1 00:00:01,120 --> 00:00:04,170 Hi, and welcome to module 9.2 of digital signal processing. 2 00:00:04,170 --> 00:00:07,842 We are talking about digital communication systems, and in this 3 00:00:07,842 --> 00:00:13,120 module, we will talk about how to fulfil the bandwidth constraint. 4 00:00:13,120 --> 00:00:15,904 The way that we're going to do this is by introducing an operation called 5 00:00:15,904 --> 00:00:19,694 upsampling. And we will see how upsampling will allow 6 00:00:19,694 --> 00:00:23,720 us to fit the spectrum generated by the transmitter onto the band allowed for by 7 00:00:23,720 --> 00:00:27,851 the channel. Remember that our assumption is that the 8 00:00:27,851 --> 00:00:31,470 signal generated by the transmitter is a wide sequence. 9 00:00:31,470 --> 00:00:35,40 And therefore, it's spiral spectral density will be full band. 10 00:00:35,40 --> 00:00:38,90 What we need to do is to shrink the support of it's spiral spectral density 11 00:00:38,90 --> 00:00:42,350 so that it fits on the band allowed by the channel. 12 00:00:42,350 --> 00:00:45,497 The way we do this is by using multirate techniques. 13 00:00:45,497 --> 00:00:49,31 In multirate, the goal is to increase or decrease the number of samples of a 14 00:00:49,31 --> 00:00:52,454 digital signal. One way to do this is to interpolate the 15 00:00:52,454 --> 00:00:55,550 digital signal into a continuous time signal. 16 00:00:55,550 --> 00:00:58,958 And then resample the interpolation at a different sampling rate. 17 00:00:58,958 --> 00:01:02,576 However, we want to avoid the transition to discrete time, and we want to perform 18 00:01:02,576 --> 00:01:07,310 this aritficial change of sampling rate entirely in the digital domain. 19 00:01:07,310 --> 00:01:10,670 Let's consider the up sampling operation, which is really what we're interested in. 20 00:01:10,670 --> 00:01:15,224 And let's look at how to do this, going through an interpolation and resampling 21 00:01:15,224 --> 00:01:19,462 operation first. So we have a discrete time signal here, 22 00:01:19,462 --> 00:01:24,610 we interpolate with the given period Ts. We obtain a continuous time signal and 23 00:01:24,610 --> 00:01:28,150 then we sample this continuous time signal with a period that is k times 24 00:01:28,150 --> 00:01:32,230 smaller than the original interpolation sample, and we obtain another discrete 25 00:01:32,230 --> 00:01:37,473 times sequence here. Graphically, assume this is our discrete 26 00:01:37,473 --> 00:01:41,900 times signal. The interpolation to continuous time will 27 00:01:41,900 --> 00:01:45,680 give us this and the resampling with a smaller sampling period will give us a 28 00:01:45,680 --> 00:01:49,460 higher density of samples on the same curve so that the resulting upsampled 29 00:01:49,460 --> 00:01:55,424 signal would look like this. Now, we are interpolating to continuous 30 00:01:55,424 --> 00:01:59,23 time, so the choice of the sampling period is completely arbitrary for 31 00:01:59,23 --> 00:02:02,622 simplicity as per usual, which is Ts equal to 1, and here we have that the 32 00:02:02,622 --> 00:02:08,720 interplay signal is given by the standard sync interplay formula. 33 00:02:08,720 --> 00:02:12,929 When we resample, with the period that is 1 over k, k times smaller than the 34 00:02:12,929 --> 00:02:19,420 original, we are taking samples of the interplayed function at n over big K. 35 00:02:19,420 --> 00:02:24,397 And the result Is an interpolation formula, where the sync function now, is 36 00:02:24,397 --> 00:02:29,458 centered at fractional intervals, so n over big K. 37 00:02:29,458 --> 00:02:32,990 In the frequency domain, the process looks like so. 38 00:02:32,990 --> 00:02:39,770 Imagine we have a discreet time signal, whose spectrum is limited to 3 pi over 4. 39 00:02:39,770 --> 00:02:44,117 We interpolate it to continuous time, and we get an analog spectrum that looks like 40 00:02:44,117 --> 00:02:49,340 this, where our nyquist frequency is omega n, equal to pi over Ts. 41 00:02:49,340 --> 00:02:54,254 When we resample, with the sampling period which is k times smaller, that is 42 00:02:54,254 --> 00:02:59,220 equivalent to multiplying nyquist frequency by k. 43 00:02:59,220 --> 00:03:01,685 So we make it bigger and we move it over here. 44 00:03:01,685 --> 00:03:05,270 And when we plot the result in digital spectrum. 