1 00:00:00,600 --> 00:00:03,780 Hi and welcome to Module 9.4 of Digital Signal Processing. 2 00:00:03,780 --> 00:00:07,400 In the previous module we saw an interesting signalling scheme that allow 3 00:00:07,400 --> 00:00:12,060 us to increase the data rate while keeping the same probability of error for 4 00:00:12,060 --> 00:00:13,620 a given power constraint. 5 00:00:13,620 --> 00:00:17,060 The problem is that the efficient communication alphabet that we devised 6 00:00:17,060 --> 00:00:18,650 is complex valued, 7 00:00:18,650 --> 00:00:21,930 where as we know that physical channels can only handle real values. 8 00:00:21,930 --> 00:00:24,050 So in this module we'll see how to transmit and 9 00:00:24,050 --> 00:00:28,430 recover a complex valued simple stream over a real value channel. 10 00:00:28,430 --> 00:00:31,996 Then we will follow this with a concrete design example for the telephone channel. 11 00:00:31,996 --> 00:00:36,370 And finally we will compare the performance of the system to the ultimate 12 00:00:36,370 --> 00:00:40,760 limit in data rate that is given us by channel's capacity formula. 13 00:00:40,760 --> 00:00:44,930 So let's review where we stand in terms of transmitter design. 14 00:00:44,930 --> 00:00:48,540 We have the user's bitstream that comes into the system. 15 00:00:48,540 --> 00:00:51,170 This is sent through a scrambler 16 00:00:51,170 --> 00:00:55,890 that makes sure that the resulting bitstream is equiprobable. 17 00:00:55,890 --> 00:00:59,340 The mapper will split the bitstream into m bit chunks, 18 00:00:59,340 --> 00:01:03,440 any chunk will be associated to a complex valued symbol. 19 00:01:03,440 --> 00:01:06,220 This will create a complex value sequence a[n]. 20 00:01:06,220 --> 00:01:09,960 And to fit that over the bandwidth prescribed by the channel 21 00:01:09,960 --> 00:01:11,550 we have to upsample it, 22 00:01:11,550 --> 00:01:15,550 which means inserting K minus one zeros after each symbol of the sequence. 23 00:01:15,550 --> 00:01:20,960 And then low passing the sequence with a filter with cutoff frequency pi over K. 24 00:01:20,960 --> 00:01:25,500 Now the sequence b[n], the upsample sequence that now fits 25 00:01:25,500 --> 00:01:29,190 the bandwidth constraint, is a complex valued baseband signal. 26 00:01:29,190 --> 00:01:30,540 Graphically we can show this. 27 00:01:30,540 --> 00:01:34,670 If this is the bandwidth constraint of the channel, positive and 28 00:01:34,670 --> 00:01:35,870 negative frequencies. 29 00:01:35,870 --> 00:01:40,360 The sequence B of N, is a baseband sequence whose bandwidth 30 00:01:40,360 --> 00:01:43,160 fits the bandwidth prescribed by the channel. 31 00:01:43,160 --> 00:01:44,170 Now the question is, 32 00:01:44,170 --> 00:01:49,800 how do we modulate this while also making the transmitted signal real? 33 00:01:49,800 --> 00:01:53,180 And making it real in a way that we can retrieve it and 34 00:01:53,180 --> 00:01:56,510 reconstruct the baseband complex signal at the receiver. 35 00:01:56,510 --> 00:01:58,753 So the way it's done is very straight forward. 36 00:01:58,753 --> 00:02:03,690 We just take the real part of the baseband signal, multiplied by e to the j omega c 37 00:02:03,690 --> 00:02:08,630 n where omega c is the center frequency of the bandwidth of the channel. 38 00:02:08,630 --> 00:02:10,440 If we work it out, the math is very simple. 39 00:02:10,440 --> 00:02:15,241 So we have the product, of course, of the real part and imaginary part of 40 00:02:15,241 --> 00:02:20,141 the baseband signal multiplied by cosine of omega cn + j sine of omega cn. 41 00:02:20,141 --> 00:02:24,899 And the real part of this equation is simply the real parts of the baseband 42 00:02:24,899 --> 00:02:29,111 signal multiplied by cosine of omega c n minus the imaginary part 43 00:02:29,111 --> 00:02:33,630 of the baseband signal multiplied by sine of omega c n. 44 00:02:33,630 --> 00:02:37,420 Now, you see here, we're multiplying each component, 45 00:02:37,420 --> 00:02:42,360 the real imaginary part of the baseband signal, by a carrier and 46 00:02:42,360 --> 00:02:45,990 the two carriers are orthogonal to each other. 