1 00:00:00,920 --> 00:00:04,650 Hi, welcome to module 9.3 of digital signal processing. 2 00:00:04,650 --> 00:00:07,550 We're still talking about digital communication systems. 3 00:00:07,550 --> 00:00:10,660 In the previous module we addressed the bandwidth constraint, 4 00:00:10,660 --> 00:00:13,530 and in this module we will tackle the power constraint. 5 00:00:13,530 --> 00:00:15,220 So first we will introduce the concept of 6 00:00:15,220 --> 00:00:18,260 noise and probability of error in a communication system. 7 00:00:18,260 --> 00:00:21,930 We will look at signalling alphabet and the related power. 8 00:00:21,930 --> 00:00:25,010 And finally we'll introduce QAM signalling. 9 00:00:25,010 --> 00:00:26,430 So, we have seen the transmitter 10 00:00:26,430 --> 00:00:30,370 sends a sequence of symbols, a of n, created by the mapper. 11 00:00:30,370 --> 00:00:32,470 Now we take the receiver into account. 12 00:00:32,470 --> 00:00:34,730 We don't yet know how, but it's safe to assume 13 00:00:34,730 --> 00:00:38,075 that the receiver in the end, will obtain an estimation. 14 00:00:38,075 --> 00:00:41,208 Hat a of n, of the original transmitted symbol sequence. 15 00:00:41,208 --> 00:00:44,410 It's an estimation.because even if there is no distortion 16 00:00:44,410 --> 00:00:47,690 introduced by the channel, even if nothing bad happens. 17 00:00:47,690 --> 00:00:49,050 There will always be a certain amount 18 00:00:49,050 --> 00:00:52,880 of noise that will corrupt the original sequence. 19 00:00:52,880 --> 00:00:55,710 When noise is very large, our estimate for the transmitted 20 00:00:55,710 --> 00:00:59,120 symbol will be off and will incur a decoding error. 21 00:00:59,120 --> 00:01:01,770 Now this probability of error will depend on the power of 22 00:01:01,770 --> 00:01:04,660 the noise with respect to the power of the signal and 23 00:01:04,660 --> 00:01:07,350 will also depend on the decoding strategies that we put in 24 00:01:07,350 --> 00:01:11,400 place, how smart we are in circumvent and defects of the noise. 25 00:01:11,400 --> 00:01:15,230 One way we can maximize the probabilty of 26 00:01:15,230 --> 00:01:18,010 correctly guessing the transmitted symbol is by using 27 00:01:18,010 --> 00:01:19,640 suitable alphabets. 28 00:01:19,640 --> 00:01:22,090 And so we'll see in more detail what that means. 29 00:01:22,090 --> 00:01:24,140 Remember the scheme for the transmitter. 30 00:01:24,140 --> 00:01:29,100 We have a bitstream coming in and then we have the scrambler, 31 00:01:30,440 --> 00:01:35,240 and then the mapper. And here we have a sequence of symbols 32 00:01:35,240 --> 00:01:39,650 a of n. These symbols will have to be sent over 33 00:01:39,650 --> 00:01:44,270 the channel and to do so we up sample and we interpolate and 34 00:01:44,270 --> 00:01:45,080 then we transmit. 35 00:01:46,510 --> 00:01:50,190 Now, how do we go from bitstreams to samples in more detail. 36 00:01:50,190 --> 00:01:52,840 In other words, how does the mapper work? 37 00:01:52,840 --> 00:01:57,110 The mapper will split the incoming bitstreams into chunks. 38 00:01:57,110 --> 00:02:02,270 And will assign a symbol, a of n, from a finite alphabet to each chunk. 39 00:02:02,270 --> 00:02:05,880 The alphabet we will decide later what it is composed of. 40 00:02:05,880 --> 00:02:09,340 To undo the mapping operation and recover the bitstream. 41 00:02:09,340 --> 00:02:12,250 The receiver will perform a slicing operation. 42 00:02:12,250 --> 00:02:15,850 So, the receiver will receive a value of hat a of n where 43 00:02:15,850 --> 00:02:21,720 hat indicates the fact that noise has leaked into the value of the signal. 