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Hello everyone and welcome to our first class on Laplace Transforms.

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So first of all, a quick look at what we're going to cover today.

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We'll start with a brief introduction on Laplace Transforms, then motivation as to why we are

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studying Laplace Transforms and follow that up with some definitions that we'll need throughout

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the classes on Laplace. Then we'll move on and look at calculating Laplace Transforms from

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the definition. Then we look at calculating Laplace Transforms from the tables and that's

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the method we'll use predominantly in this course. We'll do very little as regards calculating Laplace

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Transforms from the definition. We'll use the tables for nearly all our work. And after we've

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looked at how to calculate Laplace Transforms from basic functions, we'll move on and look at

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the first shifting theorem, which is a method for calculating Transforms of special types of

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functions. Okay, so Laplace Transforms, well they date back to the time of the French mathematician

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Pierre Simon Laplace, who's around from the mid 18th to the early 19th century. He also worked in the

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areas of engineering, physics and astronomy. And Laplace Transforms is what's known as an

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integral transform and is useful in solving physical problems found in engineering, maths and physics.

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The integral transform or integral transforms in general enable us to convert what are complicated

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problems into ones that are easier to solve. And other examples of integral transforms that are

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widely used in, for example, signal processing and the Fourier transform, Hilbert transform and

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the wavelet transform. However, we don't look at any of these in this module. We're only going to focus

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on Laplace Transforms. So as you can see, this slide is titled Laplace Transforms and ODEs. And

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ultimately that is what we are going to be using Laplace Transforms for. That's for solving ordinary

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differential equations. And as we know, ODEs can be used to describe the dynamical behaviour

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of systems, mechanical, electrical systems and so on. And traditional calculus methods, as we've seen

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previously, are used to solve ODEs for a range of different inputs. We saw when we're looking at

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second order differential equations, the inputs were the functions on the right hand side of the

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equations. We saw situations where we had constant terms, linear terms, quadratics,

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exponentials, sines and cosines. And we used the method of undetermined coefficients to solve these

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equations to obtain the output that we would get from the differential equation. However,

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when signals become more complicated, the calculus approach can become very difficult indeed. And

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the methods that we've seen up to now really don't work very well. In fact, sometimes they can't be

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used to solve certain types of differential equations. So what we're going to do is we're going to use

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Laplace transforms and that will enable us to convert these time domain calculus problems,

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such as differential equations, to problems that only need algebra to solve them. And that's what

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the Laplace transform does for us. So let's just summarise this in this diagram, this diagram here.

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So we're over here in the time domain, we've got a calculus problem, which could be a differential

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equation. We take the Laplace transform of the time domain problem of the differential equation

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and we convert in so do it by taking Laplace transforms of the ODE, we convert it from a

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calculus problem to an algebra problem. We then solve the algebra problem, which tends to be quite

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straightforward. And we have a transform solution in the Laplace domain, but we need to get back to

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the time domain because our original problem was a time domain problem and we do that by taking

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inverse Laplace transforms. And this is the place or the step that people find the most difficult.

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And it's not because of this new mass involved. It tends to be because of the different

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types of algebra that we need to know. There's nothing new in this, but it can get quite tricky

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at times. So that's really schematically what's going on. Calculus problem in the time domain

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transformed to an algebra problem in the Laplace domain. We solve the algebra problem,

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get our transform solution, and then we invert it by taking inverse Laplace transforms to obtain

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solution. As I said earlier, we would ideally like to be able to go straight from the calculus

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problem to the solution, but it can be either very, very difficult to do or even impossible

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using the methods that we know for solving time domain problems, differential equations.

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Okay, so let's have a definition of the Laplace transform. Well, so we're only going to consider

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positive time t greater than zero. So an independent variable from now on will be t time.

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So there are Laplace transforms of a function f of t. It's defined by the improper integral.

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Now an improper integral, one situation when an integral is improper is when you get infinity

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plus infinity or minus infinity or even both as the limits. And you can see that this integral,

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we have positive infinity as one of the limits. So how do we get the Laplace transform from the

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definition of a function f of t? Well, we take our function f of t, we multiply it by e to the minus

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s t. Now in here, s is actually a complex variable, but we're not really going to do anything with

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that. We just bear in mind that it is a complex variable, but we're able just to

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work this as we would do most other integrals. So we multiply our function f of t by the

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exponential e to the minus s t, and then we integrate this product from zero to infinity.

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And that will take the notation here, the L means Laplace and curly brackets are function f of t.

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This is how we calculate it, and the output will be a function of s. And it's standard practice to

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denote that with a capital F. If f of t, little lowercase f of t is a function, the output of Laplace

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transform will be uppercase f of s. So we feed in a function of time of t time, the time domain,

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and out of that comes a function of s in the Laplace domain or the s domain, as it's also called.

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So that's what we're saying down here. So the function of t, it's quite often some form of

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time signal, and we talk about moving from the time domain or t domain to the Laplace

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domain, that's the s domain, when we're performing a Laplace transformation.

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So let's have a look at an example of using that definition to calculate the Laplace transform

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of a function f of t. And I'm going to take a very simple function, f of t is equal to one,

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just the constant function that's equal to one everywhere from t greater than zero onwards.

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So that's what it looks like. The red line is just the function f of t is equal to one.

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So we want to calculate the Laplace transform of this using our definition. So the definition

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is given here in the green box, the function f of t is equal to one. So in here, I've just

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replaced a one. So the notation here is that I want the Laplace transform of the function one. So

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it's L curly brackets for the one inside. And that's equal to the function f of t is equal to one

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times e to the minus st integrated between zero and infinity. Now to perform this integration,

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there's no such number as infinity. And it's really not correct to substitute that in as one

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of the limits directly. What we do is we introduce a w variable u and we let u tend to infinity.

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That's how we handle these improper integrals. So the function is just one. So that just means

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we've just got to integrate this exponential e to the minus st with respect to t. And we know that

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when we do that, when we integrate an exponential like over here in the red type, what happens is

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it'll just be one over whatever's in front of the t. So one over k e to the minus kt. And in

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our case, what's in front of the t is the minus s. So we're going to have minus one over s e to the

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minus st. I've taken the minus one over s outside because it doesn't depend on t. So we can move

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it outside of this bracket just to make things a little easier for her. I'm going to take the limit

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as u tends to infinity between t equals zero and t equals u. Low limit is zero. As you can see,

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the upper limit is u because we've introduced this w variable. So we just substitute in these

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values. u goes in at the upper limit and zero goes in at the lower limit. So we get this expression

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here. Now e to the zero is one. So that's where that comes from. e to the minus s u. So what's

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happening here as u goes off to infinity? Well, I can write e to the minus s u as one over e to the

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s u. And I'm going to let u tend to infinity. I'm going to let u become massive, become huge.

