WEBVTT 00:00:00.090 --> 00:00:02.490 The following content is provided under a Creative 00:00:02.490 --> 00:00:04.030 Commons license. 00:00:04.030 --> 00:00:06.330 Your support will help MIT OpenCourseWare 00:00:06.330 --> 00:00:10.720 continue to offer high quality educational resources for free. 00:00:10.720 --> 00:00:13.320 To make a donation or view additional materials 00:00:13.320 --> 00:00:17.280 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.280 --> 00:00:18.450 at ocw.mit.edu. 00:00:21.371 --> 00:00:23.570 DENNIS FREEMAN: OK. 00:00:23.570 --> 00:00:28.730 So last time, the idea was to think through a different way, 00:00:28.730 --> 00:00:32.030 or in fact, several different ways, 00:00:32.030 --> 00:00:34.460 to think about discrete-time systems. 00:00:34.460 --> 00:00:40.820 Today is a crash course to do the same thing in CT systems. 00:00:40.820 --> 00:00:42.860 So last time, for the last week, we've 00:00:42.860 --> 00:00:45.440 been looking at different kinds of representations 00:00:45.440 --> 00:00:48.170 for DT systems. 00:00:48.170 --> 00:00:52.640 Difference equations, because they're concise and precise. 00:00:52.640 --> 00:00:56.090 Block diagrams, because they let us visualize the signal flow 00:00:56.090 --> 00:00:57.426 paths. 00:00:57.426 --> 00:00:59.300 And operator expressions, because it lets you 00:00:59.300 --> 00:01:01.940 treat systems like polynomials. 00:01:01.940 --> 00:01:05.450 What we'll see today is that that same kind of strategy 00:01:05.450 --> 00:01:06.980 works precisely. 00:01:06.980 --> 00:01:09.740 The strategy, not the answers. 00:01:09.740 --> 00:01:13.730 The same strategy works precisely for CT. 00:01:13.730 --> 00:01:16.490 We'll have a concise, precise representation 00:01:16.490 --> 00:01:18.740 in terms of differential equations, which I'm sure you 00:01:18.740 --> 00:01:22.370 already know all about. 00:01:22.370 --> 00:01:24.890 But we will also develop the notion of a block diagram 00:01:24.890 --> 00:01:28.220 so you can visualize signal flow paths. 00:01:28.220 --> 00:01:31.340 And we'll have an analogous operator, the A operator, 00:01:31.340 --> 00:01:34.620 that will let us treat the systems as polynomials. 00:01:34.620 --> 00:01:36.710 So that's the overview. 00:01:36.710 --> 00:01:39.170 Because it's so similar, we'll be 00:01:39.170 --> 00:01:41.510 able to do all of this in one lecture. 00:01:41.510 --> 00:01:44.990 So that's what I mean by a crash course. 00:01:44.990 --> 00:01:50.660 So this is Introduction to CT in 50 minutes. 00:01:50.660 --> 00:01:53.690 So you already know a lot about how you'd represent systems 00:01:53.690 --> 00:01:55.130 as differential equations. 00:01:55.130 --> 00:01:58.370 You've seen this in other classes. 00:01:58.370 --> 00:02:00.710 You should have seen this sort of approach in physics. 00:02:00.710 --> 00:02:05.012 You should have seen this sort of approach in math, in 1803. 00:02:05.012 --> 00:02:06.470 So we're to assume that you already 00:02:06.470 --> 00:02:09.334 know how to think about differential equations. 00:02:09.334 --> 00:02:11.000 And what we're going to do is skip ahead 00:02:11.000 --> 00:02:15.650 and think about the alternative representations instead. 00:02:15.650 --> 00:02:19.160 So just like in DT, where we thought 00:02:19.160 --> 00:02:21.400 about block diagrams, which gave us 00:02:21.400 --> 00:02:23.360 a way of thinking about signal flow paths, 00:02:23.360 --> 00:02:25.305 we'll have the same sort of idea in CT. 00:02:27.920 --> 00:02:30.530 The big difference, you can anticipate. 00:02:30.530 --> 00:02:33.620 I mean, if the CT [INAUDIBLE]. 00:02:33.620 --> 00:02:38.180 In the difference equation, the fundamental operation 00:02:38.180 --> 00:02:42.590 in the difference equation was a delay operation. 00:02:42.590 --> 00:02:44.380 If you contrast that to a CT system, 00:02:44.380 --> 00:02:47.960 a fundamental operation is not delay. 00:02:47.960 --> 00:02:49.760 You know from 1803, you know for physics, 00:02:49.760 --> 00:02:52.040 you know from lots of other exposures, 00:02:52.040 --> 00:02:56.130 that the fundamental operation is differentiation. 00:02:56.130 --> 00:02:59.750 And so the blocks will not be built out of delays, 00:02:59.750 --> 00:03:04.160 but will instead be built out of integrators. 00:03:04.160 --> 00:03:06.050 Apart from, that the block diagram structure 00:03:06.050 --> 00:03:09.480 looks extremely similar. 00:03:09.480 --> 00:03:11.990 And we'll be able to apply the same idea that we 00:03:11.990 --> 00:03:17.090 did in DT, to simplify the representation of the block 00:03:17.090 --> 00:03:20.820 diagram by thinking in terms of operators. 00:03:20.820 --> 00:03:23.180 The operator has to similarly change. 00:03:23.180 --> 00:03:24.620 So the new operator, the operator 00:03:24.620 --> 00:03:26.830 that we use to represent a CT system, 00:03:26.830 --> 00:03:33.260 we'll call the A operator, A is intended to mean accumulator. 00:03:33.260 --> 00:03:37.130 The thing that's in my mind when I say the A operator 00:03:37.130 --> 00:03:40.200 is this tank. 00:03:40.200 --> 00:03:42.680 This tank accumulates water. 00:03:42.680 --> 00:03:44.510 And you can think about the relationship 00:03:44.510 --> 00:03:47.870 between the height of the water, and the input and output rates 00:03:47.870 --> 00:03:50.630 by some differential equation. 00:03:50.630 --> 00:03:55.070 But the point is that the tank itself, the system, 00:03:55.070 --> 00:03:56.330 accumulates. 00:03:56.330 --> 00:03:58.370 And that's what we want to have in our heads 00:03:58.370 --> 00:04:01.430 when we think about the A operator. 00:04:01.430 --> 00:04:05.650 So the operator is going to be like the r operator. 00:04:05.650 --> 00:04:09.830 You will apply the r operator to a discrete time signal 00:04:09.830 --> 00:04:12.320 to generate a new discrete time signal 00:04:12.320 --> 00:04:13.970 that we shifted to the right. 00:04:13.970 --> 00:04:17.510 That's what r meant, r was right. 00:04:17.510 --> 00:04:19.800 So right shift operator. 00:04:19.800 --> 00:04:23.190 Here, you apply the A operator to the X signal. 00:04:23.190 --> 00:04:27.560 X is now CT, continuous time. 00:04:27.560 --> 00:04:31.520 And when you apply the A operator to a signal X, 00:04:31.520 --> 00:04:33.950 it generates a whole new signal Y 00:04:33.950 --> 00:04:37.550 that is everywhere equal to the indefinite integral 00:04:37.550 --> 00:04:39.050 of the input signal. 00:04:39.050 --> 00:04:44.016 So the A operator is, start with the input signal X. 00:04:44.016 --> 00:04:46.640 And integrate it with regard to time starting at minus infinity 00:04:46.640 --> 00:04:52.220 and going up to t, that's the t value of the output signal. 00:04:54.730 --> 00:04:57.470 Simple. 00:04:57.470 --> 00:05:01.000 So let's see how simple. 00:05:01.000 --> 00:05:05.350 So here's a bunch of block diagrams written in terms of A. 00:05:05.350 --> 00:05:08.110 Here's a bunch of differential equations. 00:05:08.110 --> 00:05:12.400 Figure out the correspondence, if any. 00:05:12.400 --> 00:05:15.920 OK, traditionally, it seems that people are quiet. 00:05:15.920 --> 00:05:19.430 So before you start, say something that has nothing 00:05:19.430 --> 00:05:21.350 to do with 003 to you next-door neighbor. 00:05:21.350 --> 00:05:24.710 It has to have nothing to do with 003. 00:05:24.710 --> 00:05:28.100 And then figure out which of these bottom diagrams, number 00:05:28.100 --> 00:05:30.335 1, 2, 3, 4, or none of them, corresponds 00:05:30.335 --> 00:05:31.460 best to the correspondence. 00:05:31.460 --> 00:05:36.314 [SIDE CONVERSATION] 00:06:26.940 --> 00:06:31.098 OK, you've got 30 seconds to answer the question, now. 00:06:31.098 --> 00:06:34.570 [SIDE CONVERSATION] 00:06:41.520 --> 00:06:43.166 Well, that killed everything. 