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 Well hello everyone and welcome to our fifth class on Laplace Transforms.

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 In our last class we introduced the unit step function,

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 a function that can be used to model a signal that suddenly switches on at a specific time and stays on indefinitely.

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 Today we're going to look at how to take the Laplace Transform of a unit step function

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 and this will be extremely useful when we come to solving certain initial value problems.

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 We'll then move on and introduce the second shifting theorem

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 and this will enable us to take Laplace Transforms and the inverse Laplace Transforms of functions that involve step functions

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 and again it's particularly useful for solving ODE's and initial value problems.

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 Okay so just a quick recap on the unit step function.

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 So we discussed how in real-world applications we often come across functions whose values change suddenly at specific values of time

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 and an example of that would be a voltage that's switched on or off in an electrical circuit at a given time

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 and this switching process can be described mathematically by a unit step function.

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 And solving differential equations involving step functions is made considerably easier using Laplace Transforms.

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 So we therefore need to know how to take the Laplace Transform of a step function and related functions.

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 It's functions that involve step functions just like we saw last time when we saw various products

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 and sums and differences of unit step functions.

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 And these will typically appear as floating terms in initial value problems.

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 So the unit step function as we know suddenly switches on at a given value and stays on forever.

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 So in this example my unit step function is zero all the way up to time t equals zero

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 and then it suddenly switches on, takes the value one and holds that value forever and ever.

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 So that's our unit step function and it's described mathematically here u of t is equal to zero for t less than zero

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 and is equal to one for t greater than zero.

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 Now as we saw that unit step function switched on at time t equals zero

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 it doesn't necessarily have to switch on at that value of time, it can switch on for any positive value of time.

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 And this is an example of we might call the delayed or shifted unit step function.

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 So it switches on at a time a little later than t equals zero.

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 So in this case our unit step function has switched on at time t equals capital T

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 and we call this the critical value of the step function.

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 For the previous case the critical value was capital T is equal to zero.

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 In this case the unit step function switches on at t is equal to capital T

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 and mathematically we describe it like so u of t minus t that's our critical value given by the capital T.

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 So it's zero for t less than capital T and it's one for t greater than capital T.

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 And you can see that in the diagram here and we saw that of course in our last lecture.

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 So we would now like to know how to take the Laplace transform of a unit step function

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 if we're going to use these types of functions in differential equations to model voltages, forces and so on.

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 So let's have a look at how we do that.

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 Now we're going to do it here from the definition.

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 You will not be asked to reproduce this but it's still useful to know where this comes from.

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 So if you remember the definition for the Laplace transform,

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 we calculate it from first principles, we've got Laplace transform of function f of t is the integral from zero to infinity.

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 It's an improper integral because infinity is one of the limits.

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 The integral from zero to infinity of f of t times e to the minus st dt.

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 In this case our function f of t is given by our step function u of t minus capital T.

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 So that gets replaced under the integral sign and it's times e to the minus st and we integrate this with respect to t between zero and infinity.

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 Now here's again a plot of our step function and we note that nothing happens until we hit the critical value at t equals zero.

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 So we're going to split our integral into two separate intervals.

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 We're going to calculate it over two separate intervals.

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 The first being from zero to the critical value zero to capital T and then from capital T out to infinity.

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 Representing where the unit step has now switched on.

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 So in the first interval between zero and capital T we can see that our unit step function has the value zero.

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 So that's what I've done here.

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 So my integral from zero to capital T, the step function has the value zero times e to the minus st dt.

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 Now we know that this just returns a value of zero.

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 So when you integrate zero you get zero.

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 The second integral here takes into consideration the interval that goes from capital T out to infinity.

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 And when we evaluate these improper integrals we usually introduce a dummy variable u for example and we have u as the upper limit of the integral and let that tend to infinity.

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 Now the value of the unit step function over this interval as you can see is just one.

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 So my f of t here which is the unit step function is equal to one between t out to infinity.

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 There it is. You can see that from the diagram times e to the minus st dt.

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 Now I can integrate this exponential and I get minus one over s e to the minus st.

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 Now if I evaluate that between the limits the upper limit will be u will go in here in the place of t and we'll let u tend to infinity.

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 It's a negative exponential so we'll have exponential decay.

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 So this will all go off to zero.

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 At the lower limit we substitute in t and this will evaluate to zero minus one over s e to the minus st dt.

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 And that will work out to give me one over s e to the minus s times capital T.

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 And that is the Laplace transform of the unit step function.

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 Okay whose critical value is located at t.

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 We've also called this as I said earlier the shifted or delayed unit step function.

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 Now if we were to take a value of capital T equal to zero that being the first case that we saw where the critical value of the step function is equal to zero.

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 We can see that this Laplace transform of the unit step function with critical value zero would be one over s e to the minus s times zero.

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 In other words one over s e to the zero e to the zero is one so just one over s.

