WEBVTT 00:00:08.000 --> 00:00:14.000 Well, today is the last day on Laplace transform and the first 00:00:12.000 --> 00:00:18.000 day before we start the rest of the term, which will be spent on 00:00:16.000 --> 00:00:22.000 the study of systems. I would like to spend it on one 00:00:20.000 --> 00:00:26.000 more type of input function which, in general, 00:00:23.000 --> 00:00:29.000 your teachers in other courses will expect you to have had some 00:00:28.000 --> 00:00:34.000 acquaintance with. It is the kind associated with 00:00:32.000 --> 00:00:38.000 an impulse, so an input consisted of what is sometimes 00:00:36.000 --> 00:00:42.000 called a unit impulse. Now, what's an impulse? 00:00:40.000 --> 00:00:46.000 It covers actually a lot of things. 00:00:43.000 --> 00:00:49.000 It covers a situation where you withdraw from a bank account. 00:00:48.000 --> 00:00:54.000 For example, take half your money out of a 00:00:52.000 --> 00:00:58.000 bank account one day. It also would be modeled the 00:00:56.000 --> 00:01:02.000 same way. But the simplest way to 00:00:59.000 --> 00:01:05.000 understand it the first time through is as an impulse, 00:01:03.000 --> 00:01:09.000 if you know what an impulse is. If you have a variable force 00:01:08.000 --> 00:01:14.000 acting over time, and we will assume it is acting 00:01:12.000 --> 00:01:18.000 along a straight line so I don't have to worry about it being a 00:01:17.000 --> 00:01:23.000 vector, then the impulse, according to physicists, 00:01:20.000 --> 00:01:26.000 the physical definition, the impulse of f of t 00:01:25.000 --> 00:01:31.000 over some time interval. Let's say the time interval 00:01:30.000 --> 00:01:36.000 running from a to b is, by definition, 00:01:33.000 --> 00:01:39.000 the integral from a to b of f of t dt. 00:01:38.000 --> 00:01:44.000 Actually, I am going to do the 00:01:42.000 --> 00:01:48.000 most horrible thing this period. I will assume the force is 00:01:46.000 --> 00:01:52.000 actually a constant force. So, in that case, 00:01:50.000 --> 00:01:56.000 I wouldn't even have to bother with the integral at all. 00:01:55.000 --> 00:02:01.000 If f of t is a constant, let's say capital F, 00:01:58.000 --> 00:02:04.000 then the impulse is -- Well, that integral is simply 00:02:04.000 --> 00:02:10.000 the product of the two, the impulse over that time 00:02:08.000 --> 00:02:14.000 interval is simply F times b minus a. 00:02:12.000 --> 00:02:18.000 Just the product of those two. The force times the length of 00:02:17.000 --> 00:02:23.000 time for which it acts. Now, that is what I want to 00:02:21.000 --> 00:02:27.000 calculate, want to consider in connection with our little mass 00:02:26.000 --> 00:02:32.000 system. So, once again, 00:02:29.000 --> 00:02:35.000 I think this is probably the last time you'll see the little 00:02:34.000 --> 00:02:40.000 spring. Let's bid a tearful farewell to 00:02:37.000 --> 00:02:43.000 it. There is our little mass on 00:02:39.000 --> 00:02:45.000 wheels. And let's make it an undamped 00:02:42.000 --> 00:02:48.000 mass. It has an equilibrium point and 00:02:45.000 --> 00:02:51.000 all the other little things that go with the picture. 00:02:49.000 --> 00:02:55.000 And when I apply an impulse, what I mean is applying a 00:02:54.000 --> 00:03:00.000 constant force to this over a definite time interval. 00:03:00.000 --> 00:03:06.000 And that is what I mean by applying an impulse over that 00:03:03.000 --> 00:03:09.000 time interval. Now, what is the picture of 00:03:06.000 --> 00:03:12.000 such a thing? Well, the force is only going 00:03:09.000 --> 00:03:15.000 to be applied, in other words, 00:03:11.000 --> 00:03:17.000 I am going to push on the mass or pull on the mass with a 00:03:15.000 --> 00:03:21.000 constant force. With a little electromagnet 00:03:18.000 --> 00:03:24.000 here, we will assume, there is a pile of iron filings 00:03:22.000 --> 00:03:28.000 or something inside there. I turn on the electromagnet. 00:03:26.000 --> 00:03:32.000 It pulls with a constant force just between time zero and time 00:03:30.000 --> 00:03:36.000 two seconds. And then I stop. 00:03:34.000 --> 00:03:40.000 That is going to change the motion of the thing. 00:03:37.000 --> 00:03:43.000 First it is going to start pulling it toward the thing. 00:03:40.000 --> 00:03:46.000 And then, when it lets go, it will zoom back and there 00:03:44.000 --> 00:03:50.000 will be a certain motion after that. 00:03:46.000 --> 00:03:52.000 What the question is, if I want to solve that problem 00:03:49.000 --> 00:03:55.000 of the motion of that in terms of the Laplace transform, 00:03:53.000 --> 00:03:59.000 how am I going to model this force? 00:03:55.000 --> 00:04:01.000 Well, let's draw a picture of it first. 00:03:59.000 --> 00:04:05.000 It starts here. It is zero for t, 00:04:02.000 --> 00:04:08.000 let's say the force is applied between time zero to time h. 00:04:08.000 --> 00:04:14.000 And then its force is turned on, it stays constant and then 00:04:14.000 --> 00:04:20.000 it is turned off. And those vertical lines 00:04:18.000 --> 00:04:24.000 shouldn't be there. But, since in practice, 00:04:23.000 --> 00:04:29.000 it takes a tiny bit of time to turn a force on and off. 00:04:30.000 --> 00:04:36.000 It is, in practice, not unrealistic to suppose that 00:04:34.000 --> 00:04:40.000 there are approximately vertical lines there. 00:04:38.000 --> 00:04:44.000 They are slightly slanted but not too much. 00:04:41.000 --> 00:04:47.000 Now, I want it to be unit impulse. 00:04:44.000 --> 00:04:50.000 This is the force access and this is the time access. 00:04:49.000 --> 00:04:55.000 Since the impulse is the area under this curve, 00:04:53.000 --> 00:04:59.000 if I want that to be one, then if this is h, 00:04:56.000 --> 00:05:02.000 the height to which I -- In other words, 00:05:01.000 --> 00:05:07.000 the magnitude of the force must be one over h in order 00:05:05.000 --> 00:05:11.000 that the area be one, in order, in other words, 00:05:09.000 --> 00:05:15.000 that this integral be one, the area under that curve be 00:05:13.000 --> 00:05:19.000 one. So the unit impulse looks like 00:05:15.000 --> 00:05:21.000 that. The narrower it is here, 00:05:17.000 --> 00:05:23.000 the higher it has to be that way. 00:05:20.000 --> 00:05:26.000 The bigger the force must be if you want the end result to be a 00:05:24.000 --> 00:05:30.000 unit impulse. Now, to solve a problem, 00:05:27.000 --> 00:05:33.000 a typical problem, then, would be a spring. 