00:07:44.460 --> 00:07:47.090 OK, everybody raise your hand and tell me 00:07:47.090 --> 00:07:50.390 which of the five available answers, 1 through 5, 00:07:50.390 --> 00:07:52.562 best illustrates the correspondence. 00:07:52.562 --> 00:07:53.270 Raise your hands. 00:07:53.270 --> 00:07:56.460 Let me see how many of you figured it out. 00:07:56.460 --> 00:08:00.640 Remember, if you're wrong, you can blame it on your partner. 00:08:00.640 --> 00:08:03.107 OK, it's more than 95% correct. 00:08:03.107 --> 00:08:04.190 How do you think about it? 00:08:04.190 --> 00:08:04.981 What do I do first? 00:08:06.732 --> 00:08:09.510 AUDIENCE: [INAUDIBLE] 00:08:09.510 --> 00:08:11.120 DENNIS FREEMAN: [INAUDIBLE] So what do 00:08:11.120 --> 00:08:12.564 you want me to look at first? 00:08:12.564 --> 00:08:13.730 This diagram, that equation? 00:08:13.730 --> 00:08:14.896 This diagram, that equation? 00:08:14.896 --> 00:08:16.676 AUDIENCE: [INAUDIBLE] 00:08:16.676 --> 00:08:18.050 DENNIS FREEMAN: Write an equation 00:08:18.050 --> 00:08:19.091 for each of the diagrams. 00:08:19.091 --> 00:08:22.016 How would I figure out an equation for that diagram? 00:08:22.016 --> 00:08:22.516 [INAUDIBLE] 00:08:22.516 --> 00:08:26.019 AUDIENCE: [INAUDIBLE] 00:08:26.019 --> 00:08:28.560 DENNIS FREEMAN: So what's coming in and out of the plus sign. 00:08:28.560 --> 00:08:29.390 That's very good . 00:08:29.390 --> 00:08:31.520 So let's focus on [? that ?] coming out first. 00:08:31.520 --> 00:08:31.640 Us. 00:08:31.640 --> 00:08:32.931 What's the name of that signal? 00:08:35.797 --> 00:08:37.231 AUDIENCE: [INAUDIBLE] 00:08:37.231 --> 00:08:39.612 DENNIS FREEMAN: Well, I can think of two names. 00:08:39.612 --> 00:08:41.070 That's why it's a good idea to look 00:08:41.070 --> 00:08:44.310 at what's coming in and out. 00:08:44.310 --> 00:08:46.950 One way that I could name this would be with reference 00:08:46.950 --> 00:08:49.232 to the A operator. 00:08:49.232 --> 00:08:51.190 If I know that what comes out of the A operator 00:08:51.190 --> 00:08:54.660 is capital Y, what goes into the A operator? 00:08:54.660 --> 00:08:56.640 AUDIENCE: [INAUDIBLE]. 00:08:56.640 --> 00:08:58.860 DENNIS FREEMAN: The derivative. 00:08:58.860 --> 00:09:01.850 So if I know that what comes out is Y, then what goes in 00:09:01.850 --> 00:09:02.740 must have been Y dot. 00:09:06.200 --> 00:09:08.540 But that [? also ?] has to be the output of the adder. 00:09:08.540 --> 00:09:10.682 So what were the two inputs to the adder? 00:09:10.682 --> 00:09:12.155 AUDIENCE: [INAUDIBLE] 00:09:12.155 --> 00:09:13.265 DENNIS FREEMAN: X and-- 00:09:18.070 --> 00:09:19.780 py. 00:09:19.780 --> 00:09:24.160 So y dot, the thing that goes into the accumulator, 00:09:24.160 --> 00:09:26.860 has to be the same as X plus py. 00:09:26.860 --> 00:09:29.230 So y dot is X plus py, so there's a correspondence 00:09:29.230 --> 00:09:29.980 between these two. 00:09:32.500 --> 00:09:35.240 Everybody sort of get the idea? 00:09:35.240 --> 00:09:36.970 We can do the same sort of thing here. 00:09:36.970 --> 00:09:38.261 What's the name of that signal? 00:09:42.370 --> 00:09:43.330 AUDIENCE: [INAUDIBLE] 00:09:43.330 --> 00:09:44.770 DENNIS FREEMAN: [? Shout. ?] 00:09:44.770 --> 00:09:46.910 AUDIENCE: Y dot over p. 00:09:46.910 --> 00:09:48.140 DENNIS FREEMAN: y dot over p. 00:09:48.140 --> 00:09:48.639 Exactly. 00:09:48.639 --> 00:09:51.410 So that one is going to be y dot over p. 00:09:51.410 --> 00:09:56.030 And that's going to have to be the same as x plus y. 00:09:56.030 --> 00:09:58.500 So if you clear the fraction here, 00:09:58.500 --> 00:10:00.507 you would find that y dot is px plus py. 00:10:00.507 --> 00:10:02.090 So there is a correspondence that way. 00:10:06.520 --> 00:10:10.107 And you can do the same sort of thing down here. 00:10:10.107 --> 00:10:11.440 It looks a little bit different. 00:10:11.440 --> 00:10:11.940 What's this? 00:10:11.940 --> 00:10:16.930 This says that the output of the adder, y, the output 00:10:16.930 --> 00:10:19.240 of the adder here is y. 00:10:19.240 --> 00:10:21.640 So y must have been x plus what? 00:10:24.610 --> 00:10:27.580 AUDIENCE: [INAUDIBLE] 00:10:27.580 --> 00:10:33.710 DENNIS FREEMAN: So p times the integral of y. 00:10:33.710 --> 00:10:36.300 And I'd like to write that in a differential form, 00:10:36.300 --> 00:10:38.900 so I could differentiate term by term 00:10:38.900 --> 00:10:42.560 to say that that's the same thing as y dot is 00:10:42.560 --> 00:10:46.040 x dot plus py. 00:10:46.040 --> 00:10:48.190 So y dot is x dot plus py, so there's 00:10:48.190 --> 00:10:49.190 correspondence that way. 00:10:49.190 --> 00:10:50.491 So the answer is 1. 00:10:50.491 --> 00:10:50.990 OK? 00:10:50.990 --> 00:10:53.180 Everybody's happy with them that? 00:10:53.180 --> 00:10:57.220 So the point is that the differential operators 00:10:57.220 --> 00:10:58.427 look very much the same. 00:10:58.427 --> 00:11:00.010 You can use the same kind of reasoning 00:11:00.010 --> 00:11:03.460 that you did for difference equations with r. 00:11:03.460 --> 00:11:06.490 Of course, the operators are different. 00:11:06.490 --> 00:11:11.050 And just like r, you can think about these things 00:11:11.050 --> 00:11:13.070 as polynomials. 00:11:13.070 --> 00:11:16.780 So if you think about this feedforward system that's 00:11:16.780 --> 00:11:19.840 now CT, not DT, the feedforward system 00:11:19.840 --> 00:11:25.870 says that I can construct W by taking X plus A X, X plus A X, 00:11:25.870 --> 00:11:31.310 where that's X plus the the integral of X. 00:11:31.310 --> 00:11:33.310 Then I could think about the Y signal 00:11:33.310 --> 00:11:38.350 as W plus the integral of W. So that's this one, 00:11:38.350 --> 00:11:40.420 y is the w plus the integral w. 00:11:40.420 --> 00:11:43.945 And now I can substitute instances of this into here. 00:11:46.820 --> 00:11:48.690 So this w expression could go in there. 00:11:48.690 --> 00:11:50.990 And that gives me this. 00:11:50.990 --> 00:11:53.529 If I integrated this w expression once, 00:11:53.529 --> 00:11:55.070 then I would get the integral of that 00:11:55.070 --> 00:11:58.940 plus the double integral of that, so that gives me these. 00:11:58.940 --> 00:12:02.540 And I think about those things in terms of A's, and I get 00:12:02.540 --> 00:12:06.260 precisely the same expression. 00:12:06.260 --> 00:12:09.010 So the idea is, just like r, you can 00:12:09.010 --> 00:12:14.050 manipulate operator expressions in r, 00:12:14.050 --> 00:12:15.670 like they were polynomials. 00:12:15.670 --> 00:12:21.750 You can manipulate operator expressions in A, 00:12:21.750 --> 00:12:24.720 just like they were polynomials. 00:12:24.720 --> 00:12:28.260 And the reason I can say that is that, if I think through all 00:12:28.260 --> 00:12:30.540 the properties of polynomials, I can draw 00:12:30.540 --> 00:12:34.080 an isomorphism between the way the system would have behaved 00:12:34.080 --> 00:12:36.310 and the way the polynomial would have behaved. 00:12:36.310 --> 00:12:39.210 And in particular, here's three statements 00:12:39.210 --> 00:12:42.230 that have to be true. 00:12:42.230 --> 00:12:43.940 If the operator expressions were supposed 00:12:43.940 --> 00:12:46.610 to behave like polynomials, polynomials 00:12:46.610 --> 00:12:51.470 have the property that the polynomials in A 00:12:51.470 --> 00:12:53.600 should commute. 00:12:53.600 --> 00:12:55.567 Polynomials should have the-- the A 00:12:55.567 --> 00:12:57.150 should have the distributive property, 00:12:57.150 --> 00:13:00.560 multiplication over addition. 00:13:00.560 --> 00:13:03.530 And polynomials should associate. 00:13:03.530 --> 00:13:07.000 And if you think about each of those in detail, 00:13:07.000 --> 00:13:08.800 the operations that would correspond 00:13:08.800 --> 00:13:10.650 to what you've put in the block diagram 00:13:10.650 --> 00:13:14.740 have precisely those same relations. 