Final topic and Laplace land is something called convolution. This is something which operates on two functions that I'll call F and G, and produces a third function that I'll call h. We write it as f star g. And it's defined by an integral. That integral has a function of t. It's a little hard to interpret at first. But one of the useful interpretations is that it produces a sliding, moving average of values of F, where the weight to the average are given by the function g. For example, if I let F be this sawtooth wave here that has sharp corners in it. Let's say I define g as this function that's got a little hump around 1.5. So f star g. Well, first of all, round off the corners a bit. But also these corners occur about a 2.5th in the future compared to where they ought to have been. Because that's where g is centered. By manipulating that integral and the definition, we can prove a number of properties that in fact make convolution a lot like multiplication. So it is commutative. It's associative. It distributes over addition. F star 0 is 0 where we're talking about the 0 function here. And here's one curveball f convolved with the delta function, the impulse equals f. So this does have a multiplicative identity if you want, but it's not the function. Now we come to the big thing that makes convolution important. H is f star g if and only if transform of h is the product of F of S and G of S. So not only is convolution a lot like multiplication in time, it's literally multiplication and transform space. Here's one of the implications of that. Let's say we're solving one of our second-order linear problems with 0 initial conditions. We know from experience that the transform of the solution is the transform of the forcing function divided by the characteristic polynomial. In the special case where the forcing function is an impulse, we know that the transform of Delta t is just 1. So in that case the solution is just one over the characteristic polynomial. So this is the transfer function for second-order problems. Not only that, but the x that satisfies this particular initial value problem is the impulse response. It's the same x on the left and the right except connected by transform. So the Laplace transform of the impulse response is the transfer function. Beyond that though, if we go back and look at the general forcing problem, the transform of the solution is the product of the transform of the forcing with that transfer function. That means by the convolution theorem, the solution in this first problem is the convolution of f and g, where again, g is the impulse response, the inverse transform of the transfer function. That means with just the impulse response, we could solve any initial value problem. For example, let's say we want to find a particular solution of this problem. I'm going to do it by finding the impulse response. The transfer function is one over the characteristic polynomial. And I'm gonna wanna take the inverse transform of that. So I'll need to use partial fractions. We do the usual thing to find these coefficients. If I use the pole at three, I get a. If I use the pole at one, I get B. That means that when I take the inverse transform, I get 1.5. And the one over S minus 3 gives me e to the three t minus a half e to the one t. Then we can use the convolution theorem to say that a solution of the original force problem comes from the convolution of f with g. And if we like, we can use trig identities for the difference of angles on the cosine. First I get a cosine t, cosine tau. And the cosine t doesn't depend on the integration variable, so I can pull it through. And then I get minus sine t, sine tau. And so on. Of course then you have to do the actual integrals. But let me say one more thing about this. We didn't need to use Laplace transforms at all. We could just go ahead and find the impulse response. Remember we already saw that all an impulse does in a second-order problem is cause a jump in the derivative of the solution. So the impulse response where the impulses at time 0 is equivalent to an initial value of 0 at an initial slope of one. And then this is a problem we solve with the characteristic polynomial. When we put in the initial conditions, will get the exact same impulse response as before. But now there are no Laplace transforms at all. In fact, there is a variation of parameters method for second-order problems. And it would give you these exact formulas.
V.4 Convolution
From Tobin Driscoll March 31, 2021
150 plays
150
0 comments
0
You unliked the media.