1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,720 continue to offer high quality educational resources for free. 5 00:00:10,720 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,280 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,280 --> 00:00:18,450 at ocw.mit.edu. 8 00:00:55,797 --> 00:00:57,880 ALAN V. OPPENHEIM: During the last three lectures, 9 00:00:57,880 --> 00:01:01,230 we considered the design of infinite impulse response 10 00:01:01,230 --> 00:01:03,150 digital filters. 11 00:01:03,150 --> 00:01:07,860 The two main classes of design procedures that we discussed 12 00:01:07,860 --> 00:01:10,620 involved mapping continuous time designs 13 00:01:10,620 --> 00:01:12,240 to discrete time designs. 14 00:01:12,240 --> 00:01:13,920 That was the first class. 15 00:01:13,920 --> 00:01:16,740 And the second class was the use of algorithmic design 16 00:01:16,740 --> 00:01:18,760 procedures. 17 00:01:18,760 --> 00:01:21,600 In this lecture, I would like to discuss 18 00:01:21,600 --> 00:01:25,890 the design of finite impulse response digital filters. 19 00:01:25,890 --> 00:01:29,760 And I remind you that as we've stressed 20 00:01:29,760 --> 00:01:33,840 in a number of the lectures, one of the very important aspects 21 00:01:33,840 --> 00:01:36,570 of finite impulse response digital filters 22 00:01:36,570 --> 00:01:39,450 is the fact that they can be implemented 23 00:01:39,450 --> 00:01:42,090 to have exactly linear phase. 24 00:01:42,090 --> 00:01:44,730 So finite impulse response filters, 25 00:01:44,730 --> 00:01:47,790 as opposed to infinite impulse response filters, 26 00:01:47,790 --> 00:01:51,720 are useful for problems where linear phase is 27 00:01:51,720 --> 00:01:54,120 an important constraint. 28 00:01:54,120 --> 00:01:55,330 It turns out also-- 29 00:01:55,330 --> 00:01:57,360 and this isn't a point that we'll be touching on 30 00:01:57,360 --> 00:01:58,590 in these lectures-- 31 00:01:58,590 --> 00:02:02,130 but it turns out also that finite impulse response 32 00:02:02,130 --> 00:02:05,730 digital filters have some important advantages 33 00:02:05,730 --> 00:02:08,130 over infinite impulse response filters 34 00:02:08,130 --> 00:02:10,500 with regard to round off noise, that is, 35 00:02:10,500 --> 00:02:15,470 with regard to the effects of finite register length. 36 00:02:15,470 --> 00:02:18,150 Now in general, the implementation 37 00:02:18,150 --> 00:02:21,120 of finite impulse response filters 38 00:02:21,120 --> 00:02:25,230 is considerably simpler for discrete time systems 39 00:02:25,230 --> 00:02:29,070 than it is for continuous time systems. 40 00:02:29,070 --> 00:02:33,840 Consequently, it turns out that there are not really available, 41 00:02:33,840 --> 00:02:37,770 or a wide variety of continuous time design procedures 42 00:02:37,770 --> 00:02:40,140 for finite impulse response filters 43 00:02:40,140 --> 00:02:43,350 as there were for infinite impulse response. 44 00:02:43,350 --> 00:02:48,540 And consequently, the continuous time to discrete time mapping, 45 00:02:48,540 --> 00:02:52,410 as a design procedure, is not particularly 46 00:02:52,410 --> 00:02:55,810 applicable to the design of finite impulse response 47 00:02:55,810 --> 00:02:57,060 digital filters. 48 00:02:57,060 --> 00:03:01,890 So this includes, for example, the impulse and variant 49 00:03:01,890 --> 00:03:05,580 technique, and the use of the bilinear transformation. 50 00:03:05,580 --> 00:03:09,390 That is, those techniques are not particularly useful 51 00:03:09,390 --> 00:03:13,230 when we are talking about the design of finite impulse 52 00:03:13,230 --> 00:03:15,480 response filters. 53 00:03:15,480 --> 00:03:18,810 As an additional interesting side point, 54 00:03:18,810 --> 00:03:24,030 if we were to begin with a finite impulse response 55 00:03:24,030 --> 00:03:26,640 analog or continuous time filter, 56 00:03:26,640 --> 00:03:29,340 and map it to a digital filter using the bilinear 57 00:03:29,340 --> 00:03:32,580 transformation, then in fact the resulting filter 58 00:03:32,580 --> 00:03:37,560 would not be a finite impulse response filter. 59 00:03:37,560 --> 00:03:39,480 The algorithmic procedures that we 60 00:03:39,480 --> 00:03:43,110 talked about in the previous lectures, of course, still 61 00:03:43,110 --> 00:03:47,850 do apply to the design of finite impulse response filters. 62 00:03:47,850 --> 00:03:51,150 And there also are some other algorithmic procedures 63 00:03:51,150 --> 00:03:53,160 that are discussed in the text, and that I 64 00:03:53,160 --> 00:03:58,800 will touch on just very briefly in today's lecture. 65 00:03:58,800 --> 00:04:05,460 Now for the design of finite impulse response filters, 66 00:04:05,460 --> 00:04:09,600 there are basically three classes, 67 00:04:09,600 --> 00:04:13,140 or three types of design techniques 68 00:04:13,140 --> 00:04:15,810 that we'll be discussing. 69 00:04:15,810 --> 00:04:18,750 And in this lecture, what I intend to do 70 00:04:18,750 --> 00:04:22,380 is present the three of these, primarily 71 00:04:22,380 --> 00:04:26,550 with an eye toward giving you an appreciation for what 72 00:04:26,550 --> 00:04:29,550 the style of the design techniques is. 73 00:04:29,550 --> 00:04:33,520 That is, to give you an overview of these design techniques. 74 00:04:33,520 --> 00:04:38,470 But I won't dwell on many of the details. 75 00:04:38,470 --> 00:04:42,540 So in the design of finite impulse response 76 00:04:42,540 --> 00:04:46,110 digital filters, then, we of course 77 00:04:46,110 --> 00:04:49,350 are talking about a unit sample response 78 00:04:49,350 --> 00:04:51,570 that is a finite length. 