44 00:02:21,720 --> 00:02:24,850 And the receiver will decide which symbol from the alphabet 45 00:02:24,850 --> 00:02:28,590 which is known to the receiver as well is closest 46 00:02:28,590 --> 00:02:30,640 to the received symbol, and from there, it will be 47 00:02:30,640 --> 00:02:34,950 extremely easy to piece back the original bitstream as an example, 48 00:02:34,950 --> 00:02:37,050 let's look at simple two level signalling. 49 00:02:37,050 --> 00:02:39,620 This generates signals of the kind we has seen 50 00:02:39,620 --> 00:02:43,380 in the examples so far alternating between two levels. 51 00:02:43,380 --> 00:02:48,450 The way the mapper works is by splitting the incoming bitstream into single bits. 52 00:02:49,570 --> 00:02:55,280 And the output symbol sequence uses an alphabet composed of two symbols 53 00:02:55,280 --> 00:03:00,070 G and minus G, and associates G to bit value 54 00:03:00,070 --> 00:03:03,070 one, and minus G to a bit of value zero. 55 00:03:03,070 --> 00:03:06,670 At the receiver the slicer, looks at the sign of 56 00:03:06,670 --> 00:03:11,100 the incoming symbol sequence which has been corrupted by noise 57 00:03:11,100 --> 00:03:14,620 and decides that the nth bit will be one, if 58 00:03:14,620 --> 00:03:18,790 the sign of the nth symbol is positive and zero otherwise. 59 00:03:18,790 --> 00:03:22,060 Let's look at an example, let's assume g equal to 1, 60 00:03:22,060 --> 00:03:25,633 so the two level signal will alternate between plus 1 and 61 00:03:25,633 --> 00:03:26,460 minus 1. 62 00:03:26,460 --> 00:03:32,440 And suppose we have an input bit sequence that gives rise to this signal here. 63 00:03:32,440 --> 00:03:36,110 After transmission and after decoding at the receiver. 64 00:03:36,110 --> 00:03:40,710 The resulting symbol sequence will look like this, where each 65 00:03:40,710 --> 00:03:43,322 symbol has been corrupted by a varying amount of noise. 66 00:03:43,322 --> 00:03:46,510 If we now slice this sequence by thresholding, as 67 00:03:46,510 --> 00:03:50,410 shown before, we recover a symbol sequence like this. 68 00:03:50,410 --> 00:03:50,660 Where we 69 00:03:50,660 --> 00:03:56,010 have indicated in red the errors incurred by the slicer because of the noise. 70 00:03:56,010 --> 00:03:59,990 So if you want to analyze in more detail what the probability of error is. 71 00:03:59,990 --> 00:04:03,850 We have to make some hypothesis on the signals involved in this toy experiment. 72 00:04:05,150 --> 00:04:07,620 Assume that each received symbol can be modelled 73 00:04:07,620 --> 00:04:11,230 as the original symbol plus a noise sample. 74 00:04:11,230 --> 00:04:14,530 Assume also that the bits in the bits stream are equiprobable. 75 00:04:14,530 --> 00:04:15,766 So zero and one 76 00:04:15,766 --> 00:04:19,200 appear with probability 50% each. 77 00:04:19,200 --> 00:04:22,370 Assume that the noise and the signal are independent and assume that the 78 00:04:22,370 --> 00:04:24,880 noise is additive white Gaussian noise with 79 00:04:24,880 --> 00:04:28,470 zero mean and known variance, sigma zero. 80 00:04:28,470 --> 00:04:32,680 With this hypothesis the probability of error can be written out as follows. 81 00:04:32,680 --> 00:04:34,130 First of all, we split the probability 82 00:04:34,130 --> 00:04:37,580 of errors into two conditional probabilities conditioned by 83 00:04:37,580 --> 00:04:41,880 whether the n-th bit is equal to one or the n-th bit is equal to zero. 84 00:04:41,880 --> 00:04:43,790 In the first case when the n-th bit is equal 85 00:04:43,790 --> 00:04:47,680 to one, remember the produce symbol will be equal to G. 86 00:04:47,680 --> 00:04:49,940 So the probability of error is equal to the probability 87 00:04:49,940 --> 00:04:53,210 for the noise sample to be less than minus G. 88 00:04:53,210 --> 00:04:57,870 Because only in this case, the sum of the sample plus the noise, will be negative. 