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 00:00:21,371 --> 00:00:23,570 DENNIS FREEMAN: OK. 9 00:00:23,570 --> 00:00:28,730 So last time, the idea was to think through a different way, 10 00:00:28,730 --> 00:00:32,030 or in fact, several different ways, 11 00:00:32,030 --> 00:00:34,460 to think about discrete-time systems. 12 00:00:34,460 --> 00:00:40,820 Today is a crash course to do the same thing in CT systems. 13 00:00:40,820 --> 00:00:42,860 So last time, for the last week, we've 14 00:00:42,860 --> 00:00:45,440 been looking at different kinds of representations 15 00:00:45,440 --> 00:00:48,170 for DT systems. 16 00:00:48,170 --> 00:00:52,640 Difference equations, because they're concise and precise. 17 00:00:52,640 --> 00:00:56,090 Block diagrams, because they let us visualize the signal flow 18 00:00:56,090 --> 00:00:57,426 paths. 19 00:00:57,426 --> 00:00:59,300 And operator expressions, because it lets you 20 00:00:59,300 --> 00:01:01,940 treat systems like polynomials. 21 00:01:01,940 --> 00:01:05,450 What we'll see today is that that same kind of strategy 22 00:01:05,450 --> 00:01:06,980 works precisely. 23 00:01:06,980 --> 00:01:09,740 The strategy, not the answers. 24 00:01:09,740 --> 00:01:13,730 The same strategy works precisely for CT. 25 00:01:13,730 --> 00:01:16,490 We'll have a concise, precise representation 26 00:01:16,490 --> 00:01:18,740 in terms of differential equations, which I'm sure you 27 00:01:18,740 --> 00:01:22,370 already know all about. 28 00:01:22,370 --> 00:01:24,890 But we will also develop the notion of a block diagram 29 00:01:24,890 --> 00:01:28,220 so you can visualize signal flow paths. 30 00:01:28,220 --> 00:01:31,340 And we'll have an analogous operator, the A operator, 31 00:01:31,340 --> 00:01:34,620 that will let us treat the systems as polynomials. 32 00:01:34,620 --> 00:01:36,710 So that's the overview. 33 00:01:36,710 --> 00:01:39,170 Because it's so similar, we'll be 34 00:01:39,170 --> 00:01:41,510 able to do all of this in one lecture. 35 00:01:41,510 --> 00:01:44,990 So that's what I mean by a crash course. 36 00:01:44,990 --> 00:01:50,660 So this is Introduction to CT in 50 minutes. 37 00:01:50,660 --> 00:01:53,690 So you already know a lot about how you'd represent systems 38 00:01:53,690 --> 00:01:55,130 as differential equations. 39 00:01:55,130 --> 00:01:58,370 You've seen this in other classes. 40 00:01:58,370 --> 00:02:00,710 You should have seen this sort of approach in physics. 41 00:02:00,710 --> 00:02:05,012 You should have seen this sort of approach in math, in 1803. 42 00:02:05,012 --> 00:02:06,470 So we're to assume that you already 43 00:02:06,470 --> 00:02:09,334 know how to think about differential equations. 44 00:02:09,334 --> 00:02:11,000 And what we're going to do is skip ahead 45 00:02:11,000 --> 00:02:15,650 and think about the alternative representations instead. 46 00:02:15,650 --> 00:02:19,160 So just like in DT, where we thought 47 00:02:19,160 --> 00:02:21,400 about block diagrams, which gave us 48 00:02:21,400 --> 00:02:23,360 a way of thinking about signal flow paths, 49 00:02:23,360 --> 00:02:25,305 we'll have the same sort of idea in CT. 50 00:02:27,920 --> 00:02:30,530 The big difference, you can anticipate. 51 00:02:30,530 --> 00:02:33,620 I mean, if the CT [INAUDIBLE]. 52 00:02:33,620 --> 00:02:38,180 In the difference equation, the fundamental operation 53 00:02:38,180 --> 00:02:42,590 in the difference equation was a delay operation. 54 00:02:42,590 --> 00:02:44,380 If you contrast that to a CT system, 55 00:02:44,380 --> 00:02:47,960 a fundamental operation is not delay. 56 00:02:47,960 --> 00:02:49,760 You know from 1803, you know for physics, 57 00:02:49,760 --> 00:02:52,040 you know from lots of other exposures, 58 00:02:52,040 --> 00:02:56,130 that the fundamental operation is differentiation. 59 00:02:56,130 --> 00:02:59,750 And so the blocks will not be built out of delays, 60 00:02:59,750 --> 00:03:04,160 but will instead be built out of integrators. 61 00:03:04,160 --> 00:03:06,050 Apart from, that the block diagram structure 62 00:03:06,050 --> 00:03:09,480 looks extremely similar. 63 00:03:09,480 --> 00:03:11,990 And we'll be able to apply the same idea that we 64 00:03:11,990 --> 00:03:17,090 did in DT, to simplify the representation of the block 65 00:03:17,090 --> 00:03:20,820 diagram by thinking in terms of operators. 66 00:03:20,820 --> 00:03:23,180 The operator has to similarly change. 67 00:03:23,180 --> 00:03:24,620 So the new operator, the operator 68 00:03:24,620 --> 00:03:26,830 that we use to represent a CT system, 69 00:03:26,830 --> 00:03:33,260 we'll call the A operator, A is intended to mean accumulator. 70 00:03:33,260 --> 00:03:37,130 The thing that's in my mind when I say the A operator 71 00:03:37,130 --> 00:03:40,200 is this tank. 72 00:03:40,200 --> 00:03:42,680 This tank accumulates water. 73 00:03:42,680 --> 00:03:44,510 And you can think about the relationship 74 00:03:44,510 --> 00:03:47,870 between the height of the water, and the input and output rates 75 00:03:47,870 --> 00:03:50,630 by some differential equation. 76 00:03:50,630 --> 00:03:55,070 But the point is that the tank itself, the system, 77 00:03:55,070 --> 00:03:56,330 accumulates. 78 00:03:56,330 --> 00:03:58,370 And that's what we want to have in our heads 79 00:03:58,370 --> 00:04:01,430 when we think about the A operator. 80 00:04:01,430 --> 00:04:05,650 So the operator is going to be like the r operator. 81 00:04:05,650 --> 00:04:09,830 You will apply the r operator to a discrete time signal 82 00:04:09,830 --> 00:04:12,320 to generate a new discrete time signal 83 00:04:12,320 --> 00:04:13,970 that we shifted to the right. 84 00:04:13,970 --> 00:04:17,510 That's what r meant, r was right. 85 00:04:17,510 --> 00:04:19,800 So right shift operator. 86 00:04:19,800 --> 00:04:23,190 Here, you apply the A operator to the X signal. 87 00:04:23,190 --> 00:04:27,560 X is now CT, continuous time. 88 00:04:27,560 --> 00:04:31,520 And when you apply the A operator to a signal X, 89 00:04:31,520 --> 00:04:33,950 it generates a whole new signal Y 90 00:04:33,950 --> 00:04:37,550 that is everywhere equal to the indefinite integral 91 00:04:37,550 --> 00:04:39,050 of the input signal. 92 00:04:39,050 --> 00:04:44,016 So the A operator is, start with the input signal X. 93 00:04:44,016 --> 00:04:46,640 And integrate it with regard to time starting at minus infinity 94 00:04:46,640 --> 00:04:52,220 and going up to t, that's the t value of the output signal. 95 00:04:54,730 --> 00:04:57,470 Simple. 96 00:04:57,470 --> 00:05:01,000 So let's see how simple. 97 00:05:01,000 --> 00:05:05,350 So here's a bunch of block diagrams written in terms of A. 98 00:05:05,350 --> 00:05:08,110 Here's a bunch of differential equations. 99 00:05:08,110 --> 00:05:12,400 Figure out the correspondence, if any. 100 00:05:12,400 --> 00:05:15,920 OK, traditionally, it seems that people are quiet. 101 00:05:15,920 --> 00:05:19,430 So before you start, say something that has nothing 102 00:05:19,430 --> 00:05:21,350 to do with 003 to you next-door neighbor. 103 00:05:21,350 --> 00:05:24,710 It has to have nothing to do with 003. 104 00:05:24,710 --> 00:05:28,100 And then figure out which of these bottom diagrams, number 105 00:05:28,100 --> 00:05:30,335 1, 2, 3, 4, or none of them, corresponds 106 00:05:30,335 --> 00:05:31,460 best to the correspondence. 107 00:05:31,460 --> 00:05:36,314 [SIDE CONVERSATION] 108 00:06:26,940 --> 00:06:31,098 OK, you've got 30 seconds to answer the question, now. 109 00:06:31,098 --> 00:06:34,570 [SIDE CONVERSATION] 110 00:06:41,520 --> 00:06:43,166 Well, that killed everything. 111 00:07:44,460 --> 00:07:47,090 OK, everybody raise your hand and tell me 112 00:07:47,090 --> 00:07:50,390 which of the five available answers, 1 through 5, 113 00:07:50,390 --> 00:07:52,562 best illustrates the correspondence. 114 00:07:52,562 --> 00:07:53,270 Raise your hands. 115 00:07:53,270 --> 00:07:56,460 Let me see how many of you figured it out. 116 00:07:56,460 --> 00:08:00,640 Remember, if you're wrong, you can blame it on your partner. 117 00:08:00,640 --> 00:08:03,107 OK, it's more than 95% correct. 