79 00:04:51,570 --> 00:04:56,130 And we'll choose the length, the interval over which the unit 80 00:04:56,130 --> 00:05:01,170 sample response is non-zero to be in the range from 0 81 00:05:01,170 --> 00:05:03,550 to capital N minus 1. 82 00:05:03,550 --> 00:05:08,610 So the Z-transform of the finite impulse response filter 83 00:05:08,610 --> 00:05:13,860 is then given by the sum from 0 to capital N minus 1 of h of n, 84 00:05:13,860 --> 00:05:17,880 z to the minus n, which of course is a polynomial in z 85 00:05:17,880 --> 00:05:19,390 to the minus 1. 86 00:05:19,390 --> 00:05:23,250 And we know that, for this class of systems, h of z 87 00:05:23,250 --> 00:05:27,240 has only zeros, except at z equal 0. 88 00:05:27,240 --> 00:05:31,620 That is, the poles of h of z can only fall at z equals 0, 89 00:05:31,620 --> 00:05:33,930 or z equals infinity. 90 00:05:33,930 --> 00:05:39,420 And I remind you also that for linear phase finite impulse 91 00:05:39,420 --> 00:05:43,950 response filters, the constraint on the unit sample response 92 00:05:43,950 --> 00:05:48,570 is that it basically is symmetrical about its midpoint. 93 00:05:48,570 --> 00:05:50,730 In other words, as we look at the unit sample 94 00:05:50,730 --> 00:05:53,740 response starting from the left hand end 95 00:05:53,740 --> 00:05:55,760 and starting from the right hand end, 96 00:05:55,760 --> 00:06:00,420 approaching toward the middle, then we see the same values. 97 00:06:00,420 --> 00:06:02,390 In other words, there's a symmetry that's 98 00:06:02,390 --> 00:06:06,290 required for linear phase. 99 00:06:06,290 --> 00:06:10,340 The three basic design procedures 100 00:06:10,340 --> 00:06:12,650 that I'd like to talk about in this lecture, 101 00:06:12,650 --> 00:06:15,110 basically introduce in the lecture, 102 00:06:15,110 --> 00:06:17,900 are first of all, a design technique 103 00:06:17,900 --> 00:06:21,890 which is referred to as the window method. 104 00:06:21,890 --> 00:06:24,830 The second design technique that is commonly 105 00:06:24,830 --> 00:06:28,780 referred to as the frequency sampling method. 106 00:06:28,780 --> 00:06:33,440 And a third designed technique, which is an algorithmic design 107 00:06:33,440 --> 00:06:39,020 procedure which results in equiripple filters, which 108 00:06:39,020 --> 00:06:41,450 as I'll comment on again when we get 109 00:06:41,450 --> 00:06:43,940 to this point in the lecture, turn out 110 00:06:43,940 --> 00:06:46,280 to be important because they have 111 00:06:46,280 --> 00:06:48,570 some optimality properties. 112 00:06:48,570 --> 00:06:52,220 So these are the three basic design techniques, then, 113 00:06:52,220 --> 00:06:56,370 that I'd like to discuss in this lecture. 114 00:06:56,370 --> 00:07:00,800 Well, to begin with, the window method 115 00:07:00,800 --> 00:07:09,190 is basically a method based on the following strategy. 116 00:07:09,190 --> 00:07:14,650 Let's suppose that we begin the design of the finite impulse 117 00:07:14,650 --> 00:07:19,750 response filter with a desired unit sample response, h sub 118 00:07:19,750 --> 00:07:20,590 d of n. 119 00:07:20,590 --> 00:07:22,270 For example, suppose that we wanted 120 00:07:22,270 --> 00:07:26,290 to design a low pass filter. 121 00:07:26,290 --> 00:07:29,260 We could begin with a desired unit sample 122 00:07:29,260 --> 00:07:31,840 response which was the unit sample response 123 00:07:31,840 --> 00:07:34,400 of an ideal low pass filter. 124 00:07:34,400 --> 00:07:39,550 Now as you know, the unit sample response of an ideal low pass 125 00:07:39,550 --> 00:07:46,180 filter basically has a sin x over x type of envelope. 126 00:07:46,180 --> 00:07:48,610 In particular, the important characteristic 127 00:07:48,610 --> 00:07:52,400 is that it extends from minus infinity to plus infinity. 128 00:07:52,400 --> 00:07:56,410 So in fact, it doesn't correspond to a finite impulse 129 00:07:56,410 --> 00:07:57,940 response filter. 130 00:07:57,940 --> 00:08:01,780 But we could imagine obtaining the unit sample 131 00:08:01,780 --> 00:08:09,220 response of an FIR filter by perhaps truncating 132 00:08:09,220 --> 00:08:14,020 this infinitely long unit sample response so that we retain 133 00:08:14,020 --> 00:08:18,650 only the values in the interval from 0 to capital N minus 1. 134 00:08:18,650 --> 00:08:23,890 Or more generally, we can imagine multiplying the desired 135 00:08:23,890 --> 00:08:29,020 unit sample response by a window, the window being 136 00:08:29,020 --> 00:08:33,559 0 outside the interval from 0 to capital N minus 1. 137 00:08:33,559 --> 00:08:36,370 So the window method basically consists 138 00:08:36,370 --> 00:08:40,059 of beginning with a desired unit sample response, which 139 00:08:40,059 --> 00:08:42,000 in general will be infinitely long, 140 00:08:42,000 --> 00:08:45,280 or at least longer than the filter that we want to design. 141 00:08:45,280 --> 00:08:50,630 We apply to that a window, which is 0 outside the range 0 142 00:08:50,630 --> 00:08:53,041 to capital N minus 1. 143 00:08:53,041 --> 00:08:56,050 And the result then is the unit sample response 144 00:08:56,050 --> 00:09:00,020 of the finite impulse response filter. 145 00:09:00,020 --> 00:09:04,310 Well, what is the effect of this in the frequency domain? 146 00:09:04,310 --> 00:09:07,570 We know that multiplication in the time domain 147 00:09:07,570 --> 00:09:11,570 corresponds to convolution in the frequency domain. 148 00:09:11,570 --> 00:09:16,600 And consequently, the frequency response of the finite impulse 149 00:09:16,600 --> 00:09:22,510 response filter that results, is the periodic convolution 150 00:09:22,510 --> 00:09:26,440 of the desired frequency response. 