89 00:04:58,880 --> 00:05:03,930 Similarly, when the amplitude is equal to zero, we have a negative sample. 90 00:05:03,930 --> 00:05:07,030 And the only way for that to change sine is if the noise 91 00:05:07,030 --> 00:05:08,720 sample is greater than G. 92 00:05:08,720 --> 00:05:11,900 Since the probability of each occurrence is one 93 00:05:11,900 --> 00:05:14,990 half, because of the symmetry of the Gaussian 94 00:05:14,990 --> 00:05:17,950 distribution function, this is equal to the probability 95 00:05:17,950 --> 00:05:20,860 for the noise sample to be larger than G. 96 00:05:20,860 --> 00:05:25,330 And we can compute this as the integral from G to infinity of 97 00:05:25,330 --> 00:05:27,370 the probability distribution function for the 98 00:05:27,370 --> 00:05:30,820 Gaussian distribution with the known variance here. 99 00:05:30,820 --> 00:05:32,227 So what we have here is the 100 00:05:32,227 --> 00:05:36,550 tail probability of a Gaussian with standard deviation sigma zero. 101 00:05:36,550 --> 00:05:39,790 So, this is the integral that we're trying to compute. 102 00:05:39,790 --> 00:05:42,452 Usually the tail probability of a unit variance 103 00:05:42,452 --> 00:05:46,040 Gaussian is indicated by the notation Q of G. 104 00:05:46,040 --> 00:05:48,040 The Q function of G. 105 00:05:48,040 --> 00:05:51,620 Now because our variance is actually sigma zero squared, we 106 00:05:51,620 --> 00:05:56,460 have to normalize the Q function argument by the standard deviation. 107 00:05:56,460 --> 00:05:57,590 And we find that the probability 108 00:05:57,590 --> 00:06:00,980 of error is equal to the Q function of G over sigma 0. 109 00:06:00,980 --> 00:06:06,040 Since this integral is not computable exactly, if we want to calculate the Q 110 00:06:06,040 --> 00:06:08,630 function, we have to resort to numerical 111 00:06:08,630 --> 00:06:11,763 packages, or tabulated versions of this function. 112 00:06:11,763 --> 00:06:13,100 And usually, what you would find in 113 00:06:13,100 --> 00:06:15,640 numerical packages in, is a derived function 114 00:06:15,640 --> 00:06:18,250 called the error function, which is related 115 00:06:18,250 --> 00:06:21,100 to the Q function by this formula here. 116 00:06:21,100 --> 00:06:22,640 But the point that is really important 117 00:06:22,640 --> 00:06:26,210 at the end of this derivation is that the probability of error is equal to some 118 00:06:26,210 --> 00:06:29,570 function of the ratio between the amplitude of 119 00:06:29,570 --> 00:06:32,760 the signal, and the standard deviation of the noise. 120 00:06:32,760 --> 00:06:37,600 Now we can carry this analysis further by considering the transmitted power. 121 00:06:37,600 --> 00:06:42,110 We have a bi-level signal, and each level occurs with one half probability. 122 00:06:42,110 --> 00:06:44,640 So the variance of the signal, which corresponds 123 00:06:44,640 --> 00:06:47,690 to the power is equal to G squared times 124 00:06:47,690 --> 00:06:51,620 the probability of the n-th bit being equal to 1, plus G squared times the 125 00:06:51,620 --> 00:06:55,082 probability of the n-th bit being equal to 0, which is equal to G squared. 126 00:06:55,082 --> 00:06:58,970 And so if we rewrite the probability of error, we can see that it is equal to the 127 00:06:58,970 --> 00:07:02,700 Q function of the ratio between the standard deviation 128 00:07:02,700 --> 00:07:05,960 of the signal and the standard deviation of the noise. 129 00:07:05,960 --> 00:07:08,920 But this is really equivalent to saying that the probability of error is equal 130 00:07:08,920 --> 00:07:12,920 to the Q function of the square root of the signal to noise ratio of 131 00:07:12,920 --> 00:07:14,560 the transmitted signal. 