118 00:08:03,107 --> 00:08:04,190 How do you think about it? 119 00:08:04,190 --> 00:08:04,981 What do I do first? 120 00:08:06,732 --> 00:08:09,510 AUDIENCE: [INAUDIBLE] 121 00:08:09,510 --> 00:08:11,120 DENNIS FREEMAN: [INAUDIBLE] So what do 122 00:08:11,120 --> 00:08:12,564 you want me to look at first? 123 00:08:12,564 --> 00:08:13,730 This diagram, that equation? 124 00:08:13,730 --> 00:08:14,896 This diagram, that equation? 125 00:08:14,896 --> 00:08:16,676 AUDIENCE: [INAUDIBLE] 126 00:08:16,676 --> 00:08:18,050 DENNIS FREEMAN: Write an equation 127 00:08:18,050 --> 00:08:19,091 for each of the diagrams. 128 00:08:19,091 --> 00:08:22,016 How would I figure out an equation for that diagram? 129 00:08:22,016 --> 00:08:22,516 [INAUDIBLE] 130 00:08:22,516 --> 00:08:26,019 AUDIENCE: [INAUDIBLE] 131 00:08:26,019 --> 00:08:28,560 DENNIS FREEMAN: So what's coming in and out of the plus sign. 132 00:08:28,560 --> 00:08:29,390 That's very good . 133 00:08:29,390 --> 00:08:31,520 So let's focus on [? that ?] coming out first. 134 00:08:31,520 --> 00:08:31,640 Us. 135 00:08:31,640 --> 00:08:32,931 What's the name of that signal? 136 00:08:35,797 --> 00:08:37,231 AUDIENCE: [INAUDIBLE] 137 00:08:37,231 --> 00:08:39,612 DENNIS FREEMAN: Well, I can think of two names. 138 00:08:39,612 --> 00:08:41,070 That's why it's a good idea to look 139 00:08:41,070 --> 00:08:44,310 at what's coming in and out. 140 00:08:44,310 --> 00:08:46,950 One way that I could name this would be with reference 141 00:08:46,950 --> 00:08:49,232 to the A operator. 142 00:08:49,232 --> 00:08:51,190 If I know that what comes out of the A operator 143 00:08:51,190 --> 00:08:54,660 is capital Y, what goes into the A operator? 144 00:08:54,660 --> 00:08:56,640 AUDIENCE: [INAUDIBLE]. 145 00:08:56,640 --> 00:08:58,860 DENNIS FREEMAN: The derivative. 146 00:08:58,860 --> 00:09:01,850 So if I know that what comes out is Y, then what goes in 147 00:09:01,850 --> 00:09:02,740 must have been Y dot. 148 00:09:06,200 --> 00:09:08,540 But that [? also ?] has to be the output of the adder. 149 00:09:08,540 --> 00:09:10,682 So what were the two inputs to the adder? 150 00:09:10,682 --> 00:09:12,155 AUDIENCE: [INAUDIBLE] 151 00:09:12,155 --> 00:09:13,265 DENNIS FREEMAN: X and-- 152 00:09:18,070 --> 00:09:19,780 py. 153 00:09:19,780 --> 00:09:24,160 So y dot, the thing that goes into the accumulator, 154 00:09:24,160 --> 00:09:26,860 has to be the same as X plus py. 155 00:09:26,860 --> 00:09:29,230 So y dot is X plus py, so there's a correspondence 156 00:09:29,230 --> 00:09:29,980 between these two. 157 00:09:32,500 --> 00:09:35,240 Everybody sort of get the idea? 158 00:09:35,240 --> 00:09:36,970 We can do the same sort of thing here. 159 00:09:36,970 --> 00:09:38,261 What's the name of that signal? 160 00:09:42,370 --> 00:09:43,330 AUDIENCE: [INAUDIBLE] 161 00:09:43,330 --> 00:09:44,770 DENNIS FREEMAN: [? Shout. ?] 162 00:09:44,770 --> 00:09:46,910 AUDIENCE: Y dot over p. 163 00:09:46,910 --> 00:09:48,140 DENNIS FREEMAN: y dot over p. 164 00:09:48,140 --> 00:09:48,639 Exactly. 165 00:09:48,639 --> 00:09:51,410 So that one is going to be y dot over p. 166 00:09:51,410 --> 00:09:56,030 And that's going to have to be the same as x plus y. 167 00:09:56,030 --> 00:09:58,500 So if you clear the fraction here, 168 00:09:58,500 --> 00:10:00,507 you would find that y dot is px plus py. 169 00:10:00,507 --> 00:10:02,090 So there is a correspondence that way. 170 00:10:06,520 --> 00:10:10,107 And you can do the same sort of thing down here. 171 00:10:10,107 --> 00:10:11,440 It looks a little bit different. 172 00:10:11,440 --> 00:10:11,940 What's this? 173 00:10:11,940 --> 00:10:16,930 This says that the output of the adder, y, the output 174 00:10:16,930 --> 00:10:19,240 of the adder here is y. 175 00:10:19,240 --> 00:10:21,640 So y must have been x plus what? 176 00:10:24,610 --> 00:10:27,580 AUDIENCE: [INAUDIBLE] 177 00:10:27,580 --> 00:10:33,710 DENNIS FREEMAN: So p times the integral of y. 178 00:10:33,710 --> 00:10:36,300 And I'd like to write that in a differential form, 179 00:10:36,300 --> 00:10:38,900 so I could differentiate term by term 180 00:10:38,900 --> 00:10:42,560 to say that that's the same thing as y dot is 181 00:10:42,560 --> 00:10:46,040 x dot plus py. 182 00:10:46,040 --> 00:10:48,190 So y dot is x dot plus py, so there's 183 00:10:48,190 --> 00:10:49,190 correspondence that way. 184 00:10:49,190 --> 00:10:50,491 So the answer is 1. 185 00:10:50,491 --> 00:10:50,990 OK? 186 00:10:50,990 --> 00:10:53,180 Everybody's happy with them that? 187 00:10:53,180 --> 00:10:57,220 So the point is that the differential operators 188 00:10:57,220 --> 00:10:58,427 look very much the same. 189 00:10:58,427 --> 00:11:00,010 You can use the same kind of reasoning 190 00:11:00,010 --> 00:11:03,460 that you did for difference equations with r. 191 00:11:03,460 --> 00:11:06,490 Of course, the operators are different. 192 00:11:06,490 --> 00:11:11,050 And just like r, you can think about these things 193 00:11:11,050 --> 00:11:13,070 as polynomials. 194 00:11:13,070 --> 00:11:16,780 So if you think about this feedforward system that's 195 00:11:16,780 --> 00:11:19,840 now CT, not DT, the feedforward system 196 00:11:19,840 --> 00:11:25,870 says that I can construct W by taking X plus A X, X plus A X, 197 00:11:25,870 --> 00:11:31,310 where that's X plus the the integral of X. 198 00:11:31,310 --> 00:11:33,310 Then I could think about the Y signal 199 00:11:33,310 --> 00:11:38,350 as W plus the integral of W. So that's this one, 200 00:11:38,350 --> 00:11:40,420 y is the w plus the integral w. 201 00:11:40,420 --> 00:11:43,945 And now I can substitute instances of this into here. 202 00:11:46,820 --> 00:11:48,690 So this w expression could go in there. 203 00:11:48,690 --> 00:11:50,990 And that gives me this. 204 00:11:50,990 --> 00:11:53,529 If I integrated this w expression once, 205 00:11:53,529 --> 00:11:55,070 then I would get the integral of that 206 00:11:55,070 --> 00:11:58,940 plus the double integral of that, so that gives me these. 207 00:11:58,940 --> 00:12:02,540 And I think about those things in terms of A's, and I get 208 00:12:02,540 --> 00:12:06,260 precisely the same expression. 209 00:12:06,260 --> 00:12:09,010 So the idea is, just like r, you can 210 00:12:09,010 --> 00:12:14,050 manipulate operator expressions in r, 211 00:12:14,050 --> 00:12:15,670 like they were polynomials. 212 00:12:15,670 --> 00:12:21,750 You can manipulate operator expressions in A, 213 00:12:21,750 --> 00:12:24,720 just like they were polynomials. 214 00:12:24,720 --> 00:12:28,260 And the reason I can say that is that, if I think through all 215 00:12:28,260 --> 00:12:30,540 the properties of polynomials, I can draw 216 00:12:30,540 --> 00:12:34,080 an isomorphism between the way the system would have behaved 217 00:12:34,080 --> 00:12:36,310 and the way the polynomial would have behaved. 218 00:12:36,310 --> 00:12:39,210 And in particular, here's three statements 219 00:12:39,210 --> 00:12:42,230 that have to be true. 220 00:12:42,230 --> 00:12:43,940 If the operator expressions were supposed 221 00:12:43,940 --> 00:12:46,610 to behave like polynomials, polynomials 222 00:12:46,610 --> 00:12:51,470 have the property that the polynomials in A 223 00:12:51,470 --> 00:12:53,600 should commute. 224 00:12:53,600 --> 00:12:55,567 Polynomials should have the-- the A 225 00:12:55,567 --> 00:12:57,150 should have the distributive property, 226 00:12:57,150 --> 00:13:00,560 multiplication over addition. 227 00:13:00,560 --> 00:13:03,530 And polynomials should associate. 228 00:13:03,530 --> 00:13:07,000 And if you think about each of those in detail, 229 00:13:07,000 --> 00:13:08,800 the operations that would correspond 230 00:13:08,800 --> 00:13:10,650 to what you've put in the block diagram 231 00:13:10,650 --> 00:13:14,740 have precisely those same relations. 232 00:13:14,740 --> 00:13:17,751 So that's the outline of how you go about proving something 233 00:13:17,751 --> 00:13:18,250 like this. 234 00:13:18,250 --> 00:13:21,220 But the takeaway message is the operator expressions 235 00:13:21,220 --> 00:13:28,740 in A obey all the rules that you would expect from a polynomial. 236 00:13:28,740 --> 00:13:32,720 So, second question. 