151 00:09:26,440 --> 00:09:29,740 That is, the Fourier transform of the desired unit sample 152 00:09:29,740 --> 00:09:33,580 response, that convolved with the Fourier 153 00:09:33,580 --> 00:09:38,090 transform of the window, w of n 154 00:09:38,090 --> 00:09:42,460 For example, if we were talking about a rectangular window, 155 00:09:42,460 --> 00:09:45,025 and in a rectangular window, w of n 156 00:09:45,025 --> 00:09:51,650 is simply constant over the range 0 to capital N minus 1, 157 00:09:51,650 --> 00:09:56,200 the magnitude of the Fourier transform of the window 158 00:09:56,200 --> 00:10:00,100 has a form roughly as I've indicated here. 159 00:10:00,100 --> 00:10:02,710 This is drawn for the case, N equals 8. 160 00:10:02,710 --> 00:10:06,370 And it's the magnitude of h of e to the j omega-- 161 00:10:06,370 --> 00:10:09,400 I'm sorry, of w of e to the j omega. 162 00:10:09,400 --> 00:10:11,980 If we were to remove the magnitude sines, 163 00:10:11,980 --> 00:10:16,620 then every alternate lobe would be flipped over in sine. 164 00:10:16,620 --> 00:10:22,550 So in fact, this is a sin nx over sin x curve. 165 00:10:22,550 --> 00:10:27,130 So for the particular case of a rectangular window, 166 00:10:27,130 --> 00:10:33,640 we would have not the ideal low pass filter that we began with, 167 00:10:33,640 --> 00:10:37,420 but the ideal lowpass filter frequency characteristic 168 00:10:37,420 --> 00:10:42,430 convolved with the Fourier transform of the window. 169 00:10:42,430 --> 00:10:50,980 So that's depicted here where we have the ideal low pass filter. 170 00:10:50,980 --> 00:10:55,720 Superimposed on top of it is the Fourier transform 171 00:10:55,720 --> 00:10:59,380 of, in this case, a rectangular window. 172 00:10:59,380 --> 00:11:04,390 And then to obtain the resulting overall frequency 173 00:11:04,390 --> 00:11:07,750 response of the finite impulse response filter, 174 00:11:07,750 --> 00:11:13,240 that would be obtained by convolving these two together. 175 00:11:13,240 --> 00:11:16,780 And we can see that the result of that, very roughly, 176 00:11:16,780 --> 00:11:20,650 is to, in fact, at the discontinuity, 177 00:11:20,650 --> 00:11:25,630 the almost exact effect is to integrate the window frequency 178 00:11:25,630 --> 00:11:28,900 response, the frequency response of the window. 179 00:11:28,900 --> 00:11:33,400 And the resulting frequency response for the finite impulse 180 00:11:33,400 --> 00:11:37,420 response filter then, as opposed to being 181 00:11:37,420 --> 00:11:41,770 the ideal characteristic that we desired, in fact 182 00:11:41,770 --> 00:11:44,770 has some ripples in it. 183 00:11:44,770 --> 00:11:50,800 Generally, then, it has some deviation in the pass band. 184 00:11:50,800 --> 00:11:53,470 It has some deviation in the stop band. 185 00:11:53,470 --> 00:11:57,940 And it has a transition region from pass band to stop band. 186 00:11:57,940 --> 00:12:02,890 And that's introduced because of the convolution of this window 187 00:12:02,890 --> 00:12:07,000 frequency response with the desired or ideal frequency 188 00:12:07,000 --> 00:12:09,380 response. 189 00:12:09,380 --> 00:12:14,420 Now ideally, what would we like as the frequency 190 00:12:14,420 --> 00:12:16,730 response of the window? 191 00:12:16,730 --> 00:12:19,400 Well, ideally what we would like, of course, 192 00:12:19,400 --> 00:12:25,430 is we would like this function to be an impulse. 193 00:12:25,430 --> 00:12:26,990 Because if it's an impulse, then we 194 00:12:26,990 --> 00:12:29,600 convolve it with the desired frequency response, 195 00:12:29,600 --> 00:12:34,640 and the result will again be the desired frequency response. 196 00:12:34,640 --> 00:12:39,230 The overshoot that we have in the pass band and the stop band 197 00:12:39,230 --> 00:12:44,870 has arisen because of the side lobe structure of the frequency 198 00:12:44,870 --> 00:12:47,840 response of the window. 199 00:12:47,840 --> 00:12:52,190 And the transition region has arisen because, 200 00:12:52,190 --> 00:12:56,600 primarily because of the width of this central lobe. 201 00:12:56,600 --> 00:13:01,850 So, in fact, if we want to keep the deviation in the pass band 202 00:13:01,850 --> 00:13:05,300 and stop band small, then that implies 203 00:13:05,300 --> 00:13:07,010 that we would like to keep the side lobe 204 00:13:07,010 --> 00:13:11,870 structure of the window small, the window frequency 205 00:13:11,870 --> 00:13:14,090 response small. 206 00:13:14,090 --> 00:13:17,180 And if we would like to keep the transition band narrow, 207 00:13:17,180 --> 00:13:22,970 that implies that we would like to keep this main lobe narrow. 208 00:13:22,970 --> 00:13:27,170 And as I think that your intuition should 209 00:13:27,170 --> 00:13:30,710 agree with by now, generally, an attempt 210 00:13:30,710 --> 00:13:35,960 to make this function narrower in frequency 211 00:13:35,960 --> 00:13:39,410 will have the general effect of making 212 00:13:39,410 --> 00:13:41,780 the window longer in time. 213 00:13:41,780 --> 00:13:49,010 So that, in fact, as we try to make the time window, w of n, 214 00:13:49,010 --> 00:13:51,890 or equivalently, the order of the finite impulse response 215 00:13:51,890 --> 00:13:56,030 filter smaller, the tendency will 216 00:13:56,030 --> 00:14:00,950 be toward a widening transition band in the frequency 217 00:14:00,950 --> 00:14:03,110 response of the filter. 218 00:14:03,110 --> 00:14:06,920 Likewise, as we allow the order of the filter to get longer, 219 00:14:06,920 --> 00:14:09,350 we have more flexibility in the filter. 