1 00:00:00 --> 00:00:01 2 00:00:01 --> 00:00:02 The following content is provided under a Creative 3 00:00:02 --> 00:00:03 Commons license. 4 00:00:03 --> 00:00:06 Your support will help MIT OpenCourseWare continue to 5 00:00:06 --> 00:00:09 offer high-quality educational resources for free. 6 00:00:09 --> 00:00:13 To make a donation, or to view additional materials from 7 00:00:13 --> 00:00:16 hundreds of MIT courses, visit MIT OpenCourseWare 8 00:00:16 --> 00:00:20 at ocw.mit.edu. 9 00:00:20 --> 00:00:25 PROFESSOR STRANG: OK, so I feel I should have brought pizza 10 00:00:25 --> 00:00:26 instead of math problems. 11 00:00:26 --> 00:00:34 I mean, but this is our final, final, hour. 12 00:00:34 --> 00:00:37 Open for any questions, any discussion about 13 00:00:37 --> 00:00:42 Fourier topics. 14 00:00:42 --> 00:00:45 Looking toward tomorrow evening's exam. 15 00:00:45 --> 00:00:51 And, well, hoping you'll find the whole course useful at 16 00:00:51 --> 00:00:54 many times in the future. 17 00:00:54 --> 00:01:00 So, somebody emailed me that the posted solution, so I just 18 00:01:00 --> 00:01:07 posted a quick solution to the un-collected homework ten, and 19 00:01:07 --> 00:01:10 so I haven't even looked to see what did it say 20 00:01:10 --> 00:01:11 for this question. 21 00:01:11 --> 00:01:12 But what is the answer? 22 00:01:12 --> 00:01:16 If I convolve the delta function with the delta 23 00:01:16 --> 00:01:18 function, what do I get? 24 00:01:18 --> 00:01:20 Delta, right. 25 00:01:20 --> 00:01:24 If I convolve anything with delta I get delta because I'm 26 00:01:24 --> 00:01:26 over in the other space. 27 00:01:26 --> 00:01:28 I'm multiplying by one. 28 00:01:28 --> 00:01:33 Yeah, so if I use the convolution rule to go over to 29 00:01:33 --> 00:01:37 the other space, where this becomes just one times one I'm 30 00:01:37 --> 00:01:46 getting one, and transform back and I've got delta. 31 00:01:46 --> 00:01:49 Well that question won't be on tomorrow's exam. 32 00:01:49 --> 00:01:51 But I thought we could deal with that fast. 33 00:01:51 --> 00:01:56 Now I'm open to questions about any topic. 34 00:01:56 --> 00:01:56 Homework or not. 35 00:01:56 --> 00:01:57 Yeah, thanks. 36 00:01:57 --> 00:02:01 AUDIENCE: [INAUDIBLE] 37 00:02:01 --> 00:02:02 PROFESSOR STRANG: I could try. 38 00:02:02 --> 00:02:04 Did I make up the exam? 39 00:02:04 --> 00:02:05 Probably. 40 00:02:05 --> 00:02:06 Yeah, I'll recognize it. 41 00:02:06 --> 00:02:14 OK, alright, I'll read it out. 42 00:02:14 --> 00:02:25 OK, so this an exam from 2005, not that far back. 43 00:02:25 --> 00:02:31 OK, so it has some questions. 44 00:02:31 --> 00:02:33 Shall I tell you what the whole question is but it looks 45 00:02:33 --> 00:02:36 like this Part D is sort of separate from the others. 46 00:02:36 --> 00:02:41 But why not, this gives us some questions to think about. 47 00:02:41 --> 00:02:44 So if I just read out the questions. 48 00:02:44 --> 00:02:50 One was, find the coefficients of, the function is e^(-x). 49 00:02:52 --> 00:02:55 That would naturally occur to me as a function that you 50 00:02:55 --> 00:02:58 could, so from zero to 2pi. 51 00:03:00 --> 00:03:02 And then periodic. 52 00:03:02 --> 00:03:04 Always good to graph it. 53 00:03:04 --> 00:03:06 So will e to the minus x look like? 54 00:03:06 --> 00:03:12 It'll come down to, much steeper than that, 55 00:03:12 --> 00:03:14 but that would do it. 56 00:03:14 --> 00:03:14 So that's e^(-x). 57 00:03:15 --> 00:03:18 Here it's one, and here it's e^(-2pi). 58 00:03:18 --> 00:03:20 59 00:03:20 --> 00:03:23 And then repeat, repeat, repeat. 60 00:03:23 --> 00:03:27 So find its Fourier coefficients. 61 00:03:27 --> 00:03:30 I'll just say what the question is and you can ask me. 62 00:03:30 --> 00:03:32 And then what's the decay rate. 63 00:03:32 --> 00:03:34 Oh, maybe you can tell me the decay rate. 64 00:03:34 --> 00:03:37 Even before you tell me the coefficient. 65 00:03:37 --> 00:03:42 What would the decay rate be if I expressed this as a 66 00:03:42 --> 00:03:49 combination of Fourier terms, harmonics. 67 00:03:49 --> 00:03:52 How quickly will those coefficients drop off? 68 00:03:52 --> 00:03:55 So 1/k, only. 69 00:03:55 --> 00:04:02 That's right, 1/k, because this function has a jump at zero. 70 00:04:02 --> 00:04:07 Well, at zero and again at 2pi at every one of those points. 71 00:04:07 --> 00:04:11 It has a jump, I guess, of about this distance, between 72 00:04:11 --> 00:04:12 one and that number. 73 00:04:12 --> 00:04:16 Which would be the jump that we would see of course every time. 74 00:04:16 --> 00:04:19 Yeah, so the coefficients would decay like 1/k. 75 00:04:20 --> 00:04:25 Then it asked to compute the sum of those 76 00:04:25 --> 00:04:27 coefficients squared. 77 00:04:27 --> 00:04:32 OK, so how can we find the sum of those coefficients squared 78 00:04:32 --> 00:04:37 even before I know what they are? 79 00:04:37 --> 00:04:40 Well, you know the answer, right? 80 00:04:40 --> 00:04:43 How am I going to do this problem? 81 00:04:43 --> 00:04:47 Yeah, so I'm looking for the sum of the squares 82 00:04:47 --> 00:04:51 of coefficients and then there's this magic trick. 83 00:04:51 --> 00:04:56 What is it that helps me to find that sum of squares? 84 00:04:56 --> 00:05:00 How do I go about that? 85 00:05:00 --> 00:05:05 I connected to the integral of the function square. 86 00:05:05 --> 00:05:08 It can go zero to 2pi of the function square. 87 00:05:08 --> 00:05:12 And you remember the reason, and I've left a little space 88 00:05:12 --> 00:05:13 for the fudge factor. 89 00:05:13 --> 00:05:17 2pi, or 1/2pi, we can figure out what it is by 90 00:05:17 --> 00:05:19 taking a simple case. 91 00:05:19 --> 00:05:22 Yeah, why don't we take a simple case to figure out 92 00:05:22 --> 00:05:25 should there be a 2pi or a 1/2pi or what? 93 00:05:25 --> 00:05:28 What's a simple case here? 94 00:05:28 --> 00:05:29 Suppose I take f(x)=1. 95 00:05:31 --> 00:05:36 Suppose I take the function f(x)=1, then the right side. 96 00:05:36 --> 00:05:42 So if f(x) is one, the right side would be 2pi, I guess. 97 00:05:42 --> 00:05:45 The integral of one, and the left side would be what? 98 00:05:45 --> 00:05:50 What would be the Fourier coefficients of the 99 00:05:50 --> 00:05:53 constant function? 100 00:05:53 --> 00:05:56 Well just c_0, the only one we would have would be 101 00:05:56 --> 00:06:00 c_0, the constant term. 102 00:06:00 --> 00:06:00 The c_0*e^0. 103 00:06:02 --> 00:06:03 And it would be one. 104 00:06:03 --> 00:06:08 So I'd have a one there, when k is zero, and otherwise 105 00:06:08 --> 00:06:09 I wouldn't have anything. 