1 00:00:00,025 --> 00:00:02,510 [SOUND] 2 00:00:02,510 --> 00:00:06,340 Hi and welcome to module 4.5 of digital signal processing. 3 00:00:06,340 --> 00:00:09,610 In this module we will continue the exploration of the DTFT. 4 00:00:09,610 --> 00:00:13,640 And we will consider, in particular, conditions related 5 00:00:13,640 --> 00:00:17,100 to its existence, its properties, and how we 6 00:00:17,100 --> 00:00:19,290 can look at it as a special type 7 00:00:19,290 --> 00:00:24,070 of basis expansion in the space of infinite sequences. 8 00:00:25,680 --> 00:00:27,940 So existance means simply 9 00:00:27,940 --> 00:00:31,900 that the sum that defines the DTFT does not blow up. 10 00:00:31,900 --> 00:00:35,419 This is easy to prove for absolutely summable sequences, if 11 00:00:35,419 --> 00:00:39,160 you take the magnitude of the DTFT at any point omega. 12 00:00:39,160 --> 00:00:42,050 This is equal to the sum for N that goes from minus infinity to plus 13 00:00:42,050 --> 00:00:48,160 infinity of X of N times e to the minus J omega N in magnitude. 14 00:00:48,160 --> 00:00:50,180 Now every time you have absolute value of the 15 00:00:50,180 --> 00:00:53,520 sum, you know that this is maximized by the sum 16 00:00:53,520 --> 00:00:55,930 of the absolute values of the elements of the sum. 17 00:00:55,930 --> 00:01:00,830 So we do this, and because the magnitude of the complex exponential is 1, 18 00:01:00,830 --> 00:01:05,180 this is actually equal to the sum of the absolute values of the sequence. 19 00:01:05,180 --> 00:01:07,490 Since our initial hypothesis was that the sequence 20 00:01:07,490 --> 00:01:10,790 was absolutely summable, this is less than infinity. 21 00:01:10,790 --> 00:01:14,360 And therefore, the DTFT exists for all values of omega. 22 00:01:16,400 --> 00:01:20,160 Similarly, we can invert the DTFT very easily 23 00:01:20,160 --> 00:01:25,030 if we assume absolute summability of the underlying sequence. 24 00:01:25,030 --> 00:01:28,900 As we showed in the previous module, the inversion 25 00:01:28,900 --> 00:01:32,050 formula is 1 over 2 pi times the integral 26 00:01:32,050 --> 00:01:37,337 between minus pi and pi of the DTFT times e to the J omega n in the omega. 27 00:01:37,337 --> 00:01:41,477 So we replace x of e to the j omega by the definition of the 28 00:01:41,477 --> 00:01:45,797 DTFT in here, and because of the absolute summability of 29 00:01:45,797 --> 00:01:50,380 the sequence we can invert the summation and the integral. 30 00:01:51,450 --> 00:01:54,570 When we do that we have the sum for k that goes from 31 00:01:54,570 --> 00:01:58,890 minus infinity to plus infinity of x of k time this integral here. 32 00:01:58,890 --> 00:02:00,910 So each element of the sequence in the sum 33 00:02:00,910 --> 00:02:04,820 is multiplied by this integral, which depends on k. 34 00:02:04,820 --> 00:02:06,490 But now look at the numerator of this 35 00:02:06,490 --> 00:02:07,530 fraction, here. 36 00:02:07,530 --> 00:02:13,810 This is a complex exponential, and if n is different than k, this will span an 37 00:02:13,810 --> 00:02:16,790 integer number of periods in the minus pi, 38 00:02:16,790 --> 00:02:20,120 pi integral, and therefore the integral will be 0. 39 00:02:20,120 --> 00:02:23,357 So what that means is that integral is 0 unless 40 00:02:23,357 --> 00:02:25,930 n is equal to k, at which point all the 41 00:02:25,930 --> 00:02:31,491 elements in the sum will be killed except for x of n, and so in the end, we have the 42 00:02:31,491 --> 00:02:33,490 result we're looking for. 43 00:02:34,620 --> 00:02:39,710 The DTFT looks exactly like an inner product in the space C infinity. 44 00:02:39,710 --> 00:02:43,570 If you take this inner product here between an 45 00:02:43,570 --> 00:02:45,920 infinite sequence and the sequence e to the j 46 00:02:45,920 --> 00:02:48,260 omega n, and you write the definition of the 47 00:02:48,260 --> 00:02:52,630 inner product, you get exactly the formulation for the DTFT. 48 00:02:52,630 --> 00:02:56,770 The problem here is that C infnity is not really a well definied vector space. 