220 00:14:09,350 --> 00:14:13,250 And consequently, we're able to control better the transition 221 00:14:13,250 --> 00:14:15,830 band and the overshoot in both the pass band 222 00:14:15,830 --> 00:14:17,111 and in the stop band. 223 00:14:20,200 --> 00:14:23,620 Well, in particular, again continuing 224 00:14:23,620 --> 00:14:28,240 with the example of a rectangular window, 225 00:14:28,240 --> 00:14:31,060 if we look at the frequency-- the unit sample 226 00:14:31,060 --> 00:14:38,980 response for the case of an ideal low pass 227 00:14:38,980 --> 00:14:46,180 filter multiplied by a 51 point rectangular window, 228 00:14:46,180 --> 00:14:50,630 then the unit sample response we obtain is as indicated here. 229 00:14:50,630 --> 00:14:53,980 Actually, I've marked this as the desired unit sample 230 00:14:53,980 --> 00:14:55,060 response. 231 00:14:55,060 --> 00:14:58,090 It's only equal to the desired unit sample response 232 00:14:58,090 --> 00:15:01,840 in the interval from 0 to 50. 233 00:15:01,840 --> 00:15:04,220 And the desired unit sample response, of course, 234 00:15:04,220 --> 00:15:07,330 continues to be non-zero outside that range. 235 00:15:07,330 --> 00:15:09,910 Whereas multiplying by a rectangular window 236 00:15:09,910 --> 00:15:14,530 truncates it to be 0 outside this range. 237 00:15:14,530 --> 00:15:20,230 The resulting frequency response for this particular example 238 00:15:20,230 --> 00:15:22,070 I've indicated here. 239 00:15:22,070 --> 00:15:28,930 And as opposed to the previous frequency response 240 00:15:28,930 --> 00:15:32,380 that I depicted, this is shown on a logarithmic frequency 241 00:15:32,380 --> 00:15:33,310 scale. 242 00:15:33,310 --> 00:15:35,380 There, in fact, is some ripple in the pass band, 243 00:15:35,380 --> 00:15:38,920 although it's not evident on a logarithmic frequency scale. 244 00:15:38,920 --> 00:15:44,890 But generally, you can see the effect of a transition band, 245 00:15:44,890 --> 00:15:50,650 and then a side lobe structure in the stop band. 246 00:15:50,650 --> 00:15:56,260 And you can see also that the asymptotic behavior of this 247 00:15:56,260 --> 00:15:57,220 is-- 248 00:15:57,220 --> 00:16:00,190 in the vicinity for this particular example-- 249 00:16:00,190 --> 00:16:04,750 is in the vicinity of minus 20 dB attenuation in the stop 250 00:16:04,750 --> 00:16:06,880 band. 251 00:16:06,880 --> 00:16:11,920 So this then is an example of a low pass filter that 252 00:16:11,920 --> 00:16:15,640 could be obtained by applying a rectangular window 253 00:16:15,640 --> 00:16:19,640 to the ideal low pass filter. 254 00:16:19,640 --> 00:16:24,940 There are other possible choices for windows. 255 00:16:24,940 --> 00:16:30,460 In particular, two other choices that I'd like to mention 256 00:16:30,460 --> 00:16:34,570 are what are referred to as the Hamming window 257 00:16:34,570 --> 00:16:36,760 and the Bartlett window. 258 00:16:39,530 --> 00:16:42,170 As opposed to the rectangular window, 259 00:16:42,170 --> 00:16:46,140 the Bartlett window is a triangular window, 260 00:16:46,140 --> 00:16:48,470 which has the effect basically of making 261 00:16:48,470 --> 00:16:52,610 the transition at the ends of the unit sample response 262 00:16:52,610 --> 00:16:57,920 less abrupt than they are in the case of a rectangular window. 263 00:16:57,920 --> 00:17:04,849 The Hamming window is a window which corresponds essentially 264 00:17:04,849 --> 00:17:11,099 to a cosine bell sitting on a small pedestal. 265 00:17:11,099 --> 00:17:15,650 So for this particular example, again, the low pass filter, 266 00:17:15,650 --> 00:17:18,079 the ideal low pass filter windowed, 267 00:17:18,079 --> 00:17:21,470 we can compare the frequency response obtained 268 00:17:21,470 --> 00:17:24,020 with the Bartlett window to that obtained 269 00:17:24,020 --> 00:17:27,869 with the rectangular window, and also with the Hamming window. 270 00:17:27,869 --> 00:17:32,130 So let's take a look at that. 271 00:17:32,130 --> 00:17:35,480 Here we have the frequency response 272 00:17:35,480 --> 00:17:37,760 of the Bartlett window. 273 00:17:37,760 --> 00:17:41,870 And you can see generally that the ripples aren't 274 00:17:41,870 --> 00:17:44,810 as severe in the stop band as they 275 00:17:44,810 --> 00:17:47,600 were in the case of the rectangular window. 276 00:17:47,600 --> 00:17:51,350 And in fact, we can compare these simply 277 00:17:51,350 --> 00:17:59,600 by referring back to the rectangular window frequency 278 00:17:59,600 --> 00:18:02,430 response that we had previously. 279 00:18:02,430 --> 00:18:08,330 So the rectangular window lies slightly under, or actually 280 00:18:08,330 --> 00:18:11,630 asymptotically touches the frequency response obtained 281 00:18:11,630 --> 00:18:13,370 from the Bartlett window. 282 00:18:13,370 --> 00:18:17,810 But for the Bartlett window, the ripples in the stop band 283 00:18:17,810 --> 00:18:23,690 aren't as severe as they are for the rectangular window. 284 00:18:23,690 --> 00:18:29,150 Now to compare both of these to the Hamming window, 285 00:18:29,150 --> 00:18:34,460 we can put on top of this the frequency response 286 00:18:34,460 --> 00:18:36,590 for the Hamming window. 287 00:18:40,400 --> 00:18:46,030 And in this case, you can see that the frequency response 288 00:18:46,030 --> 00:18:50,410 obtained with the Hamming window applied to a low pass filter 289 00:18:50,410 --> 00:18:54,670 has considerably better stop band attenuation 290 00:18:54,670 --> 00:18:57,490 than either the rectangular or the Bartlett window. 291 00:18:57,490 --> 00:19:01,840 In particular, it comes down to about minus 30 or minus 32 292 00:19:01,840 --> 00:19:04,300 or so dB. 293 00:19:04,300 --> 00:19:06,550 There is a slight price paid for this. 294 00:19:06,550 --> 00:19:11,320 And the price is, as you notice, that the transition band 295 00:19:11,320 --> 00:19:15,340 is slightly wider than it is for either of the other two 296 00:19:15,340 --> 00:19:16,900 windows. 