00:13:14.740 --> 00:13:17.751 So that's the outline of how you go about proving something 00:13:17.751 --> 00:13:18.250 like this. 00:13:18.250 --> 00:13:21.220 But the takeaway message is the operator expressions 00:13:21.220 --> 00:13:28.740 in A obey all the rules that you would expect from a polynomial. 00:13:28.740 --> 00:13:32.720 So, second question. 00:13:32.720 --> 00:13:34.640 Think about these two systems, and think 00:13:34.640 --> 00:13:35.960 about the notion of equivalent. 00:13:35.960 --> 00:13:41.150 Equivalent here is going to be just the same as it was in DT. 00:13:41.150 --> 00:13:45.590 In DT, equivalent [? man, ?] if all of the right shift 00:13:45.590 --> 00:13:49.970 operators started at rest, that is to say their output was 0, 00:13:49.970 --> 00:13:53.660 we will have the same kind of notion here. 00:13:53.660 --> 00:13:57.050 At rest is going to mean all of the integrators-- 00:13:57.050 --> 00:13:59.750 integrators have a starting value, an initial value. 00:13:59.750 --> 00:14:01.880 If those initial values are all, 0 then 00:14:01.880 --> 00:14:03.890 we say that the system started at rest. 00:14:03.890 --> 00:14:05.570 We have that same kind of proviso. 00:14:05.570 --> 00:14:08.960 Given that proviso, then all of the operator 00:14:08.960 --> 00:14:11.160 expressions behave like polynomials. 00:14:11.160 --> 00:14:15.350 So determine k1 so that these two systems are equivalent, 00:14:15.350 --> 00:14:19.500 given that definition of equivalence. 00:14:19.500 --> 00:17:02.000 [SIDE CONVERSATION] 00:17:02.000 --> 00:17:04.490 So anyone have an answer? 00:17:04.490 --> 00:17:06.410 So how should I choose k1 if I want those two 00:17:06.410 --> 00:17:07.940 systems to be equivalent? 00:17:07.940 --> 00:17:10.460 Yes? 00:17:10.460 --> 00:17:11.250 Raise your hands. 00:17:15.714 --> 00:17:18.711 OK, it was a trick question. 00:17:18.711 --> 00:17:21.210 So now, given the additional information that it was a trick 00:17:21.210 --> 00:17:24.839 question, revise your answers. 00:17:24.839 --> 00:17:26.369 Ah excellent, excellent. 00:17:26.369 --> 00:17:27.329 Instant revisions. 00:17:27.329 --> 00:17:28.230 Wonderful. 00:17:28.230 --> 00:17:31.300 OK, what was the trick? 00:17:31.300 --> 00:17:33.215 [INAUDIBLE] got it. 00:17:33.215 --> 00:17:35.494 What was a trick? 00:17:35.494 --> 00:17:35.994 Yeah. 00:17:35.994 --> 00:17:37.892 AUDIENCE: It would be minus 1.6. 00:17:37.892 --> 00:17:40.070 DENNIS FREEMAN: It should have been minus 1.6. 00:17:40.070 --> 00:17:41.210 Yes. 00:17:41.210 --> 00:17:44.062 So what do you do? 00:17:44.062 --> 00:17:46.520 What do I do first in order to answer a question like this? 00:17:50.920 --> 00:17:52.350 [INAUDIBLE] 00:17:52.350 --> 00:17:54.352 [? Right, ?] so translate the block diagrams 00:17:54.352 --> 00:17:55.560 into a differential equation. 00:17:55.560 --> 00:17:57.030 That's a good approach. 00:17:57.030 --> 00:17:59.880 Because what we'd like to do is figure out an equivalence. 00:17:59.880 --> 00:18:03.780 The equivalence is not trivial. 00:18:03.780 --> 00:18:05.390 So in fact, the equivalence is easier 00:18:05.390 --> 00:18:08.240 to think about if you do polynomials. 00:18:08.240 --> 00:18:10.490 You could try to manipulate the blocks 00:18:10.490 --> 00:18:12.380 to make them look the same. 00:18:12.380 --> 00:18:14.630 That's actually hard. 00:18:14.630 --> 00:18:16.170 It it's pretty easy to do it if you 00:18:16.170 --> 00:18:25.410 think about the block diagrams being represented as operators. 00:18:25.410 --> 00:18:28.290 So you can look at the first one and you can say, OK. 00:18:28.290 --> 00:18:34.815 W is the signal that comes out of A operating on an X minus 00:18:34.815 --> 00:18:36.670 0.7W. 00:18:36.670 --> 00:18:38.430 So that's what that's adder says. 00:18:42.030 --> 00:18:44.020 And you can similarly do this box. 00:18:44.020 --> 00:18:45.150 It looks the same. 00:18:45.150 --> 00:18:47.490 Treat them as polynomials and you can reduce it 00:18:47.490 --> 00:18:48.630 to some equivalent form. 00:18:51.610 --> 00:18:54.490 Does everybody have the gist of what's going on? 00:18:54.490 --> 00:18:55.440 Same thing down here. 00:18:55.440 --> 00:18:57.700 The equations look a little bit different. 00:18:57.700 --> 00:18:58.912 But it's the same idea. 00:18:58.912 --> 00:19:00.870 You figure out what are the constraints imposed 00:19:00.870 --> 00:19:02.520 by the blocks. 00:19:02.520 --> 00:19:05.230 The output of every A has to be the integral 00:19:05.230 --> 00:19:06.729 of the thing that went into it. 00:19:06.729 --> 00:19:09.270 The output of every adder has to be the sum of the two things 00:19:09.270 --> 00:19:10.115 that went into it. 00:19:10.115 --> 00:19:11.490 Figure out all those constraints, 00:19:11.490 --> 00:19:15.060 reduce them to polynomials, and then manipulate them 00:19:15.060 --> 00:19:17.250 as though they were polynomials. 00:19:17.250 --> 00:19:20.250 And the polynomial manipulation shows that minus k1 00:19:20.250 --> 00:19:21.720 should be 1.6, OK? 00:19:21.720 --> 00:19:24.490 [INAUDIBLE] should be a minus 1.6. 00:19:24.490 --> 00:19:26.280 Point is, it's easy. 00:19:26.280 --> 00:19:28.170 If you know polynomials, you know the answer. 00:19:28.170 --> 00:19:29.580 That's pretty good. 00:19:29.580 --> 00:19:31.650 So we're doing something completely different 00:19:31.650 --> 00:19:33.450 from polynomials. 00:19:33.450 --> 00:19:35.820 But we're able to use the intuitions 00:19:35.820 --> 00:19:39.960 that we get from polynomials in order to do the problem. 00:19:39.960 --> 00:19:43.920 There is one remaining problem. 00:19:43.920 --> 00:19:46.650 Something that is harder than it is in DT. 00:19:46.650 --> 00:19:49.200 Probably the hardest part of CT. 00:19:49.200 --> 00:19:51.240 And that involves thinking about what 00:19:51.240 --> 00:19:54.930 should be our basis signals. 00:19:54.930 --> 00:19:57.420 In DT, there is a really good candidate 00:19:57.420 --> 00:20:02.010 for that, which we call the unit sample single. 00:20:02.010 --> 00:20:05.920 The unit sample signal is the simplest possible 00:20:05.920 --> 00:20:08.730 non-trivial DT signal. 00:20:08.730 --> 00:20:12.752 By which I mean it is 0 everywhere, except one place. 00:20:12.752 --> 00:20:14.835 If it were 0 everywhere, we would call it trivial. 00:20:17.900 --> 00:20:19.630 So in order for it not to be trivial, 00:20:19.630 --> 00:20:22.354 it has to be non-zero at least some place. 00:20:22.354 --> 00:20:23.395 The place is non-trivial. 00:20:23.395 --> 00:20:27.360 It is at n equals 0. 00:20:27.360 --> 00:20:32.290 And the value that it is, given that it can't be 0, is 1. 00:20:32.290 --> 00:20:38.740 So the unit sample signal is the most simple non-trivial DT 00:20:38.740 --> 00:20:39.582 signal. 00:20:39.582 --> 00:20:41.290 So what we need is the equivalent for CT. 00:20:43.810 --> 00:20:46.120 So here it is. 00:20:46.120 --> 00:20:50.650 Define a signal in CT that is 0 everywhere except 0. 00:20:50.650 --> 00:20:56.150 And at 0, make it [? 1. ?] 00:20:56.150 --> 00:20:58.910 So is that a good choice? 00:20:58.910 --> 00:21:02.750 And I hope I've biased my presentation 00:21:02.750 --> 00:21:07.070 enough so that the answer is obviously no. 00:21:07.070 --> 00:21:10.240 So the real question is, why not? 00:21:10.240 --> 00:21:12.250 Can somebody think of a good reason 00:21:12.250 --> 00:21:15.201 why this is a poor choice for a building block signal? 00:21:15.201 --> 00:21:15.700 Yes. 00:21:15.700 --> 00:21:22.364 AUDIENCE: [INAUDIBLE] 00:21:22.364 --> 00:21:25.480 DENNIS FREEMAN: So you would need infinite gain 00:21:25.480 --> 00:21:27.310 to go from 0 to 1. 00:21:27.310 --> 00:21:30.370 That's sort of true for the r thing, too. 00:21:30.370 --> 00:21:33.726 You have to go from 0 to 1. 