1 00:00:00,012 --> 00:00:03,484 Welcome to module 5.4. The time has come to look at filters from 2 00:00:03,484 --> 00:00:07,667 the frequency domain prospective. And the starting point is this amazing 3 00:00:07,667 --> 00:00:10,357 result. If you take a complex exponential and 4 00:00:10,357 --> 00:00:13,286 process it with a linear time invariant filter. 5 00:00:13,286 --> 00:00:17,867 What you get at the output is a complex exponential at the same frequency as the 6 00:00:17,867 --> 00:00:21,195 input, only the phase and the amplitude may have changed. 7 00:00:21,195 --> 00:00:25,489 This has all sorts of interesting implications when we consider the Fourier 8 00:00:25,489 --> 00:00:29,392 representation of the elements involved in a filtering operation. 9 00:00:29,392 --> 00:00:33,375 We will derive what is called the frequency response of a filter, and we 10 00:00:33,375 --> 00:00:36,762 will use that to classify filters in the frequency domain. 11 00:00:36,762 --> 00:00:41,388 We will also revisit our friends the moving average, to linking to greater, and 12 00:00:41,388 --> 00:00:45,495 even the carpal strong algorithm and see how they fair In frequency. 13 00:00:45,495 --> 00:00:49,108 Welcome to module 5.4 of digital signal processing. 14 00:00:49,108 --> 00:00:53,952 A module in which we will look at filtering from the vantage point of the 15 00:00:53,952 --> 00:00:58,410 frequency domain. And to do so we will introduce the concept 16 00:00:58,410 --> 00:01:02,301 of Eigensequences. Which despite the fancy name are just 17 00:01:02,301 --> 00:01:06,443 complex exponential. We will derive the convolution theorem, 18 00:01:06,443 --> 00:01:10,570 after having obtained a frequency representation of filters. 19 00:01:10,570 --> 00:01:15,081 And then we will look frequency of phase response of filters we know. 20 00:01:15,081 --> 00:01:19,707 What happens if we take a linear time invariant filter h, and we use as the 21 00:01:19,707 --> 00:01:23,661 input a complex exponential of known frequency omega zero? 22 00:01:23,662 --> 00:01:28,691 Well let's write out the convolution. Then we exploit the fact the, that the 23 00:01:28,691 --> 00:01:34,073 fact the convolution is commutative, and we, exchange the order of the terms in the 24 00:01:34,073 --> 00:01:38,245 convolution product. At which point we can start writing out 25 00:01:38,245 --> 00:01:43,685 the convolution with some explicitly and we have the sum for k that goes from minus 26 00:01:43,685 --> 00:01:48,921 infinity to plus infinity of h of k that multiplies e to the j omega 0 and minus k. 27 00:01:48,921 --> 00:01:54,175 Now, we can take out the constant term, doesn't depend on k, out of the sum, and 28 00:01:54,175 --> 00:01:59,259 we have a Lydian term of e to the j omega 0 n, that multiplies the sum for k that 29 00:01:59,259 --> 00:02:04,507 goes from minus infinity to plus infinity of h of k times e to the minus j omega 0 30 00:02:04,507 --> 00:02:08,692 k. Now this guy here, we know very well what 31 00:02:08,692 --> 00:02:15,191 it is, is the DTFT of the impulse response computed in a mega zero. 32 00:02:15,191 --> 00:02:21,245 So it's big h of e to the j omega zero, which multiplies our original input e to 33 00:02:21,245 --> 00:02:25,942 the j on the omega 0 n. Hence, the name Igo sequences, just like 34 00:02:25,942 --> 00:02:31,104 Igo vector is a vector that when multiplied by a matrix, gives a scaled 35 00:02:31,104 --> 00:02:35,645 version of itself. An Igo sequences is a sequence that when 36 00:02:35,645 --> 00:02:41,597 Input to a linear time invariant filter, returns the sequence itself times a 37 00:02:41,597 --> 00:02:46,185 scaling factor. Which happens to be the value of the DTFT 38 00:02:46,185 --> 00:02:50,531 of the impulse response at the frequency of the input. 