237 00:13:32,720 --> 00:13:34,640 Think about these two systems, and think 238 00:13:34,640 --> 00:13:35,960 about the notion of equivalent. 239 00:13:35,960 --> 00:13:41,150 Equivalent here is going to be just the same as it was in DT. 240 00:13:41,150 --> 00:13:45,590 In DT, equivalent [? man, ?] if all of the right shift 241 00:13:45,590 --> 00:13:49,970 operators started at rest, that is to say their output was 0, 242 00:13:49,970 --> 00:13:53,660 we will have the same kind of notion here. 243 00:13:53,660 --> 00:13:57,050 At rest is going to mean all of the integrators-- 244 00:13:57,050 --> 00:13:59,750 integrators have a starting value, an initial value. 245 00:13:59,750 --> 00:14:01,880 If those initial values are all, 0 then 246 00:14:01,880 --> 00:14:03,890 we say that the system started at rest. 247 00:14:03,890 --> 00:14:05,570 We have that same kind of proviso. 248 00:14:05,570 --> 00:14:08,960 Given that proviso, then all of the operator 249 00:14:08,960 --> 00:14:11,160 expressions behave like polynomials. 250 00:14:11,160 --> 00:14:15,350 So determine k1 so that these two systems are equivalent, 251 00:14:15,350 --> 00:14:19,500 given that definition of equivalence. 252 00:14:19,500 --> 00:17:02,000 [SIDE CONVERSATION] 253 00:17:02,000 --> 00:17:04,490 So anyone have an answer? 254 00:17:04,490 --> 00:17:06,410 So how should I choose k1 if I want those two 255 00:17:06,410 --> 00:17:07,940 systems to be equivalent? 256 00:17:07,940 --> 00:17:10,460 Yes? 257 00:17:10,460 --> 00:17:11,250 Raise your hands. 258 00:17:15,714 --> 00:17:18,711 OK, it was a trick question. 259 00:17:18,711 --> 00:17:21,210 So now, given the additional information that it was a trick 260 00:17:21,210 --> 00:17:24,839 question, revise your answers. 261 00:17:24,839 --> 00:17:26,369 Ah excellent, excellent. 262 00:17:26,369 --> 00:17:27,329 Instant revisions. 263 00:17:27,329 --> 00:17:28,230 Wonderful. 264 00:17:28,230 --> 00:17:31,300 OK, what was the trick? 265 00:17:31,300 --> 00:17:33,215 [INAUDIBLE] got it. 266 00:17:33,215 --> 00:17:35,494 What was a trick? 267 00:17:35,494 --> 00:17:35,994 Yeah. 268 00:17:35,994 --> 00:17:37,892 AUDIENCE: It would be minus 1.6. 269 00:17:37,892 --> 00:17:40,070 DENNIS FREEMAN: It should have been minus 1.6. 270 00:17:40,070 --> 00:17:41,210 Yes. 271 00:17:41,210 --> 00:17:44,062 So what do you do? 272 00:17:44,062 --> 00:17:46,520 What do I do first in order to answer a question like this? 273 00:17:50,920 --> 00:17:52,350 [INAUDIBLE] 274 00:17:52,350 --> 00:17:54,352 [? Right, ?] so translate the block diagrams 275 00:17:54,352 --> 00:17:55,560 into a differential equation. 276 00:17:55,560 --> 00:17:57,030 That's a good approach. 277 00:17:57,030 --> 00:17:59,880 Because what we'd like to do is figure out an equivalence. 278 00:17:59,880 --> 00:18:03,780 The equivalence is not trivial. 279 00:18:03,780 --> 00:18:05,390 So in fact, the equivalence is easier 280 00:18:05,390 --> 00:18:08,240 to think about if you do polynomials. 281 00:18:08,240 --> 00:18:10,490 You could try to manipulate the blocks 282 00:18:10,490 --> 00:18:12,380 to make them look the same. 283 00:18:12,380 --> 00:18:14,630 That's actually hard. 284 00:18:14,630 --> 00:18:16,170 It it's pretty easy to do it if you 285 00:18:16,170 --> 00:18:25,410 think about the block diagrams being represented as operators. 286 00:18:25,410 --> 00:18:28,290 So you can look at the first one and you can say, OK. 287 00:18:28,290 --> 00:18:34,815 W is the signal that comes out of A operating on an X minus 288 00:18:34,815 --> 00:18:36,670 0.7W. 289 00:18:36,670 --> 00:18:38,430 So that's what that's adder says. 290 00:18:42,030 --> 00:18:44,020 And you can similarly do this box. 291 00:18:44,020 --> 00:18:45,150 It looks the same. 292 00:18:45,150 --> 00:18:47,490 Treat them as polynomials and you can reduce it 293 00:18:47,490 --> 00:18:48,630 to some equivalent form. 294 00:18:51,610 --> 00:18:54,490 Does everybody have the gist of what's going on? 295 00:18:54,490 --> 00:18:55,440 Same thing down here. 296 00:18:55,440 --> 00:18:57,700 The equations look a little bit different. 297 00:18:57,700 --> 00:18:58,912 But it's the same idea. 298 00:18:58,912 --> 00:19:00,870 You figure out what are the constraints imposed 299 00:19:00,870 --> 00:19:02,520 by the blocks. 300 00:19:02,520 --> 00:19:05,230 The output of every A has to be the integral 301 00:19:05,230 --> 00:19:06,729 of the thing that went into it. 302 00:19:06,729 --> 00:19:09,270 The output of every adder has to be the sum of the two things 303 00:19:09,270 --> 00:19:10,115 that went into it. 304 00:19:10,115 --> 00:19:11,490 Figure out all those constraints, 305 00:19:11,490 --> 00:19:15,060 reduce them to polynomials, and then manipulate them 306 00:19:15,060 --> 00:19:17,250 as though they were polynomials. 307 00:19:17,250 --> 00:19:20,250 And the polynomial manipulation shows that minus k1 308 00:19:20,250 --> 00:19:21,720 should be 1.6, OK? 309 00:19:21,720 --> 00:19:24,490 [INAUDIBLE] should be a minus 1.6. 310 00:19:24,490 --> 00:19:26,280 Point is, it's easy. 311 00:19:26,280 --> 00:19:28,170 If you know polynomials, you know the answer. 312 00:19:28,170 --> 00:19:29,580 That's pretty good. 313 00:19:29,580 --> 00:19:31,650 So we're doing something completely different 314 00:19:31,650 --> 00:19:33,450 from polynomials. 315 00:19:33,450 --> 00:19:35,820 But we're able to use the intuitions 316 00:19:35,820 --> 00:19:39,960 that we get from polynomials in order to do the problem. 317 00:19:39,960 --> 00:19:43,920 There is one remaining problem. 318 00:19:43,920 --> 00:19:46,650 Something that is harder than it is in DT. 319 00:19:46,650 --> 00:19:49,200 Probably the hardest part of CT. 320 00:19:49,200 --> 00:19:51,240 And that involves thinking about what 321 00:19:51,240 --> 00:19:54,930 should be our basis signals. 322 00:19:54,930 --> 00:19:57,420 In DT, there is a really good candidate 323 00:19:57,420 --> 00:20:02,010 for that, which we call the unit sample single. 324 00:20:02,010 --> 00:20:05,920 The unit sample signal is the simplest possible 325 00:20:05,920 --> 00:20:08,730 non-trivial DT signal. 326 00:20:08,730 --> 00:20:12,752 By which I mean it is 0 everywhere, except one place. 327 00:20:12,752 --> 00:20:14,835 If it were 0 everywhere, we would call it trivial. 328 00:20:17,900 --> 00:20:19,630 So in order for it not to be trivial, 329 00:20:19,630 --> 00:20:22,354 it has to be non-zero at least some place. 330 00:20:22,354 --> 00:20:23,395 The place is non-trivial. 331 00:20:23,395 --> 00:20:27,360 It is at n equals 0. 332 00:20:27,360 --> 00:20:32,290 And the value that it is, given that it can't be 0, is 1. 333 00:20:32,290 --> 00:20:38,740 So the unit sample signal is the most simple non-trivial DT 334 00:20:38,740 --> 00:20:39,582 signal. 335 00:20:39,582 --> 00:20:41,290 So what we need is the equivalent for CT. 336 00:20:43,810 --> 00:20:46,120 So here it is. 337 00:20:46,120 --> 00:20:50,650 Define a signal in CT that is 0 everywhere except 0. 338 00:20:50,650 --> 00:20:56,150 And at 0, make it [? 1. ?] 339 00:20:56,150 --> 00:20:58,910 So is that a good choice? 340 00:20:58,910 --> 00:21:02,750 And I hope I've biased my presentation 341 00:21:02,750 --> 00:21:07,070 enough so that the answer is obviously no. 342 00:21:07,070 --> 00:21:10,240 So the real question is, why not? 343 00:21:10,240 --> 00:21:12,250 Can somebody think of a good reason 344 00:21:12,250 --> 00:21:15,201 why this is a poor choice for a building block signal? 345 00:21:15,201 --> 00:21:15,700 Yes. 346 00:21:15,700 --> 00:21:22,364 AUDIENCE: [INAUDIBLE] 347 00:21:22,364 --> 00:21:25,480 DENNIS FREEMAN: So you would need infinite gain 348 00:21:25,480 --> 00:21:27,310 to go from 0 to 1. 349 00:21:27,310 --> 00:21:30,370 That's sort of true for the r thing, too. 350 00:21:30,370 --> 00:21:33,726 You have to go from 0 to 1. 351 00:21:33,726 --> 00:21:35,260 It's in 0 time. 352 00:21:35,260 --> 00:21:37,114 That seems hard. 353 00:21:37,114 --> 00:21:37,780 Any other ideas? 354 00:21:37,780 --> 00:21:37,850 Yeah. 355 00:21:37,850 --> 00:21:38,725 AUDIENCE: [INAUDIBLE] 356 00:21:38,725 --> 00:21:41,630 DENNIS FREEMAN: Not continuous. 