297 00:19:16,900 --> 00:19:21,970 And in general, this is a type of trade off 298 00:19:21,970 --> 00:19:27,250 that arises again and again in the design of-- actually, 299 00:19:27,250 --> 00:19:30,970 always in the design of digital or analog filters. 300 00:19:30,970 --> 00:19:33,760 In particular, in using windows, there 301 00:19:33,760 --> 00:19:40,420 is always this trade off between improved stop band attenuation 302 00:19:40,420 --> 00:19:45,070 and a narrower-- or wider, rather-- transition band. 303 00:19:45,070 --> 00:19:48,490 As you are willing to allow the transition 304 00:19:48,490 --> 00:19:51,970 band to become wider, you can generally 305 00:19:51,970 --> 00:19:56,540 obtain some advantage in terms of stop band attenuation. 306 00:19:56,540 --> 00:19:57,040 All right. 307 00:19:57,040 --> 00:20:03,400 So this is a quick view into the use of windows. 308 00:20:03,400 --> 00:20:06,760 In the text, in fact, there are quite a large number 309 00:20:06,760 --> 00:20:10,290 of windows that are discussed, besides the rectangular, 310 00:20:10,290 --> 00:20:12,580 the Bartlett, and the Hamming window. 311 00:20:12,580 --> 00:20:18,550 And similar frequency responses for filters 312 00:20:18,550 --> 00:20:21,780 designed using a variety of these windows 313 00:20:21,780 --> 00:20:22,835 is presented in the text. 314 00:20:25,740 --> 00:20:31,140 The next design procedure that I would like to discuss 315 00:20:31,140 --> 00:20:37,200 is the design procedure that is referred to as the frequency 316 00:20:37,200 --> 00:20:39,720 sampling design procedure. 317 00:20:39,720 --> 00:20:44,670 And basically, the idea behind that procedure 318 00:20:44,670 --> 00:20:51,000 is to argue that if we have a desired frequency 319 00:20:51,000 --> 00:20:55,620 response, which I've indicated by the continuous line-- 320 00:20:55,620 --> 00:20:58,260 a desired frequency response-- 321 00:20:58,260 --> 00:21:02,430 then we could think of designing a digital filter 322 00:21:02,430 --> 00:21:08,940 by specifying samples of the desired frequency 323 00:21:08,940 --> 00:21:12,480 response at equally spaced points along the frequency 324 00:21:12,480 --> 00:21:13,620 axis. 325 00:21:13,620 --> 00:21:16,110 Now we know that for a finite impulse response 326 00:21:16,110 --> 00:21:17,400 digital filter-- 327 00:21:17,400 --> 00:21:19,950 in general, for any finite length sequence-- 328 00:21:19,950 --> 00:21:23,700 we can represent it in terms of samples of its Fourier 329 00:21:23,700 --> 00:21:24,930 transform. 330 00:21:24,930 --> 00:21:28,060 That, in fact, is the discrete Fourier transform. 331 00:21:28,060 --> 00:21:30,810 So basically what we're saying is 332 00:21:30,810 --> 00:21:36,150 that we can begin with a desired frequency response 333 00:21:36,150 --> 00:21:40,770 and specify the discrete Fourier transform of the finite impulse 334 00:21:40,770 --> 00:21:45,810 response filter by specifying equally spaced samples 335 00:21:45,810 --> 00:21:48,780 of the desired frequency response. 336 00:21:48,780 --> 00:21:53,880 So we have then that the actual frequency response 337 00:21:53,880 --> 00:21:57,990 will correspond to capital N samples, equally 338 00:21:57,990 --> 00:22:02,040 spaced in frequency, of the desired frequency response 339 00:22:02,040 --> 00:22:04,690 characteristic. 340 00:22:04,690 --> 00:22:11,400 Now one could say that, or one could inquire as to whether, 341 00:22:11,400 --> 00:22:15,540 in fact, if we do that, what we're able to implement 342 00:22:15,540 --> 00:22:20,610 is the desired digital filter frequency response. 343 00:22:20,610 --> 00:22:24,120 In other words, we know that we can implement a finite impulse 344 00:22:24,120 --> 00:22:31,860 response filter by using only the values specified 345 00:22:31,860 --> 00:22:34,440 in the discrete Fourier transform. 346 00:22:34,440 --> 00:22:38,450 And here we've specified a discrete Fourier transform 347 00:22:38,450 --> 00:22:42,990 that's unity in the pass band and it's zero in the stop band. 348 00:22:42,990 --> 00:22:45,660 And one can ask, well, in fact, is 349 00:22:45,660 --> 00:22:50,440 that an ideal digital low pass filter? 350 00:22:50,440 --> 00:22:53,020 Well, the answer to that is no. 351 00:22:53,020 --> 00:22:57,930 And the reason is that it is important, very important 352 00:22:57,930 --> 00:23:04,350 to keep track of the fact that just because we specify 353 00:23:04,350 --> 00:23:08,310 the frequency response at equally spaced points 354 00:23:08,310 --> 00:23:12,390 along the frequency axis doesn't mean 355 00:23:12,390 --> 00:23:15,750 that, in between those points, the frequency 356 00:23:15,750 --> 00:23:18,900 characteristic has to be particularly well-behaved. 357 00:23:18,900 --> 00:23:24,840 That is, the Fourier transform of the unit sample response 358 00:23:24,840 --> 00:23:27,720 is a continuous function of frequency. 359 00:23:27,720 --> 00:23:30,870 We're specifying samples of that continuous function 360 00:23:30,870 --> 00:23:32,290 of frequency. 361 00:23:32,290 --> 00:23:36,390 And there is implied an interpolation 362 00:23:36,390 --> 00:23:39,620 in between those samples. 363 00:23:39,620 --> 00:23:43,730 Well, in fact, if we were to look at a unit sample 364 00:23:43,730 --> 00:23:51,810 response whose discrete Fourier transform magnitude is unity 365 00:23:51,810 --> 00:23:55,540 in the pass band and zero in the stop band, 366 00:23:55,540 --> 00:24:02,340 one possible choice, in fact, is the unit sample response 367 00:24:02,340 --> 00:24:05,790 and frequency response that I've indicated here. 368 00:24:05,790 --> 00:24:09,310 This is the unit sample response. 