39 00:02:50,531 --> 00:02:56,521 So the first fundamental property that we can glean from this derivation is that a 40 00:02:56,521 --> 00:03:02,515 linear time invariant system cannot change the frequency of a sinusoidal input. 41 00:03:02,515 --> 00:03:08,731 And so since the sinusoidal input is a pure frequency component it's clear that 42 00:03:08,731 --> 00:03:15,031 the DTFT of the impulse response fully determines the frequency characteristic of 43 00:03:15,031 --> 00:03:20,184 a filter at a given frequency. Let's examine a little bit more, what 44 00:03:20,184 --> 00:03:26,043 happens when we process a complex exponential with a linear time invariant 45 00:03:26,043 --> 00:03:29,684 filter. If we write the value of the DTFT of the 46 00:03:29,684 --> 00:03:35,852 impulse response at omega 0. As a times e to the j theta, where a is a 47 00:03:35,852 --> 00:03:44,255 real number and theta, of course is an angle between minus pi and pi if you will. 48 00:03:44,255 --> 00:03:50,178 Then we have the processed complex exponential, is the original complex 49 00:03:50,178 --> 00:03:56,774 exponential, which has been scaled by the amplitude a, and has been delayed or 50 00:03:56,774 --> 00:04:02,422 advanced, by a phase term theta. So if a is larger than 1, then we have an 51 00:04:02,422 --> 00:04:05,970 amplification. If a is less than 1, then we have an 52 00:04:05,970 --> 00:04:11,317 attenuation of the input sinusoid. And the phase shift again, if it's bigger 53 00:04:11,317 --> 00:04:14,970 than 0 then we have an advancement of the sinusoid. 54 00:04:14,970 --> 00:04:18,010 And if it's less than 0 then we have a delay. 55 00:04:18,011 --> 00:04:23,962 The convolution theorem tries to generalize this result by asking the 56 00:04:23,962 --> 00:04:29,606 question, what is the DTFT of the convolution of the two sequences. 57 00:04:29,606 --> 00:04:35,126 Or in another words what are the DTFT of the output of a filter? 58 00:04:35,126 --> 00:04:41,788 The intuition here is that the DTFT reconstruction formula tells us that any 59 00:04:41,788 --> 00:04:48,548 signal is made up of infinitely many sinusoidal components of the form x to the 60 00:04:48,548 --> 00:04:54,983 e to the j omega times e to the j omega n. And if I were to process a single 61 00:04:54,983 --> 00:05:02,848 sinusoidal component independently by filter h, I would get a value like so, h 62 00:05:02,848 --> 00:05:09,941 to the j omega, e to the j omega n. So I wouldn't be surprised if the DTFT of 63 00:05:09,941 --> 00:05:17,650 the convolution of two sequences was just a product of the Fourier Transform. 64 00:05:17,650 --> 00:05:21,549 So let's go see how we can drive this more formally. 65 00:05:21,549 --> 00:05:27,021 If we write out the DTFT of the convolution of 2 sequences well, we can 66 00:05:27,021 --> 00:05:33,165 write out initially the formula for the DTFT which is the sum from n from minus 67 00:05:33,165 --> 00:05:39,309 infinity to infinity of the value of the convolution in n times e to the minus j 68 00:05:39,309 --> 00:05:43,118 omega n. Then we expand the formula for the 69 00:05:43,118 --> 00:05:50,287 convolution inside of the DTFT n summation and we get a double summation over the 70 00:05:50,287 --> 00:05:57,966 indices n and k of the convolution product here times this complex exponential here. 71 00:05:57,966 --> 00:06:04,392 In the next step we do a little trick where by we add and subtract k from the 72 00:06:04,392 --> 00:06:08,599 argument n. So instead of putting n, we write n minus 73 00:06:08,599 --> 00:06:14,333 k plus k, and in so doing we managed to split the complex exponential into 2 74 00:06:14,333 --> 00:06:19,476 components here that we will distribute across the summations. 