00:05:32.000 --> 00:05:38.000 The mass is traveling on the track. 00:05:34.000 --> 00:05:40.000 Let's suppose the spring constant is one, 00:05:38.000 --> 00:05:44.000 so there would be a differential equation. 00:05:41.000 --> 00:05:47.000 And the right-hand side would be this f of t. 00:05:45.000 --> 00:05:51.000 Well, let's give it its name, the name I gave it before. 00:05:49.000 --> 00:05:55.000 Remember, I called the unit box function the thing which was one 00:05:55.000 --> 00:06:01.000 between zero and h and zero everywhere else. 00:06:00.000 --> 00:06:06.000 The notation we used for that was u, and then it had a double 00:06:04.000 --> 00:06:10.000 subscript from the starting point and the finishing point. 00:06:07.000 --> 00:06:13.000 So oh-- u(oh) of t. 00:06:10.000 --> 00:06:16.000 This much represents the thing if it only rose to the high one. 00:06:14.000 --> 00:06:20.000 But if it, instead, rises to the height one over h 00:06:17.000 --> 00:06:23.000 in order to make that area one, I have to multiply it 00:06:21.000 --> 00:06:27.000 by the factor one over h. Now, if you want to solve this 00:06:25.000 --> 00:06:31.000 by the Laplace transform. In other words, 00:06:28.000 --> 00:06:34.000 see what the motion of that mass is as I apply this unit 00:06:32.000 --> 00:06:38.000 impulse to it over that time interval. 00:06:34.000 --> 00:06:40.000 You have to take the Laplace transform, if that is the way we 00:06:38.000 --> 00:06:44.000 are doing it. Now, the left-hand side is just 00:06:41.000 --> 00:06:47.000 routine and would involve the initial conditions. 00:06:44.000 --> 00:06:50.000 The whole interest is taking the Laplace transform of the 00:06:48.000 --> 00:06:54.000 right-hand side. And that is what I want to do 00:06:51.000 --> 00:06:57.000 now. The problem is what is the 00:06:52.000 --> 00:06:58.000 Laplace transform of this guy? 00:07:02.000 --> 00:07:08.000 Well, remember, to do everything else, 00:07:04.000 --> 00:07:10.000 you do everything by writing in terms of the unit step function? 00:07:08.000 --> 00:07:14.000 This function that we are talking about is one over h 00:07:12.000 --> 00:07:18.000 times what you get by first stepping up to one. 00:07:16.000 --> 00:07:22.000 That is the unit step function, which goes up by one and tries 00:07:20.000 --> 00:07:26.000 to stay at one ever after. And then, when it gets to h, 00:07:24.000 --> 00:07:30.000 it has got to step down. Well, the way you make it step 00:07:27.000 --> 00:07:33.000 down is by subtracting off the function, which is the unit step 00:07:31.000 --> 00:07:37.000 function but where the step takes place, not at time zero 00:07:35.000 --> 00:07:41.000 but at time h. In other words, 00:07:38.000 --> 00:07:44.000 I translate the unit step function of course with, 00:07:42.000 --> 00:07:48.000 I don't think I have to draw that picture again. 00:07:45.000 --> 00:07:51.000 The unit step function looks like zing. 00:07:47.000 --> 00:07:53.000 And if you translate it to the right by h it looks like zing. 00:07:51.000 --> 00:07:57.000 And then make it negative to subtract it off. 00:07:54.000 --> 00:08:00.000 And what you will get is this box function. 00:07:56.000 --> 00:08:02.000 So we want to take the Laplace transform of this thing. 00:08:01.000 --> 00:08:07.000 Well, let's assume, for the sake of argument that 00:08:05.000 --> 00:08:11.000 you didn't remember. Well, you had to use the 00:08:08.000 --> 00:08:14.000 formula at 2:00 AM this morning and, therefore, 00:08:12.000 --> 00:08:18.000 you do remember it. [LAUGHTER] So I don't have to 00:08:16.000 --> 00:08:22.000 recopy the formula onto the board. 00:08:19.000 --> 00:08:25.000 Maybe if there is room there. All right, let's put it up 00:08:24.000 --> 00:08:30.000 there. It says that u of t minus a 00:08:26.000 --> 00:08:32.000 times f, any f, so let's call it g so 00:08:31.000 --> 00:08:37.000 you won't confuse it with this particular one, 00:08:34.000 --> 00:08:40.000 times g translated. If you translate a function 00:08:39.000 --> 00:08:45.000 from t, if you translate it to the right by a then its Laplace 00:08:44.000 --> 00:08:50.000 transform is e to the minus a s times whatever the 00:08:48.000 --> 00:08:54.000 old Laplace transform was, g of s. 00:08:51.000 --> 00:08:57.000 Multiply by an exponential on the right. 00:08:54.000 --> 00:09:00.000 On the left that corresponds to translation. 00:08:58.000 --> 00:09:04.000 Except you must remember to put in that factor u for a secret 00:09:02.000 --> 00:09:08.000 reason which I spent half of Wednesday explaining. 00:09:05.000 --> 00:09:11.000 What do we have here? The Laplace transform of u of t, 00:09:09.000 --> 00:09:15.000 that is easy. That is simply one over s. 00:09:13.000 --> 00:09:19.000 The Laplace transform of this 00:09:15.000 --> 00:09:21.000 other guy we get from the formula. 00:09:17.000 --> 00:09:23.000 It is basically one over s. No, the Laplace transform of u 00:09:21.000 --> 00:09:27.000 of t. But because it has been 00:09:24.000 --> 00:09:30.000 translated to the right by h, I have to multiply it by that 00:09:28.000 --> 00:09:34.000 factor e to the minus h times s. 00:09:38.000 --> 00:09:44.000 That is the answer. And, if you want to solve 00:09:40.000 --> 00:09:46.000 problems, this is what you would feed into the equation. 00:09:44.000 --> 00:09:50.000 And you would calculate and calculate and calculate it. 00:09:47.000 --> 00:09:53.000 But that is not what I want to do now because that was 00:09:51.000 --> 00:09:57.000 Wednesday and this is Friday. You have the right to expect 00:09:55.000 --> 00:10:01.000 something new. Here is what I am going to do 00:09:57.000 --> 00:10:03.000 new. I am going to let h go to zero. 00:10:01.000 --> 00:10:07.000 As h goes to zero, this function gets narrower and 00:10:06.000 --> 00:10:12.000 narrower, but it also has to get higher and higher because its 00:10:12.000 --> 00:10:18.000 area has to stay one. What I am interested in, 00:10:16.000 --> 00:10:22.000 first of all, is what happens to the Laplace 00:10:20.000 --> 00:10:26.000 transform as h goes to zero. In other words, 00:10:25.000 --> 00:10:31.000 what is the limit, as h goes to zero of -- 00:10:30.000 --> 00:10:36.000 Well, what is that function? One minus e to the negative hs 00:10:34.000 --> 00:10:40.000 divided by hs. 00:10:37.000 --> 00:10:43.000 Well, this is an 18.01 problem, an ordinary calculus problem, 00:10:42.000 --> 00:10:48.000 but let's do it nicely. You see, the nice way to do it 00:10:46.000 --> 00:10:52.000 is to make a substitution. We will change h s to u 00:10:50.000 --> 00:10:56.000 because it is occurring as a unit in both cases. 