47 00:02:45,990 --> 00:02:48,910 So, we have a cosine carrier and a sine carrier. 48 00:02:48,910 --> 00:02:53,140 Now, these two carriers are othogonal because they're shifted by a phase of 49 00:02:53,140 --> 00:02:54,140 90 degrees. 50 00:02:54,140 --> 00:02:58,200 Now, when two things are 90 degrees apart, they're said to be in quadrature. 51 00:02:58,200 --> 00:03:02,730 And so the first component of the signal will be called the in phase part and 52 00:03:02,730 --> 00:03:07,095 the second component, the one modulated by the sine, will be the quadrature part. 53 00:03:07,095 --> 00:03:11,440 And this is the nomenclature behind the acronym QIM. 54 00:03:11,440 --> 00:03:17,640 Okay, so now this signal s of n is clearly real and 55 00:03:17,640 --> 00:03:22,255 our next step will be to show that if we receive this at the receiver 56 00:03:22,255 --> 00:03:26,195 we will be able to recover b of n, the complex baseband signal. 57 00:03:26,195 --> 00:03:27,157 But before we do that, 58 00:03:27,157 --> 00:03:29,901 let's look at the modulation process in the frequency domain. 59 00:03:29,901 --> 00:03:34,090 Because the intuition will help us understand why we can recover the baseband 60 00:03:34,090 --> 00:03:36,760 signal exactly at the receiver. 61 00:03:36,760 --> 00:03:41,700 So in the next few diagrams, we will show the spectra of br and bi, 62 00:03:41,700 --> 00:03:44,890 the in-phase and quadrature components of the baseband signal. 63 00:03:44,890 --> 00:03:49,169 Let's assume for the sake of convenience that these spectra are purely real. 64 00:03:49,169 --> 00:03:52,376 And we will indicate our real quantities with shades of blue and 65 00:03:52,376 --> 00:03:55,380 purely imaginary quantities with shades of pink. 66 00:03:55,380 --> 00:03:56,870 The math would stay the same for 67 00:03:56,870 --> 00:04:02,250 arbitrary spectra but this assumption will allow us to draw a simpler picture. 68 00:04:02,250 --> 00:04:07,625 So if we start by plotting the spectrum of the real part of the baseband signal, 69 00:04:07,625 --> 00:04:10,080 let's assume it has a shape like this. 70 00:04:10,080 --> 00:04:13,590 Then we plot the spectrum of the imaginary part of the baseband signal. 71 00:04:13,590 --> 00:04:14,830 And let's assume that, again, 72 00:04:14,830 --> 00:04:18,400 it's a purely real spectrum except that it has a slightly different shape. 73 00:04:18,400 --> 00:04:20,130 And these shapes are completely immaterial, 74 00:04:20,130 --> 00:04:24,120 they just help us see what happens during the modulation process. 75 00:04:24,120 --> 00:04:28,030 The real part is multiplied by a cosine of omega cn and 76 00:04:28,030 --> 00:04:31,040 therefore a cosine modulation takes place. 77 00:04:31,040 --> 00:04:35,378 It is shifted left or right and centered in omega c. 78 00:04:35,378 --> 00:04:37,590 For the imaginary part of the baseband signal, 79 00:04:37,590 --> 00:04:42,200 the modulation takes place with a sine, and there is a change of sign involved. 80 00:04:42,200 --> 00:04:47,460 So the resulting spectrum will be purely imaginary, and it will be shifted at omega 81 00:04:47,460 --> 00:04:51,710 c and minus omega c, with a change of sign in the negative part of the spectrum. 82 00:04:51,710 --> 00:04:55,420 And so this is a spectrum of the signal that we actually send over 83 00:04:55,420 --> 00:04:56,470 the real channel. 