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And as u becomes bigger and bigger and bigger, this quantity here will get closer and closer

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and closer to zero because you're dividing one by something that's getting increasingly larger. So

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this term as u goes to infinity, this term here, which is equivalent to this term, will go to

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zero. And all that's left is minus one over s times minus one, which is one over s. So the Laplace

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transform of our constant function f of t is equal to one is one over s. So that's our time

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domain function f of t equals one gets fed into the Laplace integral. And the output is a function

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of s, big f of s is equal to one over s. Now that was relatively straightforward,

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but it doesn't take very much of big change in the function that we're calculating in a Laplace

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transform of for things to get quite tricky. So just imagine, you can think of one as t to the zero,

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just ended the power zero, this one. What happens if we rather than having t to zero, we've got t

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to the power one, we've got t there. So we want the Laplace transform of t, but we would need to

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use integration by parts. Things start to get a bit messy. If we want the Laplace of t squared,

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we would need integration by parts twice. So calculating from the definition can lead to

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an awful lot of work. And thankfully, all that has been done for us. And we've got a set of a table

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of Laplace transforms for a range of functions. It just depends on which course you're doing,

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which book you're looking at, how many functions are given. We've got maybe about two or three

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pages of functions that we supply the Laplace transforms for. But as I said, some books will

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give you more, some books will give you less. So we start off with one which is t to zero. We've

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just seen that the Laplace of that is one over s. So I should say that in this table here, I number

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all of these down the left hand side in the first column. In the next column, I've got my function

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at time. That's the function that I want to take the Laplace transform of. And in the third column,

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I've got the actual Laplace transform, big F of s, which is just the Laplace transform of ft. So

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we've already seen that one over s is the Laplace transform of one. For the Laplace transform of t,

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it's one over s squared. For t squared, it's two over s cubed. And then we could carry on,

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we could have t cubed, t to four, but you've got to stop somewhere. And what they do is they just

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give a general expression for t to the power n. The Laplace of that is n factorial over s to the

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n plus one. And I will see an example of that shortly. Then we've got some exponentials that

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we've got the Laplace transform of. And we'll be seeing them a lot because exponentials, as you know,

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feature a lot in the solutions of differential equations. So these ones will be quite common.

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We've got the trig functions then, sine and cosine and variations of that. And the table goes on,

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we've got the hyperbolic functions here, the hyperbolic sine, the hyperbolic cosine. We don't

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use them so much in our course. And then there's a range of other functions here. Interestingly,

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here, 2425 are the Laplace transforms of derivatives. And that's what's going to help us when we're

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going to be solving differential equations using Laplace transforms. And there's some other

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functions here, which will come across in due course, such as the unit step function

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and the delta impulse function. Okay, so before we move on to look at some examples,

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we should just get familiar with one of the properties of the Laplace transform.

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And that's the linearity property. Now, this is something you've actually seen before in both

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integration and differentiation. So what's it saying? It says that, for example, the Laplace

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transform, C1 and C2 here are constants. The Laplace transform of a sum of functions is equal

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to the sum of the individual Laplace transforms. So that's what I'm saying here. The Laplace transform

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of the sum or difference or two or more time domain functions is equal to the sum or difference of

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the Laplace transform of the individual functions. So for example, if you remember when you were

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doing differentiation, if we had a function t squared plus sine t, well, if you wanted to

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differentiate that d by dt of that, you would say that that's just d by dt of t squared plus d by dt

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of sine t. So this is all I'm saying here with Laplace. It's exactly the same as this. So the

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derivative of a sum is the sum of the individual derivatives. So you'd get 2t plus cos t. So in

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other words, you just go along, do differentiating term by term. We did the same thing with integration

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and we'll do the same thing with Laplace. When we've got terms added or subtracted, we just go

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and do them one at a time and add or subtract the result. So we'll see some examples shortly

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of that. So here's a range of functions. We've got eight different examples here to highlight

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different properties. So let's just get started. So our first function that we want to take the Laplace

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transform is 7t to the power 6. So this is the general notation. Quite often they write the L

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as a curly L like that. And then curly bracket is 7t to the 6. Now we know from the linearity

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property we just saw that the constant can be taken outside. You don't have to, but it just

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makes things a little easier. So out comes the constant and then we've got to take the Laplace

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transform of 6. Now if we look at our tables, look at this column in our tables, here's our function of

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time f of t is t to the 6. Well we run out after t squared, but we've got a general expression

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for t to any power. t to the power n, the Laplace transform of that is n factorial over s to the

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n plus 1. So all I have to do here is identify the value of n, which is pretty obvious. N is equal

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to 6. So I'll use number 4 with n equal to 6 to give me 6 factorial over s to the 6 plus 1 or 6

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factorial over s to the power 7. So that's what I want. And don't forget of course we took the 7

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outside so that must feature there. So 7 times 6 factorial over s to the 6 plus 1, s to the 7 in

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other words. Now 6 factorial, the factorial, what does function, what does that do? Well basically

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all that saying is whatever number you take the factorial of you just multiply it by every number

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smaller than it down to 1. So for example 6 factorial, so just get the pen, 6 factorial will be 6 times

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5 times 4 times 3 times 2 times 1 and if you work that out you find that it's 720. And then of course

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we've got a 7 on the outside so we must multiply that and 5 0 4 0 over s to the power 7 is the Laplace

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transform of 7t to the power 6. Okay move on now. 6 e to the minus 8t. Well we could go, we would

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go through our tables and we saw that all the way 1 2 and 3, 1 2 3 and 4, none of these involve the

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exponential function. None of these involve the exponential function but then we hit number 5.

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Number 5 is e to the minus alpha t. Well we've got e to the minus 8t so let's just start. So I want to

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take the Laplace transform of this function. I can take the constant outside, the 6 can get taken

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outside. Again you don't have to but it's a little easier to work with, a little neater. Out comes the 6

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and I want the Laplace transform of e to the minus 8t. So then I just need to identify which one we've

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already said. It's going to be number 5 in the tables and I've got to compare what I've got. I've

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got e to the minus 8t and in the table, in the table I've got e to the minus alpha t. Okay so that

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would mean that minus 8t must correspond to minus alpha t. The t's can cancel so minus 8 must correspond

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to minus alpha so in other words alpha must equal 8. Okay so that's my value of alpha identified.