75 00:06:19,476 --> 00:06:23,811 And now here we collect the terms that do not depend on n. 76 00:06:23,811 --> 00:06:29,512 And here we put the rest, this first term here is the DTFT of x of n. 77 00:06:29,512 --> 00:06:36,620 That's really easy to see, the 2nd term, looks like a DTFT, we might be confused by 78 00:06:36,620 --> 00:06:42,176 the fact that the indices here and here or n minus k instead of n. 79 00:06:42,176 --> 00:06:48,530 But the fact that the index n ranges from minus infinity to plus infinity makes this 80 00:06:48,530 --> 00:06:53,867 term completely irrelevant. And so, indeed we have the product of 2 81 00:06:53,867 --> 00:06:57,498 DTFTs. The product of the DTFT of the impulse 82 00:06:57,498 --> 00:07:03,454 response times the DTFT of that impulse. The Fourier Transform of the impulse 83 00:07:03,454 --> 00:07:09,146 response is called the frequency response. And just like in the case of a single 84 00:07:09,146 --> 00:07:14,392 complex exponential, we can split the significance of the product in the 85 00:07:14,392 --> 00:07:18,664 frequency domain. By separating the effects of the magnitude 86 00:07:18,664 --> 00:07:23,093 and those of the phrase. So the magnitude of the frequency response 87 00:07:23,093 --> 00:07:28,228 will determine weather certain frequencies of the input signal is going to be 88 00:07:28,228 --> 00:07:33,521 amplified or attenuated according to whether the frequency response is larger 89 00:07:33,521 --> 00:07:37,912 than 1 or smaller than 1 in magnitude in certain frequency bands. 90 00:07:37,912 --> 00:07:43,645 Defects of the phase are a little bit more difficult to qualify right now but it will 91 00:07:43,645 --> 00:07:49,372 be clearer later on that the phase will determine whether the signal will conserve 92 00:07:49,372 --> 00:07:53,499 its shape in the time domain or its shape will be altered. 93 00:07:53,499 --> 00:07:58,682 Let's go back now to our friend the moving average and compute it's frequency 94 00:07:58,682 --> 00:08:01,929 response. The moving average has an impulse 95 00:08:01,929 --> 00:08:07,714 response, which is just the indicator sequence for the interval 0 to capital M 96 00:08:07,714 --> 00:08:13,499 minus 1 divided by capital M, we have already computed the DTFT of such a signal 97 00:08:13,499 --> 00:08:18,135 in module 4.7. If we consider now the magnitude of the 98 00:08:18,135 --> 00:08:25,131 DTFT we find out that it's just 1 over capital M multiplies the absolute value of 99 00:08:25,131 --> 00:08:30,124 sin of omega over 2 times m divided by sign of omega over 2. 100 00:08:30,124 --> 00:08:37,950 An interesting thing to remember is that the magnitude response is going to be 0 at 101 00:08:37,950 --> 00:08:45,321 multiples of 2 pi over m except in 0. So here you see that you have indeed, M 102 00:08:45,321 --> 00:08:52,746 minus 1 0s along the frequency axis. We can plot the magnitude of the response 103 00:08:52,746 --> 00:08:59,816 for increasing values of M so for instance here you have the frequency response for m 104 00:08:59,816 --> 00:09:06,381 equal to 20 and you can count the 19 0s along the frequency axis and here you have 105 00:09:06,381 --> 00:09:12,598 the frequency response for M equal to 100. And you can see that the frequency 106 00:09:12,598 --> 00:09:16,271 response is very concentrated around the origin. 107 00:09:16,271 --> 00:09:19,637 We will see the significance of that later on. 108 00:09:19,637 --> 00:09:25,298 Let's go back to our denoising example that we studied in the time domain in the 109 00:09:25,298 --> 00:09:29,054 previous module. And look at its development in the 110 00:09:29,054 --> 00:09:33,734 frequency domain. So remember we had azimuth signal here 111 00:09:33,734 --> 00:09:38,019 we're talking time domain. So we had azimuth signal that was 112 00:09:38,019 --> 00:09:42,866 corrupted by some additive noise that we represent here in orange. 