49 00:02:56,770 --> 00:02:59,255 There are sequences that do not converge in C infinity. 50 00:02:59,255 --> 00:03:04,536 Nonetheless, if we manage to establish a formal parallel between a change of basis 51 00:03:04,536 --> 00:03:10,092 and the DTFT, it will mean that everything that we discovered about a 52 00:03:10,092 --> 00:03:14,310 DFT which is a well defined entity, will apply to the DTFT as 53 00:03:14,310 --> 00:03:16,201 well and all our intuition about the 54 00:03:16,201 --> 00:03:19,060 frequency domain will translate to the DTFT. 55 00:03:20,220 --> 00:03:21,850 Now, the basis here that we're 56 00:03:21,850 --> 00:03:25,860 talking about is not really a basis, because it's an infinite 57 00:03:25,860 --> 00:03:31,140 and uncountable set of vectors, indexed by a real value variable omega. 58 00:03:31,140 --> 00:03:32,350 So something breaks down, really. 59 00:03:32,350 --> 00:03:38,040 We start with sequences, but we end up landing in the space of functions. 60 00:03:38,040 --> 00:03:41,070 On top of it all, although we just proved existence 61 00:03:41,070 --> 00:03:47,190 and invertability for absolutely summable sequences, in reality the DTFT exists, 62 00:03:47,190 --> 00:03:52,800 for all square summable sequences, which is a larger set of sequences. 63 00:03:52,800 --> 00:03:55,700 But in that case the proofs we just gave, become 64 00:03:55,700 --> 00:03:59,070 much more technical, and so we will skip them here. 65 00:03:59,070 --> 00:04:01,390 Let's sum up the situation so far. 66 00:04:01,390 --> 00:04:06,382 For finite length signals, we start in CN, and via a change of 67 00:04:06,382 --> 00:04:11,854 basis, we compute their representation, the frequency domain, 68 00:04:11,854 --> 00:04:14,910 which as well, lives in CN. 69 00:04:14,910 --> 00:04:19,110 We can go back to the original sequence via the inversion formula. 70 00:04:19,110 --> 00:04:22,780 And the basis that allows us to go from the 71 00:04:22,780 --> 00:04:26,650 time domain to the frequency domain is the DFT basis. 72 00:04:26,650 --> 00:04:31,984 The Fourier basis for the DFT, which is a countable set of N Fourier basis vectors. 73 00:04:33,280 --> 00:04:37,250 The DFS is exactly the same. The expansion and reconstruction 74 00:04:37,250 --> 00:04:39,770 formulas are the same, except that in this 75 00:04:39,770 --> 00:04:43,670 case, we assume that everything is periodic underneath. 76 00:04:45,330 --> 00:04:47,000 And now we have the DTFT. 77 00:04:47,000 --> 00:04:49,904 We start from the space of a square summable 78 00:04:49,904 --> 00:04:54,090 sequences, and via a formal change of basis, so basis 79 00:04:54,090 --> 00:04:56,630 here is in quote, we end up in the space 80 00:04:56,630 --> 00:05:00,710 of square integrable functions, on the integral minus 5 pi. 81 00:05:00,710 --> 00:05:02,548 By looking at the DTFT as a formal 82 00:05:02,548 --> 00:05:05,770 basis of expansion the linearity property follows 83 00:05:05,770 --> 00:05:09,180 easily from the linearity of the inner product. 84 00:05:09,180 --> 00:05:11,620 So, the DTFT of a linear combination of two 85 00:05:11,620 --> 00:05:15,110 sequences will be the linear combination of the DTFTs. 86 00:05:16,680 --> 00:05:19,344 A second property that is easy to prove from 87 00:05:19,344 --> 00:05:23,090 the definition of the DTFT is the time shift property. 88 00:05:23,090 --> 00:05:28,180 So if we take a sequence and we shift it in time by big M samples, 89 00:05:28,180 --> 00:05:32,500 the DTFT of this shifted sequence is equal of the DTFT of 90 00:05:32,500 --> 00:05:37,560 the original sequence, times a delay factor e to the minus J omega 91 00:05:37,560 --> 00:05:40,650 big M, which is very similar to what we obtained in the 92 00:05:40,650 --> 00:05:44,630 case of the DFS, when we took the shift of a periodic sequence. 