132 00:07:14,560 --> 00:07:18,680 If we plot this as a function of the signal noise to ratio in dBs and I remind 133 00:07:18,680 --> 00:07:24,320 here that dBs here mean that we compute 10 times 134 00:07:24,320 --> 00:07:30,810 the log in base 10 of the power of the signal divided by the power of the noise. 135 00:07:30,810 --> 00:07:33,440 And since we are in a log, log scale. 136 00:07:33,440 --> 00:07:38,040 We can see that the probability of error decays exponentially with the signal to 137 00:07:38,040 --> 00:07:38,970 noise ratio. 138 00:07:38,970 --> 00:07:43,190 This exponentially decay is quite a norm in communication systems 139 00:07:43,190 --> 00:07:46,820 and while the absolutely rate of decay might change in terms 140 00:07:46,820 --> 00:07:50,210 of the linear constants involved in the curve, the trend 141 00:07:50,210 --> 00:07:53,830 will stay the same even for more complex signal and schemes. 142 00:07:53,830 --> 00:07:57,307 So the lesson that we learned from the simple example is that in order 143 00:07:57,307 --> 00:07:59,320 to reduce the probability of error, we 144 00:07:59,320 --> 00:08:01,939 should increase G, the amplitude of the signal. 145 00:08:01,939 --> 00:08:02,732 But of course, 146 00:08:02,732 --> 00:08:06,331 increasing G also increases the power of the transmitted signal, and 147 00:08:06,331 --> 00:08:10,200 we know that we cannot go above the channel's power constraint. 148 00:08:10,200 --> 00:08:15,050 And so that's how the power constraint limits the reliability of transmission. 149 00:08:15,050 --> 00:08:18,570 The bi-level signal is keen, is very instructive, but it's also very 150 00:08:18,570 --> 00:08:23,030 limited in the sense that we're sending just one bit per output symbol. 151 00:08:23,030 --> 00:08:25,390 So to increase the throughput, to increase the number 152 00:08:25,390 --> 00:08:27,940 of bits per second that we send over a channel, 153 00:08:27,940 --> 00:08:29,700 we can use multilevel signaling. 154 00:08:29,700 --> 00:08:32,930 There are very many ways to do so, we'll just look at a few. 155 00:08:32,930 --> 00:08:37,270 But the fundamental idea is that we take now, larger chunks of 156 00:08:37,270 --> 00:08:42,080 bits, and therefore we have alpha bits that have a higher cardinality. 157 00:08:42,080 --> 00:08:44,940 So more values in the alpha bit means more 158 00:08:44,940 --> 00:08:47,730 bits per symbol and therefore a higher data rate. 159 00:08:47,730 --> 00:08:49,570 But, not to give the ending away. 160 00:08:49,570 --> 00:08:53,060 We will see that the power of the signal will also be dependent on the size of the 161 00:08:53,060 --> 00:08:57,140 alphabet and so in order not exceed in the probability of error, given the 162 00:08:57,140 --> 00:09:01,880 channels power of constraint, we will not be able to grow the alphabet indefinitely. 163 00:09:01,880 --> 00:09:04,410 But we can be smart in a way we build this alphabet. 164 00:09:04,410 --> 00:09:06,120 And so we will look in some examples. 165 00:09:06,120 --> 00:09:09,600 The first example is PAM, Pulse-amplitude Modulation. 166 00:09:09,600 --> 00:09:12,560 We split the incoming bitstream into chunks of M bits. 167 00:09:12,560 --> 00:09:17,610 So that the each chunk corresponds to an integer between 0 and 2 to the m minus 1. 168 00:09:17,610 --> 00:09:18,590 We can 169 00:09:18,590 --> 00:09:22,680 call this sequence of integers, k of n and this sequence is mapped onto a 170 00:09:22,680 --> 00:09:28,090 sequence of symbols a of n like so There's a gain factor G, like always. 171 00:09:28,090 --> 00:09:32,430 And then we use 2 to the M minus 1, odd integers around 0. 172 00:09:32,430 --> 00:09:38,084 So for instance, if M is equal to 2, we have 0, 1, 2, 173 00:09:38,084 --> 00:09:43,850 and 3 as potential items for k of n. 174 00:09:43,850 --> 00:09:49,130 And a of n will be either lets assume 175 00:09:49,130 --> 00:09:54,642 G is equal to 1 will be either minus 3 or minus or 1 or 3. 