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So what do I want to do now? Okay so I've identified that alpha is equal to 8 so the Laplace transform

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of e to the minus 8t alpha is equal to 8 is 1 over s plus 8 and that's what I've got here and don't

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forget the 6 on the outside. We must multiply by that to give me 6 over s plus 8 as the Laplace

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transform of f of t equal to 6e to the minus 8t. So there's our time domain function. We applied the

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Laplace transform I want to do s domain or Laplace domain function here. Okay moving on let's have

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another well here's another exponential and again we just compare it with number 5. I can take

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4 outside of course. 4 can get taken outside and I've got to calculate the Laplace transform

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e to the 9t. Just got to be a little bit careful here because I've got e to the 9t but in the table

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number 5 is e to the minus alpha t. So in other words 9 must correspond to minus alpha

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or in other words alpha is equal to minus 9. Okay so alpha is equal to minus 9 so I can use

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number 5 and that would give me

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that the Laplace transform of e to the minus e to the 9t is 1 over s minus 9.

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Don't forget I've got that 4 on the outside that I can multiply by so I get 4 over s minus 9.

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And that's the Laplace transform of 4 e to the 9t. Okay so that's fine we can move on to

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look at another example though. So this again is a different type of function f of t equals 9 cos 4 t

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minus 4 sin 9t. So this involves trig functions cos and sin. Again now we could go back

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numbers 1 2 3 and 4 are only involved powers of t so they won't work. Number 5 6 7 well they

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involve the exponential functions they're not going to work either but when I reach number 8 and 9

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they involve sin and cos so that would tell me that these are the ones to use. So here I go I'm

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setting up my problem here I want to take the Laplace transform of this the difference of these

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two functions and if you remember the linearity property said that I can go in and just take the

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Laplace transform of the individual functions and then do the subtraction at the end. So I'm going to

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do that I'm going to use the linearity property so I will take the Laplace transform of this first

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term minus the Laplace transform of the second term and in both cases I've got a constant which I

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can take outside. Again you don't have to if you don't want to but I find that it just it's a little

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bit a little less messy to work with than you take the constant outside. So that just leads to define

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the Laplace of cos 4t okay so cos 4t now it's this one with omega equal to 4 so the Laplace

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transform of cos 4t that's this one here will be s over s squared plus 4 squared there we are

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down here and of course don't forget you got a 9 multiplying that term. Then we've got minus

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4 times the Laplace transform of sin 90 so omega here is equal to 9 so I'm going to have the Laplace

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transform of sin 90 would be 9 over s squared plus 9 squared so there you are it's that term there

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and if I simplify this I get the solution at the bottom 9s over s squared plus 16 minus 36

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over s squared plus 81 so that is the Laplace transform of this function f of t here.

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One thing before we move on I'll say here these two have exactly the same denominator it's s squared

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plus omega squared whatever your omega value was and this one it's 4 and this one it's 9. The way

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the Laplace of the sin and cosine differ is that for the sin in the numerator we've just got the

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constant value whatever omega is that's what goes in the numerator for the cosine however the numerator

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includes the Laplace variable s that's the difference between the Laplace of sin and cos

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that will be useful when we come to calculate inverse Laplace transforms later on.

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Okay moving on to another problem we've now got one of the hyperbolic functions

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and we don't treat this any differently we just basically look in our table in this column of

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our table to see if we can find a cos of 6t or cos of some number times t we won't find this

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exactly with a 6 but we'll find some general expression in our table so if we do that we'll

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reach down number 18 which is the hyperbolic sign so that's close but not not the one we want this

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one isn't the one we want either but then when we reach number 20 we find that we've got the

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hyperbolic cosine or cos of beta t so beta can match up with our 6 so we've just got this one here

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with beta equal to 6 and that'll be fairly easy because we just replace beta with 6 so the Laplace

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transform of cos 6t will just be s over s squared minus 6 squared which gives me in the blue box

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here s over s squared minus 36 so that's all that's to it really now another point to note here is if

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we look at our regular trig functions sin and cos as we saw on the previous slide both functions

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have a Laplace transform with s squared plus omega squared in the denominator that is the Laplace

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transform of both sin and cos the denominators are the same with s squared plus omega squared the

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tops differ with sin having just the omega the constant value but cos having the Laplace variable

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s in the numerator how do these compare with the hyperbolic functions well if you look at number 18

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and number 20 you can see that something very similar happens here the two of these 18 and 20

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have the same denominator and once again the sine version or the hyperbolic sine has a constant

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just as the regular sine did while the hyperbolic cosine like the regular cos had the variable

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the Laplace variable s in the numerator now where the regular trig functions sine and cos

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differ from the hyperbolic since if you look at the denominator the regular trig functions

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have s squared plus omega squared s squared plus some number whereas the hyperbolic functions have

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s squared minus some number so the only real difference between them is the hyperbolic have

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a minus in the denominator whereas the regular trig functions have a plus and again that will come

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in useful when we come to calculating inversions later on okay so moving on we've got a different

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thing here now we've got a product of two functions we've got an exponential multiplying a trig

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f of t is equal to e to the 8t sine 2t now we'll see later on in this class that when we've got this

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type of product involving an exponential function and some other function that we use the first

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shift in theorem but this particular case is actually given in our tables this is actually

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given in our tables so we don't need to use the first shift in theorem for this one but

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depending on what tables you're issued with you may very well have to your table tables from

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l squared might not involve might not include this type of function in the tables so let's

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have a quick look at how we do this so can we find n that looks like it we would look down

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this column in our table and eventually we would hit number 14 e to the minus alpha t

243
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sin omega t number 15 is the same or similar except and it's got the cosine instead of the

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sign so we've decided it's number 14 it doesn't matter that that's a positive and that's a negative

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we can handle that we saw that already when we're using number five in probably questions two and

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part b and c i think possibly we can handle that that's a positive eight and the table has a minus

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alpha there that's that's not a problem for us and we know from comparison that omega will be equal

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to two so i'm going to use number 14 with we've got to figure out what alpha is and we know that

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omega is equal to two so let's just compare this we know from our tables we've got e to the minus

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00:30:44.880 --> 00:30:53.520
alpha t and here we've got e to the eight so these two things must be equivalent e to the minus alpha

251
00:30:53.520 --> 00:31:05.600
t must equal e to the eight t right so in other words minus alpha t is equal to eight t and then

252
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if we go up to the the green type here that must mean that minus alpha is equal to eight

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or alpha is equal to minus eight you might be able to spot that whether you're having to work all

254
00:31:17.200 --> 00:31:25.440
this out but there's no harm in doing this so alpha is equal to minus eight so i'm going to use

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number 14 with alpha equal to minus eight and omega omega is equal to two because it's e to the minus

256
00:31:35.040 --> 00:31:43.120
alpha t sine omega t i've got e to the at alpha minus eight so that works it's going to be e to the