84 00:04:56,470 --> 00:04:59,590 As you can see the signal is real and indeed 85 00:04:59,590 --> 00:05:03,950 the spectrum has Hermitian symmetry in the sense that its real part is symmetric and 86 00:05:03,950 --> 00:05:06,053 the imaginary part is anti symmetric. 87 00:05:06,053 --> 00:05:09,890 Okay so now let's assume that the transmission goes well and 88 00:05:09,890 --> 00:05:14,240 our job at the receiver is to recover the complex baseband signal. 89 00:05:14,240 --> 00:05:18,030 In either approach we try the usual method which is multiplying the signal 90 00:05:18,030 --> 00:05:19,220 by the carrier. 91 00:05:19,220 --> 00:05:22,270 Well in this case we have two carriers, the cosine and sine. 92 00:05:22,270 --> 00:05:24,997 So let's start by multiplying by the cosine. 93 00:05:24,997 --> 00:05:29,781 And so if we take s[n] which is the transmitted and then received signal, and 94 00:05:29,781 --> 00:05:32,040 we multiply it by cosine of omega c n. 95 00:05:32,040 --> 00:05:36,322 What we obtain is b r, the real part of the baseband signal, 96 00:05:36,322 --> 00:05:41,459 multiplied by cosine square of omega c n minus b i, the imaginary part of 97 00:05:41,459 --> 00:05:48,060 the baseband signal, multiplied by sine of omega c n times cosine of omega c n. 98 00:05:48,060 --> 00:05:52,650 Now we use some really basic trigonometry to express these functions 99 00:05:52,650 --> 00:05:55,530 as functions of 2 times omega c n. 100 00:05:55,530 --> 00:06:01,640 So cosine squared of omega c n becomes 1 + cosine of 2 omega c n over 2. 101 00:06:01,640 --> 00:06:07,440 And sine of omega c n cosine of omega c n becomes sine of 2 omega cn over 2. 102 00:06:07,440 --> 00:06:11,470 And if we rearrange the terms of this equation, 103 00:06:11,470 --> 00:06:16,912 we obtain one-half times the real part of the baseband signal, 104 00:06:16,912 --> 00:06:22,983 plus one-half of a term that only contains carriers at twice omega c n. 105 00:06:22,983 --> 00:06:27,956 Which is really similar to the situation we had in the simple cosine modulation of 106 00:06:27,956 --> 00:06:30,730 the modulation for a real signal. 107 00:06:30,730 --> 00:06:34,137 Again, it's much more intuitive if we look at this in the frequency domain. 108 00:06:34,137 --> 00:06:38,280 And so, this was the signal that was transmitted and then received. 109 00:06:38,280 --> 00:06:43,514 If we multiply this by cosine of omega c n, we obtain one-half 110 00:06:43,514 --> 00:06:48,940 of the spectrum of the real part of the baseband signal. 111 00:06:48,940 --> 00:06:54,240 Plus spurious components at twice the modulation frequency. 112 00:06:55,410 --> 00:06:57,930 Now we know what to do to get rid of the spurious components, 113 00:06:57,930 --> 00:07:00,180 we just use a low pass filter. 114 00:07:00,180 --> 00:07:03,490 In this case we can use again a raised cosine. 115 00:07:03,490 --> 00:07:08,080 When we use this we have what is called a matched filter configuration where we use 116 00:07:08,080 --> 00:07:11,970 the same filter at the receiver that we used at the transmitter. 117 00:07:11,970 --> 00:07:16,210 This filter will eliminate the components out of the baseband. 118 00:07:16,210 --> 00:07:21,030 And we will recover the original, real part of the baseband signal. 119 00:07:21,030 --> 00:07:23,780 We can do exactly the same for the quadrature component, for 120 00:07:23,780 --> 00:07:25,660 the complex part of the baseband signal. 121 00:07:25,660 --> 00:07:30,800 And in this case, we multiply the received signal by sine of omega c n. 122 00:07:30,800 --> 00:07:34,638 And if we work out the math with simple trigonometry once again, 123 00:07:34,638 --> 00:07:39,331 we obtain minus one-half times the imaginary part of the baseband signal plus 124 00:07:39,331 --> 00:07:42,906 out of band components at twice the modulation frequency. 