176 00:09:54,642 --> 00:09:58,870 We will see why we used the odd integers in just a second. 177 00:09:58,870 --> 00:10:02,660 At the receiver, the slicer will work by simply associating to the 178 00:10:02,660 --> 00:10:08,080 received symbol, the closest odd integer, always taking the gain into account. 179 00:10:08,080 --> 00:10:09,540 So graphically again, 180 00:10:09,540 --> 00:10:12,740 PAM for M equal to 2 and G equal to 1 will look like this. 181 00:10:12,740 --> 00:10:14,530 Here are the odd integers. 182 00:10:14,530 --> 00:10:17,720 The distance between two transmitted points, or 183 00:10:17,720 --> 00:10:21,530 transmitted symbols, is 2G, right here G 184 00:10:21,530 --> 00:10:25,100 is equal to 1, but it would be in general, 2 times the gain. 185 00:10:25,100 --> 00:10:28,600 And using odd integers creates a zero-mean sequence. 186 00:10:28,600 --> 00:10:31,880 If we assume that each symbol is equally probable, which 187 00:10:31,880 --> 00:10:35,150 is likely, given that we've used a scrambler in the transmitter, 188 00:10:35,150 --> 00:10:36,910 then the resulting mean is zero. 189 00:10:36,910 --> 00:10:40,320 The analysis of the probability of error for PAM is 190 00:10:40,320 --> 00:10:43,770 very similar to what we carried out for bi-level signaling. 191 00:10:43,770 --> 00:10:48,280 As a matter of fact, bi-level signaling is simply PAM with m equal to 1. 192 00:10:48,280 --> 00:10:51,830 The end result is very similar and its an exponential decaying of the 193 00:10:51,830 --> 00:10:55,880 ratio between the power of the signal and the power of the noise. 194 00:10:55,880 --> 00:10:57,760 The reason why we don't analyze this further 195 00:10:57,760 --> 00:10:59,060 is because we have an improvement in store. 196 00:10:59,060 --> 00:11:01,540 And the improvement 197 00:11:01,540 --> 00:11:05,180 is aimed at increasing the throughput, increasing the numbers of bit per 198 00:11:05,180 --> 00:11:07,120 symbol that we can send without 199 00:11:07,120 --> 00:11:09,700 necessarily increasing the probability of error. 200 00:11:09,700 --> 00:11:12,000 So here's a wild idea, let's use complex 201 00:11:12,000 --> 00:11:15,810 numbers and build a complex valued transmission system. 202 00:11:15,810 --> 00:11:18,400 This requires certain suspension of disbelief for the time 203 00:11:18,400 --> 00:11:20,980 being, but believe me, it will work in the end. 204 00:11:20,980 --> 00:11:23,420 The name for this complex valued mapping scheme 205 00:11:23,420 --> 00:11:26,610 is QAM, which is an acronym for quadrature 206 00:11:26,610 --> 00:11:29,460 amplitude modulation. And it works like so. 207 00:11:29,460 --> 00:11:32,180 The mapper takes the incoming bitstream and splits 208 00:11:32,180 --> 00:11:35,220 it into chunks of M bits, with M even. 209 00:11:36,320 --> 00:11:39,280 And then it uses half of the bits to define 210 00:11:39,280 --> 00:11:42,570 a PAM sequence which we call a of r of n. 211 00:11:42,570 --> 00:11:45,270 And the remaining m over two bits to define 212 00:11:45,270 --> 00:11:48,520 an another independent PAN sequence a i of n. 213 00:11:48,520 --> 00:11:51,920 The final symbol sequence is a sequence of complex numbers 214 00:11:51,920 --> 00:11:55,790 where the real part is the first PAM sequence 215 00:11:55,790 --> 00:11:58,570 and the imaginary part is a second PAM sequence. 216 00:11:58,570 --> 00:12:01,600 And of course in front we have again factor G. 217 00:12:01,600 --> 00:12:05,640 So the transmission of alphabet A is given by points 218 00:12:05,640 --> 00:12:10,470 in the complex plane with odd valued coordinates around the origins. 