93 00:05:45,732 --> 00:05:48,580 To dual of this property is the modulation property of the 94 00:05:48,580 --> 00:05:53,340 DTFT, so if we take a sequence and we multiply this by 95 00:05:53,340 --> 00:05:58,410 a complex exponential at frequency omega 0, what happens in frequency 96 00:05:58,410 --> 00:06:02,570 is that we have a shift of the spectrum by omega 0. 97 00:06:02,570 --> 00:06:04,755 The time reversal property tells us that the 98 00:06:04,755 --> 00:06:08,670 Fourier transform of a time reverse sequence, a sequence 99 00:06:08,670 --> 00:06:11,770 where we flip the values across the origin 100 00:06:11,770 --> 00:06:16,820 will be equal to a frequency reversed Fourier transform. 101 00:06:16,820 --> 00:06:18,470 And the conjugation property says 102 00:06:18,470 --> 00:06:21,780 that if you conjugate every value of the sequence 103 00:06:21,780 --> 00:06:26,510 the Fourier transform will be both conjugated and frequency reversed. 104 00:06:27,580 --> 00:06:29,460 Now, some particular cases that are very 105 00:06:29,460 --> 00:06:32,400 useful to remember because they appear often. 106 00:06:33,520 --> 00:06:35,410 First of all, if the sequence is 107 00:06:35,410 --> 00:06:38,980 symmetric, then the DTFT is symmetric as well. 108 00:06:38,980 --> 00:06:43,810 If the sequence is real, then the DTFT is Hermitian-symmetric. 109 00:06:43,810 --> 00:06:48,472 The reality of the sequence can be expressed mathematically by saying that 110 00:06:48,472 --> 00:06:51,960 x of n is equal to the conjugate of x of n. 111 00:06:51,960 --> 00:06:57,220 This infrequency implies that the Fourier transform of the sequence is equal to 112 00:06:57,220 --> 00:07:03,000 the conjugate and Frequency reversed version of the Fourier transform. 113 00:07:03,000 --> 00:07:06,380 A simple corollary of this property is the fact that if 114 00:07:06,380 --> 00:07:09,580 x of n is real then the magnitude of the DTFT 115 00:07:09,580 --> 00:07:10,630 is symmetric. 116 00:07:10,630 --> 00:07:15,050 You can verify this simply by taking the magnitude of both terms of this equation. 117 00:07:16,248 --> 00:07:19,850 And an even more special case states that if x of n is 118 00:07:19,850 --> 00:07:25,720 real and symmetric then x of j omega is also real and symmetric. 119 00:07:25,720 --> 00:07:27,270 So this all looks nice and fine. 120 00:07:27,270 --> 00:07:30,620 It looks like we have a full fletched basis expansion, 121 00:07:30,620 --> 00:07:34,310 and that the DTFT is just another version of the DFT. 122 00:07:34,310 --> 00:07:35,850 And indeed, 123 00:07:35,850 --> 00:07:38,080 some things would lead us to believe so. 124 00:07:38,080 --> 00:07:40,770 For instance, if you take the DFT of the 125 00:07:40,770 --> 00:07:43,330 delta function, you remember you have the constant 1. 126 00:07:43,330 --> 00:07:48,630 And similarly, the DTFT of the delta function expressed as the inner 127 00:07:48,630 --> 00:07:53,900 product between the pseudo basis function and the delta function is again 1. 128 00:07:53,900 --> 00:07:56,630 However, some things are not okay at all. 129 00:07:56,630 --> 00:08:01,180 The DFT of the constant 1 is very well defined, and it's equal to 130 00:08:01,180 --> 00:08:04,570 n times the delta function in frequency. 131 00:08:04,570 --> 00:08:09,320 But the DTFT of 1 is, by definition, the sum from n that goes 132 00:08:09,320 --> 00:08:13,580 from minus infinity to plus infinity of e to the minus j omega n. 133 00:08:13,580 --> 00:08:15,990 Now to see that there is a problem with this sum, just 134 00:08:15,990 --> 00:08:20,270 put omega equal to 0, and you see that the sum diverges. 135 00:08:20,270 --> 00:08:21,430 The problem is that there are too 136 00:08:21,430 --> 00:08:24,800 many interesting sequences that are not square summable. 