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 And that makes sense because this function here for t greater than zero is just the constant one.

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 It's just equal to one and we know that the Laplace transform of one is just one over s.

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 It's the first transform that's given in our tables if you wanted to check that.

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 Okay so just remind yourselves some other step functions.

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 Here we've got a step function whose critical value is t equals four and it's got amplitude six.

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 So we just basically multiply the unit step function u t minus four by six to represent this function.

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 And we can take the Laplace transform of that quite easily.

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 We can take the constant outside with the Laplace transform of u t minus four and we can go to our tables for that actually.

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 It's number 26 in our tables with capital T equal to four.

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 So if I evaluate that I get six times one over s e to the minus four s or in other words six over s e to the minus four s.

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 Now here's our step function, our unit step function whose critical value is at the origin.

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 It switches on at time t equals zero.

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 We can either write that as u of t or we can write as u of t minus zero.

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 This represents the capital T value.

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 So if again we follow the number 26 in our table we would find that the Laplace transform would be one over s e to the minus s times zero.

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 Or one over s e to the zero which is just one over s.

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 And that's as I said earlier that's exactly what we would expect for this function that takes the value one from zero onwards.

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 Okay so let's now move on and look at the second shifting theorem.

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 If you remember a couple of lectures ago we introduced the first shifting theorem and that told us that if you've got the product,

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 we'll just get the tablet, if you've got the product of an exponential e to the minus alpha t times some function f of t then this would correspond to a shift in the estimate.

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 So this is a multiplication by an exponential in the time domain.

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 Multiplication of some function f of t by an exponential in the time domain corresponds to a shift by alpha in the s domain.

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 So basically we have the Laplace transform of f of t, our function here, f of s into the Laplace transform of this function.

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 And that's shifted by an amount alpha in the estimate where the alpha is given here.

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 So that was the first shifting theorem. Today we're going to look at the second shifting theorem.

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 Now the second shifting theorem enables us to solve initial value problems that will model time delayed systems by including the delay or time shift directly in the Laplace transform.

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 I mean you saw the idea of a delayed function earlier on in these for example.

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 So this is a delayed unit step function. It's basically switched off up until the value t equals 4.

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 So that's kind of an example of a delayed or shifted function.

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 Okay so suppose now we've got a function f of t, we can write it as f of t times u of t.

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 Basically it switches on the time t equals 0. Remember that's what the step function can be used for.

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 Use can be used to switch on and off other functions.

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 We saw that in the last lecture when we looked at products of functions and the step function.

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 We looked at the graphs of these functions.

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 So suppose we've got a function f of t times step function u of t and it's been delayed or shifted by capital t units in the time domain.

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 The corresponding delayed function would be given like so.

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 f of t minus t times the step function is critical value is capital t.

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 So this function because of the minus it gets shifted to the right.

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 So we'll discuss that a little more later on but be careful here.

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 It's not a left shift. It's a right shift. This function gets shifted to the right and it gets set to 0 all the way up to the critical value.

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 And you'll see that we'll be calling these shifted and truncated functions.

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 Truncated in the sense that it gets killed off or set to 0 all the way up to the critical value t.

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 So the second shifting theorem states that if we take the Laplace transform of this function here.

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 Shifted and truncated function if you like.

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 Then the Laplace transform of this time delayed function is a Laplace transform of the function without the delay.

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 That's the function without the delays f of t and its Laplace transform is f of s multiplied by e to the minus st where t is capital t represented the time delay.

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 So that's what I say down here. This function f of t minus capital t times its f function with critical value t.

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 It's often referred to as shifted and truncated function.

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 And I'll just go once more over that.

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 So the Laplace transform of a shifted and truncated function is simply the Laplace transform of the unshifted function.

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 That's f of s because the unshifted function is simply f of t Laplace transform of f of t is f of s.

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 So we take the Laplace transform f of s of the unshifted function.

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 We multiply that by e to the minus st where t represents the delay and it's equal to the shift and it's equal to the critical value in the step function.

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 So here's just another example of a shift and truncated function.

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 So basically I say here to obtain this shifted and truncated function f of t minus capital t times our step function with critical value t.

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 The original function f of t times u of t.

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 There it is.

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 Just basically f of t for positive values of t because the effect of this step function whose critical value is zero.

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 It's basically to switch everything off right up to t equals zero.

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 So it's just the function f of t for positive values of t.

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 So if we want to shift and truncate this function, we can shift f function by capital t units to the left.

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 That's given by the minus that shows that it's a shift to the right and we multiplied by the step function with critical value t.

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 So my function here has been shifted by capital t units to the right and it's been truncated.

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 In other words, it's been set to zero all the way up to this critical value simply by multiplying by unit step function with critical value t.

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 Okay, so here's just another example of a shift and truncated function.

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 My original function here is f of t is equal to t just the 45 degree line.

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 I'm going to shift this function two units to the right and I get my blue function which I call g of t.