257
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minus minus eight or e to the at just as i've got omega equal to two will give me sine two t and that's

258
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exactly what my question has so if that if we're going to use this one then we have omega so i'm

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going to have two over s minus eight all squared because alpha is equal to eight minus eight plus

260
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two squared because omega is equal to two multiple or yeah that just two squared is four

261
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and then i can take another step and open out these brackets to obtain my solution at the foot of

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the slide here so this is actually given in the tables but we're shortly going to see how we handle

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this sort of product of two functions where one is an exponential we use the first shift theorem

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we'll see that shortly okay let's have a look at another example of calculating the past transforms

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from the tables now this one f of t is equals cos squared t minus sine squared t well we could look

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in our tables and we wouldn't find cos squared or sine squared so we've got the problem of how are

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we going to manipulate this into a form that's given in the tables now you may know this relationship

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this is a trig identity cos squared t minus sine squared t you could actually write this if i just

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get my tablet if i write this out in full you know this is equal to cos t cos t minus sine t

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okay and that is actually equal to using the trig identity it's cos of t plus t because remember

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when you've got cos of a plus b or t plus t it's cos cos minus sine sine the sine changes when we're

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using this identity and that's just cos of cos of 2t you know that's a fairly well known trig

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identity when we've got cos squared t minus sine squared t it's equivalent to cos of 2t so whereas

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we were not able to use our tables directly for this function here we were able to apply

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a well known trig identity to rewrite cos squared t minus sine squared t as cos 2t and then if i look

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at my table when i reach number nine i've got i've got cos of omega t and this is just cos of omega t

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with omega equal to two so it's a straightforward matter of using number nine with omega equal to two

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so this is what i'm saying here that the Laplace of the function as it's given cos

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00:35:01.520 --> 00:35:07.760
squared t minus sine squared t is the same as the Laplace of cos 2t i use number nine with omega equal

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00:35:07.760 --> 00:35:18.320
to two i get s or s squared plus two squared or s over s squared plus four so that's an example of

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using a trig identity to manipulate the expression we were given and that doesn't appear in our tables

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00:35:25.680 --> 00:35:33.280
to a form that actually does appear in our tables and we are able then quite easily to obtain

283
00:35:33.280 --> 00:35:41.440
its Laplace transform okay so here's another one again you know we could look in our tables we need

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something that's got the product sine t cos t but we're not going to find this this doesn't appear

285
00:35:46.960 --> 00:35:55.520
in our tables we may then remember the trig identity that says two sine t cos t is equal to

286
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sin 2t and that's if i can just write here that's because if i write sine 2t

287
00:36:05.760 --> 00:36:17.360
if i write sine 2t i'll just get the pen going sine 2t is equal to sine t plus t

288
00:36:18.160 --> 00:36:29.920
and if you remember this trig identity that's sine t cos t plus cos t sine t and all that is

289
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it's the same thing sine t cos t plus cos t sine t is just two sine t cos t and our trig identity

290
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as we know it says that that's equal to sine 2t so i can replace the sin t cos t by sine 2t but

291
00:36:51.840 --> 00:37:00.880
remember i need a two here so my sine t cos t would just be a half sine t you know there's my

292
00:37:00.880 --> 00:37:08.080
trig identity so i've got to divide both sides by two to get an expression for sine t cos t because

293
00:37:08.080 --> 00:37:15.040
that's what we're given in the question okay so i've been able to identify a trig identity that

294
00:37:15.040 --> 00:37:23.360
i can use here so let's have a look at it so i have the Laplace transform of sine t cos t

295
00:37:25.120 --> 00:37:31.760
using my trig identity i know that that's the same as the Laplace of a half sine 2t

296
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i can take the half outside if i want to i don't have to but again it's just things are a little

297
00:37:38.480 --> 00:37:44.800
less cluttered so it's just a direct Laplace transform of sine 2t it's number eight with omega

298
00:37:44.800 --> 00:37:53.280
equal to two so that becomes a half don't forget the half on the outside times omega over s squared

299
00:37:53.280 --> 00:38:02.240
plus omega squared it's uh a half times two over s squared plus two squared that two cancels with

300
00:38:02.240 --> 00:38:09.840
that one and the answer would be one over s squared plus four so that's the Laplace transform of our

301
00:38:09.840 --> 00:38:21.200
function here so one important point that i that i should make here is that in this example

302
00:38:22.000 --> 00:38:28.720
our function was a product of two time domain functions of sine t and cos t

303
00:38:29.280 --> 00:38:37.760
this did not appear in our tables so we had to find a method of doing some sort of manipulation

304
00:38:37.760 --> 00:38:45.760
of this term into a form that's given in the tables and we were able to do that using the

305
00:38:45.760 --> 00:38:53.600
trig identity that sine 2t is equal to two sine t cos t and a little bit of rearrangement gave us

306
00:38:54.160 --> 00:39:00.560
this expression so we could directly calculate the Laplace of this term now one thing here that you

307
00:39:00.560 --> 00:39:08.080
do not do this is a product and just the same way as when we did differentiation when we're

308
00:39:08.080 --> 00:39:14.640
differentiating a product you do not do the product of the first thing the derivative of the first

309
00:39:14.640 --> 00:39:19.760
thing times the derivative of the second so you know if i just go to the next slide that's what i'm

310
00:39:19.840 --> 00:39:25.200
saying here an important point to notice while the function f of t is the product of two functions

311
00:39:25.200 --> 00:39:31.920
sin t and cos t it's not true that the Laplace transform of our function f of t

312
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is the product of sine it's the product of the Laplace of sine t and the Laplace of cos t so

313
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in other words this here does not hold that Laplace it's not the Laplace of sine t times the Laplace

314
00:39:48.640 --> 00:39:57.280
of cos t so in our case we used the trig identity we'll go to shortly see some functions that involve

315
00:39:57.280 --> 00:40:02.320
products we've already seen one of them i was seeing already seen an example but we're able to use

316
00:40:02.320 --> 00:40:08.000
the tables on it because it was given in the tables that was e to the at i can't remember what it was

317
00:40:08.000 --> 00:40:15.120
was cos t or something like that but we'll soon see the first shifting theorem now i mentioned

318
00:40:15.680 --> 00:40:23.680
uh about differentiation if you remember when we had a function f of t was equal to t squared

319
00:40:24.320 --> 00:40:32.080
sin t when we went to calculate the derivative of that f dashed of t we did not just differentiate

320
00:40:32.080 --> 00:40:38.880
the first term and multiply it by the second term that's completely wrong that's completely wrong

321
00:40:39.440 --> 00:40:49.120
the derivative of a product is not the product of the individual derivatives if you remember