113 00:09:42,867 --> 00:09:48,245 In the frequency domain, the spectrum of the smooth signal looks like this. 114 00:09:48,245 --> 00:09:54,201 Most of its energy is concentrated around low frequencies so it's low-pass signal, 115 00:09:54,201 --> 00:09:59,896 and the Fourier Transform of denoise looks like noise in the frequency domain as 116 00:09:59,896 --> 00:10:03,192 well. So when we put things together we have the 117 00:10:03,192 --> 00:10:07,947 spectrum of the smooth signal. We have the spectrum of noise and the sum 118 00:10:07,947 --> 00:10:10,744 is denoise spectrum of the measured signal. 119 00:10:10,744 --> 00:10:15,925 When we use the moving average to filter this noisy signal, in the frequency domain 120 00:10:15,925 --> 00:10:21,100 we're multiplying this spectrum by the frequency response of the moving average, 121 00:10:21,100 --> 00:10:26,180 and in magnitude it looks like this. So here we're using, for instance, a nine 122 00:10:26,180 --> 00:10:31,636 point average as you can see from the number of zeroes along the frequency axis. 123 00:10:31,636 --> 00:10:36,602 So the product in magnitude determines a very deep attenuation of the high 124 00:10:36,602 --> 00:10:42,202 frequency and therefore most of the noise that was contained in these bands has been 125 00:10:42,202 --> 00:10:45,622 eliminated. At the same time though, if you compare 126 00:10:45,622 --> 00:10:50,722 the result of the filtering operation with the original spectrum you see that the 127 00:10:50,722 --> 00:10:55,822 filtering has eliminated parts of the original spectrum that actually had every 128 00:10:55,822 --> 00:10:59,507 right to be there. So that proves that there's no free long 129 00:10:59,507 --> 00:11:04,002 signal processing as well. And in order to remove the noise sometimes 130 00:11:04,002 --> 00:11:07,306 you have to remove some of the good part with it as well. 131 00:11:07,306 --> 00:11:12,140 Before we move on let's look once again at the same denoising operation in the time 132 00:11:12,140 --> 00:11:16,109 domain now that we know what's going on and the frequency domain. 133 00:11:16,110 --> 00:11:20,814 And we can see that when we use a moving average of say, length 12, we are indeed 134 00:11:20,814 --> 00:11:25,559 smoothing out the highest components of the noise but not so much the ones that 135 00:11:25,559 --> 00:11:30,012 are slower moving and that we are ready, starting to alter a little bit the 136 00:11:30,012 --> 00:11:34,536 curvature of the signal, as explained by the spectrum we just derived. 137 00:11:34,536 --> 00:11:39,364 So what about the phase? The best way to understand the effects of 138 00:11:39,364 --> 00:11:44,498 the phase on a signal is to distinguish three different cases. 139 00:11:44,498 --> 00:11:48,622 The first one is zero phase, which means the spectrum is real. 140 00:11:48,622 --> 00:11:54,148 The second case is linear phase where the phase is proportional to the frequency by 141 00:11:54,148 --> 00:11:57,812 a, a real factor d. And the third case is nonlinear phase 142 00:11:57,812 --> 00:12:04,402 which covers all other possibilities. To understand what phase does to a signal, 143 00:12:04,402 --> 00:12:11,442 let's take a very simple discreet time signal made up of the sum of 2 sinusoid x 144 00:12:11,442 --> 00:12:18,295 of n is equal to 1 half sine of omega 0 times n plus cosine of 2 omega 0 times n. 145 00:12:18,295 --> 00:12:23,127 So it's 2 sinusoid at frequencies 1 double of the other. 146 00:12:23,127 --> 00:12:27,714 And the sum looks like this signal that we plot here in this graph. 