369 00:24:09,310 --> 00:24:12,770 This is the corresponding frequency response. 370 00:24:12,770 --> 00:24:17,400 The discrete Fourier transform of h of n, in fact, 371 00:24:17,400 --> 00:24:21,720 is exactly unity in the pass band and zero in the stop band. 372 00:24:21,720 --> 00:24:24,960 But the interpolation in between those points 373 00:24:24,960 --> 00:24:28,710 is not unity in the pass band and zero in the stop band. 374 00:24:28,710 --> 00:24:33,840 In fact, by simply putting the previous view graph 375 00:24:33,840 --> 00:24:43,800 on top of this one, we can see that this frequency response 376 00:24:43,800 --> 00:24:50,580 characteristic indeed goes through the specified frequency 377 00:24:50,580 --> 00:24:52,050 points. 378 00:24:52,050 --> 00:24:56,760 But it is not particularly well constrained 379 00:24:56,760 --> 00:24:59,020 in between those points. 380 00:24:59,020 --> 00:25:03,390 So, in fact, for one example, if we chose this 381 00:25:03,390 --> 00:25:06,180 as the unit sample response corresponding 382 00:25:06,180 --> 00:25:08,580 to this set of frequency samples, 383 00:25:08,580 --> 00:25:12,540 the resulting frequency response of the finite impulse response 384 00:25:12,540 --> 00:25:16,920 low pass filter would be as we've indicated 385 00:25:16,920 --> 00:25:20,770 with this continuous curve. 386 00:25:20,770 --> 00:25:23,730 Well, we know that there are other unit sample 387 00:25:23,730 --> 00:25:27,630 responses whose discrete Fourier transforms are 388 00:25:27,630 --> 00:25:29,332 exactly the same. 389 00:25:29,332 --> 00:25:31,290 The magnitude of the discrete Fourier transform 390 00:25:31,290 --> 00:25:33,960 is exactly the same as this one. 391 00:25:33,960 --> 00:25:37,840 In particular, we know that if we were to circularly rotate 392 00:25:37,840 --> 00:25:43,470 h of n, circularly shift h of n, the magnitude 393 00:25:43,470 --> 00:25:45,690 of the discrete Fourier transform 394 00:25:45,690 --> 00:25:47,920 would remain exactly the same. 395 00:25:47,920 --> 00:25:51,420 So in fact, another unit sample response 396 00:25:51,420 --> 00:25:56,370 that would correspond to a frequency response, which 397 00:25:56,370 --> 00:26:00,690 likewise goes through the same frequency samples, 398 00:26:00,690 --> 00:26:04,995 is the one that we have here. 399 00:26:07,580 --> 00:26:10,880 This unit sample response is simply the previous one, 400 00:26:10,880 --> 00:26:15,170 circularly shifted by approximately n over 2, 401 00:26:15,170 --> 00:26:17,330 or n minus 1 over 2. 402 00:26:17,330 --> 00:26:20,780 And in this case, the frequency response characteristic 403 00:26:20,780 --> 00:26:24,750 is even worse than it was in the previous example. 404 00:26:24,750 --> 00:26:27,290 So here we have a frequency response 405 00:26:27,290 --> 00:26:31,790 that has very wild, large fluctuations in the pass band, 406 00:26:31,790 --> 00:26:36,350 and even in the stop band, but still goes 407 00:26:36,350 --> 00:26:39,410 through the same frequency samples 408 00:26:39,410 --> 00:26:42,210 that we had specified previously. 409 00:26:42,210 --> 00:26:46,160 And we can see that by, again, putting 410 00:26:46,160 --> 00:26:48,090 these two on top of each other. 411 00:26:54,240 --> 00:26:57,560 And you can see, in this case again, 412 00:26:57,560 --> 00:27:00,770 that this frequency characteristic 413 00:27:00,770 --> 00:27:05,330 goes through the specified frequency samples. 414 00:27:05,330 --> 00:27:10,171 But it has a very wild behavior in between. 415 00:27:10,171 --> 00:27:10,670 All right. 416 00:27:10,670 --> 00:27:14,750 So the frequency sampling method, then, is basically 417 00:27:14,750 --> 00:27:19,550 a method that consists of specifying samples 418 00:27:19,550 --> 00:27:24,650 of the frequency response, and then 419 00:27:24,650 --> 00:27:28,850 accepting the interpolation in between those samples 420 00:27:28,850 --> 00:27:34,040 that is dictated by the form of the unit sample response. 421 00:27:34,040 --> 00:27:38,780 And I stress that one has to be careful in the phase that's 422 00:27:38,780 --> 00:27:40,550 applied to this, or equivalently, 423 00:27:40,550 --> 00:27:43,400 the circular shift that's applied to the unit sample 424 00:27:43,400 --> 00:27:46,910 response, since as we've seen in the last several view 425 00:27:46,910 --> 00:27:49,940 graphs, that very seriously affects 426 00:27:49,940 --> 00:27:53,810 the interpolation in between these frequency points. 427 00:27:53,810 --> 00:27:57,170 Furthermore, an additional point has been made, 428 00:27:57,170 --> 00:28:03,110 and that is that it is not-- even though we can implement 429 00:28:03,110 --> 00:28:06,080 a finite impulse response filter using 430 00:28:06,080 --> 00:28:11,440 the discrete Fourier transform, specifying the discrete Fourier 431 00:28:11,440 --> 00:28:14,000 transform to be unity in the pass band 432 00:28:14,000 --> 00:28:17,060 and zero in the stop band does not 433 00:28:17,060 --> 00:28:22,690 mean that we've implemented an ideal low pass filter. 434 00:28:22,690 --> 00:28:23,190 All right. 435 00:28:23,190 --> 00:28:26,720 Well, we had-- this, of course, is not a unit sample 436 00:28:26,720 --> 00:28:30,920 response that we would be particularly anxious to use. 437 00:28:30,920 --> 00:28:33,980 Clearly, the behavior in the pass band and stop band 438 00:28:33,980 --> 00:28:36,830 is very severe, and in fact, too severe. 439 00:28:36,830 --> 00:28:41,210 So we could at least consider as somewhat reasonable 440 00:28:41,210 --> 00:28:43,790 the unit sample response and frequency 441 00:28:43,790 --> 00:28:48,560 response that we had previously, in this case, 442 00:28:48,560 --> 00:28:49,910 relatively well-behaved. 