137 00:08:24,800 --> 00:08:26,200 And of course the constant 138 00:08:26,200 --> 00:08:27,520 1 is one of them. 139 00:08:27,520 --> 00:08:28,850 So, in order to be able to keep 140 00:08:28,850 --> 00:08:32,490 using the change of basis paradigm, even for sequences 141 00:08:32,490 --> 00:08:35,520 that are not square summable, we have to introduce 142 00:08:35,520 --> 00:08:39,080 a little mathematically trick called the direct delta function. 143 00:08:39,080 --> 00:08:41,080 This little animal here, which is usually 144 00:08:41,080 --> 00:08:45,170 indicated by the symbol delta, but now delta 145 00:08:45,170 --> 00:08:47,360 of the real variable t, not the 146 00:08:47,360 --> 00:08:51,360 delta sequence, is defined by the sifting property, 147 00:08:51,360 --> 00:08:54,500 that looks like this. If we take a delta function. 148 00:08:55,690 --> 00:09:00,160 We center it in s, where s is a variable in r. 149 00:09:01,320 --> 00:09:07,170 And then we multiply this delta functional by any function of a real variable t. 150 00:09:07,170 --> 00:09:10,730 And then we take the integral from minus infinity to plus infinity. 151 00:09:10,730 --> 00:09:15,820 Then what we get is the value of f in the point s. 152 00:09:15,820 --> 00:09:21,210 Graphically, we usually represent delta functional as an upwards pointing arrow, 153 00:09:21,210 --> 00:09:25,628 we centered this in s, we multiply this by any function of 154 00:09:25,628 --> 00:09:29,080 a real variable t and then we integrate from minus infinity to 155 00:09:29,080 --> 00:09:33,860 plus infinity and we get the value of the function in s. 156 00:09:33,860 --> 00:09:37,030 In order to develop some intuitions for the properties of the 157 00:09:37,030 --> 00:09:41,580 direct delta functional, let's consider a family of so called localizing 158 00:09:41,580 --> 00:09:44,140 functions, r k of t. 159 00:09:44,140 --> 00:09:49,050 Where k is an interger index, and t is a real valued variable. 160 00:09:49,050 --> 00:09:52,770 The properties of this family of functions are two. 161 00:09:52,770 --> 00:09:57,060 The support of each function is inversely proportional to the index k. 162 00:09:57,060 --> 00:10:01,540 But regardless of an index k, the area of each function, the 163 00:10:01,540 --> 00:10:04,980 integral from minus infinity to plus infinity of each function, is constant. 164 00:10:06,080 --> 00:10:07,050 As an example, 165 00:10:07,050 --> 00:10:10,690 take direct function. Direct function is a classic indicator 166 00:10:10,690 --> 00:10:19,020 function that is equal to 1 from minus one half to one half, and 0 everywhere else. 167 00:10:19,020 --> 00:10:23,055 So this function has a support of 1, and an area of 1. 168 00:10:24,310 --> 00:10:28,960 We can use this function to build a family of localizing functions like so. 169 00:10:28,960 --> 00:10:32,160 We multiply the rect by a factor k, 170 00:10:32,160 --> 00:10:35,920 and we shrink the support of the rect by a factor k. 171 00:10:35,920 --> 00:10:39,770 So r k of t in this case will have a support that goes from 172 00:10:39,770 --> 00:10:45,110 minus 1 over 2k to 1 over 2k, so the support is 1 over k. 173 00:10:45,110 --> 00:10:46,328 And the area is 1. 174 00:10:46,328 --> 00:10:51,040 If we plot some functions in this family, this is what we get for k equal to 1. 175 00:10:51,040 --> 00:10:52,190 The value here is 1. 176 00:10:52,190 --> 00:10:57,050 For k equal to 5, the support has shrunk to 1 over 5. 177 00:10:57,050 --> 00:11:01,265 And the area is still 1 because the value here is 5. 178 00:11:01,265 --> 00:11:03,830 We go to 15, it will look like this. 179 00:11:03,830 --> 00:11:07,700 And to 40 it will go like this, and we could on to infinity. 180 00:11:07,700 --> 00:11:12,100 Now, consider the integral between minus infinity and plus infinity of the product 181 00:11:12,100 --> 00:11:17,180 between rk of t, and any function f of a real variable t. 182 00:11:17,180 --> 00:11:22,416 Rk of t is non-zero only between minus 1 over 2k and 183 00:11:22,416 --> 00:11:24,040 1 over 2k. 184 00:11:24,040 --> 00:11:26,800 So these are the new integration image of this product. 