322
00:40:49.120 --> 00:40:56.160
we had to use the product rule we had to use the product rule where you know depending you could

323
00:40:56.160 --> 00:41:05.440
use maybe u and v for example that f dashed was equal to u dashed v plus u v dashed that was the

324
00:41:05.440 --> 00:41:10.160
product rule and similar things happen with integration as well although it's a bit more

325
00:41:10.160 --> 00:41:14.560
complicated there's no one rule that works there sometimes integration by part sometimes

326
00:41:14.560 --> 00:41:20.000
integration by substitution and there are other other methods which you know we wouldn't have seen but

327
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you know the whole purpose of me saying this is that really in general the Laplace transform of a

328
00:41:29.440 --> 00:41:36.000
product is not equal to the product of the individual Laplace transforms and that's what this

329
00:41:36.000 --> 00:41:43.600
is saying here if you take Laplace of f of t times g of t it's not the Laplace of f of t times the

330
00:41:43.600 --> 00:41:52.000
Laplace of g of t so let's have a look at one special case and what we do when we come across it

331
00:41:52.000 --> 00:41:58.000
so suppose we've got some function f of t and we're multiplying it by an exponential function

332
00:41:58.880 --> 00:42:08.560
then as I've said earlier we use the first shift in theorem so here's an example suppose or here's

333
00:42:08.560 --> 00:42:13.840
the definition sorry of the first shift in theorem which is also called the frequency shift theorem

334
00:42:14.480 --> 00:42:21.040
fst for sure suppose the function f of t has Laplace transform f of s now we know all this you

335
00:42:21.040 --> 00:42:27.440
know we've seen this f of t was the second column you know in our tables we've numbered the Laplace

336
00:42:28.480 --> 00:42:33.200
transforms in the first column the second column gives the time domain functions we've been

337
00:42:34.000 --> 00:42:39.680
using that up to now and in our third column we had f of s which was just the Laplace transform

338
00:42:39.680 --> 00:42:47.280
of f of t so if a function f of t is Laplace transform f of s then if we multiply this function f

339
00:42:47.280 --> 00:42:58.160
of t by an exponential then what this corresponds to is it corresponds to what's called a shift in

340
00:42:58.880 --> 00:43:08.000
the s domain will either be adding alpha or subtracting alpha from the s in here and the alpha

341
00:43:08.000 --> 00:43:15.520
is this value in the original function we'll crack in their Laplace transform of so basically what we

342
00:43:15.520 --> 00:43:24.480
do is we we we ignore the exponential to start with we take f of t and we find its Laplace transform

343
00:43:24.480 --> 00:43:35.040
big f of s and then wherever we see an s in the Laplace in big f of s we replace it by s plus or

344
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minus alpha so that's what the first shift in theorem does and I've summarized that here in three

345
00:43:46.080 --> 00:43:56.480
steps so that's what's involved in it but easier to remember I've got this diagram here that will

346
00:43:56.480 --> 00:44:04.160
hopefully help us so we've been given a function the bottom corner here f of t multiplied by some

347
00:44:04.720 --> 00:44:12.160
exponential so the first thing we do is we ignore the exponential to leave us f of t

348
00:44:13.760 --> 00:44:18.560
we can calculate the Laplace transform of f of t using the tables might need to do a little

349
00:44:18.560 --> 00:44:24.480
manipulation beforehand but we can certainly calculate the Laplace of f of t from the tables

350
00:44:24.480 --> 00:44:33.120
and then as I said earlier we take a big f of s that we've got from the tables and wherever we see an

351
00:44:33.120 --> 00:44:42.480
s we replace it by s plus alpha where alpha is the value that appears here in our original function

352
00:44:44.240 --> 00:44:52.160
so that's what we're going to have a look at now okay so and again it's summarized here I think I'll

353
00:44:52.160 --> 00:45:01.200
say one one more time and this diagram here gives you something hopefully to remember what this

354
00:45:01.200 --> 00:45:08.240
appear is saying so this says that if the Laplace transform of a function f of t is f of s then if

355
00:45:08.240 --> 00:45:14.320
you multiply that function f of t by an exponential then the Laplace transform of this new function

356
00:45:15.120 --> 00:45:21.520
will be the Laplace transform of the raw function f of t that's the Laplace of that is big f of s

357
00:45:21.520 --> 00:45:28.320
where s is replaced by s plus alpha okay so let's have a look at that in operation then

358
00:45:31.920 --> 00:45:38.320
so here's the question I've been given determine the Laplace transform of e to the minus 2 t

359
00:45:39.280 --> 00:45:49.600
times t cubed okay so um the first thing we have to do is we have to decide how

360
00:45:50.320 --> 00:45:58.000
we're going to go about doing this problem um suppose we didn't know that we had to use the

361
00:45:58.000 --> 00:46:06.480
first shifting theorem um well if I look at this function I can see there's two separate functions

362
00:46:06.480 --> 00:46:14.480
once an exponential and the other's a power of t so I've got t cubed multiplied by an exponential

363
00:46:14.480 --> 00:46:22.080
function now that tells me that I need the first shifting theorem I've got some function in this

364
00:46:22.080 --> 00:46:28.560
case t cubed and we've got it's been multiplied by an exponential function and that's the key for

365
00:46:29.280 --> 00:46:34.800
identifying when to use the first shifting theorem if you've got some time domain function like a power

366
00:46:34.800 --> 00:46:40.640
of t here it could be sin t it could be cos 4t something like that but if it's multiplied by an

367
00:46:40.640 --> 00:46:50.160
exponential we use the first shifting theorem so um to to use the first shifting theorem we need to

368
00:46:50.160 --> 00:46:55.120
check of course that it's in the form e to the minus alpha t f of t well yes it is there's the

369
00:46:55.120 --> 00:47:01.680
exponential and f of t is equal to t cubed so I've been able to identify that f of t is equal to t

370
00:47:01.680 --> 00:47:09.280
cubed now what about alpha what's the value of alpha well in this box here if I look here's the

371
00:47:09.280 --> 00:47:15.280
general expression from the definition of the first shifting theorem e to the minus alpha t f of t

372
00:47:15.280 --> 00:47:21.520
and here's what I've got e to the minus 2t t cubed well as we've said we can easily identify that f

373
00:47:21.520 --> 00:47:30.320
of t is t cubed alpha I think it's pretty obvious must be equal to 2 it's not minus 2 if alpha was

374
00:47:30.320 --> 00:47:36.720
minus 2 you would have e to the minus minus 2 and you would have e to the 2t if you just compare

375
00:47:36.800 --> 00:47:46.080
this term directly with this term you can see that alpha must equal 2 and that's what I say here at