125 00:07:42,906 --> 00:07:45,610 Which we can filter out with a low pass filter. 126 00:07:45,610 --> 00:07:49,832 So, we have proven that we can recover the baseband signal even when 127 00:07:49,832 --> 00:07:52,512 the original baseband signal is complex. 128 00:07:52,512 --> 00:07:56,700 The final design for the transmitter is the following. 129 00:07:56,700 --> 00:08:01,653 Bitstream, scrambler, mapper, up sampler. 130 00:08:01,653 --> 00:08:05,750 We multiply the up sample signal by e to the j omega c n. 131 00:08:07,190 --> 00:08:10,030 We obtain a complex passband signal. 132 00:08:10,030 --> 00:08:14,270 We take the real part, and finally, we have the sequence of samples 133 00:08:14,270 --> 00:08:17,942 that we can send to the D/A converter, and then over onto the channel. 134 00:08:19,100 --> 00:08:23,590 The receiver will receive the analog signal over the channel. 135 00:08:23,590 --> 00:08:26,970 We'll sample it at the proper sampling rate, and then demodulate it. 136 00:08:26,970 --> 00:08:30,540 So the signal is split into two parts that are identical. 137 00:08:30,540 --> 00:08:34,272 The first copy will be multiplied by cosine of omega cn. 138 00:08:34,272 --> 00:08:38,280 And the second copy will be multiplied by sine of omega cn. 139 00:08:38,280 --> 00:08:42,580 Both will go through a low pass filter that is really the matched filter 140 00:08:42,580 --> 00:08:46,250 of the upsampling filter we used at the transmitter. 141 00:08:46,250 --> 00:08:51,340 And this will give us the real part of the original baseband signal and 142 00:08:51,340 --> 00:08:53,595 the imaginary part of the original baseband signal. 143 00:08:53,595 --> 00:08:57,010 We multiply this by j and we sum them together and 144 00:08:57,010 --> 00:09:01,330 now we have the baseband signal, which will go through a down sampler. 145 00:09:01,330 --> 00:09:03,167 We keep one sample out of K. 146 00:09:03,167 --> 00:09:09,900 And this will give us the estimated symbol sequence created by the transmitter. 147 00:09:09,900 --> 00:09:13,080 The Slicer will find the closest alphabet symbol, and 148 00:09:13,080 --> 00:09:16,288 recover the chunk of mbits associated to the symbol. 149 00:09:16,288 --> 00:09:19,750 And finally, the Descrambler will undo the randomization and 150 00:09:19,750 --> 00:09:22,240 recover the original user data stream. 151 00:09:22,240 --> 00:09:26,200 So let's see how we can put everything we've learned so far together and 152 00:09:26,200 --> 00:09:29,630 design a practical system to send data over the telephone channel. 153 00:09:30,650 --> 00:09:35,041 Suppose that the bandwidth constraint for the telephone channel 154 00:09:35,041 --> 00:09:39,919 stipulates that we can only transmit data from 450Hz to 2850Hz. 155 00:09:39,919 --> 00:09:43,183 This gives us a usable bandwidth, W, 156 00:09:43,183 --> 00:09:48,797 of 2,400Hz with a center frequency Fc of 1,650 hertz. 157 00:09:48,797 --> 00:09:53,410 Now remember, in our all digital paradigm, we have to pick 158 00:09:53,410 --> 00:09:57,830 a sampling frequency that is at least twice the highest frequency there. 159 00:09:57,830 --> 00:10:01,960 So, F max * 2. 160 00:10:01,960 --> 00:10:07,228 This would give us a sampling frequency of at least 5,700 hertz. 161 00:10:07,228 --> 00:10:11,270 But also, we're going to use an upsampling factor, which is an integer. 162 00:10:11,270 --> 00:10:13,720 And the trick is to pick a sampling frequency 163 00:10:13,720 --> 00:10:16,870 which is an integer multiple of the bandwidth. 164 00:10:16,870 --> 00:10:21,930 So if we pick the multiple equal to 3, then we get the sampling frequency 165 00:10:21,930 --> 00:10:26,430 of 7200Hz, which of course satisfies Nyquist criterion. 