00:21:33.726 --> 00:21:35.260 It's in 0 time. 00:21:35.260 --> 00:21:37.114 That seems hard. 00:21:37.114 --> 00:21:37.780 Any other ideas? 00:21:37.780 --> 00:21:37.850 Yeah. 00:21:37.850 --> 00:21:38.725 AUDIENCE: [INAUDIBLE] 00:21:38.725 --> 00:21:41.630 DENNIS FREEMAN: Not continuous. 00:21:41.630 --> 00:21:43.988 That's also true. 00:21:43.988 --> 00:21:44.906 Yes. 00:21:44.906 --> 00:21:47.172 AUDIENCE: [INAUDIBLE] 00:21:47.172 --> 00:21:48.880 DENNIS FREEMAN: The integral is always 0. 00:21:48.880 --> 00:21:51.860 That's a problem. 00:21:51.860 --> 00:21:57.700 So it's just not going to be a very useful signal. 00:21:57.700 --> 00:21:58.880 OK, what am I doing? 00:21:58.880 --> 00:22:02.440 I'm trying to figure out a basis signal that I 00:22:02.440 --> 00:22:03.640 would use in block diagrams. 00:22:03.640 --> 00:22:07.050 My basic block diagram is this integrator thing. 00:22:07.050 --> 00:22:11.920 Mine is infinity to t on tau. 00:22:11.920 --> 00:22:14.496 So I'd like to put something in here. 00:22:14.496 --> 00:22:16.370 But if I put that signal, what did I call it? 00:22:16.370 --> 00:22:19.930 W. If I put W into an iterator, what's 00:22:19.930 --> 00:22:23.830 the output of the integrator after you integrate W? 00:22:23.830 --> 00:22:24.760 0. 00:22:24.760 --> 00:22:26.260 Well, that's a problem. 00:22:26.260 --> 00:22:28.150 It's not a very good basis function 00:22:28.150 --> 00:22:33.010 if the very first operator I go through turns it into 0. 00:22:33.010 --> 00:22:34.480 Everyone see that? 00:22:34.480 --> 00:22:36.970 That's a problem. 00:22:36.970 --> 00:22:39.820 This is the problem in CT. 00:22:39.820 --> 00:22:42.840 If you guys figure out the next slide, you're done. 00:22:42.840 --> 00:22:44.800 CT's easy. 00:22:44.800 --> 00:22:52.250 So what we do is think about defining a signal that 00:22:52.250 --> 00:22:55.580 would have the property that it behaved as though it 00:22:55.580 --> 00:22:57.377 had those desirable properties. 00:22:57.377 --> 00:22:58.460 What would have to happen? 00:22:58.460 --> 00:23:01.830 We'd like a signal that had some amount of area. 00:23:01.830 --> 00:23:03.320 Because that's what this thing is. 00:23:03.320 --> 00:23:06.390 This is an area operator. 00:23:06.390 --> 00:23:09.570 We would like a signal that didn't have a 0 area. 00:23:09.570 --> 00:23:11.460 The W signal has 0 area. 00:23:11.460 --> 00:23:13.410 Bad. 00:23:13.410 --> 00:23:15.960 We'd like a signal that doesn't have zero area. 00:23:15.960 --> 00:23:18.560 In fact, if we wanted it to be the simplest possible, 00:23:18.560 --> 00:23:22.820 then you should have an area of 1. 00:23:22.820 --> 00:23:27.010 But we'd like it to be zero almost everywhere. 00:23:27.010 --> 00:23:29.380 How do you do that? 00:23:29.380 --> 00:23:32.560 OK, well, the way you can do it, and the way we will do it, 00:23:32.560 --> 00:23:37.720 is think about building a signal [? with ?] a limit. 00:23:37.720 --> 00:23:40.900 What if I had a pulse signal that 00:23:40.900 --> 00:23:44.470 was 2 epsilon wide and 1 over 2 epsilon high? 00:23:44.470 --> 00:23:49.330 The area would be 1, regardless of the value of epsilon. 00:23:49.330 --> 00:23:56.507 And if I shrunk epsilon to be a very small number, say 0, 00:23:56.507 --> 00:23:58.090 if I think about the limit, as epsilon 00:23:58.090 --> 00:24:00.610 gets smaller and smaller and smaller and smaller, 00:24:00.610 --> 00:24:05.410 then I approach the condition I'd like to have. 00:24:05.410 --> 00:24:08.440 The signal is 0 everywhere, except 0. 00:24:08.440 --> 00:24:10.610 Now, it has a horrendous value at 0. 00:24:10.610 --> 00:24:13.510 But, oh well, I'll sweep that under the rug. 00:24:13.510 --> 00:24:16.030 The important thing is that, when I integrate it, 00:24:16.030 --> 00:24:17.920 I'd like it to give me one. 00:24:17.920 --> 00:24:20.062 That's the important thing. 00:24:20.062 --> 00:24:20.770 Everyone with me? 00:24:20.770 --> 00:24:22.895 That's the hardest part in this part of the course. 00:24:22.895 --> 00:24:26.530 If you get this, CT systems are a breeze. 00:24:26.530 --> 00:24:30.880 So this is a different kind of a signal. 00:24:30.880 --> 00:24:35.600 The idea is to incorporate in it two seemingly contradictory 00:24:35.600 --> 00:24:36.100 things. 00:24:36.100 --> 00:24:43.440 It should be 0 almost everywhere and the integral should be 1. 00:24:43.440 --> 00:24:46.110 So there is no such thing. 00:24:46.110 --> 00:24:48.125 This is an idealized signal. 00:24:48.125 --> 00:24:50.250 This is something that we think about in the limit. 00:24:50.250 --> 00:24:52.740 It turns out that we can get a lot of insight 00:24:52.740 --> 00:24:55.650 by thinking about this signal. 00:24:55.650 --> 00:24:58.680 But it is important to realize that it is only 00:24:58.680 --> 00:25:01.540 defined in a limit. 00:25:01.540 --> 00:25:04.110 So we will write it. 00:25:04.110 --> 00:25:06.800 Since it's so useful, we will write it this way. 00:25:06.800 --> 00:25:08.430 We'll draw a little arrow. 00:25:08.430 --> 00:25:10.530 The arrow is supposed to connote in your mind, 00:25:10.530 --> 00:25:13.490 this thing goes really high. 00:25:13.490 --> 00:25:16.890 It goes so high that telling you how high it is is meaningless. 00:25:16.890 --> 00:25:21.892 So instead, to the side, we tell you what's the area of it. 00:25:21.892 --> 00:25:23.600 What would come out if you integrated it? 00:25:23.600 --> 00:25:24.100 Yes. 00:25:24.100 --> 00:25:26.580 AUDIENCE: [INAUDIBLE] DT [INAUDIBLE] 00:25:26.580 --> 00:25:29.800 decompose an [INAUDIBLE] signal into a series 00:25:29.800 --> 00:25:32.342 of [INAUDIBLE] Can you do the same thing with [INAUDIBLE]? 00:25:32.342 --> 00:25:33.175 DENNIS FREEMAN: Yes. 00:25:33.175 --> 00:25:35.770 And that will be something that we spend a lot of time on, 00:25:35.770 --> 00:25:40.060 because that decomposition is not trivial. 00:25:40.060 --> 00:25:42.700 But it is possible. 00:25:42.700 --> 00:25:46.720 So the question was, can we decompose arbitrary signals 00:25:46.720 --> 00:25:48.220 into sums of these things? 00:25:48.220 --> 00:25:51.450 And the answer is absolutely yes. 00:25:51.450 --> 00:25:54.720 That's why we like it. 00:25:54.720 --> 00:25:58.930 OK, so the first thing to think about 00:25:58.930 --> 00:26:00.760 is what happens if you put, now, instead 00:26:00.760 --> 00:26:09.274 of W, let's put delta of t, the unit impulse function. 00:26:09.274 --> 00:26:10.940 What happens if you put the unit impulse 00:26:10.940 --> 00:26:13.770 function into an integrator? 00:26:13.770 --> 00:26:19.500 Well, if you integrate the delta of t, so if my input x of t 00:26:19.500 --> 00:26:25.180 is delta of t, and if I want to think about y of t, 00:26:25.180 --> 00:26:28.370 which is going to A operating on x? 00:26:28.370 --> 00:26:32.430 So think about what would be the answer to this integral 00:26:32.430 --> 00:26:34.544 for times less than 0? 00:26:34.544 --> 00:26:36.710 Well, the system started the rest and the integrator 00:26:36.710 --> 00:26:39.050 turned on at 0. 00:26:39.050 --> 00:26:43.190 What's the value of the answer of the integrator here? 00:26:43.190 --> 00:26:43.810 0. 00:26:43.810 --> 00:26:46.690 It started at 0 back at minus infinity some time. 00:26:46.690 --> 00:26:48.610 Started at rest. 00:26:48.610 --> 00:26:51.340 There was no input. 00:26:51.340 --> 00:26:52.240 So the answer is 0. 00:26:52.240 --> 00:26:56.920 So the answer is going to be 0 up until something happens. 00:26:56.920 --> 00:27:02.230 During that brief epsilon of time, a unit of area went in. 00:27:02.230 --> 00:27:06.540 So what's the output of the integrator become? 00:27:06.540 --> 00:27:08.310 1. 00:27:08.310 --> 00:27:10.020 So at that time it became 1. 00:27:13.230 --> 00:27:16.880 And then what's the value as you go forward in time? 00:27:16.880 --> 00:27:18.170 1. 