185 00:11:27,850 --> 00:11:32,130 The value of rk of t over the integration interval is k. 186 00:11:32,130 --> 00:11:35,400 So we can bring this outside of the interval. 187 00:11:35,400 --> 00:11:40,240 And inside we have simply the integral of the function over this interval. 188 00:11:40,240 --> 00:11:42,460 Now we invoke the mean value theorem. 189 00:11:42,460 --> 00:11:47,470 You can go back to your calculus textbook to revise its proof, 190 00:11:47,470 --> 00:11:53,570 and the mean value theorem says that the value of this integral here will be 191 00:11:53,570 --> 00:11:59,940 equal to f of gamma for some point gamma within the integration integral. 192 00:11:59,940 --> 00:12:02,500 Now, we don't know where gamma is inside 193 00:12:02,500 --> 00:12:04,899 this integral, but we do know that it exists. 194 00:12:06,330 --> 00:12:12,820 Now as k goes to infinity, the support of the indicator function becomes smaller and 195 00:12:12,820 --> 00:12:18,070 smaller and so gamma which is somewhere in the integration interval, 196 00:12:18,070 --> 00:12:23,740 will be sandwiched between interval limits that grow closer and closer. 197 00:12:23,740 --> 00:12:28,650 And in the limit, f of gamma will be f of 0, 198 00:12:28,650 --> 00:12:33,590 because the width of the interval has shrunk down to an infinitesimal width. 199 00:12:35,890 --> 00:12:40,460 So the delta functional is really a shorthand for this limiting operation. 200 00:12:40,460 --> 00:12:43,130 Instead of writing the limit of the inegral for a family 201 00:12:43,130 --> 00:12:46,700 of localizing functions, We just use the delta notation, and what is 202 00:12:46,700 --> 00:12:49,940 interesting is that the shape of the base function that we use 203 00:12:49,940 --> 00:12:53,530 to build the family of a localizing function, is not really critical. 204 00:12:53,530 --> 00:12:55,710 We can use pretty much any shape, and as long 205 00:12:55,710 --> 00:13:00,840 as two properties of shrinking support and constant area are satisfied, 206 00:13:00,840 --> 00:13:04,850 the limit will converge to the point-wise value of the function. 207 00:13:04,850 --> 00:13:11,150 Okay, so now a last technicality before we understand why we're doing all this. 208 00:13:11,150 --> 00:13:15,230 We will be using the direct delta functional in the frequency domain. 209 00:13:15,230 --> 00:13:19,610 Now we know that all DTFT spectra are 2 pi periodic. 210 00:13:19,610 --> 00:13:23,410 So if we want to use this tool in the frequency domain, we have to periodize it. 211 00:13:23,410 --> 00:13:25,930 The periodic version of Dirac delta functional is 212 00:13:25,930 --> 00:13:29,640 called the pulse train and it is built by placing copies of the 213 00:13:29,640 --> 00:13:34,560 delta function every 2 pi and by scaling the whole signal by 2 pi. 214 00:13:34,560 --> 00:13:38,870 So if you w to represent that in the frequency domain, with the usual 215 00:13:38,870 --> 00:13:44,360 upward arrow notation, we see that there will be a pulse every 2 pi. 216 00:13:44,360 --> 00:13:46,730 Okay, now we can let the show begin. 217 00:13:46,730 --> 00:13:51,132 Let's consider the inverse DTFT of the pulse train. 218 00:13:51,132 --> 00:13:56,150 Well, we apply the definition, and we have 1 over 2 pi times the integral between 219 00:13:56,150 --> 00:14:01,990 minus pi and pi, of the pulse train times e to the j omega n into omega. 220 00:14:01,990 --> 00:14:07,046 So, the first thing to remark is that since the periodized delta is scaled by 221 00:14:07,046 --> 00:14:09,100 a factor of 2 pi, this cancels 222 00:14:09,100 --> 00:14:13,340 the normalization factor in front of the integral. 