357 00:21:41,630 --> 00:21:43,988 That's also true. 358 00:21:43,988 --> 00:21:44,906 Yes. 359 00:21:44,906 --> 00:21:47,172 AUDIENCE: [INAUDIBLE] 360 00:21:47,172 --> 00:21:48,880 DENNIS FREEMAN: The integral is always 0. 361 00:21:48,880 --> 00:21:51,860 That's a problem. 362 00:21:51,860 --> 00:21:57,700 So it's just not going to be a very useful signal. 363 00:21:57,700 --> 00:21:58,880 OK, what am I doing? 364 00:21:58,880 --> 00:22:02,440 I'm trying to figure out a basis signal that I 365 00:22:02,440 --> 00:22:03,640 would use in block diagrams. 366 00:22:03,640 --> 00:22:07,050 My basic block diagram is this integrator thing. 367 00:22:07,050 --> 00:22:11,920 Mine is infinity to t on tau. 368 00:22:11,920 --> 00:22:14,496 So I'd like to put something in here. 369 00:22:14,496 --> 00:22:16,370 But if I put that signal, what did I call it? 370 00:22:16,370 --> 00:22:19,930 W. If I put W into an iterator, what's 371 00:22:19,930 --> 00:22:23,830 the output of the integrator after you integrate W? 372 00:22:23,830 --> 00:22:24,760 0. 373 00:22:24,760 --> 00:22:26,260 Well, that's a problem. 374 00:22:26,260 --> 00:22:28,150 It's not a very good basis function 375 00:22:28,150 --> 00:22:33,010 if the very first operator I go through turns it into 0. 376 00:22:33,010 --> 00:22:34,480 Everyone see that? 377 00:22:34,480 --> 00:22:36,970 That's a problem. 378 00:22:36,970 --> 00:22:39,820 This is the problem in CT. 379 00:22:39,820 --> 00:22:42,840 If you guys figure out the next slide, you're done. 380 00:22:42,840 --> 00:22:44,800 CT's easy. 381 00:22:44,800 --> 00:22:52,250 So what we do is think about defining a signal that 382 00:22:52,250 --> 00:22:55,580 would have the property that it behaved as though it 383 00:22:55,580 --> 00:22:57,377 had those desirable properties. 384 00:22:57,377 --> 00:22:58,460 What would have to happen? 385 00:22:58,460 --> 00:23:01,830 We'd like a signal that had some amount of area. 386 00:23:01,830 --> 00:23:03,320 Because that's what this thing is. 387 00:23:03,320 --> 00:23:06,390 This is an area operator. 388 00:23:06,390 --> 00:23:09,570 We would like a signal that didn't have a 0 area. 389 00:23:09,570 --> 00:23:11,460 The W signal has 0 area. 390 00:23:11,460 --> 00:23:13,410 Bad. 391 00:23:13,410 --> 00:23:15,960 We'd like a signal that doesn't have zero area. 392 00:23:15,960 --> 00:23:18,560 In fact, if we wanted it to be the simplest possible, 393 00:23:18,560 --> 00:23:22,820 then you should have an area of 1. 394 00:23:22,820 --> 00:23:27,010 But we'd like it to be zero almost everywhere. 395 00:23:27,010 --> 00:23:29,380 How do you do that? 396 00:23:29,380 --> 00:23:32,560 OK, well, the way you can do it, and the way we will do it, 397 00:23:32,560 --> 00:23:37,720 is think about building a signal [? with ?] a limit. 398 00:23:37,720 --> 00:23:40,900 What if I had a pulse signal that 399 00:23:40,900 --> 00:23:44,470 was 2 epsilon wide and 1 over 2 epsilon high? 400 00:23:44,470 --> 00:23:49,330 The area would be 1, regardless of the value of epsilon. 401 00:23:49,330 --> 00:23:56,507 And if I shrunk epsilon to be a very small number, say 0, 402 00:23:56,507 --> 00:23:58,090 if I think about the limit, as epsilon 403 00:23:58,090 --> 00:24:00,610 gets smaller and smaller and smaller and smaller, 404 00:24:00,610 --> 00:24:05,410 then I approach the condition I'd like to have. 405 00:24:05,410 --> 00:24:08,440 The signal is 0 everywhere, except 0. 406 00:24:08,440 --> 00:24:10,610 Now, it has a horrendous value at 0. 407 00:24:10,610 --> 00:24:13,510 But, oh well, I'll sweep that under the rug. 408 00:24:13,510 --> 00:24:16,030 The important thing is that, when I integrate it, 409 00:24:16,030 --> 00:24:17,920 I'd like it to give me one. 410 00:24:17,920 --> 00:24:20,062 That's the important thing. 411 00:24:20,062 --> 00:24:20,770 Everyone with me? 412 00:24:20,770 --> 00:24:22,895 That's the hardest part in this part of the course. 413 00:24:22,895 --> 00:24:26,530 If you get this, CT systems are a breeze. 414 00:24:26,530 --> 00:24:30,880 So this is a different kind of a signal. 415 00:24:30,880 --> 00:24:35,600 The idea is to incorporate in it two seemingly contradictory 416 00:24:35,600 --> 00:24:36,100 things. 417 00:24:36,100 --> 00:24:43,440 It should be 0 almost everywhere and the integral should be 1. 418 00:24:43,440 --> 00:24:46,110 So there is no such thing. 419 00:24:46,110 --> 00:24:48,125 This is an idealized signal. 420 00:24:48,125 --> 00:24:50,250 This is something that we think about in the limit. 421 00:24:50,250 --> 00:24:52,740 It turns out that we can get a lot of insight 422 00:24:52,740 --> 00:24:55,650 by thinking about this signal. 423 00:24:55,650 --> 00:24:58,680 But it is important to realize that it is only 424 00:24:58,680 --> 00:25:01,540 defined in a limit. 425 00:25:01,540 --> 00:25:04,110 So we will write it. 426 00:25:04,110 --> 00:25:06,800 Since it's so useful, we will write it this way. 427 00:25:06,800 --> 00:25:08,430 We'll draw a little arrow. 428 00:25:08,430 --> 00:25:10,530 The arrow is supposed to connote in your mind, 429 00:25:10,530 --> 00:25:13,490 this thing goes really high. 430 00:25:13,490 --> 00:25:16,890 It goes so high that telling you how high it is is meaningless. 431 00:25:16,890 --> 00:25:21,892 So instead, to the side, we tell you what's the area of it. 432 00:25:21,892 --> 00:25:23,600 What would come out if you integrated it? 433 00:25:23,600 --> 00:25:24,100 Yes. 434 00:25:24,100 --> 00:25:26,580 AUDIENCE: [INAUDIBLE] DT [INAUDIBLE] 435 00:25:26,580 --> 00:25:29,800 decompose an [INAUDIBLE] signal into a series 436 00:25:29,800 --> 00:25:32,342 of [INAUDIBLE] Can you do the same thing with [INAUDIBLE]? 437 00:25:32,342 --> 00:25:33,175 DENNIS FREEMAN: Yes. 438 00:25:33,175 --> 00:25:35,770 And that will be something that we spend a lot of time on, 439 00:25:35,770 --> 00:25:40,060 because that decomposition is not trivial. 440 00:25:40,060 --> 00:25:42,700 But it is possible. 441 00:25:42,700 --> 00:25:46,720 So the question was, can we decompose arbitrary signals 442 00:25:46,720 --> 00:25:48,220 into sums of these things? 443 00:25:48,220 --> 00:25:51,450 And the answer is absolutely yes. 444 00:25:51,450 --> 00:25:54,720 That's why we like it. 445 00:25:54,720 --> 00:25:58,930 OK, so the first thing to think about 446 00:25:58,930 --> 00:26:00,760 is what happens if you put, now, instead 447 00:26:00,760 --> 00:26:09,274 of W, let's put delta of t, the unit impulse function. 448 00:26:09,274 --> 00:26:10,940 What happens if you put the unit impulse 449 00:26:10,940 --> 00:26:13,770 function into an integrator? 450 00:26:13,770 --> 00:26:19,500 Well, if you integrate the delta of t, so if my input x of t 451 00:26:19,500 --> 00:26:25,180 is delta of t, and if I want to think about y of t, 452 00:26:25,180 --> 00:26:28,370 which is going to A operating on x? 453 00:26:28,370 --> 00:26:32,430 So think about what would be the answer to this integral 454 00:26:32,430 --> 00:26:34,544 for times less than 0? 455 00:26:34,544 --> 00:26:36,710 Well, the system started the rest and the integrator 456 00:26:36,710 --> 00:26:39,050 turned on at 0. 457 00:26:39,050 --> 00:26:43,190 What's the value of the answer of the integrator here? 458 00:26:43,190 --> 00:26:43,810 0. 459 00:26:43,810 --> 00:26:46,690 It started at 0 back at minus infinity some time. 460 00:26:46,690 --> 00:26:48,610 Started at rest. 461 00:26:48,610 --> 00:26:51,340 There was no input. 462 00:26:51,340 --> 00:26:52,240 So the answer is 0. 463 00:26:52,240 --> 00:26:56,920 So the answer is going to be 0 up until something happens. 464 00:26:56,920 --> 00:27:02,230 During that brief epsilon of time, a unit of area went in. 465 00:27:02,230 --> 00:27:06,540 So what's the output of the integrator become? 466 00:27:06,540 --> 00:27:08,310 1. 467 00:27:08,310 --> 00:27:10,020 So at that time it became 1. 468 00:27:13,230 --> 00:27:16,880 And then what's the value as you go forward in time? 469 00:27:16,880 --> 00:27:18,170 1. 470 00:27:18,170 --> 00:27:20,220 It got stuck at 1. 471 00:27:20,220 --> 00:27:24,650 So the most primitive signal that we'll think about 472 00:27:24,650 --> 00:27:26,020 is the unit impulse function. 