00:27:18.170 --> 00:27:20.220 It got stuck at 1. 00:27:20.220 --> 00:27:24.650 So the most primitive signal that we'll think about 00:27:24.650 --> 00:27:26.020 is the unit impulse function. 00:27:26.020 --> 00:27:29.226 We'll denote that by delta of t. 00:27:29.226 --> 00:27:31.850 And it has the property that it goes through our most primitive 00:27:31.850 --> 00:27:33.810 block. 00:27:33.810 --> 00:27:35.970 Delta goes to a unit-step And unit-step 00:27:35.970 --> 00:27:38.670 is so useful that we'll give it a special name, u of t. 00:27:38.670 --> 00:27:40.462 U means unit-step. 00:27:44.160 --> 00:27:47.380 And with that, all of these systems 00:27:47.380 --> 00:27:50.740 now make sense, in the same sense that DT systems did. 00:27:50.740 --> 00:27:55.750 So for example, if I have a feedforward system, 00:27:55.750 --> 00:27:59.680 I can think about that as having several signal flow 00:27:59.680 --> 00:28:02.250 paths going forward. 00:28:02.250 --> 00:28:04.020 And the output of the signal, when 00:28:04.020 --> 00:28:08.460 stimulated with a delta function, 00:28:08.460 --> 00:28:10.380 is the sum of all those different signal 00:28:10.380 --> 00:28:12.970 paths that I can think of. 00:28:12.970 --> 00:28:15.630 So I can express the input-output relationship 00:28:15.630 --> 00:28:18.840 by dysfunctional relationship. 00:28:18.840 --> 00:28:21.287 Everybody sees that? 00:28:21.287 --> 00:28:23.370 Or I can just think about all the different signal 00:28:23.370 --> 00:28:24.330 flow passed through here. 00:28:24.330 --> 00:28:26.371 What would happen if I had a delta function here? 00:28:26.371 --> 00:28:28.420 Well, here's a signal flow path. 00:28:28.420 --> 00:28:29.530 Here's another one. 00:28:29.530 --> 00:28:32.030 Here's another one. 00:28:32.030 --> 00:28:33.710 Here's one. 00:28:33.710 --> 00:28:36.350 For signal flow paths, I have to think 00:28:36.350 --> 00:28:39.136 about ways the input could turn into the output. 00:28:39.136 --> 00:28:40.760 If the input went through the top half, 00:28:40.760 --> 00:28:44.630 the output would be delta. 00:28:44.630 --> 00:28:52.520 If the input went through, the path the output would be u. 00:28:52.520 --> 00:28:58.000 If the input went through this path, u. 00:28:58.000 --> 00:29:01.707 If the input went through this, it turns into u, 00:29:01.707 --> 00:29:03.790 and then it goes through another accumulator, what 00:29:03.790 --> 00:29:06.730 happens when you integrate u? 00:29:06.730 --> 00:29:07.230 tu. 00:29:10.670 --> 00:29:17.732 If this is u, and if you put that in through an integrator, 00:29:17.732 --> 00:29:19.690 that integrator, like all the other integrators 00:29:19.690 --> 00:29:21.900 we'll talk about, unless we tell you otherwise, 00:29:21.900 --> 00:29:22.930 is initially at rest. 00:29:22.930 --> 00:29:25.870 Therefore, the output started at 0. 00:29:25.870 --> 00:29:27.850 Because u is 0 for t less than 0, 00:29:27.850 --> 00:29:32.700 it persists at 0 until something happens at t equals 0. 00:29:32.700 --> 00:29:34.200 At which point, the input becomes 1, 00:29:34.200 --> 00:29:37.520 and the integral of 1 is t. 00:29:37.520 --> 00:29:42.730 So we will usually write that this way, meaning it's t. 00:29:42.730 --> 00:29:45.700 So we will write as a shorthand times u. 00:29:45.700 --> 00:29:47.620 That's a strange thing to do. 00:29:47.620 --> 00:29:50.254 It's just very convenient. 00:29:50.254 --> 00:29:51.670 It's convenient, because normally, 00:29:51.670 --> 00:29:57.370 when you have a function, so that last function, this one, 00:29:57.370 --> 00:30:00.400 is a signal that does this. 00:30:03.270 --> 00:30:07.150 So we will call that tu of t because, obviously, t 00:30:07.150 --> 00:30:08.940 is that signal. 00:30:08.940 --> 00:30:10.040 We don't mean that. 00:30:10.040 --> 00:30:13.740 We want to lop off the part that came before 0. 00:30:13.740 --> 00:30:15.320 Multiplying by u of t is a quick way 00:30:15.320 --> 00:30:18.710 of writing lop off the stuff that came before 0. 00:30:18.710 --> 00:30:20.510 That's all that means. 00:30:20.510 --> 00:30:23.186 So we'll write that that way. 00:30:23.186 --> 00:30:25.310 And you can see that, by thinking about signal flow 00:30:25.310 --> 00:30:28.790 paths, and by thinking about how this basic signal, this basis 00:30:28.790 --> 00:30:30.920 function, goes through the system, 00:30:30.920 --> 00:30:34.130 it's easy to figure out the response of such a system. 00:30:37.160 --> 00:30:42.670 Just like in DT, feedback is different. 00:30:42.670 --> 00:30:44.880 So now, we have to think about what would happen 00:30:44.880 --> 00:30:47.140 if there's a feedback loop. 00:30:47.140 --> 00:30:48.660 So it was crucial, when I was doing 00:30:48.660 --> 00:30:54.060 this kind of an illustration, that it was feedforward only. 00:30:54.060 --> 00:30:58.260 I could decompose all of the signal flow paths 00:30:58.260 --> 00:31:01.240 into a finite number of forward-going paths. 00:31:01.240 --> 00:31:02.150 This is harder. 00:31:05.320 --> 00:31:06.610 This happened in DT, too. 00:31:06.610 --> 00:31:09.040 We had to figure out how the DT system was going to work 00:31:09.040 --> 00:31:10.390 when we had a feedback path. 00:31:10.390 --> 00:31:11.860 And it was more complicated, and we 00:31:11.860 --> 00:31:13.720 had to fall back on thinking about 00:31:13.720 --> 00:31:16.515 sample-by-sample propagation of the signal through the system. 00:31:16.515 --> 00:31:17.890 And we'll do the same thing here. 00:31:20.530 --> 00:31:23.430 In order to figure out how to think about signal flow 00:31:23.430 --> 00:31:29.220 through such a diagram, through a diagram that has feedback, 00:31:29.220 --> 00:31:32.912 we'll think about falling back to 1803. 00:31:32.912 --> 00:31:33.870 Something we can trust. 00:31:33.870 --> 00:31:36.750 Something we all loved. 00:31:36.750 --> 00:31:37.560 Nod your heads yes. 00:31:37.560 --> 00:31:39.720 Make me-- reassures me. 00:31:39.720 --> 00:31:42.340 Fond memories of 1803. 00:31:42.340 --> 00:31:44.910 So it's pretty simple to think about how 00:31:44.910 --> 00:31:50.020 you would solve that system by using an 1803 type of approach. 00:31:50.020 --> 00:31:52.040 Convert the system into a differential equation. 00:31:52.040 --> 00:31:53.206 We've already done that one. 00:31:56.050 --> 00:32:00.250 Then from the form, we will usually, in this class, 00:32:00.250 --> 00:32:04.490 use the general method of solving differential equations, 00:32:04.490 --> 00:32:06.451 which is-- 00:32:06.451 --> 00:32:11.050 What's the general method for solving differential equations? 00:32:11.050 --> 00:32:13.410 They may not have told you in 1803. 00:32:13.410 --> 00:32:17.110 What's the general method of solving differential equations? 00:32:17.110 --> 00:32:17.820 Guess! 00:32:17.820 --> 00:32:19.430 Yes. 00:32:19.430 --> 00:32:23.260 So the general method is guess. 00:32:23.260 --> 00:32:27.450 If you can guess, plug it in and it works, you're done. 00:32:27.450 --> 00:32:29.220 The kinds of systems that we will look at 00:32:29.220 --> 00:32:30.990 will be linear systems. 00:32:30.990 --> 00:32:33.060 They will have a single unique solution. 00:32:33.060 --> 00:32:35.610 If you can find it by guessing, you're done. 00:32:35.610 --> 00:32:37.620 And we will prove later in the course 00:32:37.620 --> 00:32:40.380 that all systems of this class, where I haven't really 00:32:40.380 --> 00:32:42.810 defined what this class is, but if you made a system out 00:32:42.810 --> 00:32:46.020 of adders, gains, and integrators, 00:32:46.020 --> 00:32:52.640 and only those parts, then that system will always be linear. 00:32:52.640 --> 00:32:55.970 And the solutions can always be written 00:32:55.970 --> 00:32:58.370 as complex exponentials. 00:32:58.370 --> 00:32:59.420 We'll prove that later. 