00:10:55.000 --> 00:11:01.000 This is going to be the same as the limit as u goes to zero. 00:11:00.000 --> 00:11:06.000 I think there are too many u's 00:11:04.000 --> 00:11:10.000 here already. I cannot use u, 00:11:07.000 --> 00:11:13.000 you cannot use t, v is velocity, 00:11:10.000 --> 00:11:16.000 w is wavefunction. There is no letter. 00:11:13.000 --> 00:11:19.000 All right, u. It is one minus e to the 00:11:17.000 --> 00:11:23.000 negative u over u. 00:11:20.000 --> 00:11:26.000 So what is the answer? Well, either you know the 00:11:25.000 --> 00:11:31.000 answer or you replace this by, say, the first couple of terms 00:11:30.000 --> 00:11:36.000 of the Taylor series. But I think most of you would 00:11:35.000 --> 00:11:41.000 use L'Hopital's rule, so let's do that. 00:11:38.000 --> 00:11:44.000 The derivative of the top is zero here. 00:11:40.000 --> 00:11:46.000 The derivative by the chain rule of e to the negative u is e 00:11:44.000 --> 00:11:50.000 to the negative u times minus one. 00:11:47.000 --> 00:11:53.000 And that minus one cancels that minus. 00:11:49.000 --> 00:11:55.000 So the derivative of the top is simply e to the negative u and 00:11:53.000 --> 00:11:59.000 the derivative of the bottom is one. 00:11:55.000 --> 00:12:01.000 So, as u goes to zero, that limit is one. 00:12:00.000 --> 00:12:06.000 Interesting. Let's draw a picture this way. 00:12:03.000 --> 00:12:09.000 I will draw it schematically. Up here is the function one 00:12:08.000 --> 00:12:14.000 over h times u zero h of t, our box function, 00:12:14.000 --> 00:12:20.000 except it has the height one over h instead of the 00:12:19.000 --> 00:12:25.000 height one. We have just calculated that 00:12:23.000 --> 00:12:29.000 its Laplace transform is that funny thing, one minus e to the 00:12:28.000 --> 00:12:34.000 minus hs divided by hs. 00:12:34.000 --> 00:12:40.000 That is the top line. All this is completely kosher, 00:12:39.000 --> 00:12:45.000 but now I am going to let h go to zero. 00:12:43.000 --> 00:12:49.000 And the question is what do we get now? 00:12:47.000 --> 00:12:53.000 Well, I just calculated for you that this thing approaches one, 00:12:53.000 --> 00:12:59.000 has the limit one. And now, let's fill in the 00:12:58.000 --> 00:13:04.000 picture. What does this thing approach? 00:13:03.000 --> 00:13:09.000 Well, it approaches a function which is zero everywhere. 00:13:10.000 --> 00:13:16.000 As h approaches zero, this green box turns into a box 00:13:17.000 --> 00:13:23.000 which is zero everywhere except at zero. 00:13:22.000 --> 00:13:28.000 And there, it is infinitely high. 00:13:26.000 --> 00:13:32.000 So, keep going up. 00:13:35.000 --> 00:13:41.000 Now, of course, that is not a function. 00:13:38.000 --> 00:13:44.000 People call it a function but it isn't. 00:13:41.000 --> 00:13:47.000 Mathematicians call it a generalized function, 00:13:45.000 --> 00:13:51.000 but that is not a function either. 00:13:48.000 --> 00:13:54.000 It is just a way of making you feel comfortable by talking 00:13:53.000 --> 00:13:59.000 about something which isn't really a function. 00:13:56.000 --> 00:14:02.000 It was given the name, introduced formally into 00:14:00.000 --> 00:14:06.000 mathematics by a physicist, Dirac. 00:14:05.000 --> 00:14:11.000 And he, looking ahead to the future, did what many people do 00:14:09.000 --> 00:14:15.000 who introduce something into the literature, a formula or a 00:14:13.000 --> 00:14:19.000 function or something which they think is going to be important. 00:14:17.000 --> 00:14:23.000 They never name it directly after themselves, 00:14:20.000 --> 00:14:26.000 but they always use as the symbol for it the first letter 00:14:24.000 --> 00:14:30.000 of their name. I cannot tell you how often 00:14:26.000 --> 00:14:32.000 that has happened. Maybe even Euler called e for 00:14:31.000 --> 00:14:37.000 that reason, although he claims it was in Latin because it has 00:14:37.000 --> 00:14:43.000 to do with exponentials. Well, luckily his name began 00:14:42.000 --> 00:14:48.000 with an E, too. That is Paul Dirac's delta 00:14:45.000 --> 00:14:51.000 function. I won't dignify it by the name 00:14:49.000 --> 00:14:55.000 function by writing that out, by putting the world function 00:14:54.000 --> 00:15:00.000 here, too, but it is called the delta function. 00:15:00.000 --> 00:15:06.000 From this point on, the entire rest of the lecture 00:15:03.000 --> 00:15:09.000 has a slight fictional element. The entire rest of the lecture 00:15:08.000 --> 00:15:14.000 is in figurative quotation marks, so you are not entirely 00:15:12.000 --> 00:15:18.000 responsible for anything I say. This is a non-function, 00:15:16.000 --> 00:15:22.000 but you put it in there and call it a function. 00:15:19.000 --> 00:15:25.000 And you naturally want to complete, if it's a function 00:15:23.000 --> 00:15:29.000 then it must have a Laplace transform, even though it 00:15:27.000 --> 00:15:33.000 doesn't, so the diagram is completed that way. 00:15:32.000 --> 00:15:38.000 And its Laplace transform is declared to be one. 00:15:35.000 --> 00:15:41.000 So let's start listing the properties of this weird thing. 00:15:54.000 --> 00:16:00.000 The delta function, its Laplace transform is one. 00:16:05.000 --> 00:16:11.000 Now, one of the things is we have not yet expressed the fact 00:16:09.000 --> 00:16:15.000 that it is a unit impulse. In other words, 00:16:13.000 --> 00:16:19.000 since the areas of all of these boxes, they all have areas one 00:16:17.000 --> 00:16:23.000 as they are shrunk this way they get higher that way. 00:16:22.000 --> 00:16:28.000 By convention, one says that the area under 00:16:25.000 --> 00:16:31.000 the orange curve also remains one in the limit. 00:16:30.000 --> 00:16:36.000 Now, how am I going to express that? 00:16:32.000 --> 00:16:38.000 Well, it is done by the following formula that the 00:16:35.000 --> 00:16:41.000 integral, the total impulse of the delta function should be 00:16:39.000 --> 00:16:45.000 one. Now, where do I integrate? 