219 00:12:10,470 --> 00:12:13,520 And the receiver just lies through works by finding the 220 00:12:13,520 --> 00:12:17,560 symbol in the alphabet which is closest in Euclidean distance 221 00:12:17,560 --> 00:12:20,980 to the received symbol. Let's look at this graphically. 222 00:12:21,980 --> 00:12:26,440 This is the set of points for QAM transmission with M equal to two, which 223 00:12:26,440 --> 00:12:33,790 corresponds to two bi-level PAM signals on the real axis and on the imaginary axis. 224 00:12:33,790 --> 00:12:36,630 So that results into four points. 225 00:12:36,630 --> 00:12:41,030 If we increase the number of bits per symbol, we set M equal to four. 226 00:12:41,030 --> 00:12:42,800 That corresponds to two PAM 227 00:12:42,800 --> 00:12:47,180 signals with two bits each. Which makes for a constellation. 228 00:12:47,180 --> 00:12:51,920 This is how these arrangement of points in the complex plane are called. 229 00:12:51,920 --> 00:12:54,402 A constellation of four by four points at 230 00:12:54,402 --> 00:12:58,000 the odd valued coordinates in the complex plane. 231 00:12:58,000 --> 00:13:01,670 If we increase M to 8, then we have 232 00:13:01,670 --> 00:13:06,760 a 256 point constellation with 16 points per side. 233 00:13:06,760 --> 00:13:08,400 Lets look at what happens 234 00:13:08,400 --> 00:13:10,970 when a symbol is received and how we 235 00:13:10,970 --> 00:13:13,790 derive an expression from the probability of the error. 236 00:13:13,790 --> 00:13:16,828 If this is the nominal constellation the transmitter will 237 00:13:16,828 --> 00:13:20,031 choose one of these values for transmission, say this one. 238 00:13:20,031 --> 00:13:21,858 And this value will be corrupted by 239 00:13:21,858 --> 00:13:25,290 noise in the transmission and the receiving process. 240 00:13:25,290 --> 00:13:27,420 And will appear somewhere in the complex plane, 241 00:13:27,420 --> 00:13:31,050 not necessarily exactly on the point it originates from. 242 00:13:32,050 --> 00:13:33,910 The way the slicer operates 243 00:13:33,910 --> 00:13:38,490 is by defining decision regions around each point in the constellation. 244 00:13:38,490 --> 00:13:41,730 So suppose for this point here, the transmitted point, the decision 245 00:13:41,730 --> 00:13:48,680 region is a square of side, 2G, centered around the transmitted point. 246 00:13:48,680 --> 00:13:51,540 So what happens, is that when we receive 247 00:13:51,540 --> 00:13:54,150 symbols, they will not fall on the original 248 00:13:54,150 --> 00:13:56,230 point, but as long as they fall within 249 00:13:56,230 --> 00:13:58,750 the decision region, they will be decoded correctly. 250 00:13:58,750 --> 00:14:01,860 So for instance, here. We will decode this correctly. 251 00:14:01,860 --> 00:14:04,780 Here we will decode this correctly, same here. 252 00:14:04,780 --> 00:14:07,990 But this point for instance falls outside of the decision region 253 00:14:07,990 --> 00:14:11,180 and therefore it will be associated to a different constellation point. 254 00:14:11,180 --> 00:14:12,560 Thereby, causing an error. 255 00:14:13,590 --> 00:14:17,480 To quantify the probability of error, we assume as per usual that each 256 00:14:17,480 --> 00:14:23,050 received symbol is the sum of the transmitted symbol plus a noise sample. 257 00:14:23,050 --> 00:14:23,740 Inter of n. 258 00:14:23,740 --> 00:14:29,020 And we further assume that this noise is a complex value 259 00:14:29,020 --> 00:14:33,330 Gaussian noise of equal variance in the complex and real components. 260 00:14:34,330 --> 00:14:37,168 We're working on a completely digital 261 00:14:37,168 --> 00:14:41,610 system that operates with complex valued quantities. 262 00:14:41,610 --> 00:14:44,020 So we're making a new model for the noise. 