00:32:59.420 --> 00:33:01.130 I'm not going to bother with it now. 00:33:01.130 --> 00:33:02.494 For now, you would say, OK. 00:33:02.494 --> 00:33:04.160 First-order linear differential equation 00:33:04.160 --> 00:33:05.990 with constant coefficiency, the answer 00:33:05.990 --> 00:33:09.140 is obviously an exponential function. 00:33:09.140 --> 00:33:11.242 So you plug that into the differential equation. 00:33:11.242 --> 00:33:12.700 You figure out the constraints that 00:33:12.700 --> 00:33:15.860 would have to be solved for this to be true. 00:33:15.860 --> 00:33:19.480 And the answer is that y should be e to the pt [? of ?] u of t. 00:33:19.480 --> 00:33:21.820 Should start at 0. 00:33:21.820 --> 00:33:24.670 And there should be an exponential after that. 00:33:24.670 --> 00:33:31.790 And the p shows up as the exponent in the exponential. 00:33:31.790 --> 00:33:34.880 What we'd like to do is develop an alternative way 00:33:34.880 --> 00:33:37.490 of thinking about that, using the operator approach. 00:33:41.030 --> 00:33:43.630 So a completely different way to solve that problem 00:33:43.630 --> 00:33:46.330 would be to think about, write an operator expression 00:33:46.330 --> 00:33:48.480 for this. 00:33:48.480 --> 00:33:51.260 OK, so Y is A times the signal that 00:33:51.260 --> 00:33:52.970 results when you add x to py. 00:33:59.100 --> 00:34:03.300 Solve for the ratio Y by X, and you get an operator expression, 00:34:03.300 --> 00:34:05.240 a over 1 minus pA. 00:34:08.900 --> 00:34:11.090 Just like in DT, we have to figure out 00:34:11.090 --> 00:34:14.070 how we would think about that. 00:34:14.070 --> 00:34:15.790 It's an implicit operation. 00:34:15.790 --> 00:34:18.300 It's telling me that if I knew the answer, 00:34:18.300 --> 00:34:21.330 the answer is the signal that, when operated on by 1 minus p 00:34:21.330 --> 00:34:24.544 is the integral of the input. 00:34:24.544 --> 00:34:25.710 But I don't know the answer. 00:34:25.710 --> 00:34:27.300 So that reasoning doesn't quite work. 00:34:27.300 --> 00:34:33.179 The reasoning that I used to solve the feedforward question 00:34:33.179 --> 00:34:34.366 doesn't quite work. 00:34:34.366 --> 00:34:36.449 So I have to think of a different way of doing it, 00:34:36.449 --> 00:34:39.350 just like we did in DT. 00:34:39.350 --> 00:34:41.489 And the same solution works. 00:34:41.489 --> 00:34:43.110 Not surprisingly. 00:34:43.110 --> 00:34:44.909 A behaves like a polynomial. 00:34:44.909 --> 00:34:46.409 R behaves like a polynomial. 00:34:46.409 --> 00:34:49.820 After you turn it into a polynomial, 00:34:49.820 --> 00:34:53.000 you can't tell if you started with a or r. 00:34:53.000 --> 00:34:55.190 It's a polynomial. 00:34:55.190 --> 00:34:58.501 So for that reason, exactly the same stuff that you did in DT 00:34:58.501 --> 00:34:59.000 works. 00:34:59.000 --> 00:35:02.680 That's why we can solve everything in 50 minutes. 00:35:02.680 --> 00:35:05.440 It's all the same. 00:35:05.440 --> 00:35:12.100 So in DT, we thought about this as being a series. 00:35:12.100 --> 00:35:15.220 You could figure out an ascending series, a power 00:35:15.220 --> 00:35:17.530 series, that's equivalent by thinking about something 00:35:17.530 --> 00:35:20.980 like synthetic division, Taylor series, whatever 00:35:20.980 --> 00:35:22.040 you're comfortable with. 00:35:22.040 --> 00:35:24.280 But whatever you're comfortable with. 00:35:24.280 --> 00:35:29.140 This, A over 1 minus pA, take the A out front. 00:35:29.140 --> 00:35:32.390 Then you're left with 1 over 1 minus pA. 00:35:32.390 --> 00:35:34.030 Think about that is Taylor series. 00:35:34.030 --> 00:35:36.160 That's 1 plus pA, plus p squared A squared, 00:35:36.160 --> 00:35:38.171 plus p cubed A cubed, etc. 00:35:38.171 --> 00:35:38.670 Done. 00:35:43.440 --> 00:35:46.560 Now I know what the system functional looks like. 00:35:46.560 --> 00:35:50.550 Now I've got a feedforward system. 00:35:50.550 --> 00:35:54.210 So I can use the technique from the other side. 00:35:54.210 --> 00:35:58.470 What happens if you put delta into a system whose functional 00:35:58.470 --> 00:36:00.888 representation is A? 00:36:05.750 --> 00:36:07.010 So you get a u. 00:36:07.010 --> 00:36:08.170 Let me skip that one. 00:36:08.170 --> 00:36:10.160 That's on the next slide. 00:36:10.160 --> 00:36:16.180 What happens if you put in-- 00:36:16.180 --> 00:36:17.042 no, that's right. 00:36:17.042 --> 00:36:17.750 I did that right. 00:36:17.750 --> 00:36:19.610 OK 1 u. 00:36:19.610 --> 00:36:22.120 Sorry, I'm confusing myself. 00:36:22.120 --> 00:36:23.710 This A turned into that u. 00:36:26.420 --> 00:36:28.700 What happens if you put it into pA squared? 00:36:32.260 --> 00:36:34.670 So the first A turned delta into u. 00:36:34.670 --> 00:36:40.900 The second A turns u into tu. 00:36:40.900 --> 00:36:42.460 And so I get this term. 00:36:42.460 --> 00:36:47.480 So the pA squared turns into ptu. 00:36:47.480 --> 00:36:52.890 The p squared A cubed turns into a half p squared t 00:36:52.890 --> 00:36:56.410 squared u, etc. 00:36:56.410 --> 00:36:58.420 So what I've done is I've thought 00:36:58.420 --> 00:37:02.710 of a way of constructing this output signal from the input 00:37:02.710 --> 00:37:04.725 signal without ever using calculus. 00:37:08.410 --> 00:37:10.240 I started with a solution based on 1803 00:37:10.240 --> 00:37:12.760 that's strictly calculus. 00:37:12.760 --> 00:37:16.870 I just redid the whole problem and didn't use any calculus. 00:37:16.870 --> 00:37:20.830 And the method works just the same reason 00:37:20.830 --> 00:37:23.071 that the our method worked. 00:37:23.071 --> 00:37:25.570 In the r method, when we were thinking 00:37:25.570 --> 00:37:28.750 about the simple feedback with an r system, 00:37:28.750 --> 00:37:33.370 every loop around the feedback loop generated one new sample. 00:37:33.370 --> 00:37:35.620 Here, every loop around the CT loop 00:37:35.620 --> 00:37:39.040 generates one more contribution to the output. 00:37:39.040 --> 00:37:42.530 You put a delta function in, it gets integrated once, 00:37:42.530 --> 00:37:45.950 and you get a u function out. 00:37:45.950 --> 00:37:48.550 [INAUDIBLE] [? first ?] in this symbolic representations. 00:37:48.550 --> 00:37:52.750 The first thing that comes out is A times the input. 00:37:52.750 --> 00:37:56.390 Second thing that comes out is A squared p times the input. 00:37:56.390 --> 00:38:00.740 Third thing that comes out, p squared a cubed, etc. 00:38:00.740 --> 00:38:02.660 Every time you go through the loop, 00:38:02.660 --> 00:38:04.790 you pick up one more turn. 00:38:04.790 --> 00:38:08.300 Now, think about it in a time domain. 00:38:08.300 --> 00:38:10.560 You put in a delta function. 00:38:10.560 --> 00:38:12.350 The first thing that comes out is a step. 00:38:17.070 --> 00:38:20.290 The second thing that comes out, and gets 00:38:20.290 --> 00:38:23.350 added to the first thing that comes out, 00:38:23.350 --> 00:38:27.010 is integrate, multiply by p, and integrate again. 00:38:29.620 --> 00:38:32.510 So that's this term, p times t. 00:38:32.510 --> 00:38:37.050 So we get the unit step from the first term, 00:38:37.050 --> 00:38:38.980 which looks like this. 00:38:38.980 --> 00:38:42.210 We get the linear increase from the second term. 00:38:42.210 --> 00:38:46.260 We get a squared term, we get a cubed' term. 00:38:46.260 --> 00:38:52.200 And voila, we get the series expansion for e to the pt. 00:38:52.200 --> 00:38:53.760 No calculus. 00:38:53.760 --> 00:38:55.470 It's purely thinking about how you 00:38:55.470 --> 00:39:01.440 would think about constructing this system by a series 00:39:01.440 --> 00:39:04.050 representation for spinning around the feedback loop. 00:39:07.430 --> 00:39:11.870 So that's an insight that comes from thinking about the system 00:39:11.870 --> 00:39:17.540 as a functional representation, as a polynomial. 