106 00:06:09 --> 00:06:12 So this would be one squared, this would be a one. 107 00:06:12 --> 00:06:14 So I do need to divide by 2pi. 108 00:06:16 --> 00:06:19 Do you agree with that? 109 00:06:19 --> 00:06:26 To make it correct I'd better have a 1/2pi to get that. 110 00:06:26 --> 00:06:27 To be right. 111 00:06:27 --> 00:06:33 OK, and then this gives me the answer then, I just do this 112 00:06:33 --> 00:06:37 integral, e^(-2x), then f is e^(-x). 113 00:06:38 --> 00:06:43 So f squared would be e^(-2x), which I can certainly integrate 114 00:06:43 --> 00:06:45 and get some expression. 115 00:06:45 --> 00:06:46 Yeah, OK. 116 00:06:46 --> 00:06:49 So that's a straightforward part. 117 00:06:49 --> 00:06:53 And now no to the Part D, which you asked about 118 00:06:53 --> 00:06:56 which, is oh wow. 119 00:06:56 --> 00:07:05 OK, so now I'm asking about an equation u, du/dx, plus u(x) 120 00:07:05 --> 00:07:08 equals a train of deltas. 121 00:07:08 --> 00:07:12 So delta(x) made periodic. 122 00:07:12 --> 00:07:14 Can I write it that way? 123 00:07:14 --> 00:07:16 So this is the periodic case. 124 00:07:16 --> 00:07:20 So I'm looking for a periodic answer. 125 00:07:20 --> 00:07:26 So I'm in the Fourier series world when I take, I'll be 126 00:07:26 --> 00:07:30 looking for coefficients, not the integral transform. 127 00:07:30 --> 00:07:33 OK. 128 00:07:33 --> 00:07:36 So alright, so I'm making it clear. 129 00:07:36 --> 00:07:40 So how do I proceed now, with such a problem? 130 00:07:40 --> 00:07:44 So I've mentioned this morning that a problem 131 00:07:44 --> 00:07:45 like this could appear. 132 00:07:45 --> 00:07:48 So I'm happy to discuss it. 133 00:07:48 --> 00:07:50 What do I do? 134 00:07:50 --> 00:07:51 Transform. 135 00:07:51 --> 00:07:52 Absolutely. 136 00:07:52 --> 00:07:54 Take Fourier transform. 137 00:07:54 --> 00:07:58 OK, so what happens when I take Fourier transforms of u? 138 00:07:58 --> 00:08:01 Well, I guess so we're doing periodic, so we're going to 139 00:08:01 --> 00:08:03 go to coefficients, right? 140 00:08:03 --> 00:08:10 I'll call them, shall I call, shall I think of u(x), shall I 141 00:08:10 --> 00:08:12 just call its coefficient c? 142 00:08:12 --> 00:08:14 It's the only function and I have around so I might 143 00:08:14 --> 00:08:17 as well use c for it. 144 00:08:17 --> 00:08:22 So here's the function, and Fourier takes this to the 145 00:08:22 --> 00:08:25 set of coefficients, OK. 146 00:08:25 --> 00:08:30 And the point is that we can separate - I mean this is 147 00:08:30 --> 00:08:33 the whole reason why Fourier's so great. 148 00:08:33 --> 00:08:37 That we can do each frequency separately. 149 00:08:37 --> 00:08:38 So let's see then. 150 00:08:38 --> 00:08:44 So this has coefficient c_k, and what are the coefficients, 151 00:08:44 --> 00:08:50 what's the k'th coefficient, so this is all the e^(ikx) term. 152 00:08:50 --> 00:09:00 What's the k'th coefficient of the derivative? ikc_k, right? 153 00:09:00 --> 00:09:04 And what's the k'th coefficient of the delta function? 154 00:09:04 --> 00:09:06 One, right? 155 00:09:06 --> 00:09:07 Is that right or is there 1/2pi? 156 00:09:07 --> 00:09:09 157 00:09:09 --> 00:09:10 Is it 1/2pi? 158 00:09:10 --> 00:09:11 OK, 1/2pi. 159 00:09:13 --> 00:09:14 Yeah, I guess that's right. 160 00:09:14 --> 00:09:21 Because the coefficient, I would have to take the integral 161 00:09:21 --> 00:09:24 times, yeah, the integral with the delta function will give me 162 00:09:24 --> 00:09:26 the one, and then there's a 1/2pi. 163 00:09:27 --> 00:09:28 Right, OK. 164 00:09:28 --> 00:09:33 So now I've got the answer. 165 00:09:33 --> 00:09:37 Now I have this equation at each k. 166 00:09:37 --> 00:09:40 Following each eigenvector, following each 167 00:09:40 --> 00:09:42 separate frequency. 168 00:09:42 --> 00:09:44 And it's easy to do. 169 00:09:44 --> 00:09:45 It's just 1/2pi. 170 00:09:46 --> 00:09:53 And there's one plus ik, I think. 171 00:09:53 --> 00:09:57 OK, and now those are the c_k's. 172 00:09:57 --> 00:10:00 So I think maybe this is it. 173 00:10:00 --> 00:10:05 Sum from minus infinity, I know what the c_k is. 174 00:10:05 --> 00:10:10 2pi(1+ik), times e^(ikx). 175 00:10:10 --> 00:10:15 176 00:10:15 --> 00:10:18 That's my u(x), I think. 177 00:10:18 --> 00:10:19 Had to be, right? 178 00:10:19 --> 00:10:26 So I just did the sort of automatic steps. 179 00:10:26 --> 00:10:28 When I made up this exam, I don't know whether I was 180 00:10:28 --> 00:10:44 thinking I could find out can I actually do that sum to produce 181 00:10:44 --> 00:10:47 some function u(x) that I am familiar with. 182 00:10:47 --> 00:10:52 I'm not sure. 183 00:10:52 --> 00:10:56 Anybody reading the question, so the question says solve 184 00:10:56 --> 00:10:58 this differential equation. 185 00:10:58 --> 00:11:02 Doesn't say what it means by solve. 186 00:11:02 --> 00:11:03 Anyway, I've done it. 187 00:11:03 --> 00:11:05 I guess. 188 00:11:05 --> 00:11:13 Shall we agree that I passed on this question? 189 00:11:13 --> 00:11:18 If we knew some tricky way to do this addition, but I 190 00:11:18 --> 00:11:19 don't immediately see it. 191 00:11:19 --> 00:11:21 So that's good. 192 00:11:21 --> 00:11:23 OK, you're welcome. 193 00:11:23 --> 00:11:24 Thanks. 194 00:11:24 --> 00:11:27 OK, so that was a very good question to ask. 195 00:11:27 --> 00:11:34 Has it got, it's got the method of how do you deal with 196 00:11:34 --> 00:11:37 a differential equation. 197 00:11:37 --> 00:11:41 By the way, if this could be a difference equation, too. 198 00:11:41 --> 00:11:42 That could be u(x+h)-u(x-h). 199 00:11:42 --> 00:11:46 200 00:11:46 --> 00:11:51 Shall I just emphasize that that wouldn't have been any 201 00:11:51 --> 00:11:59 harder? u(x+h)-u(x-h)/2h, let me take, for example, 202 00:11:59 --> 00:12:03 plus u equals delta. 203 00:12:03 --> 00:12:09 This is like I've taken a center difference there, 204 00:12:09 --> 00:12:10 instead of a derivative. 205 00:12:10 --> 00:12:12 How would you write a solution to that? 206 00:12:12 --> 00:12:18 Same method. 207 00:12:18 --> 00:12:19 So what would the method be? 208 00:12:19 --> 00:12:21 I'd take transforms. 209 00:12:21 --> 00:12:25 So here this would have coefficient c_k, this would 210 00:12:25 --> 00:12:26 have coefficients 1/2pi. 211 00:12:28 --> 00:12:30 And what would be the coefficients of these guys? 212 00:12:30 --> 00:12:31 Well, there's a 1/2h. 213 00:12:33 --> 00:12:35 So what are the Fourier coefficients of u(x+h)? 