223 00:14:13,340 --> 00:14:16,440 Then, we are integrating only between minus pi and pi, 224 00:14:16,440 --> 00:14:18,290 in this interval we only have one 225 00:14:18,290 --> 00:14:23,000 pulse, so we can remove the implicit periodization. 226 00:14:23,000 --> 00:14:25,550 And then, because of the sifting property of the 227 00:14:25,550 --> 00:14:29,930 delta functional, this integral will be just the value 228 00:14:29,930 --> 00:14:33,010 of this function of the real valued variable omega 229 00:14:33,010 --> 00:14:36,400 in 0, because this delta is centered in 0. 230 00:14:36,400 --> 00:14:41,700 And so the value of e to the j omega n for omega equal to 0 is equal to 1. 231 00:14:41,700 --> 00:14:47,140 This is formally similar to the fact that the inverse DFT 232 00:14:47,140 --> 00:14:51,660 of N delta of k is 233 00:14:51,660 --> 00:14:55,710 actually equal to 1. So, by using the delta functional, we have 234 00:14:55,710 --> 00:15:01,670 established another formal parallel between the DFT and DTFT. 235 00:15:01,670 --> 00:15:06,840 So if the inverse DTFT of the delta functional is 1, 236 00:15:06,840 --> 00:15:11,530 then it means that the direct DTFT, the forward Fourier transform 237 00:15:11,530 --> 00:15:15,060 of the constant 1 is formally equal to the pulse train. 238 00:15:17,590 --> 00:15:19,060 Does this make sense? 239 00:15:19,060 --> 00:15:22,450 Well, we could try to compute numerically the partial sums that 240 00:15:22,450 --> 00:15:26,890 are involved in the computation of the DTFT of the constant 1. 241 00:15:26,890 --> 00:15:30,770 So define Sk of omega as the sum that goes from 242 00:15:30,770 --> 00:15:35,580 minus k to k of e to the minus j omega n. 243 00:15:35,580 --> 00:15:38,240 As k goes to infinity as k of omega, 244 00:15:38,240 --> 00:15:41,540 should converge to the DTFT of the constant one. 245 00:15:41,540 --> 00:15:42,950 So if we plot these partial 246 00:15:42,950 --> 00:15:47,010 sums in magnitude for increasing values of k, we have something 247 00:15:47,010 --> 00:15:49,960 like this for k equal to 5 we have this shape. 248 00:15:49,960 --> 00:15:53,820 And as we increase, the index, we see that this 249 00:15:53,820 --> 00:15:58,665 family of partial sums looks like a family of localizing functions. 250 00:15:58,665 --> 00:16:06,420 So, the support gets narrow and the area stays constant. 251 00:16:06,420 --> 00:16:08,040 So, it really makes sense to 252 00:16:08,040 --> 00:16:10,490 say that in the limit, this partial 253 00:16:10,490 --> 00:16:14,010 sums will converge to the direct delta function. 254 00:16:14,010 --> 00:16:16,860 With this fundamental result in our pocket, we can now proceed 255 00:16:16,860 --> 00:16:22,500 to derive some other interesting DTFT pairs for non-square summable sequences. 256 00:16:22,500 --> 00:16:27,420 With the same technique we used before, we can show that the inverse DTFT 257 00:16:27,420 --> 00:16:32,640 of a shifted pulse train a pulse train shifted by a frequency omega 0. 258 00:16:32,640 --> 00:16:33,540 Gives a 259 00:16:33,540 --> 00:16:38,360 complex exponential of frequency omega 0 in the time domain. 260 00:16:38,360 --> 00:16:42,370 So if the DTFT of 1 is the pulse train centered in 0. 261 00:16:42,370 --> 00:16:46,878 The DTFT of an arbitrary complex exponential of frequency 262 00:16:46,878 --> 00:16:51,650 omega 0, is the pulse train shifted by omega 0. 263 00:16:51,650 --> 00:16:55,290 By using Euler's relation and the linearity of the 264 00:16:55,290 --> 00:16:59,030 DTFT, we can derive the DTFT of the cosine 265 00:16:59,030 --> 00:17:03,772 of omega 0 n which is just one half times the sum of 2 266 00:17:03,772 --> 00:17:08,842 pulse trains, one centered in omega 0 and the other one centered in minus 267 00:17:08,842 --> 00:17:14,110 omega 0. And the DTFT of the sine of omega 0 n, 268 00:17:14,110 --> 00:17:19,610 which is minus j over 2, times a pulse train centered in the omega 0, and another 269 00:17:19,610 --> 00:17:25,387 pulse train centered in minus omega 0.