00:39:17.540 --> 00:39:19.580 Polynomials can be expanded in Taylor series. 00:39:19.580 --> 00:39:23.920 Therefore, systems can be expanded in Taylor series. 00:39:23.920 --> 00:39:27.330 If you change the sign of p, you'll 00:39:27.330 --> 00:39:29.205 notice that that previous example exploded. 00:39:31.730 --> 00:39:33.824 So we got to an increasing exponential. 00:39:33.824 --> 00:39:35.990 If you change the sign of p, it's not too surprising 00:39:35.990 --> 00:39:39.670 that what will happen is that it will converge. 00:39:39.670 --> 00:39:43.040 So here, symbolically, the first signal 00:39:43.040 --> 00:39:45.980 is a unit-step just like before. 00:39:45.980 --> 00:39:47.980 But now because of the minus sign, 00:39:47.980 --> 00:39:51.195 there is a negative on this pt, so you get sloping downward. 00:39:54.680 --> 00:39:56.790 In the squared term, it's still positive. 00:39:56.790 --> 00:39:59.750 So that breaks it up. 00:39:59.750 --> 00:40:03.170 But the cubic is now downward again. 00:40:03.170 --> 00:40:05.480 And you add an infinite number of those 00:40:05.480 --> 00:40:10.750 and you get another exponential, this time a convergent one. 00:40:10.750 --> 00:40:12.520 So the point is that it works extremely 00:40:12.520 --> 00:40:15.880 like the way the DT stuff did, except now the basis 00:40:15.880 --> 00:40:17.260 functions are different. 00:40:17.260 --> 00:40:20.110 The basis functions are derived from this thing, which 00:40:20.110 --> 00:40:24.270 is the unit impulse function. 00:40:24.270 --> 00:40:26.610 And the things that come out of the integrator 00:40:26.610 --> 00:40:32.160 are steps and ramps and parabolas and stuff like that. 00:40:32.160 --> 00:40:39.760 So in r, the basic input signal was the unit sample signal. 00:40:39.760 --> 00:40:44.260 And what came out on each revolution around-- 00:40:44.260 --> 00:40:47.490 for each cycle in a feedback system, what came out 00:40:47.490 --> 00:40:49.410 was a successive delay. 00:40:49.410 --> 00:40:54.162 Here, we get successive integration. 00:40:54.162 --> 00:40:55.620 Delay was the fundamental operator. 00:40:55.620 --> 00:40:57.450 Integration is the fundamental operator. 00:40:57.450 --> 00:41:01.020 Successive delays, successive integrations. 00:41:01.020 --> 00:41:04.200 And we get the idea that we have convergence and divergence, 00:41:04.200 --> 00:41:05.490 just like we did in DT. 00:41:05.490 --> 00:41:08.442 Except now, the shapes of the regions are different. 00:41:11.540 --> 00:41:15.870 So if the p, which we will later call the pole, 00:41:15.870 --> 00:41:18.390 by complete analogy to what we did in DT, 00:41:18.390 --> 00:41:20.250 if the pole is in the right half plane, 00:41:20.250 --> 00:41:23.520 then the system's response is divergent. 00:41:23.520 --> 00:41:25.350 If the pole is in the left half plane, 00:41:25.350 --> 00:41:28.890 the system's responses is convergent. 00:41:28.890 --> 00:41:31.560 So it looks just like DT. 00:41:31.560 --> 00:41:33.780 We think about this as signal flow paths. 00:41:33.780 --> 00:41:35.940 We think about how feedback gives rise 00:41:35.940 --> 00:41:39.610 to an infinite number of those. 00:41:39.610 --> 00:41:44.580 In DT, each time through, it gave one unit of delay in CT. 00:41:44.580 --> 00:41:48.900 It gave one unit of integration. 00:41:48.900 --> 00:41:54.730 The method is the same, the answer is different. 00:41:54.730 --> 00:41:57.480 So what we got is tremendous similarities. 00:41:57.480 --> 00:41:59.740 Differential equation compared to difference equation, 00:41:59.740 --> 00:42:02.280 block diagram compared to block diagram. 00:42:02.280 --> 00:42:05.250 Integrators instead of delays. 00:42:05.250 --> 00:42:09.530 Functional functional, A instead of r. 00:42:09.530 --> 00:42:10.315 Pole, pole. 00:42:13.170 --> 00:42:14.460 Mode, mode. 00:42:14.460 --> 00:42:16.620 Fundamental mode. 00:42:16.620 --> 00:42:19.920 Here, the fundamental mode was a geometric sequence. 00:42:19.920 --> 00:42:22.830 Here, it's an exponential function of time. 00:42:22.830 --> 00:42:25.080 And we've got different kinds of regional convergence. 00:42:25.080 --> 00:42:29.410 Signal converges for poles in the left half plane. 00:42:29.410 --> 00:42:33.310 Signals converge for poles inside the unit circle. 00:42:33.310 --> 00:42:38.770 Same idea, but some differences. 00:42:38.770 --> 00:42:46.100 OK, so [INAUDIBLE] have time think 00:42:46.100 --> 00:42:50.030 about this for 30 seconds. 00:42:50.030 --> 00:42:52.280 We now have two representations, R polynomials 00:42:52.280 --> 00:42:55.000 and A polynomials. 00:42:55.000 --> 00:42:58.240 I'm giving you four examples to think through, 00:42:58.240 --> 00:43:00.580 and your job is to figure out which 00:43:00.580 --> 00:43:06.610 of those functionals correspond to convergent responses 00:43:06.610 --> 00:43:10.240 when excited by a unit sample or a unit impulse signal. 00:44:26.910 --> 00:44:28.280 So, how do I think about this? 00:44:28.280 --> 00:44:31.140 Is this convergent or divergent? 00:44:31.140 --> 00:44:33.380 What do I think about? 00:44:33.380 --> 00:44:35.770 Step 1. 00:44:35.770 --> 00:44:36.270 Convergent. 00:44:36.270 --> 00:44:37.870 How did you get that? 00:44:37.870 --> 00:44:38.370 Yeah. 00:44:38.370 --> 00:44:39.242 It's this one. 00:44:39.242 --> 00:44:41.552 AUDIENCE: [INAUDIBLE] 00:44:41.552 --> 00:44:43.050 DENNIS FREEMAN: So factor. 00:44:43.050 --> 00:44:45.450 It's polynomial. 00:44:45.450 --> 00:44:49.670 You get a pole at minus 1/2 and a pole at plus 1/2. 00:44:49.670 --> 00:44:51.720 They're both inside [? the unit ?] circle. 00:44:51.720 --> 00:44:54.310 They both converge. 00:44:54.310 --> 00:44:56.500 How about this one? 00:44:56.500 --> 00:45:00.000 Same equation, same answer. 00:45:00.000 --> 00:45:02.450 Kind of. 00:45:02.450 --> 00:45:07.030 Same poles, minus 1/2 and plus 1/2. 00:45:07.030 --> 00:45:09.180 Convergent or divergent? 00:45:09.180 --> 00:45:12.700 Divergent, because the regions are different. 00:45:12.700 --> 00:45:15.570 One of the poles is in the right half plane. 00:45:15.570 --> 00:45:16.830 Divergent. 00:45:16.830 --> 00:45:19.660 How about this one. 00:45:19.660 --> 00:45:22.430 Where were the poles? 00:45:22.430 --> 00:45:24.260 Minus 1/2, minus 3/2. 00:45:27.240 --> 00:45:29.590 Outside unit circle? 00:45:29.590 --> 00:45:31.530 One of the poles is outside the unit circle. 00:45:31.530 --> 00:45:34.660 Diverges, because of that pole. 00:45:34.660 --> 00:45:38.590 Minus 1/2 and minus 3/2, same poles. 00:45:38.590 --> 00:45:41.070 Both in the left half lane. 00:45:41.070 --> 00:45:41.850 Convergence. 00:45:41.850 --> 00:45:43.470 That's the point. 00:45:43.470 --> 00:45:44.920 So they worked very similar. 00:45:44.920 --> 00:45:46.980 But there are differences. 00:45:46.980 --> 00:45:49.940 And now in the last three minutes, 00:45:49.940 --> 00:45:51.990 the last thing I want to tell you about 00:45:51.990 --> 00:45:54.820 is that even complex numbers work just the same as in DT. 00:45:57.330 --> 00:46:00.690 So think about a system that's slightly harder. 00:46:00.690 --> 00:46:04.310 Here, I can represent this by this kind of relationship. 00:46:04.310 --> 00:46:10.100 So F equals Kx, but F is MA. 00:46:10.100 --> 00:46:12.660 So 801. 00:46:12.660 --> 00:46:15.330 And so I generate a block diagram or a differential 00:46:15.330 --> 00:46:18.090 equation or however I want to think about it. 00:46:18.090 --> 00:46:19.470 And I can solve for the answer. 00:46:19.470 --> 00:46:21.011 What I'd like to do is think about it 00:46:21.011 --> 00:46:23.800 in terms of the functional representation. 00:46:23.800 --> 00:46:25.467 So I just think about, take this system 00:46:25.467 --> 00:46:26.550 that you're familiar with. 00:46:26.550 --> 00:46:28.350 We did this on the first lecture. 