263 00:14:44,020 --> 00:14:45,670 And we will see later how to translate 264 00:14:45,670 --> 00:14:49,020 the physical real noise into a complex variable. 265 00:14:50,020 --> 00:14:50,950 With this assumptions. 266 00:14:50,950 --> 00:14:54,420 The probability of error is equal to the probability 267 00:14:54,420 --> 00:14:56,680 that the real part of the noise is larger than 268 00:14:56,680 --> 00:15:00,060 G in magnitude, plus the probability that the imaginary 269 00:15:00,060 --> 00:15:03,470 part of the noise is larger than G in magnitude. 270 00:15:03,470 --> 00:15:06,200 We assume that real and imaginary components of the noise are 271 00:15:06,200 --> 00:15:09,840 independent, and that's why we can split the probability like so. 272 00:15:09,840 --> 00:15:11,550 Now, if you remember the shape of 273 00:15:11,550 --> 00:15:15,105 the decision region, this condition is equivalent to 274 00:15:15,105 --> 00:15:20,010 saying that the noise is pushing the real part of the point 275 00:15:20,010 --> 00:15:26,100 outside of the decision region in either direction and same for the majority part. 276 00:15:26,100 --> 00:15:30,990 Now if we develop this, this is to equal 1 minus the probability the real part of 277 00:15:30,990 --> 00:15:34,920 the noise is less than G and the imaginary part of the noise is less than G. 278 00:15:34,920 --> 00:15:38,750 This is the complimentary condition to what we just wrote above. 279 00:15:38,750 --> 00:15:40,380 And so this is equal to 1 280 00:15:40,380 --> 00:15:43,840 minus the integral over the decision region, d, of 281 00:15:43,840 --> 00:15:47,370 the complex valued probability density function for the noise. 282 00:15:47,370 --> 00:15:50,610 In order to compute this integral, we're going to approximate 283 00:15:50,610 --> 00:15:55,110 the shape of the decision region with the inbound circle. 284 00:15:55,110 --> 00:15:56,735 So instead of using the square. 285 00:15:56,735 --> 00:15:59,840 We're going to a circle centered around the transmission point. 286 00:15:59,840 --> 00:16:04,620 When the conciliation is very dense, this approximation is quite accurate. 287 00:16:04,620 --> 00:16:05,550 With this approximation, 288 00:16:05,550 --> 00:16:09,320 we can compute the integral exactly for a Gaussian distribution. 289 00:16:09,320 --> 00:16:12,170 And if we assume that the variance of the noise is 290 00:16:12,170 --> 00:16:17,920 sigma 0 squared over 2 in each component, real and imaginary. 291 00:16:17,920 --> 00:16:19,980 It turns out that the probability of error is equal 292 00:16:19,980 --> 00:16:23,540 to e to the minus g square over sigma 0 square. 293 00:16:23,540 --> 00:16:26,360 Now to obtain a probability of error as a function of the signal 294 00:16:26,360 --> 00:16:31,170 to noise ratio, we have to compute the power of the transmitted signal. 295 00:16:31,170 --> 00:16:34,550 So if all symbols are equiprobable and independent. 296 00:16:34,550 --> 00:16:37,180 It turns out that the variance of the signal is G 297 00:16:37,180 --> 00:16:40,410 square times 1 over 2 to the power of M, which 298 00:16:40,410 --> 00:16:43,840 is a probability of each symbol times the sum of over 299 00:16:43,840 --> 00:16:47,632 all symbols in the alphabet of the magnitude of the symbol squared. 300 00:16:47,632 --> 00:16:53,670 Now it's a little bit tedious but we can solve it exactly for M 301 00:16:53,670 --> 00:16:56,280 and it turns out that the power of the of the transmitted signal is G 302 00:16:56,280 --> 00:16:59,520 square, 2 3rds, 2 the M minus 1. 303 00:16:59,520 --> 00:17:02,960 Now if we plug this in to the formula for the probability of error, that we've 304 00:17:02,960 --> 00:17:07,040 seen before, we get that the result is 305 00:17:07,040 --> 00:17:11,400 an exponential function where the argument is minus 3. 