214 00:12:35 --> 00:12:38 215 00:12:38 --> 00:12:42 Just to see that all these things, I've got constant 216 00:12:42 --> 00:12:44 coefficients here. 217 00:12:44 --> 00:12:46 Why shouldn't Fourier work? 218 00:12:46 --> 00:12:53 So if it's u(x+h), so there's an e to the something 219 00:12:53 --> 00:12:54 factor, right? 220 00:12:54 --> 00:12:58 When I've shifted a function then in the transform I 221 00:12:58 --> 00:13:00 see an e to the i thing. 222 00:13:00 --> 00:13:01 So let's just see. 223 00:13:01 --> 00:13:14 If u(x), so if u(x) is the sum of c_k*e^(ikx), then u(x+h) 224 00:13:14 --> 00:13:21 would be the sum of c_k*e^ik(x+h), right? 225 00:13:21 --> 00:13:21 No. 226 00:13:21 --> 00:13:23 Nothing surprising there. 227 00:13:23 --> 00:13:25 So now I'm seeing the factor. 228 00:13:25 --> 00:13:29 There's c_k, it's multiplied by the factor e^(ikh). 229 00:13:29 --> 00:13:34 230 00:13:34 --> 00:13:43 So this is the same c_k, but an e^(ikh), from that one. 231 00:13:43 --> 00:13:46 And this would be the minus, and this would shift 232 00:13:46 --> 00:13:47 the other way. 233 00:13:47 --> 00:13:50 So it would be the same c_k, and probably e^(-ikh). 234 00:13:52 --> 00:13:54 If I did that right. 235 00:13:54 --> 00:13:55 Yeah. 236 00:13:55 --> 00:13:59 Anyway, once again we have still linear problem. 237 00:13:59 --> 00:14:04 This is c_k times something equals 1/2pi. 238 00:14:04 --> 00:14:08 239 00:14:08 --> 00:14:11 By the way, this thing is what I would call the 240 00:14:11 --> 00:14:12 transfer function. 241 00:14:12 --> 00:14:14 You see that language? 242 00:14:14 --> 00:14:16 We haven't used that language much. 243 00:14:16 --> 00:14:21 Or at all, it's sort of more of a systems theory, 244 00:14:21 --> 00:14:23 control theory word. 245 00:14:23 --> 00:14:29 But hey, we're always doing the same correct thing. 246 00:14:29 --> 00:14:35 That's the thing that transfers the input to the output. 247 00:14:35 --> 00:14:38 So maybe the transfer function is one over. 248 00:14:38 --> 00:14:41 Maybe I include the division in the transfer function. 249 00:14:41 --> 00:14:48 So finally I get c_k is this 1/2pi guy, divided by whatever 250 00:14:48 --> 00:14:53 that was, 1/2h times that number, minus 1/2h 251 00:14:53 --> 00:14:55 of that plus one. 252 00:14:55 --> 00:14:57 Whatever is multiplying c_k there. 253 00:14:57 --> 00:14:59 It's a transfer function. 254 00:14:59 --> 00:14:59 Yeah. 255 00:14:59 --> 00:15:04 So there's another word which we could have, and should 256 00:15:04 --> 00:15:06 have, used before today. 257 00:15:06 --> 00:15:09 But here it is. 258 00:15:09 --> 00:15:12 OK, so that's some thoughts about that question. 259 00:15:12 --> 00:15:14 Any other direction to go? 260 00:15:14 --> 00:15:15 AUDIENCE: [INAUDIBLE] 261 00:15:15 --> 00:15:17 PROFESSOR STRANG: Yes. go ahead. 262 00:15:17 --> 00:15:18 Yeah. 263 00:15:18 --> 00:15:22 AUDIENCE: [INAUDIBLE] 264 00:15:22 --> 00:15:24 PROFESSOR STRANG: Lab problem 12, OK. 265 00:15:24 --> 00:15:28 4.5, Fourier integrals, number 12. 266 00:15:28 --> 00:15:30 Oh yeah, maybe that was also. 267 00:15:30 --> 00:15:33 A question was raised about was it right on 268 00:15:33 --> 00:15:42 the posted solution. 269 00:15:42 --> 00:15:45 So maybe not. 270 00:15:45 --> 00:15:46 Let's see what we want to do. 271 00:15:46 --> 00:15:50 Yeah I just thought that was pretty - I could remember the 272 00:15:50 --> 00:15:51 day I thought of this equation. 273 00:15:51 --> 00:15:55 Integral of u minus derivative of u equal delta(x). 274 00:15:55 --> 00:16:02 I thought that's kind of a cool looking equation. 275 00:16:02 --> 00:16:05 But I don't know that I've ever solved it, before. 276 00:16:05 --> 00:16:15 So integral of u, actually yeah, I look cheerful, but so 277 00:16:15 --> 00:16:18 I'm finishing the fourth edition of the linear 278 00:16:18 --> 00:16:19 algebra book. 279 00:16:19 --> 00:16:21 Introduction to Linear Algebra. 280 00:16:21 --> 00:16:23 So I love writing. 281 00:16:23 --> 00:16:27 I've done all the writing and now comes the only horrible 282 00:16:27 --> 00:16:30 part, solving all those stupid exercises. 283 00:16:30 --> 00:16:36 So that, you may think of me in sunny Singapore or somewhere, 284 00:16:36 --> 00:16:40 but what I'll be doing is not on the beach. 285 00:16:40 --> 00:16:47 It's in an office somewhere, doing Problem 1.1.1, and 286 00:16:47 --> 00:16:50 two, three, it's not life. 287 00:16:50 --> 00:16:55 I mean, it's. 288 00:16:55 --> 00:16:57 But, OK. 289 00:16:57 --> 00:16:59 Anyway, that looks like a pretty good equation. 290 00:16:59 --> 00:17:03 And certainly I'm going to transform it. 291 00:17:03 --> 00:17:06 So I can see a little question coming up. 292 00:17:06 --> 00:17:09 So am in the Fourier integral world, yeah. 293 00:17:09 --> 00:17:15 So I should you use this notation. u hat of k, right? 294 00:17:15 --> 00:17:22 Over ik, is that the right thing when I've integrated? 295 00:17:22 --> 00:17:23 Yeah. 296 00:17:23 --> 00:17:24 That's made it smoother. 297 00:17:24 --> 00:17:28 That's dragging the function down. 298 00:17:28 --> 00:17:32 And I guess the one point that's a little tricky 299 00:17:32 --> 00:17:38 there is always, are you dividing by zero. 300 00:17:38 --> 00:17:41 Because zero is one of the frequencies. 301 00:17:41 --> 00:17:46 And so I sort of need u hat of zero to be zero for this. 302 00:17:46 --> 00:17:50 So it's looking slightly dangerous at k=0, but otherwise 303 00:17:50 --> 00:17:55 it's certainly improving things by speeding up the decay rate. 304 00:17:55 --> 00:17:59 And then this would be minus ik u hat of k, as we just said. 305 00:17:59 --> 00:18:04 And this would be the one, maybe it's a one in the Fourier 306 00:18:04 --> 00:18:07 integral world, where the 2pi went in a different point. 307 00:18:07 --> 00:18:11 So I think that's, yeah. 308 00:18:11 --> 00:18:14 So I would now just, the usual thing was the 309 00:18:14 --> 00:18:15 transfer function. 310 00:18:15 --> 00:18:21 1/(ik)-ik, so that's my transfer function. 311 00:18:21 --> 00:18:23 That's multiplying u hat. 312 00:18:23 --> 00:18:27 Maybe I can simplify that by putting it all over 313 00:18:27 --> 00:18:30 ik, one minus ik. 314 00:18:30 --> 00:18:35 Is it plus k? 315 00:18:35 --> 00:18:41 Does that look good, or does that look bad? k squared? 316 00:18:41 --> 00:18:44 Does that look better? 317 00:18:44 --> 00:18:53 So I'm wondering, so over here is k squared over ik, is that 318 00:18:53 --> 00:18:54 the same as the minus ik? 319 00:18:54 --> 00:18:57 320 00:18:57 --> 00:18:58 Yeah. 321 00:18:58 --> 00:19:01 Yeah, bring it up and the minus i squared is one 322 00:19:01 --> 00:19:02 and I've got k squared. 