473 00:27:26,020 --> 00:27:29,226 We'll denote that by delta of t. 474 00:27:29,226 --> 00:27:31,850 And it has the property that it goes through our most primitive 475 00:27:31,850 --> 00:27:33,810 block. 476 00:27:33,810 --> 00:27:35,970 Delta goes to a unit-step And unit-step 477 00:27:35,970 --> 00:27:38,670 is so useful that we'll give it a special name, u of t. 478 00:27:38,670 --> 00:27:40,462 U means unit-step. 479 00:27:44,160 --> 00:27:47,380 And with that, all of these systems 480 00:27:47,380 --> 00:27:50,740 now make sense, in the same sense that DT systems did. 481 00:27:50,740 --> 00:27:55,750 So for example, if I have a feedforward system, 482 00:27:55,750 --> 00:27:59,680 I can think about that as having several signal flow 483 00:27:59,680 --> 00:28:02,250 paths going forward. 484 00:28:02,250 --> 00:28:04,020 And the output of the signal, when 485 00:28:04,020 --> 00:28:08,460 stimulated with a delta function, 486 00:28:08,460 --> 00:28:10,380 is the sum of all those different signal 487 00:28:10,380 --> 00:28:12,970 paths that I can think of. 488 00:28:12,970 --> 00:28:15,630 So I can express the input-output relationship 489 00:28:15,630 --> 00:28:18,840 by dysfunctional relationship. 490 00:28:18,840 --> 00:28:21,287 Everybody sees that? 491 00:28:21,287 --> 00:28:23,370 Or I can just think about all the different signal 492 00:28:23,370 --> 00:28:24,330 flow passed through here. 493 00:28:24,330 --> 00:28:26,371 What would happen if I had a delta function here? 494 00:28:26,371 --> 00:28:28,420 Well, here's a signal flow path. 495 00:28:28,420 --> 00:28:29,530 Here's another one. 496 00:28:29,530 --> 00:28:32,030 Here's another one. 497 00:28:32,030 --> 00:28:33,710 Here's one. 498 00:28:33,710 --> 00:28:36,350 For signal flow paths, I have to think 499 00:28:36,350 --> 00:28:39,136 about ways the input could turn into the output. 500 00:28:39,136 --> 00:28:40,760 If the input went through the top half, 501 00:28:40,760 --> 00:28:44,630 the output would be delta. 502 00:28:44,630 --> 00:28:52,520 If the input went through, the path the output would be u. 503 00:28:52,520 --> 00:28:58,000 If the input went through this path, u. 504 00:28:58,000 --> 00:29:01,707 If the input went through this, it turns into u, 505 00:29:01,707 --> 00:29:03,790 and then it goes through another accumulator, what 506 00:29:03,790 --> 00:29:06,730 happens when you integrate u? 507 00:29:06,730 --> 00:29:07,230 tu. 508 00:29:10,670 --> 00:29:17,732 If this is u, and if you put that in through an integrator, 509 00:29:17,732 --> 00:29:19,690 that integrator, like all the other integrators 510 00:29:19,690 --> 00:29:21,900 we'll talk about, unless we tell you otherwise, 511 00:29:21,900 --> 00:29:22,930 is initially at rest. 512 00:29:22,930 --> 00:29:25,870 Therefore, the output started at 0. 513 00:29:25,870 --> 00:29:27,850 Because u is 0 for t less than 0, 514 00:29:27,850 --> 00:29:32,700 it persists at 0 until something happens at t equals 0. 515 00:29:32,700 --> 00:29:34,200 At which point, the input becomes 1, 516 00:29:34,200 --> 00:29:37,520 and the integral of 1 is t. 517 00:29:37,520 --> 00:29:42,730 So we will usually write that this way, meaning it's t. 518 00:29:42,730 --> 00:29:45,700 So we will write as a shorthand times u. 519 00:29:45,700 --> 00:29:47,620 That's a strange thing to do. 520 00:29:47,620 --> 00:29:50,254 It's just very convenient. 521 00:29:50,254 --> 00:29:51,670 It's convenient, because normally, 522 00:29:51,670 --> 00:29:57,370 when you have a function, so that last function, this one, 523 00:29:57,370 --> 00:30:00,400 is a signal that does this. 524 00:30:03,270 --> 00:30:07,150 So we will call that tu of t because, obviously, t 525 00:30:07,150 --> 00:30:08,940 is that signal. 526 00:30:08,940 --> 00:30:10,040 We don't mean that. 527 00:30:10,040 --> 00:30:13,740 We want to lop off the part that came before 0. 528 00:30:13,740 --> 00:30:15,320 Multiplying by u of t is a quick way 529 00:30:15,320 --> 00:30:18,710 of writing lop off the stuff that came before 0. 530 00:30:18,710 --> 00:30:20,510 That's all that means. 531 00:30:20,510 --> 00:30:23,186 So we'll write that that way. 532 00:30:23,186 --> 00:30:25,310 And you can see that, by thinking about signal flow 533 00:30:25,310 --> 00:30:28,790 paths, and by thinking about how this basic signal, this basis 534 00:30:28,790 --> 00:30:30,920 function, goes through the system, 535 00:30:30,920 --> 00:30:34,130 it's easy to figure out the response of such a system. 536 00:30:37,160 --> 00:30:42,670 Just like in DT, feedback is different. 537 00:30:42,670 --> 00:30:44,880 So now, we have to think about what would happen 538 00:30:44,880 --> 00:30:47,140 if there's a feedback loop. 539 00:30:47,140 --> 00:30:48,660 So it was crucial, when I was doing 540 00:30:48,660 --> 00:30:54,060 this kind of an illustration, that it was feedforward only. 541 00:30:54,060 --> 00:30:58,260 I could decompose all of the signal flow paths 542 00:30:58,260 --> 00:31:01,240 into a finite number of forward-going paths. 543 00:31:01,240 --> 00:31:02,150 This is harder. 544 00:31:05,320 --> 00:31:06,610 This happened in DT, too. 545 00:31:06,610 --> 00:31:09,040 We had to figure out how the DT system was going to work 546 00:31:09,040 --> 00:31:10,390 when we had a feedback path. 547 00:31:10,390 --> 00:31:11,860 And it was more complicated, and we 548 00:31:11,860 --> 00:31:13,720 had to fall back on thinking about 549 00:31:13,720 --> 00:31:16,515 sample-by-sample propagation of the signal through the system. 550 00:31:16,515 --> 00:31:17,890 And we'll do the same thing here. 551 00:31:20,530 --> 00:31:23,430 In order to figure out how to think about signal flow 552 00:31:23,430 --> 00:31:29,220 through such a diagram, through a diagram that has feedback, 553 00:31:29,220 --> 00:31:32,912 we'll think about falling back to 1803. 554 00:31:32,912 --> 00:31:33,870 Something we can trust. 555 00:31:33,870 --> 00:31:36,750 Something we all loved. 556 00:31:36,750 --> 00:31:37,560 Nod your heads yes. 557 00:31:37,560 --> 00:31:39,720 Make me-- reassures me. 558 00:31:39,720 --> 00:31:42,340 Fond memories of 1803. 559 00:31:42,340 --> 00:31:44,910 So it's pretty simple to think about how 560 00:31:44,910 --> 00:31:50,020 you would solve that system by using an 1803 type of approach. 561 00:31:50,020 --> 00:31:52,040 Convert the system into a differential equation. 562 00:31:52,040 --> 00:31:53,206 We've already done that one. 563 00:31:56,050 --> 00:32:00,250 Then from the form, we will usually, in this class, 564 00:32:00,250 --> 00:32:04,490 use the general method of solving differential equations, 565 00:32:04,490 --> 00:32:06,451 which is-- 566 00:32:06,451 --> 00:32:11,050 What's the general method for solving differential equations? 567 00:32:11,050 --> 00:32:13,410 They may not have told you in 1803. 568 00:32:13,410 --> 00:32:17,110 What's the general method of solving differential equations? 569 00:32:17,110 --> 00:32:17,820 Guess! 570 00:32:17,820 --> 00:32:19,430 Yes. 571 00:32:19,430 --> 00:32:23,260 So the general method is guess. 572 00:32:23,260 --> 00:32:27,450 If you can guess, plug it in and it works, you're done. 573 00:32:27,450 --> 00:32:29,220 The kinds of systems that we will look at 574 00:32:29,220 --> 00:32:30,990 will be linear systems. 575 00:32:30,990 --> 00:32:33,060 They will have a single unique solution. 576 00:32:33,060 --> 00:32:35,610 If you can find it by guessing, you're done. 577 00:32:35,610 --> 00:32:37,620 And we will prove later in the course 578 00:32:37,620 --> 00:32:40,380 that all systems of this class, where I haven't really 579 00:32:40,380 --> 00:32:42,810 defined what this class is, but if you made a system out 580 00:32:42,810 --> 00:32:46,020 of adders, gains, and integrators, 581 00:32:46,020 --> 00:32:52,640 and only those parts, then that system will always be linear. 582 00:32:52,640 --> 00:32:55,970 And the solutions can always be written 583 00:32:55,970 --> 00:32:58,370 as complex exponentials. 584 00:32:58,370 --> 00:32:59,420 We'll prove that later. 585 00:32:59,420 --> 00:33:01,130 I'm not going to bother with it now. 586 00:33:01,130 --> 00:33:02,494 For now, you would say, OK. 587 00:33:02,494 --> 00:33:04,160 First-order linear differential equation 588 00:33:04,160 --> 00:33:05,990 with constant coefficiency, the answer 589 00:33:05,990 --> 00:33:09,140 is obviously an exponential function. 