45 00:03:05,270 --> 00:03:09,650 We map, as per usual, the nyquist frequency to pi, which corresponds to a 46 00:03:09,650 --> 00:03:14,694 contraction of the frequency spectrum by a factor of K. 47 00:03:14,694 --> 00:03:18,750 If here, we choose k, which is equal to 3. 48 00:03:18,750 --> 00:03:22,404 We get that what was the highest frequency of the spectrum, 3 pi over 4, 49 00:03:22,404 --> 00:03:27,562 now becomes pi over 4. Can we do this completely in the digital 50 00:03:27,562 --> 00:03:30,380 domain? Well the idea is that we need to increase 51 00:03:30,380 --> 00:03:35,262 the number of samples by a factor of K. And obviously the sample sequence will 52 00:03:35,262 --> 00:03:40,294 have to coincide with the original values when the index of the up sample sequence 53 00:03:40,294 --> 00:03:45,254 is a multiple of K. There are several reasons why this is so, 54 00:03:45,254 --> 00:03:48,558 but probably the most intuitive one is that if we then discard the extra 55 00:03:48,558 --> 00:03:53,580 samples, we should be able to obtain the original sequence again. 56 00:03:53,580 --> 00:03:57,606 So for lack of a better strategy, we can start by building a sequence where we put 57 00:03:57,606 --> 00:04:01,571 the original samples, every K samples and then we put zeros everywhere else in 58 00:04:01,571 --> 00:04:06,90 between. So for example, for k equal to 3, the 59 00:04:06,90 --> 00:04:10,430 upsample sequence, say this is m equal to 0, will be equal to x0 and m equal to 60 00:04:10,430 --> 00:04:14,310 zero. Then we put two zeros, then we put x1, 61 00:04:14,310 --> 00:04:18,400 then we put two zero, then we put x2, and then we put two zeroes. 62 00:04:18,400 --> 00:04:22,54 We can see this in the time domain start with the same sequence that we showed 63 00:04:22,54 --> 00:04:25,625 before. And what we are doing, we're simply 64 00:04:25,625 --> 00:04:31,436 introducing zeroes between each stamp. With this choice, the Fourier transform 65 00:04:31,436 --> 00:04:34,820 of the upsample sequence is rather easy to compute. 66 00:04:34,820 --> 00:04:40,524 We just write out the standard DTFT formula, but now here, we remember that 67 00:04:40,524 --> 00:04:49,80 xu of m will be equal to zero every time that m is not a multiple of K. 68 00:04:49,80 --> 00:04:53,790 And so with this, we can simplify the sum and use only the known zero terms. 69 00:04:53,790 --> 00:04:57,450 And we get the sum from n that goes to minus infinity to plus infinity of x of 70 00:04:57,450 --> 00:05:03,380 n, which is our original sequence, that multiplies e to the minis j, omega nK. 71 00:05:03,380 --> 00:05:08,70 And so, this is simply a scaling of the frequency axis by a factor of K. 72 00:05:08,70 --> 00:05:11,486 Graphically, we can plot the digital spectrum and we know that now, since 73 00:05:11,486 --> 00:05:15,238 we're multiply the frequency access be a factor of K, there will be a shrinkage of 74 00:05:15,238 --> 00:05:20,138 the frequency access like this. But we should never forget that the 75 00:05:20,138 --> 00:05:24,734 digital spectrum is 2 pi periodic. So lets plot this explicitly, for minus 5 76 00:05:24,734 --> 00:05:28,264 pi to 5 pi. If we choose k equal to 3, we're mapping 77 00:05:28,264 --> 00:05:33,810 the interval from minus 3 pi to 3 pi back onto the minus pi, pi interval. 78 00:05:33,810 --> 00:05:36,982 And when we do that, we get something that is very close to what we obtained 79 00:05:36,982 --> 00:05:41,728 going through the analog domain. In the sense that this frequency here is 80 00:05:41,728 --> 00:05:46,205 again pi over 4. But, we have extra copies that have crop 81 00:05:46,205 --> 00:05:51,198 in the main frequency interval. Now, we know what to do in this cases, we 82 00:05:51,198 --> 00:05:55,50 apply some drastic low pass filter to get rid of them. 83 00:05:55,50 --> 00:05:59,673 We choose an ideal low-pass filter with cutoff frequency pi over K, because this 84 00:05:59,673 --> 00:06:05,520 is where the original pi in the frequency spectrum would be mapped to. 85 00:06:05,520 --> 00:06:08,692 And this will get rid of the extra copies, and leave us with a spectrum that 86 00:06:08,692 --> 00:06:14,220 is identical to what we obtain using an interpolator followed by a sampler. 87 00:06:14,220 --> 00:06:17,30 So now let's look at the procedure back in the time domain. 88 00:06:17,30 --> 00:06:21,766 So the first step is to insert K minus 1 zeros after each sample, followed by an 89 00:06:21,766 --> 00:06:26,802 ideal low-pass filter. And we choose the cutoff frequency for 90 00:06:26,802 --> 00:06:31,250 the filter to be pi over K as we saw in the previous graph. 