147 00:12:27,714 --> 00:12:33,032 We can call this signal, a zero phase signal, because the phase associated to 148 00:12:33,032 --> 00:12:37,876 each sinusoidal function is zero. Now, let's add a phase term to each 149 00:12:37,876 --> 00:12:43,420 component, and make sure that the phase is proportional to the frequency of the 150 00:12:43,420 --> 00:12:46,818 sinusiod. So what we add is theta 0 to the first 151 00:12:46,818 --> 00:12:52,041 component at frequency omega zero. And we add 2 theta zero to the second 152 00:12:52,041 --> 00:12:55,271 component, whose frequency is 2 omega zero. 153 00:12:55,271 --> 00:13:00,523 The value of theta zero is actually irrelevant, but we can see that when we 154 00:13:00,523 --> 00:13:03,896 add a phase term to the sinusoidal components. 155 00:13:03,896 --> 00:13:08,146 That it's proportional the, the frequency of this sinusoid. 156 00:13:08,146 --> 00:13:12,678 The net effect in the time domain is just a shift of the signal. 157 00:13:12,679 --> 00:13:16,696 The shape of the signal remains exactly the same. 158 00:13:16,696 --> 00:13:22,222 If now we add a phase that is non proportional to the frequency, so in this 159 00:13:22,222 --> 00:13:28,386 case, we change the phase to the first term to zero and we'll leave a phase term 160 00:13:28,386 --> 00:13:34,826 of two theta zero to the second term while the frequencies of the two components have 161 00:13:34,826 --> 00:13:41,301 not changed, the shape of the signal in the time domain has changed significantly. 162 00:13:41,301 --> 00:13:45,627 Please note that in all 3 cases, the spectrum of x of n remains exactly the 163 00:13:45,627 --> 00:13:49,368 same in magnitude. To understand that the linear face term is 164 00:13:49,368 --> 00:13:53,896 simply a delay in time domain. Consider the following system, where we 165 00:13:53,896 --> 00:13:58,221 have a filter d that simply produces a delayed version of its input. 166 00:13:58,221 --> 00:14:01,286 The input output relationship is this one here. 167 00:14:01,286 --> 00:14:06,902 And in the frequency domain, we can take Fourier Transforms left to right and we 168 00:14:06,902 --> 00:14:12,002 obtain that big Y of e to the j omega is equal to e to the minus j omega d times 169 00:14:12,002 --> 00:14:16,652 big X of e to the j omega. This is really equivalent to saying that 170 00:14:16,652 --> 00:14:22,436 the frequency response of this delaying filter is simply e to the minus j omega d. 171 00:14:22,436 --> 00:14:28,247 So, again, we have a linear phased term. Where the phase introduced by the filter 172 00:14:28,247 --> 00:14:31,696 is proportional to the frequency via a factor d. 173 00:14:31,696 --> 00:14:35,770 Which actually represents the delay in time domain. 174 00:14:35,770 --> 00:14:41,572 In general, if we can split the frequency response of the filter into the product of 175 00:14:41,572 --> 00:14:46,696 a pure real term and a pure phase term, it means that the filter operates by 176 00:14:46,696 --> 00:14:52,576 combining the action of a zero phase and, therefore, zero delay component that only 177 00:14:52,576 --> 00:14:57,821 affects the magnitude of the input followed by a delay by these samples. 178 00:14:57,821 --> 00:15:02,971 And this delay is inevitable in the case of causal filters for instance. 179 00:15:02,971 --> 00:15:09,487 Let's consider the moving average again, the frequency response is the product of a 180 00:15:09,487 --> 00:15:15,346 real term here and a pure phase term here where the delay d is exactly n minus 1 181 00:15:15,346 --> 00:15:21,586 over 2 which represents half the length of the support of the impulse response. 182 00:15:21,586 --> 00:15:26,631 And this is indeed the delay introduced by the moving average. 