00:16:41.000 --> 00:16:47.000 Well, from any place that it is zero to any place that it is 00:16:45.000 --> 00:16:51.000 zero on the other side of that vertical line. 00:16:48.000 --> 00:16:54.000 But, in order to avoid controversy, people integrate 00:16:52.000 --> 00:16:58.000 all the way from negative infinity to infinity since it 00:16:55.000 --> 00:17:01.000 doesn't hurt. Does it? 00:16:57.000 --> 00:17:03.000 It is zero practically all the time. 00:17:01.000 --> 00:17:07.000 This is the function whose Laplace transform is one. 00:17:06.000 --> 00:17:12.000 Its integral from minus infinity to infinity is one. 00:17:11.000 --> 00:17:17.000 How else can we calculate for it? 00:17:15.000 --> 00:17:21.000 Well, I would like to calculate its convolution. 00:17:20.000 --> 00:17:26.000 Here is f of t. What happens if I convolute it 00:17:26.000 --> 00:17:32.000 with the delta function? Well, if you go back to the 00:17:31.000 --> 00:17:37.000 definition of the convolution, you know, it is that funny 00:17:35.000 --> 00:17:41.000 integral, you are going to do a lot of head scratching because 00:17:40.000 --> 00:17:46.000 it is not really all that clear how to integrate with the delta 00:17:44.000 --> 00:17:50.000 function. Instead of doing that let's 00:17:46.000 --> 00:17:52.000 assume that it follows the laws of the Laplace transform. 00:17:50.000 --> 00:17:56.000 In that case, its Laplace transform would be 00:17:53.000 --> 00:17:59.000 what? Well, the whole thing of a 00:17:55.000 --> 00:18:01.000 convolution is that the Laplace transform of the convolution is 00:18:00.000 --> 00:18:06.000 the product of the two separate Laplace transforms. 00:18:05.000 --> 00:18:11.000 So that is going to be F of s times the Laplace 00:18:09.000 --> 00:18:15.000 transform of the delta function, which is one. 00:18:13.000 --> 00:18:19.000 Now, what must this thing be? Well, there is some ambiguity 00:18:18.000 --> 00:18:24.000 as to what it is for negative values of t. 00:18:21.000 --> 00:18:27.000 But if we, by brute force, decide for negative values of t 00:18:26.000 --> 00:18:32.000 it is going to have the value zero, that is the way we make 00:18:31.000 --> 00:18:37.000 things unique. In fact, why don't we make f of 00:18:35.000 --> 00:18:41.000 unique that way to start with? 00:18:38.000 --> 00:18:44.000 This is a function now that is allowed to do anything it wants 00:18:41.000 --> 00:18:47.000 on the right-hand side of zero starting at zero, 00:18:44.000 --> 00:18:50.000 but on the left-hand side of zero it is wiped away and must 00:18:48.000 --> 00:18:54.000 be zero. This is a definite thing now. 00:18:50.000 --> 00:18:56.000 Its convolution is this. And the inverse Laplace 00:18:53.000 --> 00:18:59.000 transform is -- The answer, in other words, 00:18:57.000 --> 00:19:03.000 is the same thing as what u of t f of t would be. 00:19:02.000 --> 00:19:08.000 It's the same thing, F of s. 00:19:04.000 --> 00:19:10.000 And so, the conclusion is that these are equal, 00:19:08.000 --> 00:19:14.000 since they must be unique. They have been made unique by 00:19:12.000 --> 00:19:18.000 making them zero for t negative. In other words, 00:19:16.000 --> 00:19:22.000 apply to a function, well, I won't recopy it. 00:19:19.000 --> 00:19:25.000 But the point is that delta t, for the convolution operation, 00:19:24.000 --> 00:19:30.000 is acting like an identity. If I multiply, 00:19:29.000 --> 00:19:35.000 in the sense of convolution, it is a peculiar operation. 00:19:33.000 --> 00:19:39.000 But algebraically, it has a lot of the properties 00:19:36.000 --> 00:19:42.000 of multiplication. It is communitive. 00:19:39.000 --> 00:19:45.000 It is linear in both factors. In other words, 00:19:42.000 --> 00:19:48.000 it is almost anything you would want with multiplication. 00:19:46.000 --> 00:19:52.000 It has an identity element, identity function. 00:19:49.000 --> 00:19:55.000 And the identity function is the Dirac delta function. 00:19:53.000 --> 00:19:59.000 Anything else here? Yeah, I will throw in one more 00:19:57.000 --> 00:20:03.000 thing. It would just require one more 00:20:01.000 --> 00:20:07.000 phony argument, which I won't bother giving 00:20:04.000 --> 00:20:10.000 you, but it is not totally implausible. 00:20:06.000 --> 00:20:12.000 After all, u of t, the unit step function is not 00:20:11.000 --> 00:20:17.000 differentiable, is not a differentiable 00:20:13.000 --> 00:20:19.000 function. It looks like this. 00:20:15.000 --> 00:20:21.000 Here its derivative is zero, here its derivative is zero, 00:20:19.000 --> 00:20:25.000 and in this class it is not even defined in between. 00:20:23.000 --> 00:20:29.000 But, I don't care, I will make it go straight up. 00:20:27.000 --> 00:20:33.000 The question is what's its derivative? 00:20:31.000 --> 00:20:37.000 Well, zero here, zero there and infinity at 00:20:34.000 --> 00:20:40.000 zero, so it must be the delta function. 00:20:37.000 --> 00:20:43.000 That has exactly the right properties. 00:20:40.000 --> 00:20:46.000 So the same people who will tell you this will tell you that 00:20:44.000 --> 00:20:50.000 also. And, in fact, 00:20:45.000 --> 00:20:51.000 when you use it to solve differential equations it acts 00:20:50.000 --> 00:20:56.000 as if that is true. I think I have given you an 00:20:53.000 --> 00:20:59.000 example on your homework. Let me now show you a typical 00:20:57.000 --> 00:21:03.000 example of the way the Dirac delta function would be used to 00:21:02.000 --> 00:21:08.000 solve a problem. Let's go back to our little 00:21:06.000 --> 00:21:12.000 spring, since it is the easiest thing. 00:21:09.000 --> 00:21:15.000 You are familiar with it from a physical point of view, 00:21:13.000 --> 00:21:19.000 and it is the easiest thing to illustrate on. 00:21:16.000 --> 00:21:22.000 We have our spring mass system. Where is it? 00:21:20.000 --> 00:21:26.000 Is it on the board? Up there. 00:21:22.000 --> 00:21:28.000 That one. And the differential equation 00:21:25.000 --> 00:21:31.000 we are going to solve is y double prime plus y 00:21:29.000 --> 00:21:35.000 equals -- And now, I am going to assume 00:21:33.000 --> 00:21:39.000 that the spring is kicked with impulse a. 00:21:37.000 --> 00:21:43.000 I am not going to kick it at time t equals zero, 00:21:42.000 --> 00:21:48.000 since that would get us into slight technical difficulties. 