306 00:17:11,400 --> 00:17:16,490 That multiplies 2 to the minus m plus 1, that multiplies the signal's noise ratio. 307 00:17:17,600 --> 00:17:21,490 We can plot this probability of error in a log log scale, like we did before. 308 00:17:21,490 --> 00:17:24,350 And we can parameterize the curve as a 309 00:17:24,350 --> 00:17:27,570 function of the number of points in the constellation. 310 00:17:27,570 --> 00:17:30,880 So here you have the curve for a four point constellation. 311 00:17:30,880 --> 00:17:34,680 Here's the curve for 16 points. And here's the curve for 64 points. 312 00:17:34,680 --> 00:17:37,150 Now, you can see that for a given signal to noise 313 00:17:37,150 --> 00:17:41,450 ratio, the probability of error increases with the number of points. 314 00:17:41,450 --> 00:17:42,630 Why is that? 315 00:17:42,630 --> 00:17:44,820 Well, if the signal to noise remains the same, 316 00:17:44,820 --> 00:17:46,510 and we assume that the noise is always at the 317 00:17:46,510 --> 00:17:47,505 same level. 318 00:17:47,505 --> 00:17:50,690 Then it means that the power of the signal remains constant as well. 319 00:17:50,690 --> 00:17:54,710 In that case, if the number of points increases G has to become 320 00:17:54,710 --> 00:17:59,490 smaller in order to accommodate a larger number of points for the same power. 321 00:17:59,490 --> 00:18:03,490 But if G becomes smaller then the decision reaches become smaller. 322 00:18:03,490 --> 00:18:06,010 The separation between points becomes smaller and 323 00:18:06,010 --> 00:18:08,740 the decision process becomes more vulnerable to noise. 324 00:18:10,300 --> 00:18:11,980 So in the end, here is the final recipe 325 00:18:11,980 --> 00:18:14,150 to design a QAM transmitter. 326 00:18:14,150 --> 00:18:16,628 First you pick a probability of error that you can live with. 327 00:18:16,628 --> 00:18:18,380 In general 10 to the minus 6 is 328 00:18:18,380 --> 00:18:20,930 an acceptable probability of error at the symbol level. 329 00:18:22,070 --> 00:18:23,720 Then you find out the signal to noise 330 00:18:23,720 --> 00:18:26,550 ratio that is imposed by the channel's power constraint. 331 00:18:26,550 --> 00:18:32,140 Once you have that, you can find the size of your constellation by finding m which 332 00:18:32,140 --> 00:18:37,530 based on the previous equations is the log in base 2 of 1 minus 3 over 2 333 00:18:37,530 --> 00:18:39,890 times the signal to noise ratio divided by 334 00:18:39,890 --> 00:18:42,350 the natural logarithm of the probability of error. 335 00:18:42,350 --> 00:18:45,560 Of course you will have to round this to a suitable integer value and 336 00:18:45,560 --> 00:18:49,170 potentially to an even power of 2 in order to have the square constellation. 337 00:18:49,170 --> 00:18:53,280 The final data rate of your system will be m the number of bits 338 00:18:53,280 --> 00:18:58,460 per symbol, times w which if you remember is the board rate of system. 339 00:18:58,460 --> 00:19:02,990 And corresponds to the bandwidth allowed for by the channel. 340 00:19:02,990 --> 00:19:07,820 So, we know how to fit the bandwidth constraint by upsampling. 341 00:19:07,820 --> 00:19:12,560 With QAM, we know how many bits per symbol we can use given the power constraint. 342 00:19:12,560 --> 00:19:14,310 And so we know the theoretical throughput 343 00:19:14,310 --> 00:19:16,720 of the transmitter, for a given reliability figure. 344 00:19:18,370 --> 00:19:20,560 However, the question remains, how are we going 345 00:19:20,560 --> 00:19:24,910 to send complex value symbols over a physical channel. 346 00:19:24,910 --> 00:19:28,880 It's time therefore to stop the suspension of disbelief and 347 00:19:28,880 --> 00:19:33,080 look at techniques to do complex signaling over a real value channel.