376
00:47:46.080 --> 00:47:55.920
step one so in my diagram what I've done is um I've identified I've ignored the exponential first of

377
00:47:55.920 --> 00:48:02.960
all and I identified what f of t is it's t cubed I then went in and I identified what alpha was alpha

378
00:48:02.960 --> 00:48:11.120
is equal to 2 the next step going around this diagram in a clockwise direction the next step is to find

379
00:48:11.120 --> 00:48:20.560
the Laplace transform of f of t in other words find the Laplace transform of t cubed so is that

380
00:48:20.560 --> 00:48:26.880
given my tables well no it isn't I've run out after t squared but I've got a general expression t to

381
00:48:26.880 --> 00:48:35.920
the power n whether Laplace is n factorial over s to the n plus one so that'll be with n equal to 3

382
00:48:35.920 --> 00:48:44.880
because t cubed it'll be 3 factorial over s to the 3 plus one that's s to the fourth so it'll become

383
00:48:44.880 --> 00:48:51.280
6 over s to the power 4 so that's big f of s there we are we're up in the top right hand corner

384
00:48:51.280 --> 00:48:59.760
one more step to do and that says that I must replace every occurrence of s in f of s that are

385
00:48:59.760 --> 00:49:07.120
found in step two by s plus alpha where alpha's the value I found at step one in this case alpha is

386
00:49:07.120 --> 00:49:13.520
equal to 2 so what this is telling me to do is replace every occurrence of s whether it's only

387
00:49:13.520 --> 00:49:22.880
appears once by s plus 2 because alpha was equal to 2 so s plus alpha becomes s plus 2

388
00:49:23.680 --> 00:49:34.560
so if I do that I obtain 6 over s plus 2 all to the power 4 that's my f of s plus alpha so that

389
00:49:34.560 --> 00:49:42.320
there is my solution now one thing to be very careful not to do what people sometimes do here

390
00:49:42.320 --> 00:49:49.360
is they get to they get step two correctly and then they say well I've got to replace

391
00:49:50.960 --> 00:50:00.320
it's s plus alpha so they've got to have a plus 2 in it so what they write is 6 over s to the 4

392
00:50:00.320 --> 00:50:09.360
plus 2 that's wrong what's what's been done here is that s is getting replaced by s plus 2 so if s

393
00:50:09.360 --> 00:50:15.840
was raised to the power 4 then what you're replacing it by must get raised to the power 4 and that's why

394
00:50:16.480 --> 00:50:25.840
it's s plus 2 all to the power 4 so that one is correct this one is definitely wrong so bear that

395
00:50:25.840 --> 00:50:33.440
in mind when when we're replacing s by s plus alpha that if it's raised to a power then the s

396
00:50:33.520 --> 00:50:41.600
plus alpha in this case s plus 2 must also be raised to the power and that was power 4 in this case

397
00:50:42.640 --> 00:50:50.480
okay so let's just do another example now so what do we got here well here's my little diagram to

398
00:50:50.480 --> 00:50:57.520
remind me of what the steps that are involved the three steps we go around the diagram in a clockwise

399
00:50:58.080 --> 00:51:04.320
direction here's the expression from the first shifted theorem that I need to compare the

400
00:51:04.320 --> 00:51:12.240
expression I've been given so my first step here says that I've got to identify f of t well I can do

401
00:51:12.240 --> 00:51:18.800
that just by ignoring the exponential so f of t is t to the power 6 or I can just match it up like

402
00:51:18.800 --> 00:51:27.120
this I can see f of t is t to the 6 I then must identify the value of alpha and I've got e to the

403
00:51:27.120 --> 00:51:37.120
minus alpha t must be equivalent to e to the minus 4t so clearly alpha must equal 4 okay so it's not

404
00:51:37.120 --> 00:51:44.320
minus 4 just as in the last example because it would be e to the minus minus 4t which is e to the

405
00:51:44.320 --> 00:51:52.960
4t and that's not what I've got I've got e to the minus 4t so alpha in this case just by simple

406
00:51:52.960 --> 00:52:01.760
comparison alpha is equal to 4 okay so I'm ignoring the exponential to give me f of t I know what f of

407
00:52:01.760 --> 00:52:09.040
t is it's t to the power 6 so I'm up here I now want to calculate the Laplace transform of f of t

408
00:52:10.640 --> 00:52:18.400
t to some power well we know it certainly would be one of the early ones in the table we don't have

409
00:52:18.400 --> 00:52:24.800
t to the power 6 in the table but we've got the general expression t to the end and we can take

410
00:52:24.800 --> 00:52:34.480
n is equal to 6 and the Laplace transform will be 6 factorial over s to the 6 plus 1 or 6 factorial

411
00:52:34.480 --> 00:52:43.680
over s to the power 7 so that's what I've got 6 factorial 720 because it's 6 times 5 times 4 times

412
00:52:43.760 --> 00:52:52.240
3 times 2 times 1 that's 720 so that's big f of s so there we are we've got we're up in this corner

413
00:52:52.240 --> 00:53:00.000
and I just these one last thing to do is replace every occurrence of s in f of s by s plus alpha

414
00:53:00.560 --> 00:53:08.400
alpha is equal to 4 so I must replace s by s plus 4 that's just that's this final step here

415
00:53:08.960 --> 00:53:15.840
s gets replaced by s plus alpha alpha in this case is 4 so what I'm going to have in the denominator

416
00:53:15.840 --> 00:53:23.680
is s plus 4 getting slotted in here and so if s was raised to the power 7 it's replacement

417
00:53:23.680 --> 00:53:30.880
s plus 4 must get raised to the power 7 and I end up with my solution at the foot of the slide here

418
00:53:31.280 --> 00:53:39.840
so that's the answer to that question so let's move on and have a look at another example here now

419
00:53:42.640 --> 00:53:50.320
so we've got a function here again I would look at it and I would say well okay I've got a function

420
00:53:50.320 --> 00:53:56.880
t to the power 8 I'm multiplying that by an exponential so that tells me that I need the

421
00:53:56.880 --> 00:54:05.840
first shift in theorem because I'm multiplying this power of t t to the power 8 by an exponential

422
00:54:06.640 --> 00:54:12.160
that's the first shift in theorem that's needed to obtain the Laplace transform of the whole function

423
00:54:13.280 --> 00:54:22.320
okay so let's do a usual let's get the general expression for the first shift in theorem is

424
00:54:22.320 --> 00:54:28.720
given here let's have a look we've been given this function here well we can find f of t either by

425
00:54:28.720 --> 00:54:34.880
direct comparison or simply by just ignoring the exponential that tells me that f of t is equal to