00:21:47.000 --> 00:21:53.000 Anyway, it is more fun to kick it at time pi over two. 00:21:52.000 --> 00:21:58.000 The thing is, 00:21:54.000 --> 00:22:00.000 what is happening? Well, we have got to have 00:21:58.000 --> 00:22:04.000 initial conditions. The initial conditions are 00:22:02.000 --> 00:22:08.000 going to be, let's start at time zero. 00:22:05.000 --> 00:22:11.000 We will start it at the position one. 00:22:08.000 --> 00:22:14.000 So I take my spring, I drag it to the position one, 00:22:11.000 --> 00:22:17.000 I take the little mass there and then let it go. 00:22:15.000 --> 00:22:21.000 And so it starts going birr. But right when it gets to the 00:22:19.000 --> 00:22:25.000 equilibrium point I give it a, "cha!" with unit impulse. 00:22:23.000 --> 00:22:29.000 I started it from rest. Those will be the initial 00:22:27.000 --> 00:22:33.000 conditions. And I want to say that I kicked 00:22:31.000 --> 00:22:37.000 it, not with unit impulse, but with the impulse a. 00:22:35.000 --> 00:22:41.000 Bigger. And I did that at time pi over 00:22:38.000 --> 00:22:44.000 two. So how are we going to say 00:22:41.000 --> 00:22:47.000 that? Well, kick it means delivered 00:22:43.000 --> 00:22:49.000 that impulse over an extremely short time interval, 00:22:47.000 --> 00:22:53.000 but in such a way kicked it sufficiently hard that the total 00:22:52.000 --> 00:22:58.000 impulse was a. The way to say that is kick it 00:22:56.000 --> 00:23:02.000 with the Dirac delta function. Translate it to the point time 00:23:02.000 --> 00:23:08.000 pi over two. Not at zero any longer. 00:23:06.000 --> 00:23:12.000 t minus pi over two. 00:23:09.000 --> 00:23:15.000 But that would kick it with a unit impulse. 00:23:12.000 --> 00:23:18.000 I want it to kick it with the impulse a, so I will just 00:23:16.000 --> 00:23:22.000 multiply that by the constant factor a. 00:23:20.000 --> 00:23:26.000 Let's put this over here. y of zero equals one, 00:23:24.000 --> 00:23:30.000 that's the starting value. Now we have a problem. 00:23:30.000 --> 00:23:36.000 The only thing new in solving this with the Laplace transform 00:23:33.000 --> 00:23:39.000 is I have this funny right-hand side. 00:23:36.000 --> 00:23:42.000 But it corresponds to a physical situation. 00:23:38.000 --> 00:23:44.000 Let's do it. You take the Laplace transform 00:23:41.000 --> 00:23:47.000 of both sides of the equation. Remember how to do that? 00:23:44.000 --> 00:23:50.000 You have to take account of the initial conditions. 00:23:48.000 --> 00:23:54.000 The Laplace transform of the second derivative is you 00:23:51.000 --> 00:23:57.000 multiply by s squared, and then you have to subtract. 00:23:55.000 --> 00:24:01.000 You have to use these initial conditions. 00:23:59.000 --> 00:24:05.000 This one won't give you anything, but the first one 00:24:02.000 --> 00:24:08.000 means I have to subtract one times s. 00:24:05.000 --> 00:24:11.000 That is the Laplace transform of y double prime. 00:24:09.000 --> 00:24:15.000 The Laplace transform of y, of course, is just capital Y. 00:24:13.000 --> 00:24:19.000 And how about the Laplace 00:24:16.000 --> 00:24:22.000 transform of the right-hand side. 00:24:18.000 --> 00:24:24.000 Well, we will have the constant factor a because the Laplace 00:24:22.000 --> 00:24:28.000 transform is linear. And now, the delta function 00:24:25.000 --> 00:24:31.000 would have the transform one. But when I translate it, 00:24:30.000 --> 00:24:36.000 pi over two, that means I have to use that 00:24:33.000 --> 00:24:39.000 formula. Translate it by pi over two 00:24:35.000 --> 00:24:41.000 means take the one that it would have been otherwise and multiply 00:24:40.000 --> 00:24:46.000 it by e, that exponential factor. 00:24:42.000 --> 00:24:48.000 It would be e to the minus pi over two, 00:24:46.000 --> 00:24:52.000 that is the A times s times one, which would be the g of s, 00:24:49.000 --> 00:24:55.000 the Laplace transform or the delta function before it 00:24:54.000 --> 00:25:00.000 had been translated. But I don't have to put that in 00:24:57.000 --> 00:25:03.000 because it's one. I am multiplying by one. 00:25:01.000 --> 00:25:07.000 And to do everything now is routine. 00:25:03.000 --> 00:25:09.000 Solve for the Laplace transform. 00:25:05.000 --> 00:25:11.000 Well, what is it? It is y is equal to. 00:25:08.000 --> 00:25:14.000 I put the s on the other side. That makes the right-hand side 00:25:12.000 --> 00:25:18.000 the sum of two terms. And I divide by the coefficient 00:25:15.000 --> 00:25:21.000 of y, which is s squared plus one. 00:25:18.000 --> 00:25:24.000 The s is over on the right-hand side and it is divided by s 00:25:22.000 --> 00:25:28.000 squared plus one. And the other factor is there, 00:25:25.000 --> 00:25:31.000 too. And it, too, 00:25:26.000 --> 00:25:32.000 is divided by s squared plus one. 00:25:35.000 --> 00:25:41.000 Now, we take the inverse Laplace transform of those two 00:25:38.000 --> 00:25:44.000 terms and add them up. 00:26:00.000 --> 00:26:06.000 What will we get? Well, y is equal to, 00:26:02.000 --> 00:26:08.000 the inverse Laplace transform of s over s squared plus one is 00:26:07.000 --> 00:26:13.000 cosine t. 00:26:11.000 --> 00:26:17.000 Now, for this thing we will have to use our formula. 00:26:15.000 --> 00:26:21.000 If this weren't here, the inverse Laplace transform 00:26:19.000 --> 00:26:25.000 of a over s squared plus one would 00:26:24.000 --> 00:26:30.000 be what? Well, it would be a times the 00:26:27.000 --> 00:26:33.000 sine of t. 00:26:35.000 --> 00:26:41.000 In other words, if this is the g of s 00:26:37.000 --> 00:26:43.000 then the function on the left would be basically A sine t. 00:26:41.000 --> 00:26:47.000 But because it has been 00:26:43.000 --> 00:26:49.000 multiplied by that exponential factor, e to the minus as 00:26:47.000 --> 00:26:53.000 where a is pi over two, 00:26:50.000 --> 00:26:56.000 the left-hand side has to be changed from A sine t 00:26:53.000 --> 00:26:59.000 to what it would be with the translated form. 00:26:58.000 --> 00:27:04.000 So the rest of it is u of t minus pi over two, 00:27:01.000 --> 00:27:07.000 because a is pi over two, times what it would have 00:27:05.000 --> 00:27:11.000 been just from the factor g of s itself. 