91 00:06:31,250 --> 00:06:36,780 So now, the resulting sequence is simply the convolution of the upsampled sequence 92 00:06:36,780 --> 00:06:40,836 with zeroes. And the impulse response of the filter 93 00:06:40,836 --> 00:06:45,440 that, with this cutoff frequency will be simply sink of n over K. 94 00:06:45,440 --> 00:06:49,720 And if we work out the convolution sum, we have this summation here. 95 00:06:49,720 --> 00:06:55,850 But again, we remember that of these terms, only 1 every K will be non zero. 96 00:06:55,850 --> 00:07:00,500 So we replace i with mK, and we sum over m. 97 00:07:00,500 --> 00:07:04,460 And we get the sum for m that goes to minus infinity to plus infinity of x of 98 00:07:04,460 --> 00:07:10,774 m, sync of n over K minus m. Which is exactly the same formula we got 99 00:07:10,774 --> 00:07:16,287 using an interpolator and a sample. As we've mentioned before, if we have an 100 00:07:16,287 --> 00:07:18,978 upsampled sequence we can always recover the original sequence by downsampling, 101 00:07:18,978 --> 00:07:23,270 which means we keep only one sample out of k and throw away the rest. 102 00:07:23,270 --> 00:07:27,428 Now, this is obvious in the case of an upsample sequence where we just introduce 103 00:07:27,428 --> 00:07:31,523 K minus 1 zeroes every sample, but it is also true for a filtered sequence where 104 00:07:31,523 --> 00:07:37,223 we used an ideal filter. Or any other filter that fulfills the 105 00:07:37,223 --> 00:07:41,650 interpolation properties that we have seen in module 6. 106 00:07:41,650 --> 00:07:46,258 In other words, we want impulse response to be equal to 1 for n equal to 0, and to 107 00:07:46,258 --> 00:07:52,63 be equal to 0, for all multiples of K. In general, downsampling is a more 108 00:07:52,63 --> 00:07:55,939 complex operation that upsampling, just like sampling is more complicated than 109 00:07:55,939 --> 00:07:59,550 interpolation. We have the pesky problem of aliasing, 110 00:07:59,550 --> 00:08:04,96 because we are throwing away information. We will not develop the properties of the 111 00:08:04,96 --> 00:08:08,310 downsampling operator in detail because we will not need it in the following. 112 00:08:08,310 --> 00:08:11,773 But you're encouraged to read about multirate signal processing in the book. 113 00:08:11,773 --> 00:08:15,982 So let's go back to the bandwidth constraint, as you remember the channel 114 00:08:15,982 --> 00:08:21,40 imposes that we only use frequencies between Fmin and Fmax. 115 00:08:21,40 --> 00:08:24,630 We also want to translate these requirements into the digital domain. 116 00:08:24,630 --> 00:08:28,851 So we choose a sampling frequency and Fs over two, half of our sampling frequency 117 00:08:28,851 --> 00:08:33,20 will be our nyquist frequency in the analog domain. 118 00:08:33,20 --> 00:08:36,760 Now, here's a new trick. Compute the positive bandwidth. 119 00:08:36,760 --> 00:08:40,725 Namely, the width of the channel bandwidth on the positive axis, and call 120 00:08:40,725 --> 00:08:43,690 that W. Now, pick the sample frequency so that 121 00:08:43,690 --> 00:08:47,204 two things happen. First of all, the sample frequency will 122 00:08:47,204 --> 00:08:51,488 have to be at least twice the maximum frequency that we can use in the channel 123 00:08:51,488 --> 00:08:56,110 to avoid aliasing. But then, we choose the sample frequency 124 00:08:56,110 --> 00:08:59,400 as an integer multiple of the positive bandwidth. 125 00:08:59,400 --> 00:09:04,560 So we say that Fs is equal to KW for K a positive integer. 126 00:09:04,560 --> 00:09:09,370 With this choice when we translate the analogue specifications into digital 127 00:09:09,370 --> 00:09:15,190 domain and remember the formula is always the same, 2 pi F over Fs. 128 00:09:15,190 --> 00:09:20,584 What happens is that the bandwidth in visual domain will be 2 pi W divided by 129 00:09:20,584 --> 00:09:28,152 Fs which is equal to 2 pi over K and so. We can simply upsample the symbol 130 00:09:28,152 --> 00:09:32,740 sequence by k so that it's bandwidth will move from two pi to two pi over k and 131 00:09:32,740 --> 00:09:39,702 therefore, it's width will fit on the band allowed for on the channel. 132 00:09:39,702 --> 00:09:43,306 Now, upsampling does not change the data rate, because we're creating a sequence 133 00:09:43,306 --> 00:09:46,562 of symbols. From the user data bitstream and then we 134 00:09:46,562 --> 00:09:49,322 are introducing the zeros between samples. 