443 00:28:49,910 --> 00:28:52,400 That is, it ripples somewhat in the pass band 444 00:28:52,400 --> 00:28:55,130 and in the stop band. 445 00:28:55,130 --> 00:28:58,340 But now we could inquire as to whether there 446 00:28:58,340 --> 00:29:04,700 is some way of improving the pass band ripple and the stop 447 00:29:04,700 --> 00:29:06,320 band ripple. 448 00:29:06,320 --> 00:29:09,500 We know, or we anticipate, that there's a price that we 449 00:29:09,500 --> 00:29:11,240 would have to pay for that. 450 00:29:11,240 --> 00:29:13,790 Probably the price, and indeed it 451 00:29:13,790 --> 00:29:16,400 turns out that the price we have to pay 452 00:29:16,400 --> 00:29:20,780 is the price of widening the transition band. 453 00:29:20,780 --> 00:29:23,480 Well, how could we widen the transition band? 454 00:29:23,480 --> 00:29:28,080 We could imagine beginning with this particular design, 455 00:29:28,080 --> 00:29:34,460 taking as one possibility one of the stop band constrained 456 00:29:34,460 --> 00:29:35,660 points. 457 00:29:35,660 --> 00:29:39,620 And allow it to move into the transition band 458 00:29:39,620 --> 00:29:42,980 so that, in fact, we have a wider frequency 459 00:29:42,980 --> 00:29:47,210 range that we're permitting the transition from pass band 460 00:29:47,210 --> 00:29:50,360 to stop band to go to. 461 00:29:50,360 --> 00:30:00,800 And let me illustrate that here, where now what we've done 462 00:30:00,800 --> 00:30:17,400 is to take one of the frequency samples in the stop band, 463 00:30:17,400 --> 00:30:20,790 move it up into the transition band. 464 00:30:20,790 --> 00:30:25,410 And as we do that, you can see for this particular example, 465 00:30:25,410 --> 00:30:29,550 the pass band ripple hasn't changed too much. 466 00:30:29,550 --> 00:30:31,590 But in fact, the stop band ripple 467 00:30:31,590 --> 00:30:34,380 has changed somewhat dramatically. 468 00:30:34,380 --> 00:30:38,490 As we allow this transition point to move farther up 469 00:30:38,490 --> 00:30:41,430 into the transition band, basically widening 470 00:30:41,430 --> 00:30:44,370 the general pass band region, what we'll see 471 00:30:44,370 --> 00:30:48,990 is that the pass band ripple become somewhat smaller. 472 00:30:48,990 --> 00:30:51,420 And there, in fact, is a trade off 473 00:30:51,420 --> 00:30:55,680 that is introduced between the pass band ripple and the stop 474 00:30:55,680 --> 00:30:56,760 band ripple. 475 00:30:56,760 --> 00:31:04,620 And let me illustrate that with this next example, where 476 00:31:04,620 --> 00:31:10,110 in this case, we have moved the transition point up somewhat 477 00:31:10,110 --> 00:31:10,610 further. 478 00:31:16,920 --> 00:31:19,590 So that whereas in the previous case 479 00:31:19,590 --> 00:31:22,470 the transition point was someplace down here, 480 00:31:22,470 --> 00:31:25,380 we've now moved it farther up into the transition region, 481 00:31:25,380 --> 00:31:27,870 closer to the pass band. 482 00:31:27,870 --> 00:31:32,820 And the pass band ripple has become correspondingly smaller. 483 00:31:32,820 --> 00:31:37,080 So this, in fact, is a design technique 484 00:31:37,080 --> 00:31:44,010 that can be used for the design of, generally, band frequency 485 00:31:44,010 --> 00:31:49,350 selective filters, where we first of all specify 486 00:31:49,350 --> 00:31:50,970 frequency samples. 487 00:31:50,970 --> 00:31:52,500 In fact, this can be used, really, 488 00:31:52,500 --> 00:31:56,190 for the design of any digital filter characteristic 489 00:31:56,190 --> 00:31:58,800 with transition regions. 490 00:31:58,800 --> 00:32:04,500 And then, perhaps, [AUDIO OUT] reducing the ripple in the pass 491 00:32:04,500 --> 00:32:07,650 band and/or the stop band, or in the various bands, 492 00:32:07,650 --> 00:32:12,540 by allowing an additional point to be used 493 00:32:12,540 --> 00:32:14,640 to widen the transition band. 494 00:32:14,640 --> 00:32:17,370 And as we move this point up or down, 495 00:32:17,370 --> 00:32:21,320 then, we will reduce the pass band and the stop band ripple. 496 00:32:21,320 --> 00:32:25,230 We can also, of course, consider moving not just one point 497 00:32:25,230 --> 00:32:28,170 into the transition region, but two points or three 498 00:32:28,170 --> 00:32:29,760 points or more points. 499 00:32:29,760 --> 00:32:34,260 Obviously, as we do that, we will widen the transition band 500 00:32:34,260 --> 00:32:35,250 further. 501 00:32:35,250 --> 00:32:38,670 And at the same time, that offers 502 00:32:38,670 --> 00:32:41,430 some additional flexibility, which 503 00:32:41,430 --> 00:32:45,300 permits us to reduce the pass band and stop band ripple. 504 00:32:45,300 --> 00:32:49,800 So this then is basically the strategy 505 00:32:49,800 --> 00:32:57,060 in the frequency sampling method of digital filter design. 506 00:32:57,060 --> 00:32:59,460 The final method that I would like 507 00:32:59,460 --> 00:33:04,510 to touch on just very briefly is a method 508 00:33:04,510 --> 00:33:08,970 which I referred to at the beginning of this lecture 509 00:33:08,970 --> 00:33:12,300 as an equiripple design method. 510 00:33:12,300 --> 00:33:15,690 And basically, the notion here is 511 00:33:15,690 --> 00:33:21,330 that we've seen that there is a trade off 512 00:33:21,330 --> 00:33:24,820 between the transition region on the one hand, 513 00:33:24,820 --> 00:33:28,350 and pass band and stop band ripple on the other hand. 514 00:33:28,350 --> 00:33:31,920 Well, one could inquire as to whether, in at least 515 00:33:31,920 --> 00:33:35,070 in some sense, we could design for ourselves 516 00:33:35,070 --> 00:33:37,890 an optimum class of filters. 