323 00:19:02 --> 00:19:04 Yeah, so this looks good. 324 00:19:04 --> 00:19:08 So that's then dividing by it gives me because, 325 00:19:08 --> 00:19:10 I'm dividing the one. 326 00:19:10 --> 00:19:14 So u hat k is one over this. 327 00:19:14 --> 00:19:17 So it's the transfer function. 328 00:19:17 --> 00:19:19 One plus k squared. 329 00:19:19 --> 00:19:25 OK. 330 00:19:25 --> 00:19:28 I don't know whether I had in mind to go any 331 00:19:28 --> 00:19:29 further than this. 332 00:19:29 --> 00:19:32 Did you get this far, or what? 333 00:19:32 --> 00:19:34 AUDIENCE: [INAUDIBLE] 334 00:19:34 --> 00:19:36 PROFESSOR STRANG: It said that? 335 00:19:36 --> 00:19:40 Oh jeez. 336 00:19:40 --> 00:19:43 I bitterly regret saying these things. 337 00:19:43 --> 00:19:45 Yeah, OK. 338 00:19:45 --> 00:19:46 I don't know u(x). 339 00:19:46 --> 00:19:48 Did anybody have any luck? 340 00:19:48 --> 00:19:50 You did. 341 00:19:50 --> 00:19:53 Did I say it with a derivative? 342 00:19:53 --> 00:19:56 Oh one, over one plus k squared, oh yeah. 343 00:19:56 --> 00:20:01 One over one plus k squared, we know what to do. 344 00:20:01 --> 00:20:06 And then i k will take its derivative. 345 00:20:06 --> 00:20:09 So what do we do if it's one over one plus k squared? 346 00:20:09 --> 00:20:12 That's the one where, that was the guy which was 347 00:20:12 --> 00:20:18 the e^(-x), right? 348 00:20:18 --> 00:20:20 Oh, it's just one side. 349 00:20:20 --> 00:20:22 No, two sides, isn't it? 350 00:20:22 --> 00:20:23 Two sides, yeah. 351 00:20:23 --> 00:20:25 Yeah, we just got a corner. 352 00:20:25 --> 00:20:26 Yeah, OK. 353 00:20:26 --> 00:20:28 That's looking good. 354 00:20:28 --> 00:20:35 And now maybe I need to divide by two. a is one here but 355 00:20:35 --> 00:20:37 there is a division by two. 356 00:20:37 --> 00:20:39 Yeah, that looks good. 357 00:20:39 --> 00:20:44 OK, and now I take - This function, its 358 00:20:44 --> 00:20:47 transform has that. 359 00:20:47 --> 00:20:49 And now if I want to multiply by ik I have to take 360 00:20:49 --> 00:20:50 the derivative. 361 00:20:50 --> 00:20:54 So I believe with your help here, that the derivative 362 00:20:54 --> 00:20:58 of that is the same. 363 00:20:58 --> 00:21:03 Can I just take the derivative while we're looking at it? 364 00:21:03 --> 00:21:08 I mean e^x is the most - you know, that was created to 365 00:21:08 --> 00:21:10 take its derivative, right? 366 00:21:10 --> 00:21:12 So its derivative is itself. 367 00:21:12 --> 00:21:15 The derivative of this is a minus. 368 00:21:15 --> 00:21:22 So I think that maybe that's the answer. e^x over two, 369 00:21:22 --> 00:21:25 coming up to 1/2, I guess. 370 00:21:25 --> 00:21:29 And then - oh, well the picture won't look, because 371 00:21:29 --> 00:21:30 of that minus sign. 372 00:21:30 --> 00:21:31 Yeah. 373 00:21:31 --> 00:21:31 Huh. 374 00:21:31 --> 00:21:39 Good, because that minus sign it's minus e^(-x) over two. 375 00:21:39 --> 00:21:41 So it starts at minus 1/2. 376 00:21:41 --> 00:21:43 So there's a jump of one. 377 00:21:43 --> 00:21:45 Or drop of one. 378 00:21:45 --> 00:21:51 Is that right, there should be a drop of one in u hat? 379 00:21:51 --> 00:21:53 Probably. 380 00:21:53 --> 00:21:54 Yeah, yeah. 381 00:21:54 --> 00:21:58 The integral will be smooth, at zero where the delta 382 00:21:58 --> 00:22:00 function's hitting us. 383 00:22:00 --> 00:22:02 I have a minus sign and a derivative. 384 00:22:02 --> 00:22:06 But yeah, doesn't that look good? 385 00:22:06 --> 00:22:11 If the derivative has a delta, the function has a jump. 386 00:22:11 --> 00:22:15 And with that minus sign, and with one delta the jump 387 00:22:15 --> 00:22:17 should be a drop of one. 388 00:22:17 --> 00:22:17 Which is that. 389 00:22:17 --> 00:22:18 Yeah. 390 00:22:18 --> 00:22:21 And the integral minus the derivative should 391 00:22:21 --> 00:22:23 be otherwise zero. 392 00:22:23 --> 00:22:29 So again this is not unrelated to exam questions. 393 00:22:29 --> 00:22:33 I did all this stuff, and I really should check, 394 00:22:33 --> 00:22:34 did I get it right? 395 00:22:34 --> 00:22:37 Does that solve the differential equation? 396 00:22:37 --> 00:22:39 I mean, you could say of course it does if you took 397 00:22:39 --> 00:22:40 all these steps right. 398 00:22:40 --> 00:22:42 But it's wise to check. 399 00:22:42 --> 00:22:44 Plug it in. 400 00:22:44 --> 00:22:46 OK, it has the drop of one. 401 00:22:46 --> 00:22:49 So it deals correctly with the delta because, the 402 00:22:49 --> 00:22:52 derivative has a delta. 403 00:22:52 --> 00:22:55 And things match. 404 00:22:55 --> 00:22:58 And now what else do I have to check? 405 00:22:58 --> 00:23:00 I have to check out all the other points, right? 406 00:23:00 --> 00:23:03 I've just checked that yep, this answer is 407 00:23:03 --> 00:23:05 good at the jump. 408 00:23:05 --> 00:23:06 The tricky point. 409 00:23:06 --> 00:23:09 But now what about all the other points? 410 00:23:09 --> 00:23:14 So where the delta is zero, I should just check that the 411 00:23:14 --> 00:23:18 integral of u, so what is the integral of u? 412 00:23:18 --> 00:23:24 Let me just check for the points left of the origin. 413 00:23:24 --> 00:23:29 So I'm just going to look at this part. 414 00:23:29 --> 00:23:31 Yeah, x negative. 415 00:23:31 --> 00:23:36 So its integral, what's the integral of e^x over two? 416 00:23:36 --> 00:23:42 I guess e^x over two, right? 417 00:23:42 --> 00:23:43 Minus? 418 00:23:43 --> 00:23:45 And what's the derivative of e^x? 419 00:23:45 --> 00:23:49 Oh, the derivative is also e^x over two and I get 420 00:23:49 --> 00:23:51 zero, or x negative. 421 00:23:51 --> 00:23:53 As I want to. 422 00:23:53 --> 00:23:57 Because the delta is zero in that left side. 423 00:23:57 --> 00:23:58 And similarly on the right side. 424 00:23:58 --> 00:23:59 It should be OK. 425 00:23:59 --> 00:24:03 And then the key point was that the jump was right. 426 00:24:03 --> 00:24:04 Yeah, so thank you. 427 00:24:04 --> 00:24:06 That's good. 428 00:24:06 --> 00:24:06 Yes please. 429 00:24:06 --> 00:24:09 AUDIENCE: [INAUDIBLE] 430 00:24:09 --> 00:24:11 PROFESSOR STRANG: What, sorry? 431 00:24:11 --> 00:24:14 Why did I take the derivative? 432 00:24:14 --> 00:24:18 What will be the derivative of this? 433 00:24:18 --> 00:24:19 Yes, there is. 434 00:24:19 --> 00:24:20 That's right. 435 00:24:20 --> 00:24:22 What would be the derivative of this? 436 00:24:22 --> 00:24:23 Huh, yeah, good question. 