183 00:15:26,631 --> 00:15:31,902 So we've looked at the moving average and by now you know that what comes next is 184 00:15:31,902 --> 00:15:36,050 the leaky integrator. And indeed, let's consider the impulse 185 00:15:36,050 --> 00:15:40,948 response which you remember is 1 minus lambda times lambda to the power of n 186 00:15:40,948 --> 00:15:45,189 multiplied by uni step so an exponentially decaying sequence. 187 00:15:45,189 --> 00:15:50,475 If we take the Fourier Transform of that, we've done that in detail in module 4.4 188 00:15:50,475 --> 00:15:55,254 and here is the result. Big H, lead to the j omega is 1 minus 189 00:15:55,254 --> 00:16:00,296 lambda divided by one minus lamda times e to the j omega. 190 00:16:00,296 --> 00:16:06,980 Now to find the magnitude and the phase of this animal, we need to use a little 191 00:16:06,980 --> 00:16:11,821 algebra and we need to recall this very simple result. 192 00:16:11,821 --> 00:16:18,739 If you have one over a plus jb, you can always rewrite that as a minus jb divided 193 00:16:18,739 --> 00:16:24,302 by a squared plus b squared. So if the number x is actually 1 over a 194 00:16:24,302 --> 00:16:30,463 plus jb, the magnitude square of x is 1 over a squared plus b squared and the 195 00:16:30,463 --> 00:16:34,699 phase of x is the inverse tangent of minus b over a. 196 00:16:34,699 --> 00:16:40,645 This applies to the leaky integrator in the sense that we can rewrite the 197 00:16:40,645 --> 00:16:47,080 frequency response as 1 minus lambda divided by 1 minus lambda cosine of omega 198 00:16:47,080 --> 00:16:53,812 which is the real part of the denominator, minus j times sine of omega, which is the 199 00:16:53,812 --> 00:17:00,137 imaginary part of the denominator. And by applying the formula that we just 200 00:17:00,137 --> 00:17:06,217 saw, we find out that the magnitude squared of the leaky integrator is 1 minus 201 00:17:06,217 --> 00:17:12,677 lambda squared divided by 1 minus 2 lambda cosine of omega plus lambda square, and 202 00:17:12,677 --> 00:17:18,947 the phase of the leaky integrator is the inverse tangent of lambda sine of omega 203 00:17:18,947 --> 00:17:24,721 divided by 1 minus lambda cosine of omega. So you can see that the phase is 204 00:17:24,721 --> 00:17:30,555 definitely non-linear in, in this case. If we plot the magnitude we can see that 205 00:17:30,555 --> 00:17:35,639 we have a characteristic that is very similar to that of the moving average 206 00:17:35,639 --> 00:17:41,297 although we don't have any zeros and the characteristics is monotonic rather than 207 00:17:41,297 --> 00:17:45,188 the [unknown]. As lambda goes closer to 1 we see that 208 00:17:45,188 --> 00:17:50,644 once again the frequency response concentrates around the origin and for 209 00:17:50,644 --> 00:17:55,465 lambda very, very close to 1. We have something that resembles the 210 00:17:55,465 --> 00:18:01,052 frequency response of a moving average computed over a large number of tabs. 211 00:18:01,052 --> 00:18:06,426 The phase looks like this, it's a non linear characteristic and as lambda 212 00:18:06,426 --> 00:18:12,146 changes the steepness of the transition between positive and negative phase 213 00:18:12,146 --> 00:18:16,123 increases as well. What is interesting in terms of the 214 00:18:16,123 --> 00:18:22,143 combination between the magnitude and the phase response is that what interests us 215 00:18:22,143 --> 00:18:27,647 is the part of the filter where the attenuation is not too big, because that's 216 00:18:27,647 --> 00:18:33,848 where the frequencies will pass through whereas here there will be fundamentally. 