00:46:28.350 --> 00:46:31.460 Turn it into a differential equation. 00:46:31.460 --> 00:46:33.280 Turn that into a functional representation. 00:46:33.280 --> 00:46:35.821 Now, I want to think about how you would solve the functional 00:46:35.821 --> 00:46:38.180 representation. 00:46:38.180 --> 00:46:40.840 The idea is just like before, factor. 00:46:40.840 --> 00:46:46.655 But now the trick is that the factors become complex. 00:46:49.340 --> 00:46:51.580 So one way you can think about this 00:46:51.580 --> 00:46:55.450 is to force the second order system into a canonical form. 00:46:55.450 --> 00:47:03.880 The canonical form for DT was 1 over 1 minus p not R. 00:47:03.880 --> 00:47:08.650 For CT, it's A over 1 minus p not A. 00:47:08.650 --> 00:47:10.990 So I want to make factors look like that, because I 00:47:10.990 --> 00:47:15.190 know that here, the response looks like p not to the n. 00:47:15.190 --> 00:47:18.520 Here, the response looks like e to the p, not t. 00:47:18.520 --> 00:47:19.990 If I can coerce it into that form, 00:47:19.990 --> 00:47:22.294 I already know the answer. 00:47:22.294 --> 00:47:23.710 So that's what's illustrated here. 00:47:23.710 --> 00:47:27.640 I coerced this into that form. 00:47:27.640 --> 00:47:32.440 And then I'll know what the answer looks like. 00:47:32.440 --> 00:47:35.930 Just like we could in DT, substitute R goes to 1 over z. 00:47:35.930 --> 00:47:40.390 In CT, we can substitute A to 1 over s, and solve for s. 00:47:40.390 --> 00:47:41.410 We get the same answer. 00:47:41.410 --> 00:47:45.780 Same as we did in DT, except now we call it s. 00:47:45.780 --> 00:47:50.040 So the poles to this system are plus or minus j constant. 00:47:50.040 --> 00:47:53.940 For convenience, I'll call the constant omega not. 00:47:53.940 --> 00:47:58.730 So then I have a pole at e to the j omega not. 00:47:58.730 --> 00:48:01.230 And another one-- so I have a pole at j omega 0 00:48:01.230 --> 00:48:03.820 and a second pole at minus j omega 0. 00:48:03.820 --> 00:48:07.310 By that argument, the fundamental modes 00:48:07.310 --> 00:48:13.400 are e the j omega not t and e to the minus j omega not t. 00:48:13.400 --> 00:48:19.130 Just like in DT, the complex poles gave complex modes. 00:48:19.130 --> 00:48:22.400 Here, the complex pole, j omega not, 00:48:22.400 --> 00:48:25.280 gave a complex mode, cos omega t plus j sine omega t. 00:48:29.240 --> 00:48:34.210 And just like DT, the system conspires. 00:48:34.210 --> 00:48:37.210 So that started out being mass and spring system. 00:48:37.210 --> 00:48:40.940 Obviously, it's not going to have an imaginary output. 00:48:40.940 --> 00:48:44.260 The system conspires so that the imaginary parts 00:48:44.260 --> 00:48:47.990 of the different fundamental modes kill each other off. 00:48:47.990 --> 00:48:51.510 And the answer is a real number. 00:48:51.510 --> 00:48:54.670 So even though the mode, even though the pole is complex, 00:48:54.670 --> 00:48:59.266 the multipliers are complex, this sum is real. 00:48:59.266 --> 00:49:01.640 And if you just think about what that sum is, by thinking 00:49:01.640 --> 00:49:04.422 about how complex numbers work, you 00:49:04.422 --> 00:49:06.880 get an expression that looks like omega not sine, omega not 00:49:06.880 --> 00:49:08.420 t. 00:49:08.420 --> 00:49:10.430 It's a little more fun to think about how 00:49:10.430 --> 00:49:12.770 that evolves as a series. 00:49:12.770 --> 00:49:14.720 If we do a Taylor series for this, 00:49:14.720 --> 00:49:16.010 we can represent it that way. 00:49:18.530 --> 00:49:21.284 And the series, then, the first term is t. 00:49:23.950 --> 00:49:28.690 The second term is a t cubed, which goes down. 00:49:28.690 --> 00:49:35.540 Then there's a fifth and a seventh and a ninth. 00:49:35.540 --> 00:49:38.860 And if you keep adding up the terms, amazingly enough, 00:49:38.860 --> 00:49:43.370 the series representation for the operator expression 00:49:43.370 --> 00:49:47.880 is the Taylor series expansion of the sine wave. 00:49:47.880 --> 00:49:50.070 Just like we would have expected. 00:49:50.070 --> 00:49:54.900 Again, here, the response looks like e to the p not t. 00:49:54.900 --> 00:49:56.370 If I can coerce it into that form, 00:49:56.370 --> 00:49:58.684 I already know the answer. 00:49:58.684 --> 00:50:00.100 So that's what's illustrated here. 00:50:00.100 --> 00:50:06.170 I coerce this into that form, and then I'll 00:50:06.170 --> 00:50:08.830 know what the answer looks like. 00:50:08.830 --> 00:50:12.330 Just like we could in DT, substitute R goes to 1 over z. 00:50:12.330 --> 00:50:16.780 In CT, we can substitute A goes to 1 over s, and solve for s. 00:50:16.780 --> 00:50:17.800 We get the same answer. 00:50:17.800 --> 00:50:22.170 Same as we did in DT, except now we call it s. 00:50:22.170 --> 00:50:26.430 So the poles to this system are plus or minus j constant. 00:50:26.430 --> 00:50:30.330 For convenience, I'll call the constant omega not. 00:50:30.330 --> 00:50:32.955 So then I have a pole at e to the j omega 00:50:32.955 --> 00:50:37.500 not, and another one-- so I have a pole at j omega not 00:50:37.500 --> 00:50:40.180 and a second poet minus j omega not. 00:50:40.180 --> 00:50:43.700 By that argument, the fundamental modes 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. 00:50:49.790 --> 00:50:55.580 Just like in DT, the complex poles gave complex modes. 00:50:55.580 --> 00:50:58.770 Here, the complex pole, j omega not, 00:50:58.770 --> 00:51:02.020 gave a complex mode, cos omega t plus j sine omega t. 00:51:05.640 --> 00:51:10.600 And just like DT, the system conspires. 00:51:10.600 --> 00:51:13.600 So that started out being a [? mass ?] [INAUDIBLE] system. 00:51:13.600 --> 00:51:17.330 Obviously, it's not going to have an imaginary output. 00:51:17.330 --> 00:51:20.650 The system conspires so that the imaginary parts 00:51:20.650 --> 00:51:24.380 of the different fundamental modes kill each other off, 00:51:24.380 --> 00:51:27.900 and the answer is a real number. 00:51:27.900 --> 00:51:31.320 So even though the pole is complex, 00:51:31.320 --> 00:51:35.637 the multipliers are complex, this sum is real. 00:51:35.637 --> 00:51:37.470 And if you just think about what that sum is 00:51:37.470 --> 00:51:40.625 by thinking about how complex numbers work, 00:51:40.625 --> 00:51:43.250 you get an expression that looks like omega not sine, omega not 00:51:43.250 --> 00:51:44.840 t. 00:51:44.840 --> 00:51:46.820 It's a little more fun to think about how 00:51:46.820 --> 00:51:49.160 that evolves as a series. 00:51:49.160 --> 00:51:51.110 If we do a Taylor series for this, 00:51:51.110 --> 00:51:52.400 we can represent it that way. 00:51:54.940 --> 00:51:57.672 And the series, then, the first term is t. 00:52:00.330 --> 00:52:05.060 The second term is a t cubed, which goes down. 00:52:05.060 --> 00:52:11.930 Then there's a fifth and a seventh and a ninth. 00:52:11.930 --> 00:52:15.260 And if you keep adding up the terms, amazingly enough, 00:52:15.260 --> 00:52:19.760 the series representation for the operator expression 00:52:19.760 --> 00:52:24.250 is the Taylor series expansion of the sine wave. 00:52:24.250 --> 00:52:26.130 Just like we would have expected. 00:52:26.130 --> 00:52:28.420 Again, I solved the differential equation, 00:52:28.420 --> 00:52:30.580 a second order differential equation, this time 00:52:30.580 --> 00:52:32.590 without calculus. 00:52:32.590 --> 00:52:35.290 All I did was polynomial math. 00:52:35.290 --> 00:52:38.410 And so with that, I'll finish just by saying, 00:52:38.410 --> 00:52:40.990 today, what we did was introduce, and basically 00:52:40.990 --> 00:52:43.720 finish, CT. 00:52:43.720 --> 00:52:46.780 Because there's such a strong analogy between what 00:52:46.780 --> 00:52:51.810 we did in DT, and how we'll approach thinking about CT.