437 00:24:23 --> 00:24:27 What's the derivative of this answer. 438 00:24:27 --> 00:24:33 So it's that, on the left. and then there's a delta function, 439 00:24:33 --> 00:24:36 a minus delta, right, because it's dropped down. 440 00:24:36 --> 00:24:39 And then the derivative of that is e^(-x). 441 00:24:40 --> 00:24:43 Yeah, so if I graph the derivative, cool. 442 00:24:43 --> 00:24:50 I graph the derivative, it looks like the function. 443 00:24:50 --> 00:24:56 It has a spike going down there. 444 00:24:56 --> 00:25:00 At the origin, and then the derivative here is positive. 445 00:25:00 --> 00:25:01 Right? 446 00:25:01 --> 00:25:03 This function's coming up. 447 00:25:03 --> 00:25:05 It's e(-x) over two. 448 00:25:05 --> 00:25:06 It's positive, it's there. 449 00:25:06 --> 00:25:08 Yeah. 450 00:25:08 --> 00:25:11 Oh wow, that's a nice graph. 451 00:25:11 --> 00:25:17 So that's the derivative, this is the graph of du/dx. 452 00:25:17 --> 00:25:19 Yeah, thanks. 453 00:25:19 --> 00:25:23 And similarly I could graph the integral, the integral - well, 454 00:25:23 --> 00:25:26 would you want to see the integral? 455 00:25:26 --> 00:25:28 Probably not. 456 00:25:28 --> 00:25:30 Maybe, yeah. 457 00:25:30 --> 00:25:34 The integral would probably look pretty much just like that 458 00:25:34 --> 00:25:37 but without the delta, yeah. 459 00:25:37 --> 00:25:40 Cool, isn't that nice? 460 00:25:40 --> 00:25:44 It's artistic. 461 00:25:44 --> 00:25:46 That's the derivative, and the integral is the same 462 00:25:46 --> 00:25:48 thing without the delta. 463 00:25:48 --> 00:25:52 And when you subtract, you get the delta. 464 00:25:52 --> 00:25:57 OK, you sure you guys don't want to come to Singapore 465 00:25:57 --> 00:26:01 and help with these problem solutions? 466 00:26:01 --> 00:26:03 There's plenty for everybody. 467 00:26:03 --> 00:26:04 OK. 468 00:26:04 --> 00:26:07 Alright, so another question. 469 00:26:07 --> 00:26:10 Yeah. 470 00:26:10 --> 00:26:10 2005. 471 00:26:10 --> 00:26:11 AUDIENCE: [INAUDIBLE] 472 00:26:11 --> 00:26:14 PROFESSOR STRANG: OK, alright. 473 00:26:14 --> 00:26:14 Yeah, which one? 474 00:26:14 --> 00:26:19 AUDIENCE: [INAUDIBLE] 475 00:26:19 --> 00:26:22 PROFESSOR STRANG: OK, I'll just read out the whole question. 476 00:26:22 --> 00:26:26 Yeah. 477 00:26:26 --> 00:26:32 So this is 2005, Problem 3. 478 00:26:32 --> 00:26:37 OK it's a half hat function. 479 00:26:37 --> 00:26:41 This is Fourier integral, because it's on the whole line. 480 00:26:41 --> 00:26:46 And it's half a hat for between zero and one. 481 00:26:46 --> 00:26:47 The function is coming down. 482 00:26:47 --> 00:26:53 And then otherwise it's zero. 483 00:26:53 --> 00:26:55 So its graph, done. 484 00:26:55 --> 00:26:56 Graph its derivative. 485 00:26:56 --> 00:27:01 OK, let's graph the derivative of that function. 486 00:27:01 --> 00:27:04 Derivative is certainly zero along there. 487 00:27:04 --> 00:27:07 Then the derivative is minus one here. 488 00:27:07 --> 00:27:09 And then the derivative is zero. 489 00:27:09 --> 00:27:16 So the derivative looks to me like it's that, OK. 490 00:27:16 --> 00:27:21 So yeah, that function - is that right? 491 00:27:21 --> 00:27:21 No. 492 00:27:21 --> 00:27:24 It's not right. 493 00:27:24 --> 00:27:25 There's a delta. 494 00:27:25 --> 00:27:28 That was the tricky part of the problem. 495 00:27:28 --> 00:27:33 Right, there's a delta in the derivative, right here 496 00:27:33 --> 00:27:35 because of that jump. 497 00:27:35 --> 00:27:37 There's a delta. 498 00:27:37 --> 00:27:37 OK. 499 00:27:37 --> 00:27:39 Good. 500 00:27:39 --> 00:27:42 Now, is that better? 501 00:27:42 --> 00:27:43 That's good. 502 00:27:43 --> 00:27:44 OK. 503 00:27:44 --> 00:27:47 What's its transform? 504 00:27:47 --> 00:27:52 The transform of this derivative. 505 00:27:52 --> 00:27:54 What's the Fourier transform? 506 00:27:54 --> 00:27:56 Well, I guess we got two parts. 507 00:27:56 --> 00:27:58 So this is my function u'(x). 508 00:27:59 --> 00:28:00 The derivative. 509 00:28:00 --> 00:28:04 Now, what's u' transform? 510 00:28:04 --> 00:28:07 So it should be a function of k, now. 511 00:28:07 --> 00:28:09 Alright, what do you think? 512 00:28:09 --> 00:28:12 So it's the sum of two things. 513 00:28:12 --> 00:28:17 That delta, which Fourier transforms to one. 514 00:28:17 --> 00:28:25 And this guy down below between zero and one, which was like 515 00:28:25 --> 00:28:30 the one we did today, we've got the integral from zero to one. 516 00:28:30 --> 00:28:33 Now it happens to be a minus one e^(-ikx)dx. 517 00:28:33 --> 00:28:37 518 00:28:37 --> 00:28:44 And that, of course, we can do. 519 00:28:44 --> 00:28:47 OK Oh, there's solutions here. 520 00:28:47 --> 00:28:49 AUDIENCE: [INAUDIBLE] 521 00:28:49 --> 00:28:55 PROFESSOR STRANG: Yeah. 522 00:28:55 --> 00:28:58 So OK, so I think we're doing alright. 523 00:28:58 --> 00:29:03 This expression is familiar. 524 00:29:03 --> 00:29:04 It's (1-e^(-ik))/ik. 525 00:29:04 --> 00:29:09 526 00:29:09 --> 00:29:12 Is that right? 527 00:29:12 --> 00:29:13 We have a minus. 528 00:29:13 --> 00:29:18 Is the minus, yeah, I think the minus looks good. 529 00:29:18 --> 00:29:23 And yeah, at least, that's here and it's probably got 530 00:29:23 --> 00:29:26 the minus signs correct. 531 00:29:26 --> 00:29:27 So why was it? 532 00:29:27 --> 00:29:30 I plugged in x=1, and that's why I got an e^(-ik). 533 00:29:32 --> 00:29:34 And then I plugged in x=0 and that's where 534 00:29:34 --> 00:29:36 that one came from. 535 00:29:36 --> 00:29:42 OK, so but your question was about - that was it? 536 00:29:42 --> 00:29:43 Oh, I see. 537 00:29:43 --> 00:29:45 OK. 538 00:29:45 --> 00:29:49 Right, and then the next part asked about the transform 539 00:29:49 --> 00:29:51 of the original guy. 540 00:29:51 --> 00:29:53 The transform of the original guy. 541 00:29:53 --> 00:30:00 Now, what would be the transform of the original guy? 542 00:30:00 --> 00:30:11 The original u. 543 00:30:11 --> 00:30:17 Right, OK, yeah. 544 00:30:17 --> 00:30:21 The reason I'm sort of stuttering is that if I'm going 545 00:30:21 --> 00:30:25 to integrate - I mean, what are you going to tell me? 546 00:30:25 --> 00:30:27 What's a quick way to find the transform of 547 00:30:27 --> 00:30:30 the original u now? 548 00:30:30 --> 00:30:31 Divide by ik. 549 00:30:31 --> 00:30:32 Divide by ik. 550 00:30:33 --> 00:30:36 And then I'm all ready to do that except I'm 551 00:30:36 --> 00:30:37 worried that k=0. 552 00:30:39 --> 00:30:41 But it should come out alright, now. 553 00:30:41 --> 00:30:44 So I have to hope that this thing comes out 554 00:30:44 --> 00:30:48 to be zero at k=0. 