00:27:09.000 --> 00:27:15.000 In other words, A times the sine of, 00:27:11.000 --> 00:27:17.000 again, t minus pi over two. 00:27:15.000 --> 00:27:21.000 I am applying that formula, but I am applying it in that 00:27:19.000 --> 00:27:25.000 direction. I started with this, 00:27:21.000 --> 00:27:27.000 and I want to recover the left-hand side. 00:27:24.000 --> 00:27:30.000 And that is what it must look like. 00:27:26.000 --> 00:27:32.000 The A, of course, just gets dragged along for the 00:27:29.000 --> 00:27:35.000 free ride. Now, as I emphasized to you 00:27:34.000 --> 00:27:40.000 last time, and I hope you did on your homework that you handed 00:27:38.000 --> 00:27:44.000 in, you mustn't leave it in that form. 00:27:41.000 --> 00:27:47.000 You have to make cases because people will expect you to tell 00:27:46.000 --> 00:27:52.000 them what the meaning of this is. 00:27:49.000 --> 00:27:55.000 Now, if t is less than pi over two, this is zero. 00:27:52.000 --> 00:27:58.000 And, therefore, that term does not exist. 00:27:56.000 --> 00:28:02.000 So the first part of it is just the cosine t term if 00:28:00.000 --> 00:28:06.000 t lies between zero and pi over two. 00:28:05.000 --> 00:28:11.000 If t is bigger than pi over two then this factor is 00:28:10.000 --> 00:28:16.000 one. It's the unit step function. 00:28:12.000 --> 00:28:18.000 And I, therefore, must add in this term. 00:28:16.000 --> 00:28:22.000 Now, what is that term? What is the sine of t minus pi 00:28:20.000 --> 00:28:26.000 over two? The sine of t looks 00:28:25.000 --> 00:28:31.000 like that. The sine of t, 00:28:27.000 --> 00:28:33.000 if I translate it, looks like this. 00:28:31.000 --> 00:28:37.000 If I translate it by pi over two. 00:28:33.000 --> 00:28:39.000 And let's finish it up, the pi that was over here moved 00:28:38.000 --> 00:28:44.000 into position. That curve is the curve 00:28:41.000 --> 00:28:47.000 negative cosine t. 00:28:51.000 --> 00:28:57.000 And so the answer is if t is bigger than pi over two, 00:28:55.000 --> 00:29:01.000 it is cosine t minus A times cosine t. 00:29:00.000 --> 00:29:06.000 Or, in other words, 00:29:02.000 --> 00:29:08.000 it is one minus A times cosine t. 00:29:11.000 --> 00:29:17.000 Now, do those match up? They have always got to match 00:29:13.000 --> 00:29:19.000 up, or you have made a mistake. You always have to get a 00:29:17.000 --> 00:29:23.000 continuous function when you have just discontinuities. 00:29:20.000 --> 00:29:26.000 Do we get a continuous function? 00:29:22.000 --> 00:29:28.000 Yeah, when t is pi over two the value here is 00:29:25.000 --> 00:29:31.000 zero. The value of this is also zero 00:29:27.000 --> 00:29:33.000 at pi over two. There is no conflict in the 00:29:31.000 --> 00:29:37.000 values. Values doesn't suddenly jump. 00:29:33.000 --> 00:29:39.000 The function is continuous. It is not differential but it 00:29:38.000 --> 00:29:44.000 is continuous. Well, what function does that 00:29:41.000 --> 00:29:47.000 look like? There are cases. 00:29:43.000 --> 00:29:49.000 It starts out life as the function cosine t. 00:29:47.000 --> 00:29:53.000 So it gets to here. And at t equals pi over two, 00:29:51.000 --> 00:29:57.000 the mass gets kicked and that changes the function. 00:29:55.000 --> 00:30:01.000 Now, what are the values? Well, if A is bigger than one 00:30:01.000 --> 00:30:07.000 this is a negative number and it therefore becomes 00:30:07.000 --> 00:30:13.000 the function negative cosine t. 00:30:11.000 --> 00:30:17.000 Now, negative cosine t looks like this, the blue guy. 00:30:16.000 --> 00:30:22.000 Negative cosine t is a function that looks like this. 00:30:21.000 --> 00:30:27.000 So it goes from here, it reverses direction, 00:30:26.000 --> 00:30:32.000 the mass reverses direction from what you thought it was 00:30:31.000 --> 00:30:37.000 going to do. And it does that because A is 00:30:36.000 --> 00:30:42.000 so large that that impulse was enough to make it reverse 00:30:40.000 --> 00:30:46.000 direction. Of course it might only do 00:30:42.000 --> 00:30:48.000 this, but this is what will happen if A is bigger than one. 00:30:47.000 --> 00:30:53.000 This will be A, which is a lot bigger than one. 00:30:50.000 --> 00:30:56.000 If it's not so much bigger than one it might look like that. 00:30:55.000 --> 00:31:01.000 So A is just bigger than one. How's that? 00:30:59.000 --> 00:31:05.000 Well, what if A is less than one? 00:31:01.000 --> 00:31:07.000 Well, in that case it stays positive. 00:31:04.000 --> 00:31:10.000 If A is less than one, this is now still a positive 00:31:07.000 --> 00:31:13.000 number. And, therefore, 00:31:09.000 --> 00:31:15.000 the cosine continues on its merry way. 00:31:12.000 --> 00:31:18.000 The only thing is it might be a little more sluggish or it might 00:31:17.000 --> 00:31:23.000 be very peppy and do that. Let's just go that far. 00:31:20.000 --> 00:31:26.000 This will be the case A less than one. 00:31:23.000 --> 00:31:29.000 Well, of course, the most interesting case is 00:31:26.000 --> 00:31:32.000 what happens if A is exactly equal to one? 00:31:32.000 --> 00:31:38.000 The porridge is exactly just right, I think that's the 00:31:37.000 --> 00:31:43.000 phrase. Too hot. 00:31:38.000 --> 00:31:44.000 Too cold. Just right. 00:31:40.000 --> 00:31:46.000 When A is equal to one, it is zero. 00:31:44.000 --> 00:31:50.000 It starts out as cosine t. 00:31:48.000 --> 00:31:54.000 When it gets to t, it continues on ever after as 00:31:52.000 --> 00:31:58.000 the function zero. I have a visual aid for the 00:31:57.000 --> 00:32:03.000 only time this term. It didn't work at all. 00:32:02.000 --> 00:32:08.000 I mean, on the other hand, the last hour, 00:32:06.000 --> 00:32:12.000 the people who worked it were not intrinsically baseball 00:32:11.000 --> 00:32:17.000 players, so we will use the equation of the pendulum 00:32:15.000 --> 00:32:21.000 instead. That is a lot easier than mass 00:32:18.000 --> 00:32:24.000 spring. This is a pendulum. 00:32:20.000 --> 00:32:26.000 It is undamped because I declare it to be and it swings 00:32:25.000 --> 00:32:31.000 back and forth. And here I am releasing it. 00:32:30.000 --> 00:32:36.000 The variable is not x or y but theta, the angle through. 00:32:34.000 --> 00:32:40.000 Here theta is one, let's say. 00:32:37.000 --> 00:32:43.000 That's about one radian. It starts there and swings back 00:32:41.000 --> 00:32:47.000 and forth. It is not damped, 00:32:44.000 --> 00:32:50.000 so it never loses amplitude, particularly if I swish it, 00:32:48.000 --> 00:32:54.000 if I move my hand a little bit. I want someone who knows how to 00:32:53.000 --> 00:32:59.000 bat a baseball. That was the problem last hour. 00:32:57.000 --> 00:33:03.000 Two people. One to release it. 00:33:01.000 --> 00:33:07.000 I will stand up and try to hold it here. 00:33:04.000 --> 00:33:10.000 Somebody releases it. And then somebody who has to be 00:33:08.000 --> 00:33:14.000 very skillful should apply a unit impulse of exactly one when 00:33:13.000 --> 00:33:19.000 it gets to the equilibrium point. 