426
00:54:34.880 --> 00:54:42.320
8 now I've got to be a little careful here I've got to identify alpha and again I compare this I've

427
00:54:42.320 --> 00:54:50.640
got e to the minus alpha t is equal to e to the 70 that corresponds to e to the 70 so in other words

428
00:54:50.640 --> 00:55:01.680
minus alpha t must correspond to e correspond to 70 so that's what I've got in here okay so for this

429
00:55:01.680 --> 00:55:08.240
to hold then minus alpha must equal 7 you know think of it if you're afraid think of it just cancel

430
00:55:08.240 --> 00:55:16.400
the t's out minus alpha is equal to 7 or in other words alpha is equal to minus 7 so that's what I've

431
00:55:16.400 --> 00:55:27.840
got here alpha is equal to minus 7 so that's step one so the next step well step one I've

432
00:55:27.840 --> 00:55:34.320
ignored the exponential I identified f of t at step two I've got to get the Laplace transform

433
00:55:34.320 --> 00:55:42.720
of f of t f of t was t to the power 8 well I can do that from the tables and I once again it's number

434
00:55:42.800 --> 00:55:50.240
four this time with n equal to 8 so the Laplace transform of t to the 8 is 8 factorial over

435
00:55:50.960 --> 00:55:58.720
s to the 8 plus 1 so it's 8 factorial over s to the 9 so 8 factorial is quite a large number

436
00:55:58.720 --> 00:56:04.960
and quite often it's okay just to leave it in this form or you could work it out if you prefer

437
00:56:05.440 --> 00:56:13.520
it's the factorial button is on any on most calculators anyway so big f of s is 8 factorial

438
00:56:13.520 --> 00:56:18.880
over s to the power 9 that's using number four on the table with n equal to 8 and I'm using n equal

439
00:56:18.880 --> 00:56:29.520
to 8 because we've got t to the power 8 that's what we've got here okay so I'm up here now in my table

440
00:56:29.520 --> 00:56:38.960
my diagram I've got f of s and my final step says replace s in f of s by s plus alpha alpha is minus

441
00:56:38.960 --> 00:56:46.800
seven so I've got to replace s in f of s by s minus seven there's f of s so s minus seven

442
00:56:48.240 --> 00:56:55.600
gets replaced replaces s so if s is raised to the power 9 then s minus 7 must be raised to the power

443
00:56:55.600 --> 00:57:03.760
9 so that is the final answer so just be careful it's not s to the power 9 minus 7 as I spoke about

444
00:57:03.760 --> 00:57:12.640
a couple of examples ago it's s minus 7 all to the power 9 so that's yet another example of the first

445
00:57:12.640 --> 00:57:19.520
shifted theorem now this is perhaps the example we saw earlier on this is given in the tables but in

446
00:57:19.520 --> 00:57:25.520
many many tables this will not be given and we would have to figure out how to calculate the Laplace

447
00:57:25.520 --> 00:57:34.880
Transword and again if I look at this example I would see a function a trig function sin 2t

448
00:57:34.880 --> 00:57:40.880
been multiplied by an exponential so that tells me I've got to use the first shift in theorem

449
00:57:40.880 --> 00:57:48.240
here's my diagram that I use I set up my comparison here the general expression from the first

450
00:57:48.240 --> 00:57:55.040
shift in theorem is here and the expression I've got is here f of t is easy to get either by just

451
00:57:55.040 --> 00:58:02.320
direct comparison or just by ignoring exponential that tells me f of t is equal to sin 2t what about

452
00:58:02.320 --> 00:58:13.520
alpha well alpha just like in this example we can compare and we can find that alpha is minus eight

453
00:58:13.520 --> 00:58:21.760
you know if you if I just get the tablet if you compare this you've got e to the minus alpha t e to

454
00:58:21.760 --> 00:58:34.320
the minus alpha t must match e to the 8t so in other words minus alpha t must be equivalent to

455
00:58:34.960 --> 00:58:42.400
8t I could cancel the t so if I like minus alpha is equal to 8 or alpha is minus 8 so

456
00:58:44.160 --> 00:58:50.720
that's me identified f of t is sin 2t and I've also identified that alpha is equal to minus eight

457
00:58:50.720 --> 00:58:56.400
so I can go ahead now and use the first shift in theorem following my diagram here starting in the

458
00:58:56.400 --> 00:59:02.480
bottom corner with a function I was given I first of all ignore the exponential to give me f of t

459
00:59:02.480 --> 00:59:11.760
I've done that f of t is sin 2t I get the Laplace transform of it well I've just got sin 2t so I

460
00:59:11.760 --> 00:59:18.480
would look down my second column my table till I hit sin omega t and clearly just by comparison

461
00:59:18.480 --> 00:59:28.080
omega is equal to 2 so the Laplace of sin 2t is 2 over s squared plus 2 squared or in other words

462
00:59:28.640 --> 00:59:35.840
2 over s squared plus 4 so that's what that gives me here so I'm now up in the top right of my table

463
00:59:35.840 --> 00:59:46.480
I've found f of s the final step the final step just involves me replacing any occurrence of s in f of s

464
00:59:47.360 --> 00:59:54.800
by s plus alpha so let's have a look so f of s here it is there's an s here so I've got to replace

465
00:59:54.800 --> 01:00:03.280
that by s plus alpha alpha is minus 8 so this s here must get replaced by s minus 8 and once again

466
01:00:04.240 --> 01:00:11.040
the s squared so the thing that's replacing it the s minus 8 must be squared so that's where

467
01:00:11.040 --> 01:00:19.280
that comes from so it's not s squared minus 8 it's s minus 8 all squared and I get this answer down

468
01:00:19.280 --> 01:00:24.960
here which if I wanted to I could expand the denominator the Braggs and the denominator obtain

469
01:00:24.960 --> 01:00:34.960
this solution both are perfectly correct okay so that's an example another example of the first

470
01:00:34.960 --> 01:00:45.120
shifted theorem let's do another one now slightly different or very slightly different here's my

471
01:00:45.120 --> 01:00:54.160
function e to the 60 times t minus 3 all squared again I compared with a general expression for

472
01:00:54.160 --> 01:01:00.320
the first shifted theorem e to the minus alpha t f of t some function of time f of t multiplied by

473
01:01:00.320 --> 01:01:08.320
an exponential function is that what I've got well yes it is because I've got my exponential

474
01:01:08.320 --> 01:01:19.360
function e to the 6t and this part here would be f of t okay I could easily identify f of t again

475
01:01:19.360 --> 01:01:26.720
either by direct comparison or just by ignorantly exponential in the given function t minus 3