555 00:30:48 --> 00:30:51 556 00:30:51 --> 00:30:54 Can you see that it does? 557 00:30:54 --> 00:30:57 What is, yeah? 558 00:30:57 --> 00:30:59 This is a good point. 559 00:30:59 --> 00:31:03 What does u hat at zero represent? 560 00:31:03 --> 00:31:05 Have you thought about that? 561 00:31:05 --> 00:31:08 If I look at the Fourier transform, which I've got here. 562 00:31:08 --> 00:31:10 So let me write what I've got here. 563 00:31:10 --> 00:31:17 I've got here that Fourier transform of this function. 564 00:31:17 --> 00:31:20 And if I take the Fourier transform of any function, and 565 00:31:20 --> 00:31:22 I look at zero frequency. 566 00:31:22 --> 00:31:24 What am I seeing? 567 00:31:24 --> 00:31:25 The average, right. 568 00:31:25 --> 00:31:27 I'm saying the average. 569 00:31:27 --> 00:31:32 What's the average value of that function? 570 00:31:32 --> 00:31:33 We never thought about that. 571 00:31:33 --> 00:31:35 But that's fun. 572 00:31:35 --> 00:31:37 What's the average of this guy? 573 00:31:37 --> 00:31:40 If I integrate over the whole line. 574 00:31:40 --> 00:31:42 Do I get zero? 575 00:31:42 --> 00:31:43 Yes. 576 00:31:43 --> 00:31:48 Because I get the integral of the delta part gives me a one. 577 00:31:48 --> 00:31:51 But the integral of this box part gives me a minus one. 578 00:31:51 --> 00:31:56 So the integral, so this does equal zero at k=0, and I can 579 00:31:56 --> 00:32:05 safely do that division and get, so now u hat of k is this 580 00:32:05 --> 00:32:07 expression divided by ik. 581 00:32:07 --> 00:32:11 So it's one minus this thing. 582 00:32:11 --> 00:32:12 I'm just copying. 583 00:32:12 --> 00:32:16 Minus ik over ik, all that divided by ik. 584 00:32:16 --> 00:32:20 And I can, of course, maneuver that some more 585 00:32:20 --> 00:32:23 to make it look better. 586 00:32:23 --> 00:32:27 OK, these are good questions, because it's giving me a few 587 00:32:27 --> 00:32:31 functions to do the standard rules. 588 00:32:31 --> 00:32:39 Shifting, integrating, differentiating. 589 00:32:39 --> 00:32:40 Oh, here's a Christmas present. 590 00:32:40 --> 00:32:42 What does that mean? 591 00:32:42 --> 00:32:47 Is the convolution of this function with itself 592 00:32:47 --> 00:32:52 the whole hat? 593 00:32:52 --> 00:32:53 What is that, a Christmas present? 594 00:32:53 --> 00:32:57 Is the convolution, the convolution of 595 00:32:57 --> 00:33:00 that with itself. 596 00:33:00 --> 00:33:05 So first point, we don't want to compute a convolution. 597 00:33:05 --> 00:33:07 You may have noticed, you have not computed convolutions 598 00:33:07 --> 00:33:09 or some things like this. 599 00:33:09 --> 00:33:13 It's not a lot of fun. 600 00:33:13 --> 00:33:15 Because you have that integral to do. 601 00:33:15 --> 00:33:19 And you'd have to separate out the parts where it's this and 602 00:33:19 --> 00:33:22 the part where it's this and the parts where it's that. 603 00:33:22 --> 00:33:23 It's not nice. 604 00:33:23 --> 00:33:27 Much better to multiply in the transform domain. 605 00:33:27 --> 00:33:28 Much better. 606 00:33:28 --> 00:33:34 OK, so if I multiply in the transform domain, I don't 607 00:33:34 --> 00:33:40 think I get the Fourier transform of the hat. 608 00:33:40 --> 00:33:43 Of the whole hat. 609 00:33:43 --> 00:33:47 But can you give me a convincing argument for why - 610 00:33:47 --> 00:33:51 this is here I'm in the transform domain and I'm going 611 00:33:51 --> 00:33:59 to convolve with itself, so I'm just going to square it. 612 00:33:59 --> 00:34:05 With the decay rate, what's the decay rate as it is now? 613 00:34:05 --> 00:34:09 The decay rate is, yes. 614 00:34:09 --> 00:34:11 The decay rate is a good key. 615 00:34:11 --> 00:34:14 So what is the decay rate right now? 616 00:34:14 --> 00:34:15 Here's a function with a jump. 617 00:34:15 --> 00:34:18 So I know that even though has a slightly messy looking 618 00:34:18 --> 00:34:21 transform, I know the decay rate. 619 00:34:21 --> 00:34:24 It's what, with a jump is 1/k. 620 00:34:25 --> 00:34:28 OK, so this must be, and I sort of do see a k down 621 00:34:28 --> 00:34:30 here, so that's sensible. 622 00:34:30 --> 00:34:35 OK, now, if I convolve, then I multiply in this domain. 623 00:34:35 --> 00:34:38 So that would give me the k squared, as you said. 624 00:34:38 --> 00:34:43 So could that be the transform of the whole hat? 625 00:34:43 --> 00:34:50 Does the whole hat have a - ooh, could be. 626 00:34:50 --> 00:34:53 I'm pretty sure the answer's no. 627 00:34:53 --> 00:34:54 Yeah, no way. 628 00:34:54 --> 00:34:59 Yeah, I think that that's the best answer. 629 00:34:59 --> 00:35:03 I've lost the reason. 630 00:35:03 --> 00:35:06 Because I'm getting the k squared to k rate. 631 00:35:06 --> 00:35:10 If I convolve something with itself, that would give me a k 632 00:35:10 --> 00:35:14 squared, and the decay rate for the transform of the 633 00:35:14 --> 00:35:16 hat is a k squared. 634 00:35:16 --> 00:35:18 But they wouldn't be the same, yeah. 635 00:35:18 --> 00:35:21 OK, yeah, I won't. 636 00:35:21 --> 00:35:22 Yes, please. 637 00:35:22 --> 00:35:30 AUDIENCE: [INAUDIBLE] 638 00:35:30 --> 00:35:30 PROFESSOR STRANG: Right. 639 00:35:30 --> 00:35:32 It's just convention. 640 00:35:32 --> 00:35:34 The difference was that that was the Fourier 641 00:35:34 --> 00:35:36 integral problem. 642 00:35:36 --> 00:35:39 Where we put, you might have noticed that the 2pi goes on 643 00:35:39 --> 00:35:46 the other - and this was the Fourier series problem. 644 00:35:46 --> 00:35:47 That was its only difference. 645 00:35:47 --> 00:35:50 So it's just a convention. 646 00:35:50 --> 00:35:54 And maybe other people would choose a different convention. 647 00:35:54 --> 00:36:01 So that is a slight wiggle in the presentation that the 2pi 648 00:36:01 --> 00:36:08 in the Fourier integral section got moved to the transform 649 00:36:08 --> 00:36:09 in the other direction. 650 00:36:09 --> 00:36:12 Yeah, good. 651 00:36:12 --> 00:36:13 Yes, thanks. 652 00:36:13 --> 00:36:16 AUDIENCE: [INAUDIBLE] 653 00:36:16 --> 00:36:19 PROFESSOR STRANG: Oh, OK. 654 00:36:19 --> 00:36:24 I'm really just going by this rule that if there is a jump 655 00:36:24 --> 00:36:29 in the function, and nothing worse, then the decay rate is 656 00:36:29 --> 00:36:35 1/k, and if there's a jump in the derivative, as 657 00:36:35 --> 00:36:37 there would be here. 658 00:36:37 --> 00:36:40 So that has a jump in slope, then the decay rate is one 659 00:36:40 --> 00:36:42 over k squared and so on. 660 00:36:42 --> 00:36:44 But those are the main cases. 