00:33:16.000 --> 00:33:22.000 So who can do that? Who can play baseball here? 00:33:20.000 --> 00:33:26.000 Come on. Somebody elected? 00:33:30.000 --> 00:33:36.000 All right. Come on. [APPLAUSE] 00:33:41.000 --> 00:33:47.000 Somebody release it, too. 00:33:44.000 --> 00:33:50.000 Somebody tall to handle it all. I think that will be me. 00:33:52.000 --> 00:33:58.000 Just hold it at what you would take to be one radian. 00:34:00.000 --> 00:34:06.000 He releases it. When it gets to the bottom, 00:34:04.000 --> 00:34:10.000 you will have to get way down, and maybe on this side. 00:34:11.000 --> 00:34:17.000 Are you a lefty or a righty? Rightly. 00:34:15.000 --> 00:34:21.000 Okay. Bat it what part. 00:34:17.000 --> 00:34:23.000 Give it a good swat. I will stand up higher. 00:34:22.000 --> 00:34:28.000 Help. I'm not very stable. 00:34:25.000 --> 00:34:31.000 [APPLAUSE] A trial run. Again. 00:34:30.000 --> 00:34:36.000 Okay. A little further out. 00:34:32.000 --> 00:34:38.000 First of all, you have to see where it's 00:34:36.000 --> 00:34:42.000 going. Why don't you stand, 00:34:38.000 --> 00:34:44.000 oh, you bat rightly. That's right. 00:34:41.000 --> 00:34:47.000 Okay. Let's try it again. 00:34:59.000 --> 00:35:05.000 Strike one. It's okay. It's the beginning of the baseball season. One more. The Red Sox are having trouble, too. Not bad. [APPLAUSE] 00:35:13.000 --> 00:35:19.000 If he had hit even harder it would have reversed direction 00:35:16.000 --> 00:35:22.000 and gone that way. If you hadn't hit it quite as 00:35:20.000 --> 00:35:26.000 hard it would have continued on, still at cosine t, 00:35:24.000 --> 00:35:30.000 but with less amplitude. But if you hit it exactly right 00:35:28.000 --> 00:35:34.000 -- It is fun to try to do. 00:35:31.000 --> 00:35:37.000 Toomre in our department is a master at this, 00:35:36.000 --> 00:35:42.000 but he has been practicing for years. 00:35:40.000 --> 00:35:46.000 He can take a little mallet and go blunk, and it stops 00:35:45.000 --> 00:35:51.000 absolutely dead. It is unbelievable. 00:35:49.000 --> 00:35:55.000 I should have had him give the lecture. 00:35:53.000 --> 00:35:59.000 Now, I would like to do something truly serious. 00:36:00.000 --> 00:36:06.000 Here, I guess. Because there is a certain 00:36:03.000 --> 00:36:09.000 amount of engineering lingo you have to learn. 00:36:07.000 --> 00:36:13.000 It is used by almost everybody. Not architects and biologists 00:36:12.000 --> 00:36:18.000 probably quite yet, but anybody that uses the 00:36:16.000 --> 00:36:22.000 Laplace transform will use these words in connection with it. 00:36:21.000 --> 00:36:27.000 I really think, since it is such a widespread 00:36:25.000 --> 00:36:31.000 technique, that these are things you should know. 00:36:31.000 --> 00:36:37.000 Anyway, it will be easy. It is just the enrichment of 00:36:34.000 --> 00:36:40.000 your vocabulary. It is always fun to learn new 00:36:37.000 --> 00:36:43.000 vocabulary words. So, let's just consider a 00:36:40.000 --> 00:36:46.000 general second order equation. By the way, all this applies to 00:36:45.000 --> 00:36:51.000 higher order equations, too. 00:36:47.000 --> 00:36:53.000 It applies to systems. The same words are used, 00:36:50.000 --> 00:36:56.000 but let's use something that you know. 00:36:52.000 --> 00:36:58.000 Here is a system. It could be a spring mass 00:36:55.000 --> 00:37:01.000 dashpot system. It could be an RLC circuit. 00:37:00.000 --> 00:37:06.000 Or that pendulum, a damped pendulum, 00:37:02.000 --> 00:37:08.000 anything that is modeled by that differential equation with 00:37:06.000 --> 00:37:12.000 constant coefficients, second-order. 00:37:08.000 --> 00:37:14.000 This is the input. The input can be any kind of a 00:37:11.000 --> 00:37:17.000 function. Exponential functions, 00:37:13.000 --> 00:37:19.000 sine, cosine. It could be a Dirac delta 00:37:16.000 --> 00:37:22.000 function. It could be a sum of these 00:37:18.000 --> 00:37:24.000 things. It could be a Fourier series. 00:37:21.000 --> 00:37:27.000 Anything of the sort of stuff we have been talking about 00:37:24.000 --> 00:37:30.000 throughout the last few weeks. And let's have simple initial 00:37:30.000 --> 00:37:36.000 conditions so that doesn't louse things up, the simplest possible 00:37:34.000 --> 00:37:40.000 ones. The mass starts at the 00:37:36.000 --> 00:37:42.000 equilibrium point from rest. Of course, it doesn't stay that 00:37:40.000 --> 00:37:46.000 way because there is an input that is asking it to move along. 00:37:45.000 --> 00:37:51.000 Now all I want to do is solve this in general with a Laplace 00:37:49.000 --> 00:37:55.000 transform. If I do it in general, 00:37:51.000 --> 00:37:57.000 that is always easier than doing it in particular since you 00:37:56.000 --> 00:38:02.000 don't ever have to do any calculations. 00:38:00.000 --> 00:38:06.000 It is s squared Y. There are no other terms here 00:38:05.000 --> 00:38:11.000 because the initial conditions are zero. 00:38:08.000 --> 00:38:14.000 This part will be a times s Y. 00:38:12.000 --> 00:38:18.000 Again, no other terms because the initial conditions are zero. 00:38:17.000 --> 00:38:23.000 Plus b times Y. And all that is equal to 00:38:21.000 --> 00:38:27.000 whatever the Laplace transform is of the right-hand side. 00:38:26.000 --> 00:38:32.000 So it is F of s. Next step. 00:38:31.000 --> 00:38:37.000 Boy, this is an easy problem. You solve for Y. 00:38:35.000 --> 00:38:41.000 Well, Y is F of s times one over s squared plus as plus b. 00:38:45.000 --> 00:38:51.000 Now, what is that? The next step now is to figure 00:38:50.000 --> 00:38:56.000 out what the answer to the problem is, what's the Y of t? 00:38:55.000 --> 00:39:01.000 Well, you do that by taking the 00:39:00.000 --> 00:39:06.000 inverse Laplace transform. But because these are general 00:39:04.000 --> 00:39:10.000 functions, I don't have to write down any specific answer. 00:39:09.000 --> 00:39:15.000 The only thing is to use the convolution because this is the 00:39:13.000 --> 00:39:19.000 product of two functions of s. The inverse transform will be 00:39:18.000 --> 00:39:24.000 the convolution of their respective things. 