517 00:33:37,890 --> 00:33:40,710 And indeed, you can. 518 00:33:40,710 --> 00:33:44,010 And in fact, it's possible to show 519 00:33:44,010 --> 00:33:49,980 that the optimum strategy in terms of minimizing pass band 520 00:33:49,980 --> 00:33:53,250 or stop band ripple, or in terms of minimizing a transition 521 00:33:53,250 --> 00:33:59,340 band, is to design the filter to have an equiripple frequency 522 00:33:59,340 --> 00:34:02,070 response characteristic. 523 00:34:02,070 --> 00:34:07,200 Well, one way of setting up the optimal condition-- and it 524 00:34:07,200 --> 00:34:09,659 turns out that there are a variety of ways, 525 00:34:09,659 --> 00:34:13,870 and these are discussed in quite a bit more detail in the text. 526 00:34:13,870 --> 00:34:18,360 But for example, if we pick a pass band cutoff 527 00:34:18,360 --> 00:34:22,469 frequency and a stop band cutoff frequency. 528 00:34:22,469 --> 00:34:25,860 Furthermore, if we, of course, are 529 00:34:25,860 --> 00:34:28,230 going to allow some deviation in the pass band 530 00:34:28,230 --> 00:34:32,260 and some deviation in the stop band, if in addition, 531 00:34:32,260 --> 00:34:39,030 we fix the ratio of stop band to pass band ripple, 532 00:34:39,030 --> 00:34:42,989 and if we then inquire as to what sort of characteristic 533 00:34:42,989 --> 00:34:47,159 should result if delta 1, the pass band 534 00:34:47,159 --> 00:34:49,949 ripple, or equivalently, delta 2, 535 00:34:49,949 --> 00:34:52,949 the stop band ripple is minimized-- 536 00:34:52,949 --> 00:34:55,889 this should be delta 2-- 537 00:34:55,889 --> 00:35:00,240 so if delta 1 and delta 2 are minimized, 538 00:35:00,240 --> 00:35:04,110 it turns out it can be shown that that implies 539 00:35:04,110 --> 00:35:08,190 an equiripple characteristic in the pass band, 540 00:35:08,190 --> 00:35:11,790 and an equiripple characteristic in the stop band. 541 00:35:11,790 --> 00:35:14,130 That is, the ripples in the pass band 542 00:35:14,130 --> 00:35:18,660 should alternate between the upper and lower limits. 543 00:35:18,660 --> 00:35:21,030 And the ripples in a stop band should 544 00:35:21,030 --> 00:35:24,890 alternate between the upper and lower limits. 545 00:35:24,890 --> 00:35:29,580 In the text, there are discussed several algorithmic procedures 546 00:35:29,580 --> 00:35:32,040 for actually designing equiripple filters. 547 00:35:32,040 --> 00:35:35,190 We won't go through those details in this lecture. 548 00:35:35,190 --> 00:35:37,230 Mainly, I want to emphasize the point 549 00:35:37,230 --> 00:35:41,700 that there are procedures that allow you to implement designs 550 00:35:41,700 --> 00:35:42,840 of this type. 551 00:35:42,840 --> 00:35:47,010 And that designs of this type are optimum. 552 00:35:47,010 --> 00:35:54,240 Let me just show you one example of an optimum equiripple finite 553 00:35:54,240 --> 00:35:59,400 impulse response filter that was designed when our computer 554 00:35:59,400 --> 00:36:02,880 budget was particularly big. 555 00:36:02,880 --> 00:36:08,640 This is an example of a length 251 point finite impulse 556 00:36:08,640 --> 00:36:12,450 response digital filter, designed 557 00:36:12,450 --> 00:36:16,950 using one of the algorithms that is discussed in the text. 558 00:36:16,950 --> 00:36:22,440 And you can see that, indeed, the equiripple characteristics 559 00:36:22,440 --> 00:36:26,130 are evident in the pass band. 560 00:36:26,130 --> 00:36:29,850 This curve corresponds to the frequency characteristics 561 00:36:29,850 --> 00:36:32,430 plotted in dB. 562 00:36:32,430 --> 00:36:35,160 So the pass band ripple doesn't show up. 563 00:36:35,160 --> 00:36:38,520 So with that scale expanded, we can see the ripple. 564 00:36:38,520 --> 00:36:43,260 And then we have the ripple in the stop band. 565 00:36:43,260 --> 00:36:47,230 There should be 125 ripples all together. 566 00:36:47,230 --> 00:36:49,609 And in fact, we could see if that was true. 567 00:36:49,609 --> 00:36:51,150 We could [AUDIO OUT] one, two, three. 568 00:36:51,150 --> 00:36:53,700 But let's not do that. 569 00:36:53,700 --> 00:36:57,030 Obviously, there's a very sharp transition band. 570 00:36:57,030 --> 00:37:02,130 And what turns out to be minus approximately 80-- 571 00:37:02,130 --> 00:37:07,230 approximately minus 85 dB attenuation in the stop band. 572 00:37:07,230 --> 00:37:13,230 So this is an example of a very high order equiripple optimum 573 00:37:13,230 --> 00:37:16,620 finite impulse response digital filter that 574 00:37:16,620 --> 00:37:19,675 was designed using one of the algorithms discussed 575 00:37:19,675 --> 00:37:20,175 in the text. 576 00:37:23,590 --> 00:37:24,090 OK. 577 00:37:24,090 --> 00:37:29,880 Well, this, as I indicated at the beginning of the lecture, 578 00:37:29,880 --> 00:37:36,390 was intended primarily as a very brief overview 579 00:37:36,390 --> 00:37:40,500 of finite impulse response digital filter design. 580 00:37:40,500 --> 00:37:44,850 Hopefully, it gives you some appreciation 581 00:37:44,850 --> 00:37:47,880 for the style of the design techniques. 582 00:37:47,880 --> 00:37:51,450 There are, of course, a large number of details, 583 00:37:51,450 --> 00:37:54,450 some of which are discussed in the text and some of which 584 00:37:54,450 --> 00:37:55,800 aren't. 585 00:37:55,800 --> 00:38:00,660 But in any case, this, then, concludes our discussion 586 00:38:00,660 --> 00:38:04,110 of the design of infinite impulse response 587 00:38:04,110 --> 00:38:07,500 and finite impulse response digital filters. 588 00:38:07,500 --> 00:38:12,300 In the next lecture, we will begin a new topic, specifically 589 00:38:12,300 --> 00:38:14,940 the topic involving the computation 590 00:38:14,940 --> 00:38:17,050 of the discrete Fourier transform. 591 00:38:17,050 --> 00:38:18,810 Thank you.