476
01:01:27.440 --> 01:01:36.640
all squared that's where f of t is so there's f of t we've identified it now when we come at step 2

477
01:01:36.640 --> 01:01:42.560
to calculate the Laplace transform of this I know this I will not have this in my tables but what I'm

478
01:01:42.560 --> 01:01:49.920
going to do is I'm going to expand it because I know I've got t squared I've got I've got t in the

479
01:01:49.920 --> 01:01:55.040
tables and I've got a constant in the table so if I expand this bracket I can certainly get the Laplace

480
01:01:55.040 --> 01:02:01.600
of these three terms from the table so it'll just be number three number two number one in the tables

481
01:02:02.480 --> 01:02:07.360
so that's fine I could certainly do that what's the value of alpha well again you know if we

482
01:02:07.360 --> 01:02:14.960
compare what have we got we've got in our question we've got e to the 6t let me just write that down

483
01:02:17.040 --> 01:02:23.440
and in the general expression for the first shifted theorem is e to the minus alpha t

484
01:02:24.320 --> 01:02:32.320
so in other words 6t must be equivalent to minus alpha t the t's can cancel if you like

485
01:02:32.320 --> 01:02:38.400
6 is equal to minus alpha or alpha is equal to minus 6 so that's where that comes from

486
01:02:39.600 --> 01:02:50.400
okay so we're just about ready to to start using the diagram to go around the diagram in the clockwise

487
01:02:50.960 --> 01:02:57.280
direction so that's the function we're given down in the bottom left hand corner we ignore the

488
01:02:57.280 --> 01:03:06.000
exponential to leave us f of t f of t we're already identified as this function here so that's f of t

489
01:03:06.000 --> 01:03:11.760
so I now want to take the Laplace transform of this function which involves three separate terms

490
01:03:12.320 --> 01:03:19.760
so but we know because the linearity property that the Laplace transform of this entire function

491
01:03:19.760 --> 01:03:25.920
is equivalent to just taking the Laplace transform of each individual term and adding or subtracting

492
01:03:25.920 --> 01:03:33.840
them together depending on what's required so let's do that so I've got the Laplace transform of t

493
01:03:33.840 --> 01:03:40.560
squared well that's going to be number three in the table minus well I can take the constant out

494
01:03:40.560 --> 01:03:47.520
if I like minus six times the Laplace transform of t that's just number two plus nine times the

495
01:03:47.520 --> 01:03:54.240
Laplace transform of one and that's just number one in our table so I can get these quite easy

496
01:03:54.240 --> 01:04:02.560
from the table t squared it's two over s cubed it's the Laplace transform of t squared minus six

497
01:04:02.560 --> 01:04:10.400
times the Laplace transform of t so it's minus six times one over s squared plus nine times the Laplace

498
01:04:10.400 --> 01:04:20.560
transform of one well it's just nine times one over s we've seen that before anyway and then we

499
01:04:20.560 --> 01:04:30.080
can tidy all of this up and there is the expression for big f of s that's the Laplace transform of f of t

500
01:04:30.080 --> 01:04:36.800
and I'm now up in the top right in my diagram and the last step that I have to do is I've got to replace

501
01:04:36.800 --> 01:04:45.840
every occurrence of s by s plus alpha now up to now I think we've only had one occurrence of s in f of s

502
01:04:45.840 --> 01:04:52.720
but if we look at this expression here we can see that s appears three times it appears in each of the

503
01:04:52.720 --> 01:05:01.200
denominators and that being the case we must replace every occurrence of s by s plus alpha alpha is

504
01:05:01.200 --> 01:05:08.720
minus six so each of these gets replaced by s minus six going all to the next slide that's what I've

505
01:05:08.720 --> 01:05:17.440
got for f of s I've got to replace it by s each s by s minus six so let's just do that so I'll have two

506
01:05:17.440 --> 01:05:25.760
over s minus six all cubed so s minus six is going in here for that s and if the original s was cubed

507
01:05:25.760 --> 01:05:33.280
then the thing that replaces it which was s minus six that too must be cubed similarly here s was

508
01:05:33.280 --> 01:05:40.560
squared so it's going to be s minus six all squared and simply here s gets replaced by s minus six so

509
01:05:40.560 --> 01:05:47.920
that is the Laplace transform of my original function and that's the answer to the question

510
01:05:48.640 --> 01:05:57.600
so just to summarize then today we've introduced Laplace transforms and looked at the motivation

511
01:05:57.600 --> 01:06:04.720
for using them and we've looked at the definition of the Laplace transform that it's actually

512
01:06:05.680 --> 01:06:12.640
an improper integral and it would be a lot of work to calculate if we didn't have the Laplace tables

513
01:06:13.920 --> 01:06:20.080
so we looked at some examples of calculating Laplace transforms of standard functions using the tables

514
01:06:21.040 --> 01:06:28.880
and then we looked at the first shifted theorem which was able to handle a product of two functions

515
01:06:28.880 --> 01:06:35.680
where one of them was an exponential so we saw that in that case we used the first shifted theorem

516
01:06:35.680 --> 01:06:44.000
and very importantly we saw or we know now that the Laplace transform of a function like this is not

517
01:06:44.000 --> 01:06:49.520
the Laplace transform of the first function times the Laplace transform of the second function

518
01:06:49.520 --> 01:06:56.640
that is not how to do that is incorrect so in this case because we multiplied a function by an

519
01:06:56.640 --> 01:07:04.560
exponential we use the first shifted theorem in the next class we're going to look at inverse Laplace

520
01:07:04.560 --> 01:07:13.200
transforms so if today we went from the time domain to the s domain or the Laplace domains it's also

521
01:07:13.200 --> 01:07:19.600
called in our next class we're going to look at going the other way from the s domain from the Laplace

522
01:07:19.600 --> 01:07:30.480
domain back to the time domain and that involves taking inverse Laplace transforms so after this

523
01:07:30.480 --> 01:07:36.400
class then you should be able to attempt question one from the Mobius questions and just remember

524
01:07:36.400 --> 01:07:44.880
that each individual question Mobius question has a large database of questions associated with it so

525
01:07:44.880 --> 01:07:50.960
you can keep on just try you click in the try another button and to get a different question

526
01:07:50.960 --> 01:07:56.640
of the same type so you could practice these once you've seen the Mobius questions and I also

527
01:07:56.640 --> 01:08:02.640
included here from the lecture notes which questions you should also be able to attempt

528
01:08:03.360 --> 01:08:08.720
so we'll leave it at that then for today and as I said the next time we're going to continue Laplace

529
01:08:08.720 --> 01:08:15.760
transforms looking at calculating inverse Laplace transforms