Now we want to solve problems with impulse and step forcing. There are two key transforms we need to use. The second one, I'll remind you is called the shift theorem. And is a special case. When the function f of t is equal to one, then it's transform is 1 over S. Here's an example with impulse forcing. The transform of the solution is the transform of the forcing function divided by the characteristic polynomial. I'm going to make a definition of capital Y. It will be one over S square plus 1. That leaves x sub p in the form e to the minus st times capital Y of S. Now why did I do that? Well, it's because this now is the expression that's on one side of the shift theorem. So this expression is the transform of shifted step time shifted y. That means when I take the inverse transform, I'll get the solution is just that shifted step times the shifted. Why? So everything now comes down to finding y of t. I can do that because I have the transform of y. So if this is capital Y of S, then y of t has to be sine of t. So in my solution, I just have to put sine of t minus capital T instead. If we look at the solution piecewise, it's 0 up until the impulse time. And then it's shifted sine afterwards. Solution is 0 up to time t. And then it looks like a sine wave. And if we look at the impulse time and what happens there, the solution is continuous and the slope is discontinuous, jumps up by one. Here's our next example. Find a particular solution. We take the transform of the forcing function, e to the minus four S and divide by the characteristic polynomial. Once again, I want to use the shift theorem. And again I wrote x sub p wrong. That should be e to the minus 4 s. So I want to use capital T equal to four and call this part Y of S. Then, just like last time, the shift theorem gives us the solution if we can find y of t. So we'll reset the problem now to be finding the inverse transform of this. You might want to use partial fractions. But if we look at the poles of this thing, which other characteristic roots? They're complex. So this is already as factors as we can do with real partial fractions. Multiple reach into our bag of tricks. And what will pull out is completing the square in the denominator. S squared plus two. S is the beginning of a perfect square if I have plus one. So I've added 0. But now this group here is a perfect square. I'll point out to you that the real part appears here. It's s minus that real part. And the imaginary part is square to get here, that's always going to be the case. Now I want to use this identity from our transform table. This argues that I should choose a to be negative one. Because then this thing becomes capital F of S plus 1. And then I can take the inverse transform according to that identity. So really now I have to find f of t. And we've identify what capital F of S plus 1 is. And that means that capital F just S has S in place of S plus 1. And now that is something we can take an inverse transform of that gives us f of t. Now we just have to retrace our steps. First, y of t is given by e to the minus t times f. And finally, the particular solution come from applying the shift theorem to y. Let's see what happens with a step forcing function. The transform of the solution is the transform of the forcing function divided by the characteristic polynomial. We know by now that we should factor out that e to the minus t term so that we can apply the shift theorem. I'll let everything else be called capital Y of S. So once we find y of t, the shift theorem gives us a solution. So starting with this capital Y, we use a partial fraction decomposition. Let's clear the denominators. I have one real pole at S equals 0. And from that, we immediately get the a equals one. And the other poles are imaginary. So rather than substituting more values in, I'll put in the value we have for a. I can subtract one from both sides of this identity. And from here it's clear that b plus 1 has to be 0 and c has to be 0. So I can put those into our partial fraction decomposition. And now I can inverse transform each of these terms. Whenever S is the transform of one. And the second term is the transform of cosine t. Let's write out the solution. If I plot this solution 0 up to the time of the step. And then it's a shifted negative cosine from there. And because the cosine comes in at a 0 slope, at the moment of the step here, both x and x prime are continuous. Here's a problem with a piecewise-defined forcing function. The first thing we have to do is express this algebraically using a step function. And this is not something we know how to take the transform AV right away. But we do know how to transform us something like this using the shift theorem. So we have to make it look like that. In other words, we have to pick g the right way. We want g of t minus 3, t equal t. If I let t minus 3 be a new variable, then g of u is three plus u. The transform doesn't care what I call that variable, whether it's T or U. The transform is the same. The transform of 3 is 3 over S, and the transform of t or of you is one over S squared. The transform of the solution is the transform of the forcing function divided by the characteristic polynomial. And we just use the shift theorem to find F of S. I'll factor out the exponential, put everything else in terms of one fraction. So as usual, I'll need to find the inverse transform of this part. And then the solution will come from the shift theorem. Let's reset. This time I have a double pole at 0. That means I need a linear term over that poll. And then I factor the rest into real poles. As usual, clear the denominators. I'll substitute in each of our real poles. When I put in x equals 0, everything disappears except B. When I put an S equals 1, only see survives. And when I put in x equals negative one, I get T. Now I'm fresh out of polls. But I still haven't determined a. What I'll do is I'll fill in all these values that we do know now. Looking on the right, let's just collect the x cubed terms. Not going to need the rest. This will give me a, because the coefficient of x cubed is a plus three on the right and it's 0 on the left. So a must be negative three. Now we'll break up that first fraction into two parts. Each of these terms is something I can invert. Inverse transform of 1 over S is 1. 1 over S squared gives you t. And so on. And then the solution is the shift theorem applied to y.
V.3 Steps and impulses at 2nd order
From Tobin Driscoll March 31, 2021
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