661 00:36:44 --> 00:36:50 Yeah, so I'm not using anything deep there. 662 00:36:50 --> 00:36:54 And by the way, I realize here that one MATLAB problem that I 663 00:36:54 --> 00:36:59 didn't assign this semester but it's quite fun, is to actually 664 00:36:59 --> 00:37:03 see the Gibbs phenomenon. 665 00:37:03 --> 00:37:09 We know the terms in the Fourier series, with a jump. 666 00:37:09 --> 00:37:16 And if you computed - make it the periodic case, if you've 667 00:37:16 --> 00:37:20 computed the terms in the Fourier series they would stay 668 00:37:20 --> 00:37:27 near here and then I'd see the famous Gibbs stuff here. 669 00:37:27 --> 00:37:32 So that - you know, it's quite pleasant to see the printout 670 00:37:32 --> 00:37:37 of the Gibbs phenomenon. 671 00:37:37 --> 00:37:42 Not getting any shorter, just moving closer to the jump as 672 00:37:42 --> 00:37:44 you increase the number of terms. 673 00:37:44 --> 00:37:48 Yeah, it's quite interesting. 674 00:37:48 --> 00:37:50 But couldn't do everything. 675 00:37:50 --> 00:37:51 Yeah, ready for more. 676 00:37:51 --> 00:37:54 Any questions? 677 00:37:54 --> 00:37:55 Well, these are good. 678 00:37:55 --> 00:37:57 Yes, thanks. 679 00:37:57 --> 00:37:58 AUDIENCE: [INAUDIBLE] 680 00:37:58 --> 00:37:59 PROFESSOR STRANG: With cyclic convolutions. 681 00:37:59 --> 00:38:13 OK, what could I do with cyclic convolutions? 682 00:38:13 --> 00:38:17 Let's see. 683 00:38:17 --> 00:38:20 What should I do with cyclic convolutions? 684 00:38:20 --> 00:38:30 I better make a little space. 685 00:38:30 --> 00:38:31 Let me make a little space and think. 686 00:38:31 --> 00:38:43 Anybody got a suggested problem for a cyclic convolution? 687 00:38:43 --> 00:38:50 I mean, one type of problem would certainly be if I gave 688 00:38:50 --> 00:39:02 you a cyclic convolution and asked for - I mean, the direct 689 00:39:02 --> 00:39:06 way would be to say OK, what's the cyclic convolution of 690 00:39:06 --> 00:39:14 (1, 4, 2), cyclically with (2, 1, 3) or something. 691 00:39:14 --> 00:39:17 But you could - that's just a calculation, so 692 00:39:17 --> 00:39:19 you would get that. 693 00:39:19 --> 00:39:22 I won't discuss that further. 694 00:39:22 --> 00:39:28 Now suppose I ask it with the unknown, the thing 695 00:39:28 --> 00:39:30 that I don't know here? 696 00:39:30 --> 00:39:31 So let me do that. 697 00:39:31 --> 00:39:40 Shall I say three, yeah. (3, 1, 1). 698 00:39:40 --> 00:39:49 Cyclically convolved with some unknown (x_0, x_1, x_2) equals 699 00:39:49 --> 00:39:51 some right-hand side, what should we take for the 700 00:39:51 --> 00:39:57 right-hand side? (1, 1, 1), for example? 701 00:39:57 --> 00:40:05 OK, I'll erase the top one. 702 00:40:05 --> 00:40:14 OK, how would you go with that? 703 00:40:14 --> 00:40:17 So that's three by three, three unknowns. 704 00:40:17 --> 00:40:20 It's linear, so somehow I could write that as three 705 00:40:20 --> 00:40:23 equations in three unknowns. 706 00:40:23 --> 00:40:28 The question is how do I get some insight into this. 707 00:40:28 --> 00:40:31 And when I'm seeing a convolution, what's my 708 00:40:31 --> 00:40:34 immediate thought? 709 00:40:34 --> 00:40:36 Transform. 710 00:40:36 --> 00:40:39 That's what this month is all about. 711 00:40:39 --> 00:40:42 November. 712 00:40:42 --> 00:40:46 So I'll take the DFT of everything. 713 00:40:46 --> 00:40:51 So what's the discrete Fourier transform of (1, 1, 1)? 714 00:40:51 --> 00:40:54 Up there is going to be the transform. 715 00:40:54 --> 00:40:59 So let me call it x_0 hat, x_1 hat, x_2 hat. 716 00:40:59 --> 00:41:01 Just to emphasize. 717 00:41:01 --> 00:41:07 So n is three here, in this problem. 718 00:41:07 --> 00:41:12 Or maybe c_0, c_1, c_2 you might prefer. 719 00:41:12 --> 00:41:13 Whatever. 720 00:41:13 --> 00:41:17 And now this is going to be a multiplication, so it's 721 00:41:17 --> 00:41:20 in MATLAB notation will be a dot star, right? 722 00:41:20 --> 00:41:25 And so now I have to - these are the known ones, so what's 723 00:41:25 --> 00:41:28 the discrete Fourier transform of (1, 1, 1), the cyclic 724 00:41:28 --> 00:41:31 convolution of that? 725 00:41:31 --> 00:41:33 It's (1, 0, 0) is it? 726 00:41:33 --> 00:41:35 Or it's (3, 0, 0)? 727 00:41:35 --> 00:41:43 Is it three - I guess when I asked that question, you're 728 00:41:43 --> 00:41:44 open to two answers. 729 00:41:44 --> 00:41:51 You could either be going from x space to frequency space 730 00:41:51 --> 00:41:52 or frequency to the other. 731 00:41:52 --> 00:41:57 Because I haven't told you what space we're in. 732 00:41:57 --> 00:41:59 OK, let's go with that one. 733 00:41:59 --> 00:42:07 How did you get to there? 734 00:42:07 --> 00:42:10 You multiplied by the three by three Fourier matrix, 735 00:42:10 --> 00:42:12 is that what you did? 736 00:42:12 --> 00:42:15 Aright, so now can you do that here? 737 00:42:15 --> 00:42:18 So separately I have to do a little multiplication of the 738 00:42:18 --> 00:42:20 three by three Fourier matrix. 739 00:42:20 --> 00:42:22 One, one - sorry, let me get it right. 740 00:42:22 --> 00:42:27 So I'm going to transform this. 741 00:42:27 --> 00:42:28 Make a little space here. 742 00:42:28 --> 00:42:29 Transform that. 743 00:42:29 --> 00:42:37 So it's 1, 1, 1, 1, 1 and this is w, w squared, w squared and 744 00:42:37 --> 00:42:40 w to the fourth, which is the same as w. 745 00:42:40 --> 00:42:44 OK, and now I want to multiply that by (3, 1, 1). 746 00:42:44 --> 00:42:47 And get the answer there. 747 00:42:47 --> 00:42:50 OK, so what do I have? 748 00:42:50 --> 00:42:52 Five. 749 00:42:52 --> 00:42:53 And what is that other one? 750 00:42:53 --> 00:42:54 Oh, boy. 751 00:42:54 --> 00:42:59 What is three plus w plus w squared? 752 00:42:59 --> 00:43:02 Anybody know that one? 753 00:43:02 --> 00:43:07 Well, I do know what one plus w plus w squared is. 754 00:43:07 --> 00:43:09 It is zero. 755 00:43:09 --> 00:43:12 Why is one plus w plus w squared zero? 756 00:43:12 --> 00:43:15 So I claim that if I have three plus, I think the 757 00:43:15 --> 00:43:17 answer there is two. 758 00:43:17 --> 00:43:20 And here I think I've also got three, w, w squared, I think if 759 00:43:20 --> 00:43:24 it was a one the answer would be a zero. 760 00:43:24 --> 00:43:26 But it's a three, so I still have two there. 761 00:43:26 --> 00:43:29 I think that's probably right. 