590 00:33:09,140 --> 00:33:11,242 So you plug that into the differential equation. 591 00:33:11,242 --> 00:33:12,700 You figure out the constraints that 592 00:33:12,700 --> 00:33:15,860 would have to be solved for this to be true. 593 00:33:15,860 --> 00:33:19,480 And the answer is that y should be e to the pt [? of ?] u of t. 594 00:33:19,480 --> 00:33:21,820 Should start at 0. 595 00:33:21,820 --> 00:33:24,670 And there should be an exponential after that. 596 00:33:24,670 --> 00:33:31,790 And the p shows up as the exponent in the exponential. 597 00:33:31,790 --> 00:33:34,880 What we'd like to do is develop an alternative way 598 00:33:34,880 --> 00:33:37,490 of thinking about that, using the operator approach. 599 00:33:41,030 --> 00:33:43,630 So a completely different way to solve that problem 600 00:33:43,630 --> 00:33:46,330 would be to think about, write an operator expression 601 00:33:46,330 --> 00:33:48,480 for this. 602 00:33:48,480 --> 00:33:51,260 OK, so Y is A times the signal that 603 00:33:51,260 --> 00:33:52,970 results when you add x to py. 604 00:33:59,100 --> 00:34:03,300 Solve for the ratio Y by X, and you get an operator expression, 605 00:34:03,300 --> 00:34:05,240 a over 1 minus pA. 606 00:34:08,900 --> 00:34:11,090 Just like in DT, we have to figure out 607 00:34:11,090 --> 00:34:14,070 how we would think about that. 608 00:34:14,070 --> 00:34:15,790 It's an implicit operation. 609 00:34:15,790 --> 00:34:18,300 It's telling me that if I knew the answer, 610 00:34:18,300 --> 00:34:21,330 the answer is the signal that, when operated on by 1 minus p 611 00:34:21,330 --> 00:34:24,544 is the integral of the input. 612 00:34:24,544 --> 00:34:25,710 But I don't know the answer. 613 00:34:25,710 --> 00:34:27,300 So that reasoning doesn't quite work. 614 00:34:27,300 --> 00:34:33,179 The reasoning that I used to solve the feedforward question 615 00:34:33,179 --> 00:34:34,366 doesn't quite work. 616 00:34:34,366 --> 00:34:36,449 So I have to think of a different way of doing it, 617 00:34:36,449 --> 00:34:39,350 just like we did in DT. 618 00:34:39,350 --> 00:34:41,489 And the same solution works. 619 00:34:41,489 --> 00:34:43,110 Not surprisingly. 620 00:34:43,110 --> 00:34:44,909 A behaves like a polynomial. 621 00:34:44,909 --> 00:34:46,409 R behaves like a polynomial. 622 00:34:46,409 --> 00:34:49,820 After you turn it into a polynomial, 623 00:34:49,820 --> 00:34:53,000 you can't tell if you started with a or r. 624 00:34:53,000 --> 00:34:55,190 It's a polynomial. 625 00:34:55,190 --> 00:34:58,501 So for that reason, exactly the same stuff that you did in DT 626 00:34:58,501 --> 00:34:59,000 works. 627 00:34:59,000 --> 00:35:02,680 That's why we can solve everything in 50 minutes. 628 00:35:02,680 --> 00:35:05,440 It's all the same. 629 00:35:05,440 --> 00:35:12,100 So in DT, we thought about this as being a series. 630 00:35:12,100 --> 00:35:15,220 You could figure out an ascending series, a power 631 00:35:15,220 --> 00:35:17,530 series, that's equivalent by thinking about something 632 00:35:17,530 --> 00:35:20,980 like synthetic division, Taylor series, whatever 633 00:35:20,980 --> 00:35:22,040 you're comfortable with. 634 00:35:22,040 --> 00:35:24,280 But whatever you're comfortable with. 635 00:35:24,280 --> 00:35:29,140 This, A over 1 minus pA, take the A out front. 636 00:35:29,140 --> 00:35:32,390 Then you're left with 1 over 1 minus pA. 637 00:35:32,390 --> 00:35:34,030 Think about that is Taylor series. 638 00:35:34,030 --> 00:35:36,160 That's 1 plus pA, plus p squared A squared, 639 00:35:36,160 --> 00:35:38,171 plus p cubed A cubed, etc. 640 00:35:38,171 --> 00:35:38,670 Done. 641 00:35:43,440 --> 00:35:46,560 Now I know what the system functional looks like. 642 00:35:46,560 --> 00:35:50,550 Now I've got a feedforward system. 643 00:35:50,550 --> 00:35:54,210 So I can use the technique from the other side. 644 00:35:54,210 --> 00:35:58,470 What happens if you put delta into a system whose functional 645 00:35:58,470 --> 00:36:00,888 representation is A? 646 00:36:05,750 --> 00:36:07,010 So you get a u. 647 00:36:07,010 --> 00:36:08,170 Let me skip that one. 648 00:36:08,170 --> 00:36:10,160 That's on the next slide. 649 00:36:10,160 --> 00:36:16,180 What happens if you put in-- 650 00:36:16,180 --> 00:36:17,042 no, that's right. 651 00:36:17,042 --> 00:36:17,750 I did that right. 652 00:36:17,750 --> 00:36:19,610 OK 1 u. 653 00:36:19,610 --> 00:36:22,120 Sorry, I'm confusing myself. 654 00:36:22,120 --> 00:36:23,710 This A turned into that u. 655 00:36:26,420 --> 00:36:28,700 What happens if you put it into pA squared? 656 00:36:32,260 --> 00:36:34,670 So the first A turned delta into u. 657 00:36:34,670 --> 00:36:40,900 The second A turns u into tu. 658 00:36:40,900 --> 00:36:42,460 And so I get this term. 659 00:36:42,460 --> 00:36:47,480 So the pA squared turns into ptu. 660 00:36:47,480 --> 00:36:52,890 The p squared A cubed turns into a half p squared t 661 00:36:52,890 --> 00:36:56,410 squared u, etc. 662 00:36:56,410 --> 00:36:58,420 So what I've done is I've thought 663 00:36:58,420 --> 00:37:02,710 of a way of constructing this output signal from the input 664 00:37:02,710 --> 00:37:04,725 signal without ever using calculus. 665 00:37:08,410 --> 00:37:10,240 I started with a solution based on 1803 666 00:37:10,240 --> 00:37:12,760 that's strictly calculus. 667 00:37:12,760 --> 00:37:16,870 I just redid the whole problem and didn't use any calculus. 668 00:37:16,870 --> 00:37:20,830 And the method works just the same reason 669 00:37:20,830 --> 00:37:23,071 that the our method worked. 670 00:37:23,071 --> 00:37:25,570 In the r method, when we were thinking 671 00:37:25,570 --> 00:37:28,750 about the simple feedback with an r system, 672 00:37:28,750 --> 00:37:33,370 every loop around the feedback loop generated one new sample. 673 00:37:33,370 --> 00:37:35,620 Here, every loop around the CT loop 674 00:37:35,620 --> 00:37:39,040 generates one more contribution to the output. 675 00:37:39,040 --> 00:37:42,530 You put a delta function in, it gets integrated once, 676 00:37:42,530 --> 00:37:45,950 and you get a u function out. 677 00:37:45,950 --> 00:37:48,550 [INAUDIBLE] [? first ?] in this symbolic representations. 678 00:37:48,550 --> 00:37:52,750 The first thing that comes out is A times the input. 679 00:37:52,750 --> 00:37:56,390 Second thing that comes out is A squared p times the input. 680 00:37:56,390 --> 00:38:00,740 Third thing that comes out, p squared a cubed, etc. 681 00:38:00,740 --> 00:38:02,660 Every time you go through the loop, 682 00:38:02,660 --> 00:38:04,790 you pick up one more turn. 683 00:38:04,790 --> 00:38:08,300 Now, think about it in a time domain. 684 00:38:08,300 --> 00:38:10,560 You put in a delta function. 685 00:38:10,560 --> 00:38:12,350 The first thing that comes out is a step. 686 00:38:17,070 --> 00:38:20,290 The second thing that comes out, and gets 687 00:38:20,290 --> 00:38:23,350 added to the first thing that comes out, 688 00:38:23,350 --> 00:38:27,010 is integrate, multiply by p, and integrate again. 689 00:38:29,620 --> 00:38:32,510 So that's this term, p times t. 690 00:38:32,510 --> 00:38:37,050 So we get the unit step from the first term, 691 00:38:37,050 --> 00:38:38,980 which looks like this. 692 00:38:38,980 --> 00:38:42,210 We get the linear increase from the second term. 693 00:38:42,210 --> 00:38:46,260 We get a squared term, we get a cubed' term. 694 00:38:46,260 --> 00:38:52,200 And voila, we get the series expansion for e to the pt. 695 00:38:52,200 --> 00:38:53,760 No calculus. 696 00:38:53,760 --> 00:38:55,470 It's purely thinking about how you 697 00:38:55,470 --> 00:39:01,440 would think about constructing this system by a series 698 00:39:01,440 --> 00:39:04,050 representation for spinning around the feedback loop. 699 00:39:07,430 --> 00:39:11,870 So that's an insight that comes from thinking about the system 700 00:39:11,870 --> 00:39:17,540 as a functional representation, as a polynomial. 701 00:39:17,540 --> 00:39:19,580 Polynomials can be expanded in Taylor series. 