217 00:18:33,849 --> 00:18:38,656 And in that area where the frequency response has a magnitude that is 218 00:18:38,656 --> 00:18:43,986 sufficiently close to 1, the phase is actually more or less linear so we can use 219 00:18:43,986 --> 00:18:49,152 the leaky integrator without incurring excessive phase distortion in the 220 00:18:49,152 --> 00:18:52,846 [inaudible]. Finally, let's revisit another classic 221 00:18:52,846 --> 00:18:57,846 namely the Karplus-Strong Algorithm. And let's try to analyze its behavior 222 00:18:57,846 --> 00:19:03,381 using the convolution theory. Remember the Karplus-Strong Algorithm is 223 00:19:03,381 --> 00:19:09,977 initialized with a finite support signal x of support capital M and then we use the 224 00:19:09,977 --> 00:19:16,185 feedback loop with a delay of capital M taps to produce multiple copies of the 225 00:19:16,185 --> 00:19:22,301 original finite support signal. Scaled by an exponentially decaying factor 226 00:19:22,301 --> 00:19:25,703 alpha. So for instance, if we initialize the 227 00:19:25,703 --> 00:19:31,683 algorithm with one period of a sawtooth wave, what we get in the end is multiple 228 00:19:31,683 --> 00:19:37,591 repetitions of that period scaled by an exponentially and decaying emblem. 229 00:19:37,591 --> 00:19:42,906 In the time domain we can write this out explicitly as such. 230 00:19:42,906 --> 00:19:49,174 Here we have the first period of the output signal is just a copy of the known 231 00:19:49,174 --> 00:19:54,804 0 values of the finite support signal. Followed by another copy of the no 0 232 00:19:54,804 --> 00:19:59,952 values of the finance support signal scaled by alpha, followed by another copy 233 00:19:59,952 --> 00:20:02,865 scaled by alpha square and so on and so forth. 234 00:20:02,865 --> 00:20:08,947 The key observation to analyze in the algorithm within the paradigm of the 235 00:20:08,947 --> 00:20:15,415 convolution theorem is to see that the output can be expressed as a convolution 236 00:20:15,415 --> 00:20:20,903 of the finite support input with a sequence w of n that we build in the 237 00:20:20,903 --> 00:20:25,825 following way. W of n is equal to a to the power of k for 238 00:20:25,825 --> 00:20:32,100 values of the index n. There are the k multiple of capital M and 239 00:20:32,100 --> 00:20:38,893 it's zero, otherwise. So if I was to draw the sequence it would 240 00:20:38,893 --> 00:20:43,524 look like so. So it would be one and zero, it would be 241 00:20:43,524 --> 00:20:49,972 alpha and m alpha square in 2 m, and it would be zero in between, so a series of 242 00:20:49,972 --> 00:20:54,680 exponentially decayed deltas spaced n points apart. 243 00:20:54,680 --> 00:21:01,088 With this, we can write the convolution theorem, and we say okay, the output you 244 00:21:01,088 --> 00:21:07,587 simply in frequency the product of the fully transformed of the finance support 245 00:21:07,587 --> 00:21:12,188 signal times the fully transformed of the signal w of n. 246 00:21:12,188 --> 00:21:16,875 Well, and now we can proceed exactly like in module 4.4. 247 00:21:16,875 --> 00:21:20,103 If we consider the example of a sawtooth wave. 248 00:21:20,104 --> 00:21:27,398 We know that the Fourier Transform of one period is this one, whereas the Fourier 249 00:21:27,398 --> 00:21:33,878 Transform of the sequence w of n is the rescaled Fourier Transform of the 250 00:21:33,878 --> 00:21:41,546 exponentially decaying sequence, which means we will have several peaks happening 251 00:21:41,546 --> 00:21:47,877 in the minus pi, pi interval. We put them together graphically, we have 252 00:21:47,877 --> 00:21:54,807 the Fourier Transform in magnitude of the sawtooth period; we have the Fourier 253 00:21:54,807 --> 00:22:00,507 Transform. A magnitude of the staggered exponential 254 00:22:00,507 --> 00:22:08,892 sequence, we take the product of the two and we obtain the same spectrum that we 255 00:22:08,892 --> 00:22:11,422 derived in module 4.4.