --> 01:13:09,200 So actually I want to draw this horizontal asymptote at 1186 01:13:09,200 --> 01:13:13,030 two when rho Shannon equals 2. 1187 01:13:13,030 --> 01:13:15,720 So this is my rho Shannon. 1188 01:13:15,720 --> 01:13:19,200 Now, I can now also plot rho binary from 1189 01:13:19,200 --> 01:13:21,020 this expression here. 1190 01:13:21,020 --> 01:13:23,220 I can never exceed two bits per two dimensions. 1191 01:13:23,220 --> 01:13:25,720 So I had plotted this horizontal line. 1192 01:13:25,720 --> 01:13:28,720 So as Eb N_0 goes to infinity, the most I can get is two bits 1193 01:13:28,720 --> 01:13:30,250 per two dimensions. 1194 01:13:30,250 --> 01:13:35,380 If Eb N_0 is smaller, I cannot do better so I will be doing 1195 01:13:35,380 --> 01:13:38,210 worse and worse, and I will always be below this line. 1196 01:13:38,210 --> 01:13:41,330 Ultimately, it can be shown that I will have the same 1197 01:13:41,330 --> 01:13:43,210 Shannon limit here. 1198 01:13:43,210 --> 01:13:45,020 So this is rho binary. 1199 01:13:45,020 --> 01:13:48,380 1200 01:13:48,380 --> 01:13:51,830 And the main observation here is if I'm in the power-limited 1201 01:13:51,830 --> 01:13:55,150 regime, which is say, in this part of the curve -- 1202 01:13:55,150 --> 01:13:57,650 like, say around one bit per two dimension -- the gap here 1203 01:13:57,650 --> 01:13:59,390 is quite small. 1204 01:13:59,390 --> 01:14:02,620 In fact, this gap can be shown to be at most 0.2 dB. 1205 01:14:02,620 --> 01:14:05,190 1206 01:14:05,190 --> 01:14:08,130 So if I want to achieve a certain spectral efficiency, 1207 01:14:08,130 --> 01:14:10,855 if I impose a binary signaling constraint, the additional Eb 1208 01:14:10,855 --> 01:14:15,300 N_0 that I require when the spectral efficiency is 1 dB, 1209 01:14:15,300 --> 01:14:18,360 one bit per two dimensions, is at most 0.2 dB. 1210 01:14:18,360 --> 01:14:22,280 So there is not much to lose by imposing a binary signaling 1211 01:14:22,280 --> 01:14:23,870 constraint. 1212 01:14:23,870 --> 01:14:26,220 Said in different words words if you had the best possible 1213 01:14:26,220 --> 01:14:29,150 binary linear code, followed by a binary signaling 1214 01:14:29,150 --> 01:14:33,210 constellation, the most you can lose is 0.2 dB. 1215 01:14:33,210 --> 01:14:36,070 And so this type of an architecture does make sense. 1216 01:14:36,070 --> 01:14:39,783 There are a lot of details that the next powerpoint will 1217 01:14:39,783 --> 01:14:44,680 be to kind of formalize the notion of a binary block code, 1218 01:14:44,680 --> 01:14:46,850 then specialize it to the case of binary linear 1219 01:14:46,850 --> 01:14:48,450 codes and so on. 1220 01:14:48,450 --> 01:14:51,370 Unfortunately, the first thing to do is to start with some 1221 01:14:51,370 --> 01:14:54,970 finite field theory because there's a lot of finite field 1222 01:14:54,970 --> 01:14:57,780 algebra involved in this algebraic block codes. 1223 01:14:57,780 --> 01:15:02,580 So how many of you have seen finite field theory before? 1224 01:15:02,580 --> 01:15:05,810 OK, if you haven't seen it, don't worry about it. 1225 01:15:05,810 --> 01:15:08,570 We'll be starting right from the basics. 1226 01:15:08,570 --> 01:15:11,580 But I think I will be stopping here for today because I don't 1227 01:15:11,580 --> 01:15:13,390 want to go into those details today. 1228 01:15:13,390 --> 01:15:36,017