702 00:39:19,580 --> 00:39:23,920 Therefore, systems can be expanded in Taylor series. 703 00:39:23,920 --> 00:39:27,330 If you change the sign of p, you'll 704 00:39:27,330 --> 00:39:29,205 notice that that previous example exploded. 705 00:39:31,730 --> 00:39:33,824 So we got to an increasing exponential. 706 00:39:33,824 --> 00:39:35,990 If you change the sign of p, it's not too surprising 707 00:39:35,990 --> 00:39:39,670 that what will happen is that it will converge. 708 00:39:39,670 --> 00:39:43,040 So here, symbolically, the first signal 709 00:39:43,040 --> 00:39:45,980 is a unit-step just like before. 710 00:39:45,980 --> 00:39:47,980 But now because of the minus sign, 711 00:39:47,980 --> 00:39:51,195 there is a negative on this pt, so you get sloping downward. 712 00:39:54,680 --> 00:39:56,790 In the squared term, it's still positive. 713 00:39:56,790 --> 00:39:59,750 So that breaks it up. 714 00:39:59,750 --> 00:40:03,170 But the cubic is now downward again. 715 00:40:03,170 --> 00:40:05,480 And you add an infinite number of those 716 00:40:05,480 --> 00:40:10,750 and you get another exponential, this time a convergent one. 717 00:40:10,750 --> 00:40:12,520 So the point is that it works extremely 718 00:40:12,520 --> 00:40:15,880 like the way the DT stuff did, except now the basis 719 00:40:15,880 --> 00:40:17,260 functions are different. 720 00:40:17,260 --> 00:40:20,110 The basis functions are derived from this thing, which 721 00:40:20,110 --> 00:40:24,270 is the unit impulse function. 722 00:40:24,270 --> 00:40:26,610 And the things that come out of the integrator 723 00:40:26,610 --> 00:40:32,160 are steps and ramps and parabolas and stuff like that. 724 00:40:32,160 --> 00:40:39,760 So in r, the basic input signal was the unit sample signal. 725 00:40:39,760 --> 00:40:44,260 And what came out on each revolution around-- 726 00:40:44,260 --> 00:40:47,490 for each cycle in a feedback system, what came out 727 00:40:47,490 --> 00:40:49,410 was a successive delay. 728 00:40:49,410 --> 00:40:54,162 Here, we get successive integration. 729 00:40:54,162 --> 00:40:55,620 Delay was the fundamental operator. 730 00:40:55,620 --> 00:40:57,450 Integration is the fundamental operator. 731 00:40:57,450 --> 00:41:01,020 Successive delays, successive integrations. 732 00:41:01,020 --> 00:41:04,200 And we get the idea that we have convergence and divergence, 733 00:41:04,200 --> 00:41:05,490 just like we did in DT. 734 00:41:05,490 --> 00:41:08,442 Except now, the shapes of the regions are different. 735 00:41:11,540 --> 00:41:15,870 So if the p, which we will later call the pole, 736 00:41:15,870 --> 00:41:18,390 by complete analogy to what we did in DT, 737 00:41:18,390 --> 00:41:20,250 if the pole is in the right half plane, 738 00:41:20,250 --> 00:41:23,520 then the system's response is divergent. 739 00:41:23,520 --> 00:41:25,350 If the pole is in the left half plane, 740 00:41:25,350 --> 00:41:28,890 the system's responses is convergent. 741 00:41:28,890 --> 00:41:31,560 So it looks just like DT. 742 00:41:31,560 --> 00:41:33,780 We think about this as signal flow paths. 743 00:41:33,780 --> 00:41:35,940 We think about how feedback gives rise 744 00:41:35,940 --> 00:41:39,610 to an infinite number of those. 745 00:41:39,610 --> 00:41:44,580 In DT, each time through, it gave one unit of delay in CT. 746 00:41:44,580 --> 00:41:48,900 It gave one unit of integration. 747 00:41:48,900 --> 00:41:54,730 The method is the same, the answer is different. 748 00:41:54,730 --> 00:41:57,480 So what we got is tremendous similarities. 749 00:41:57,480 --> 00:41:59,740 Differential equation compared to difference equation, 750 00:41:59,740 --> 00:42:02,280 block diagram compared to block diagram. 751 00:42:02,280 --> 00:42:05,250 Integrators instead of delays. 752 00:42:05,250 --> 00:42:09,530 Functional functional, A instead of r. 753 00:42:09,530 --> 00:42:10,315 Pole, pole. 754 00:42:13,170 --> 00:42:14,460 Mode, mode. 755 00:42:14,460 --> 00:42:16,620 Fundamental mode. 756 00:42:16,620 --> 00:42:19,920 Here, the fundamental mode was a geometric sequence. 757 00:42:19,920 --> 00:42:22,830 Here, it's an exponential function of time. 758 00:42:22,830 --> 00:42:25,080 And we've got different kinds of regional convergence. 759 00:42:25,080 --> 00:42:29,410 Signal converges for poles in the left half plane. 760 00:42:29,410 --> 00:42:33,310 Signals converge for poles inside the unit circle. 761 00:42:33,310 --> 00:42:38,770 Same idea, but some differences. 762 00:42:38,770 --> 00:42:46,100 OK, so [INAUDIBLE] have time think 763 00:42:46,100 --> 00:42:50,030 about this for 30 seconds. 764 00:42:50,030 --> 00:42:52,280 We now have two representations, R polynomials 765 00:42:52,280 --> 00:42:55,000 and A polynomials. 766 00:42:55,000 --> 00:42:58,240 I'm giving you four examples to think through, 767 00:42:58,240 --> 00:43:00,580 and your job is to figure out which 768 00:43:00,580 --> 00:43:06,610 of those functionals correspond to convergent responses 769 00:43:06,610 --> 00:43:10,240 when excited by a unit sample or a unit impulse signal. 770 00:44:26,910 --> 00:44:28,280 So, how do I think about this? 771 00:44:28,280 --> 00:44:31,140 Is this convergent or divergent? 772 00:44:31,140 --> 00:44:33,380 What do I think about? 773 00:44:33,380 --> 00:44:35,770 Step 1. 774 00:44:35,770 --> 00:44:36,270 Convergent. 775 00:44:36,270 --> 00:44:37,870 How did you get that? 776 00:44:37,870 --> 00:44:38,370 Yeah. 777 00:44:38,370 --> 00:44:39,242 It's this one. 778 00:44:39,242 --> 00:44:41,552 AUDIENCE: [INAUDIBLE] 779 00:44:41,552 --> 00:44:43,050 DENNIS FREEMAN: So factor. 780 00:44:43,050 --> 00:44:45,450 It's polynomial. 781 00:44:45,450 --> 00:44:49,670 You get a pole at minus 1/2 and a pole at plus 1/2. 782 00:44:49,670 --> 00:44:51,720 They're both inside [? the unit ?] circle. 783 00:44:51,720 --> 00:44:54,310 They both converge. 784 00:44:54,310 --> 00:44:56,500 How about this one? 785 00:44:56,500 --> 00:45:00,000 Same equation, same answer. 786 00:45:00,000 --> 00:45:02,450 Kind of. 787 00:45:02,450 --> 00:45:07,030 Same poles, minus 1/2 and plus 1/2. 788 00:45:07,030 --> 00:45:09,180 Convergent or divergent? 789 00:45:09,180 --> 00:45:12,700 Divergent, because the regions are different. 790 00:45:12,700 --> 00:45:15,570 One of the poles is in the right half plane. 791 00:45:15,570 --> 00:45:16,830 Divergent. 792 00:45:16,830 --> 00:45:19,660 How about this one. 793 00:45:19,660 --> 00:45:22,430 Where were the poles? 794 00:45:22,430 --> 00:45:24,260 Minus 1/2, minus 3/2. 795 00:45:27,240 --> 00:45:29,590 Outside unit circle? 796 00:45:29,590 --> 00:45:31,530 One of the poles is outside the unit circle. 797 00:45:31,530 --> 00:45:34,660 Diverges, because of that pole. 798 00:45:34,660 --> 00:45:38,590 Minus 1/2 and minus 3/2, same poles. 799 00:45:38,590 --> 00:45:41,070 Both in the left half lane. 800 00:45:41,070 --> 00:45:41,850 Convergence. 801 00:45:41,850 --> 00:45:43,470 That's the point. 802 00:45:43,470 --> 00:45:44,920 So they worked very similar. 803 00:45:44,920 --> 00:45:46,980 But there are differences. 804 00:45:46,980 --> 00:45:49,940 And now in the last three minutes, 805 00:45:49,940 --> 00:45:51,990 the last thing I want to tell you about 806 00:45:51,990 --> 00:45:54,820 is that even complex numbers work just the same as in DT. 807 00:45:57,330 --> 00:46:00,690 So think about a system that's slightly harder. 808 00:46:00,690 --> 00:46:04,310 Here, I can represent this by this kind of relationship. 809 00:46:04,310 --> 00:46:10,100 So F equals Kx, but F is MA. 810 00:46:10,100 --> 00:46:12,660 So 801. 811 00:46:12,660 --> 00:46:15,330 And so I generate a block diagram or a differential 812 00:46:15,330 --> 00:46:18,090 equation or however I want to think about it. 813 00:46:18,090 --> 00:46:19,470 And I can solve for the answer. 814 00:46:19,470 --> 00:46:21,011 What I'd like to do is think about it 815 00:46:21,011 --> 00:46:23,800 in terms of the functional representation. 816 00:46:23,800 --> 00:46:25,467 So I just think about, take this system 817 00:46:25,467 --> 00:46:26,550 that you're familiar with. 818 00:46:26,550 --> 00:46:28,350 We did this on the first lecture. 819 00:46:28,350 --> 00:46:31,460 Turn it into a differential equation. 