166 00:10:27,500 --> 00:10:30,180 When we translate the specs into the digital domain, 167 00:10:30,180 --> 00:10:33,930 we find that the modulating frequency is 0.458 pi. 168 00:10:35,570 --> 00:10:37,340 Now let's tackle the power constraint and 169 00:10:37,340 --> 00:10:41,850 assume that the telephone line has a maximum SNR of 22dB's. 170 00:10:41,850 --> 00:10:44,360 You have to pick a probability of error that you can live with and 171 00:10:44,360 --> 00:10:46,402 let's say that we pick 10 to the -6. 172 00:10:47,430 --> 00:10:52,272 If we use QAM, we can use the formula that we saw in the previous module to 173 00:10:52,272 --> 00:10:54,371 find the size of the alphabet or 174 00:10:54,371 --> 00:10:58,592 alternatively the number of bits per symbol that we can send. 175 00:10:58,592 --> 00:11:02,235 And the formula is this one, and when we plug in the values for 176 00:11:02,235 --> 00:11:05,378 the probability of error and signal to noise ratio, 177 00:11:05,378 --> 00:11:09,048 we find that we can send at least 4 bits per symbol. 178 00:11:09,048 --> 00:11:13,390 And with 4 bits per symbol we will have a constellation of 16 points. 179 00:11:13,390 --> 00:11:16,211 So, that would look like this. 180 00:11:19,741 --> 00:11:24,850 The final data rate, remember, is the baud rate times the bits per symbol. 181 00:11:24,850 --> 00:11:28,490 The baud rate is equal to the bandwidth, so 2,400, and so 182 00:11:28,490 --> 00:11:31,190 we have a total of 9,600 bits per second. 183 00:11:32,420 --> 00:11:38,394 This is actually an operating mode of a modem standard called V.24. 184 00:11:38,394 --> 00:11:42,510 It was popular in the 90s, and it is still sometimes used in fax machines. 185 00:11:44,180 --> 00:11:47,370 Now the question is, are we doing good with the respect to the maximum amount of 186 00:11:47,370 --> 00:11:49,400 information that we can send over this channel? 187 00:11:50,540 --> 00:11:54,180 Remember, we used very specific design choices to derive this 188 00:11:54,180 --> 00:11:56,680 figure of 9,600 bits per second. 189 00:11:56,680 --> 00:12:01,710 A specific modulation scheme, a specific probability of error and so on, so forth. 190 00:12:01,710 --> 00:12:03,410 What is the best one can do? 191 00:12:03,410 --> 00:12:05,110 Well this is a complex question and 192 00:12:05,110 --> 00:12:09,580 an exhaustive answer would require several lectures in information theory. 193 00:12:09,580 --> 00:12:14,990 But there is a formula derived by Claude Shannon in the late 1940s that states 194 00:12:14,990 --> 00:12:20,550 the capacity of a channel given its bandwidth and its signal to noise ratio. 195 00:12:20,550 --> 00:12:25,160 And this capacity specifies the amount of information that we can sell reliably, 196 00:12:25,160 --> 00:12:29,280 meaning with an arbitrarily low probably of error over a channel. 197 00:12:29,280 --> 00:12:31,389 The formula, unfortunately, is not constructed. 198 00:12:31,389 --> 00:12:35,681 It doesn't tell us how to send this data, but gives an upper bound on the amount 199 00:12:35,681 --> 00:12:39,070 of information that can be sent over the channel. 200 00:12:39,070 --> 00:12:41,860 So, for instance, for the parameters that we used before. 201 00:12:41,860 --> 00:12:47,535 The maximum capacity for that channel would be 17,500 bits per second. 202 00:12:47,535 --> 00:12:48,975 And with our design scheme, 203 00:12:48,975 --> 00:12:53,005 we're basically hitting half the capacity of the channel. 204 00:12:53,005 --> 00:12:56,575 The gap can be narrowed if we use more sophisticated modulation and 205 00:12:56,575 --> 00:12:57,925 data coding techniques. 206 00:12:57,925 --> 00:12:59,999 But as I said, to explore these topics, 207 00:12:59,999 --> 00:13:03,479 we would have to start an entirely new class in information theory.