762 00:43:29 --> 00:43:32 Can I just remind you why? 763 00:43:32 --> 00:43:38 There's one, there's w, and there's w squared 764 00:43:38 --> 00:43:40 and they add to zero. 765 00:43:40 --> 00:43:42 Yeah, yeah. 766 00:43:42 --> 00:43:47 For many reasons. 767 00:43:47 --> 00:43:51 That's a very, very, handy property. 768 00:43:51 --> 00:43:57 That the sum of the nth roots of one add to zero, OK, 769 00:43:57 --> 00:43:58 so now where I am I? 770 00:43:58 --> 00:44:04 So I got (5, 2, 2). 771 00:44:04 --> 00:44:05 So I've transformed. 772 00:44:05 --> 00:44:09 Now what do I do next? 773 00:44:09 --> 00:44:11 I divide. 774 00:44:11 --> 00:44:18 So this is telling me that this is three separate equations. 775 00:44:18 --> 00:44:23 There's three, two, x_1 hat is zero. 776 00:44:23 --> 00:44:24 Right? 777 00:44:24 --> 00:44:26 No, x_2 hat. 778 00:44:26 --> 00:44:28 And 2x_2 hat. 779 00:44:28 --> 00:44:31 Sorry, one was right. 780 00:44:31 --> 00:44:35 And 2x_2 hat is zero. 781 00:44:35 --> 00:44:40 I'm just doing the multiplication as 782 00:44:40 --> 00:44:41 I'm supposed to. 783 00:44:41 --> 00:44:45 In the other space it's at each, for each 784 00:44:45 --> 00:44:46 component separately. 785 00:44:46 --> 00:44:52 So now I know x_0 hat is 3/5, and the others are zero. 786 00:44:52 --> 00:44:55 So now I know what these numbers are. 787 00:44:55 --> 00:45:01 Oh, yeah. (3/5, 0, 0). 788 00:45:01 --> 00:45:08 And now what? 789 00:45:08 --> 00:45:11 I mean, that statement's clearly true, right? 790 00:45:11 --> 00:45:14 Point star means five times this gives me three. 791 00:45:14 --> 00:45:17 Two times zero gives me zero, two times zero gives zero. 792 00:45:17 --> 00:45:21 So I'm golden, but I'm not finished. 793 00:45:21 --> 00:45:26 So that's not x, that's its transform. 794 00:45:26 --> 00:45:29 So what do I have to do? 795 00:45:29 --> 00:45:30 Go back. 796 00:45:30 --> 00:45:36 OK, so now can you tell me what this is? 797 00:45:36 --> 00:45:41 Up to a factor of 1/n, or n, which we may have to 798 00:45:41 --> 00:45:45 figure out separately. 799 00:45:45 --> 00:45:50 So apart from this possibly appearing, and probably 800 00:45:50 --> 00:45:56 appearing, give me an idea of what's the inverse transform of 801 00:45:56 --> 00:46:01 that? (3/5, 3/5, 3/5), good. 802 00:46:01 --> 00:46:06 3/5, all three times. 803 00:46:06 --> 00:46:16 OK, and now I can check to see if it's right. 804 00:46:16 --> 00:46:22 And again, I have not attempted to get the 1/n's right, I've 805 00:46:22 --> 00:46:23 just waited to the end. 806 00:46:23 --> 00:46:26 So now I'll just do this convolution and see if I 807 00:46:26 --> 00:46:28 really do get (1, 1, 1). 808 00:46:28 --> 00:46:30 So can we do that convolution? 809 00:46:30 --> 00:46:36 What's the typical, say, this guy. 810 00:46:36 --> 00:46:40 Where does that come from in the convolution? 811 00:46:40 --> 00:46:45 It comes from this, times what? 812 00:46:45 --> 00:46:46 This. 813 00:46:46 --> 00:46:48 That's the constant, right? 814 00:46:48 --> 00:46:49 And that's the constant. 815 00:46:49 --> 00:46:51 Contributes to the constant. 816 00:46:51 --> 00:46:55 And what other products contribute to the constant? 817 00:46:55 --> 00:47:00 So we're really seeing - thank you for this question. 818 00:47:00 --> 00:47:13 So this one multiplies which one? 819 00:47:13 --> 00:47:16 The answer is, this one multiplies this guy. 820 00:47:16 --> 00:47:18 You may say don't worry me about it because 821 00:47:18 --> 00:47:19 they're the same. 822 00:47:19 --> 00:47:21 But it's nice to know. 823 00:47:21 --> 00:47:23 So this one multiplies this one. 824 00:47:23 --> 00:47:25 And this guy multiplies this guy. 825 00:47:25 --> 00:47:27 And we'll remember in a minute why. 826 00:47:27 --> 00:47:30 And then you add them up, that's what convolution is. 827 00:47:30 --> 00:47:38 So I'm getting 9/5+3/5+3/5 is 15/5, so that's three. 828 00:47:38 --> 00:47:43 So what's your conclusion here? 829 00:47:43 --> 00:47:45 I should divide by three, right? 830 00:47:45 --> 00:47:46 So I should divide by three. 831 00:47:46 --> 00:47:48 So that's easy to do. 832 00:47:48 --> 00:47:49 Those will all be one. 833 00:47:49 --> 00:47:52 Yeah, OK. 834 00:47:52 --> 00:47:54 So that's the right answer, right? 835 00:47:54 --> 00:47:59 And then I check that it's right. 836 00:47:59 --> 00:48:04 So let me just close by remembering why was it - 837 00:48:04 --> 00:48:08 so I'm doing up just a forward convolution here. 838 00:48:08 --> 00:48:11 Solving the equation was a deconvolution, 839 00:48:11 --> 00:48:12 figuring out that x. 840 00:48:12 --> 00:48:16 But now that I've got it and I'm just checking, that's just 841 00:48:16 --> 00:48:19 do it forward and see if you got the right answer. 842 00:48:19 --> 00:48:24 And so we're just remembering what it means to 843 00:48:24 --> 00:48:30 do a convolution. 844 00:48:30 --> 00:48:34 So you remember the point about convolutions. 845 00:48:34 --> 00:48:41 That if I'm convolving (a, b, c) cyclically with (d, e, 846 00:48:41 --> 00:48:45 f), that's a constant term. 847 00:48:45 --> 00:48:48 Just tell me again, now, what's the constant terms in this? 848 00:48:48 --> 00:48:55 It's ad plus b times what? 849 00:48:55 --> 00:49:01 This is the first term in the convolution. 850 00:49:01 --> 00:49:06 You remember that, the day we brought convolutions 851 00:49:06 --> 00:49:07 into the course? 852 00:49:07 --> 00:49:13 What does b multiply? f, why f. 853 00:49:13 --> 00:49:21 Because b is the coefficient of w, if I'm taking polynomials, 854 00:49:21 --> 00:49:25 this is a plus bw plus cw squared, and this corresponds 855 00:49:25 --> 00:49:29 to d plus ew plus fw squared. 856 00:49:29 --> 00:49:32 And I'm looking for the constant. 857 00:49:32 --> 00:49:37 So I get an a times ad gives me a constant. bw times 858 00:49:37 --> 00:49:43 what? times f w squared gives me the w cubed. 859 00:49:43 --> 00:49:47 Which is the one so that bf here and then there's a ce. 860 00:49:48 --> 00:49:55 Because c w squared multiplying ew gives me the w cubed, 861 00:49:55 --> 00:49:57 the one which folds in. 862 00:49:57 --> 00:50:01 Yeah, OK, so anyway. 863 00:50:01 --> 00:50:04 That's a quick reminder of cyclic convolutions. 864 00:50:04 --> 00:50:07 That was a good question to ask, yeah. 865 00:50:07 --> 00:50:12 You know, I fully realize this especially this third topic, 866 00:50:12 --> 00:50:14 lots of things like this. 867 00:50:14 --> 00:50:17 Straightforward if you have lots of time, but 868 00:50:17 --> 00:50:20 we had to keep moving. 869 00:50:20 --> 00:50:26 So remembering what cyclic convolution was is a 870 00:50:26 --> 00:50:29 good chance to do it.