820 00:46:31,460 --> 00:46:33,280 Turn that into a functional representation. 821 00:46:33,280 --> 00:46:35,821 Now, I want to think about how you would solve the functional 822 00:46:35,821 --> 00:46:38,180 representation. 823 00:46:38,180 --> 00:46:40,840 The idea is just like before, factor. 824 00:46:40,840 --> 00:46:46,655 But now the trick is that the factors become complex. 825 00:46:49,340 --> 00:46:51,580 So one way you can think about this 826 00:46:51,580 --> 00:46:55,450 is to force the second order system into a canonical form. 827 00:46:55,450 --> 00:47:03,880 The canonical form for DT was 1 over 1 minus p not R. 828 00:47:03,880 --> 00:47:08,650 For CT, it's A over 1 minus p not A. 829 00:47:08,650 --> 00:47:10,990 So I want to make factors look like that, because I 830 00:47:10,990 --> 00:47:15,190 know that here, the response looks like p not to the n. 831 00:47:15,190 --> 00:47:18,520 Here, the response looks like e to the p, not t. 832 00:47:18,520 --> 00:47:19,990 If I can coerce it into that form, 833 00:47:19,990 --> 00:47:22,294 I already know the answer. 834 00:47:22,294 --> 00:47:23,710 So that's what's illustrated here. 835 00:47:23,710 --> 00:47:27,640 I coerced this into that form. 836 00:47:27,640 --> 00:47:32,440 And then I'll know what the answer looks like. 837 00:47:32,440 --> 00:47:35,930 Just like we could in DT, substitute R goes to 1 over z. 838 00:47:35,930 --> 00:47:40,390 In CT, we can substitute A to 1 over s, and solve for s. 839 00:47:40,390 --> 00:47:41,410 We get the same answer. 840 00:47:41,410 --> 00:47:45,780 Same as we did in DT, except now we call it s. 841 00:47:45,780 --> 00:47:50,040 So the poles to this system are plus or minus j constant. 842 00:47:50,040 --> 00:47:53,940 For convenience, I'll call the constant omega not. 843 00:47:53,940 --> 00:47:58,730 So then I have a pole at e to the j omega not. 844 00:47:58,730 --> 00:48:01,230 And another one-- so I have a pole at j omega 0 845 00:48:01,230 --> 00:48:03,820 and a second pole at minus j omega 0. 846 00:48:03,820 --> 00:48:07,310 By that argument, the fundamental modes 847 00:48:07,310 --> 00:48:13,400 are e the j omega not t and e to the minus j omega not t. 848 00:48:13,400 --> 00:48:19,130 Just like in DT, the complex poles gave complex modes. 849 00:48:19,130 --> 00:48:22,400 Here, the complex pole, j omega not, 850 00:48:22,400 --> 00:48:25,280 gave a complex mode, cos omega t plus j sine omega t. 851 00:48:29,240 --> 00:48:34,210 And just like DT, the system conspires. 852 00:48:34,210 --> 00:48:37,210 So that started out being mass and spring system. 853 00:48:37,210 --> 00:48:40,940 Obviously, it's not going to have an imaginary output. 854 00:48:40,940 --> 00:48:44,260 The system conspires so that the imaginary parts 855 00:48:44,260 --> 00:48:47,990 of the different fundamental modes kill each other off. 856 00:48:47,990 --> 00:48:51,510 And the answer is a real number. 857 00:48:51,510 --> 00:48:54,670 So even though the mode, even though the pole is complex, 858 00:48:54,670 --> 00:48:59,266 the multipliers are complex, this sum is real. 859 00:48:59,266 --> 00:49:01,640 And if you just think about what that sum is, by thinking 860 00:49:01,640 --> 00:49:04,422 about how complex numbers work, you 861 00:49:04,422 --> 00:49:06,880 get an expression that looks like omega not sine, omega not 862 00:49:06,880 --> 00:49:08,420 t. 863 00:49:08,420 --> 00:49:10,430 It's a little more fun to think about how 864 00:49:10,430 --> 00:49:12,770 that evolves as a series. 865 00:49:12,770 --> 00:49:14,720 If we do a Taylor series for this, 866 00:49:14,720 --> 00:49:16,010 we can represent it that way. 867 00:49:18,530 --> 00:49:21,284 And the series, then, the first term is t. 868 00:49:23,950 --> 00:49:28,690 The second term is a t cubed, which goes down. 869 00:49:28,690 --> 00:49:35,540 Then there's a fifth and a seventh and a ninth. 870 00:49:35,540 --> 00:49:38,860 And if you keep adding up the terms, amazingly enough, 871 00:49:38,860 --> 00:49:43,370 the series representation for the operator expression 872 00:49:43,370 --> 00:49:47,880 is the Taylor series expansion of the sine wave. 873 00:49:47,880 --> 00:49:50,070 Just like we would have expected. 874 00:49:50,070 --> 00:49:54,900 Again, here, the response looks like e to the p not t. 875 00:49:54,900 --> 00:49:56,370 If I can coerce it into that form, 876 00:49:56,370 --> 00:49:58,684 I already know the answer. 877 00:49:58,684 --> 00:50:00,100 So that's what's illustrated here. 878 00:50:00,100 --> 00:50:06,170 I coerce this into that form, and then I'll 879 00:50:06,170 --> 00:50:08,830 know what the answer looks like. 880 00:50:08,830 --> 00:50:12,330 Just like we could in DT, substitute R goes to 1 over z. 881 00:50:12,330 --> 00:50:16,780 In CT, we can substitute A goes to 1 over s, and solve for s. 882 00:50:16,780 --> 00:50:17,800 We get the same answer. 883 00:50:17,800 --> 00:50:22,170 Same as we did in DT, except now we call it s. 884 00:50:22,170 --> 00:50:26,430 So the poles to this system are plus or minus j constant. 885 00:50:26,430 --> 00:50:30,330 For convenience, I'll call the constant omega not. 886 00:50:30,330 --> 00:50:32,955 So then I have a pole at e to the j omega 887 00:50:32,955 --> 00:50:37,500 not, and another one-- so I have a pole at j omega not 888 00:50:37,500 --> 00:50:40,180 and a second poet minus j omega not. 889 00:50:40,180 --> 00:50:43,700 By that argument, the fundamental modes 890 00:50:43,700 --> 00:50:49,790 are e to the j omega not t and e to the minus j omega not t. 891 00:50:49,790 --> 00:50:55,580 Just like in DT, the complex poles gave complex modes. 892 00:50:55,580 --> 00:50:58,770 Here, the complex pole, j omega not, 893 00:50:58,770 --> 00:51:02,020 gave a complex mode, cos omega t plus j sine omega t. 894 00:51:05,640 --> 00:51:10,600 And just like DT, the system conspires. 895 00:51:10,600 --> 00:51:13,600 So that started out being a [? mass ?] [INAUDIBLE] system. 896 00:51:13,600 --> 00:51:17,330 Obviously, it's not going to have an imaginary output. 897 00:51:17,330 --> 00:51:20,650 The system conspires so that the imaginary parts 898 00:51:20,650 --> 00:51:24,380 of the different fundamental modes kill each other off, 899 00:51:24,380 --> 00:51:27,900 and the answer is a real number. 900 00:51:27,900 --> 00:51:31,320 So even though the pole is complex, 901 00:51:31,320 --> 00:51:35,637 the multipliers are complex, this sum is real. 902 00:51:35,637 --> 00:51:37,470 And if you just think about what that sum is 903 00:51:37,470 --> 00:51:40,625 by thinking about how complex numbers work, 904 00:51:40,625 --> 00:51:43,250 you get an expression that looks like omega not sine, omega not 905 00:51:43,250 --> 00:51:44,840 t. 906 00:51:44,840 --> 00:51:46,820 It's a little more fun to think about how 907 00:51:46,820 --> 00:51:49,160 that evolves as a series. 908 00:51:49,160 --> 00:51:51,110 If we do a Taylor series for this, 909 00:51:51,110 --> 00:51:52,400 we can represent it that way. 910 00:51:54,940 --> 00:51:57,672 And the series, then, the first term is t. 911 00:52:00,330 --> 00:52:05,060 The second term is a t cubed, which goes down. 912 00:52:05,060 --> 00:52:11,930 Then there's a fifth and a seventh and a ninth. 913 00:52:11,930 --> 00:52:15,260 And if you keep adding up the terms, amazingly enough, 914 00:52:15,260 --> 00:52:19,760 the series representation for the operator expression 915 00:52:19,760 --> 00:52:24,250 is the Taylor series expansion of the sine wave. 916 00:52:24,250 --> 00:52:26,130 Just like we would have expected. 917 00:52:26,130 --> 00:52:28,420 Again, I solved the differential equation, 918 00:52:28,420 --> 00:52:30,580 a second order differential equation, this time 919 00:52:30,580 --> 00:52:32,590 without calculus. 920 00:52:32,590 --> 00:52:35,290 All I did was polynomial math. 921 00:52:35,290 --> 00:52:38,410 And so with that, I'll finish just by saying, 922 00:52:38,410 --> 00:52:40,990 today, what we did was introduce, and basically 923 00:52:40,990 --> 00:52:43,720 finish, CT. 924 00:52:43,720 --> 00:52:46,780 Because there's such a strong analogy between what 925 00:52:46,780 --> 00:52:51,810 we did in DT, and how we'll approach thinking about CT.