135 00:09:49,322 --> 00:09:51,662 So, we are not introducing extra information. 136 00:09:51,662 --> 00:09:55,950 So, we produce and transmit W symbols per second and then we have sampled that by K 137 00:09:55,950 --> 00:10:01,464 to achieve a sample rate which is equal to the sample of frequency. 138 00:10:01,464 --> 00:10:05,136 W, therefore is the fundamental data rate of the system and sometimes it's called 139 00:10:05,136 --> 00:10:09,562 the baud rate of the system. A golden rule for digital communication 140 00:10:09,562 --> 00:10:13,594 systems is that the baud rate will be equal to the positive bandwidth allowed 141 00:10:13,594 --> 00:10:18,588 by the channel. So here is our revised baud diagram for 142 00:10:18,588 --> 00:10:22,804 the transmitter. User data comes in as a bit stream. 143 00:10:22,804 --> 00:10:27,725 The Scrambler makes sure that we have a random sequence of symbols. 144 00:10:27,725 --> 00:10:31,998 The mapper will create, the random sequence of symbols. 145 00:10:31,998 --> 00:10:36,600 We upsample the sequence by K. We filter this with a low pass filter 146 00:10:36,600 --> 00:10:41,336 with cutoff frequency pi over K, and we obtain base band signal b of n, which is 147 00:10:41,336 --> 00:10:47,730 centered in 0, and extends from minus pi over K to pi over K. 148 00:10:47,730 --> 00:10:52,880 Now, we need to move this base band signal, to the pass band of the channel. 149 00:10:52,880 --> 00:10:56,490 And to do so, we modulated with a cosine carrier. 150 00:10:56,490 --> 00:11:00,120 This frequency is the center frequency of the channel's band. 151 00:11:00,120 --> 00:11:04,26 This pass band signal s of n can now be converted to the analog domain, before 152 00:11:04,26 --> 00:11:08,754 being transmitted over the channel. Graphically, assume these are the 153 00:11:08,754 --> 00:11:13,410 specifications dictated by the channel and translated to the digital domain. 154 00:11:13,410 --> 00:11:18,100 So here we have the positive bandwidth, and the negative bandwidth. 155 00:11:18,100 --> 00:11:22,447 The sequence of symbols generated by the mapper, is a wide sequence and therefore, 156 00:11:22,447 --> 00:11:27,990 inspires spectral density Pa of e to the j omega, is a full band signal. 157 00:11:27,990 --> 00:11:32,870 Now when we upsample this sequence by a factor of k, we reduce its spectral 158 00:11:32,870 --> 00:11:40,127 support, to a baseband signal that goes from a minus pi over K, 2pi over K. 159 00:11:40,127 --> 00:11:45,247 And then, we modulate this with a cosine carrier, to fit it onto the bands that 160 00:11:45,247 --> 00:11:50,903 are available on the channel. As a final note, since we are developing 161 00:11:50,903 --> 00:11:55,173 a completely digital transmission system, we will probably want to use FIR filters 162 00:11:55,173 --> 00:11:59,837 in its implementation. Now, we know that the sync filter that we 163 00:11:59,837 --> 00:12:03,731 have used in the upsampling operator Will be a notoriously difficult filter to 164 00:12:03,731 --> 00:12:08,371 approximate with an FIR. So what is used in practice is another 165 00:12:08,371 --> 00:12:12,924 type of filter called a raised cosine. The frequency response of the raised 166 00:12:12,924 --> 00:12:16,488 cosine is shown here in this picture, and you can see that the transition band is 167 00:12:16,488 --> 00:12:21,222 no longer a discontinuity. But it is actually a smooth transition 168 00:12:21,222 --> 00:12:24,750 from past band to star band as a matter of fact a raised cosine has a parameter 169 00:12:24,750 --> 00:12:29,450 that you can tune to have an even gentler transition then. 170 00:12:29,450 --> 00:12:33,872 Now, the raised cosine remains an ideal filter because you can see it is constant 171 00:12:33,872 --> 00:12:39,36 over the pass band, and the stop band. But it is much easier to approximate than 172 00:12:39,36 --> 00:12:41,872 the sync. Another good property of the raise 173 00:12:41,872 --> 00:12:46,20 cosine, is that it fulfills the interpolation property that we need to do 174 00:12:46,20 --> 00:12:50,581 upsampling. And the final selling pointing, is that 175 00:12:50,581 --> 00:12:56,200 the impulse response, it can be shown decays as 1 over n cubed. 176 00:12:56,200 --> 00:13:00,566 So even short FIR approximations can get a very good response. 177 00:13:00,566 --> 00:13:06,102 [BLANK_AUDIO]