00:39:21.000 --> 00:39:27.000 The answer is going to be the convolution of F of t, 00:39:26.000 --> 00:39:32.000 the input function in other words, convoluted with the 00:39:30.000 --> 00:39:36.000 inverse Laplace transform of that thing. 00:39:35.000 --> 00:39:41.000 Now, we have to have a name for that, and those are the two 00:39:39.000 --> 00:39:45.000 words I want to introduce you to because they are used 00:39:43.000 --> 00:39:49.000 everywhere. The function, 00:39:44.000 --> 00:39:50.000 on the right-hand side, this function one over s 00:39:48.000 --> 00:39:54.000 squared plus as plus b, 00:39:51.000 --> 00:39:57.000 notice it only depends upon the left-hand side of the 00:39:55.000 --> 00:40:01.000 differential equation, on the damping constant. 00:40:00.000 --> 00:40:06.000 The spring constant if you are thinking of a mass spring 00:40:03.000 --> 00:40:09.000 dashpot system. So this depends only on the 00:40:06.000 --> 00:40:12.000 system, not on what input is going into it. 00:40:08.000 --> 00:40:14.000 And it is called the transfer function. 00:40:11.000 --> 00:40:17.000 Is usually called capital W of, sometimes it is 00:40:15.000 --> 00:40:21.000 capital H of s, there are different things, 00:40:18.000 --> 00:40:24.000 but it is always called the transfer function. 00:40:27.000 --> 00:40:33.000 What we are interested in putting here its inverse Laplace 00:40:31.000 --> 00:40:37.000 transform. Well, I will call that W of t 00:40:34.000 --> 00:40:40.000 to go with the capital W of s by the usual 00:40:39.000 --> 00:40:45.000 notation. Its inverse Laplace transform, 00:40:42.000 --> 00:40:48.000 well, I cannot calculate that. I will just give it a name, 00:40:46.000 --> 00:40:52.000 W of t. And that is called the weight 00:40:49.000 --> 00:40:55.000 function of the system. This is the transfer function 00:40:53.000 --> 00:40:59.000 of the system, so put in "of the system" if 00:40:57.000 --> 00:41:03.000 you are taking notes. And so the answer is that 00:41:02.000 --> 00:41:08.000 always the solution is the convolution to this differential 00:41:06.000 --> 00:41:12.000 equation that we have been solving for the last three or 00:41:11.000 --> 00:41:17.000 four weeks. It is the convolution of that. 00:41:14.000 --> 00:41:20.000 And, therefore, the solution is expressed as a 00:41:18.000 --> 00:41:24.000 definite integral of the function of the input on the 00:41:22.000 --> 00:41:28.000 right-hand side, what is forcing the equation, 00:41:26.000 --> 00:41:32.000 times this magic function but flipped and translated by t. 00:41:32.000 --> 00:41:38.000 That says du for you guys over there. 00:41:34.000 --> 00:41:40.000 In other words, the solution to the 00:41:37.000 --> 00:41:43.000 differential equation is presented as a definite 00:41:41.000 --> 00:41:47.000 integral. Marvelous. 00:41:42.000 --> 00:41:48.000 And the only thing is the definite integral involves this 00:41:47.000 --> 00:41:53.000 funny function W of t. To understand why that is the 00:41:52.000 --> 00:41:58.000 solution, you have to understand what W of t is. 00:41:55.000 --> 00:42:01.000 Well, formally, of course, it's that. 00:42:00.000 --> 00:42:06.000 But what does it really mean? The problem is what is W of t 00:42:05.000 --> 00:42:11.000 really? Not just formally, 00:42:08.000 --> 00:42:14.000 but what does it really mean? I mean, is it real? 00:42:12.000 --> 00:42:18.000 I think the simplest way of thinking of it, 00:42:16.000 --> 00:42:22.000 once you know about the delta function is just to think of 00:42:21.000 --> 00:42:27.000 this differential equation y double prime plus a y prime plus 00:42:27.000 --> 00:42:33.000 b. Except use as the input the 00:42:32.000 --> 00:42:38.000 Dirac delta function. In other words, 00:42:35.000 --> 00:42:41.000 we are kicking the mass. The mass starts at rest, 00:42:39.000 --> 00:42:45.000 so the initial conditions are going to be what they were 00:42:43.000 --> 00:42:49.000 before. y of zero, 00:42:46.000 --> 00:42:52.000 y prime of zero. Both zero. 00:42:49.000 --> 00:42:55.000 The mass starts at rest from the equilibrium position, 00:42:53.000 --> 00:42:59.000 and it is kicked in the positive direction, 00:42:57.000 --> 00:43:03.000 I guess that's this way, with unit impulse. 00:43:02.000 --> 00:43:08.000 At time zero with unit impulse. In other words, 00:43:05.000 --> 00:43:11.000 kick it just hard enough so you impart a unit impulse. 00:43:10.000 --> 00:43:16.000 So that situation is modeled by this differential equation. 00:43:15.000 --> 00:43:21.000 The kick at time zero is modeled by this input, 00:43:19.000 --> 00:43:25.000 the Dirac delta function. And now, what happens if I 00:43:23.000 --> 00:43:29.000 solve it? Well, you see, 00:43:25.000 --> 00:43:31.000 everything in the solution is the same. 00:43:30.000 --> 00:43:36.000 The left stays the same, but on the right-hand side I 00:43:34.000 --> 00:43:40.000 should have not f of s here. 00:43:37.000 --> 00:43:43.000 Since this is the delta function, I should have one. 00:43:42.000 --> 00:43:48.000 What I get is, on the left-hand side, 00:43:45.000 --> 00:43:51.000 s squared Y plus as Y plus bY equals, 00:43:50.000 --> 00:43:56.000 for the Laplace transform of the right-hand side is simply 00:43:56.000 --> 00:44:02.000 one. And, therefore, 00:43:57.000 --> 00:44:03.000 Y is what? Y is one over exactly the 00:44:02.000 --> 00:44:08.000 transform function. And therefore its inverse 00:44:05.000 --> 00:44:11.000 Laplace transform is that weight function. 00:44:09.000 --> 00:44:15.000 That is the simplest interpretation I know of what 00:44:13.000 --> 00:44:19.000 this magic weight function is, which gives the solution to all 00:44:18.000 --> 00:44:24.000 the differential equations, no matter what the input is. 00:44:23.000 --> 00:44:29.000 The weight function is the response of the system at rest 00:44:27.000 --> 00:44:33.000 to a sharp kick at time zero with unit impulse. 00:44:33.000 --> 00:44:39.000 And read the notes because they will explain to you why this 00:44:37.000 --> 00:44:43.000 could be thought of as the superposition of a lot of sharp 00:44:42.000 --> 00:44:48.000 kicks times zero a little later. Kick, kick, kick, 00:44:46.000 --> 00:44:52.000 kick. And that's what makes the 00:44:49.000 --> 00:44:55.000 solution. Next time we start systems.