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 So g of t is t minus two because my shift is to the right.

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 And you can see that this makes sense because if you set y, let's call y is equal to t minus two.

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 So where is y equal to zero? Well, it's when t is equal to two.

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 So the minus is the correct sign here.

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 As I said, people often think that because it's a minus that the shift is to the left, it's not. The shift is to the right.

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 So from my f of t, I shifted it to two units to the right to get my new function g of t, which I've shown in blue.

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 And then I'm going to truncate this function.

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 I'm going to set it to zero all the way up to the value t equals two.

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 And to do that, I multiply by the unit step function, which remember whose critical value is equal to two,

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 which as we know will set everything equal to zero all the way up to the critical value, which is t equals two.

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 So this is what we get.

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 Our function h of t, I've called it, is zero all the way up to value two.

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 And then it's simply the graph of t minus two from there onwards.

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 And again, we saw this last time when we were looking at plots of functions that involved products of functions where one of the functions is a unit step function.

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 Okay, so there we are.

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 So the first shift in theorem, sorry, the second shift in theorem, therefore, says that the last transform of f of t, or if you prefer, f of t times u of t,

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 if that's equal to capital F of s, then the Laplace transform of the shifted and truncated function, as I said earlier on,

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 is simply the product of the Laplace transform of the unshifted function.

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 That's the Laplace transform of f of t, which is f of s, and we multiply that by e to the minus s t,

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 where t is the shift or the delay if you prefer, and it's also the critical value that appears in our step function.

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 So that's what the second shifting theorem says, and it allows us to take Laplace transforms and inverse Laplace transforms of step functions

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 and other functions that will involve possibly products of step functions with other functions.

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 Now, we're going to now show how we can apply the second shifting theorem to calculate the Laplace transforms of certain functions,

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 and we're going to do that using this diagram here.

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 So we're going to use these shifted and truncated functions in our examples.

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 So we would like to be able to go directly over here, but we can't do it right away like this.

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 So what we do is we use the diagram.

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 First of all, we ignore the step function, we ignore the shift, and that leaves us f of t.

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 We take the Laplace transform of f of t to give us f of s, and our final step, as we talked about before,

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 is we multiply f of s, the Laplace transform of the unshifted function.

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 We multiply that by e to the minus s t, where t represents the shift or the delay if you prefer,

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 and it also represents the critical value of the step function.

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 So we have one, two, three steps to go from the time domain where we have a shifted and truncated function

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 to go from the time domain to the s domain.

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 In other words, to get the Laplace transform of this shifted and truncated function.

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 Okay, so here we are then, we've got a function g of t is equal to t minus 4 ut minus 4.

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 Now, does this match the type of function that we need for applying the second shifted theorem?

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 So basically, I've got to have some function of t minus capital T,

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 multiplied by a step function whose critical value is the same as the amount of shift in our function.

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 Is that the case? Well, yes it is. The critical value of the step function is 4.

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 Our function f of t, if you like, has been shifted by 4 to the right, so that we can call it f of t minus 4.

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 I just get my tablet, so just say a couple of words about that.

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 If I've got a function f of t is equal to t, say, well, my function f of t minus 4, how do I get that?

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 Well, I get that simply by replacing t minus 4 wherever I see t in the definition of the function.

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 In this case, it's very simple. It only appears once, so f of t minus 4 is equal to t minus 4 in this case.

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 So, we can move on now and look at the calculating the last transform of this shifted and truncated function.

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 So we start off on the bottom left hand corner of the diagram in the time domain.

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 We're going to move around in a clockwise direction to end up in the estimate and thereby obtain the Laplace transform of the shifted and truncated function.

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 So that's what I say here. Start bottom left and move around clockwise.

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 So the first thing to do, step one, ignore the step function and ignore the shift to simply leave f of t is equal to t.

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 So if I ignore the step function here, that's ignored, I ignore the shift so my function f of t would be equal to t.

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 I then want to take the Laplace transform and that's simple because just number two in the tables, it's 1 over s squared.

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 So f of s, the Laplace transform of f of t is simply 1 over s squared, so there it is.

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 And my final step says multiply f of s, we just found at step two.

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 Multiply that by e to the minus s t with the t that I'm talking about as the capital t and that corresponds to the shift, which also corresponds to the critical value of the step function.

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 In this case, we can easily identify that just by looking at our question, our critical value of the step function is 4, our shift is equal to 4, so capital t is equal to 4.

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00:24:31.000 --> 00:24:46.000
 So the Laplace transform of my shifted and truncated function t minus 4 times u t minus 4 is simply 1 over s squared, that's f of s times e to the minus 4 s.

196
00:24:46.000 --> 00:24:59.000
 Because my capital t, my shift, which equals the critical value of the step function is equal to 4, so that gives me the Laplace transform of this function.

197
00:24:59.000 --> 00:25:05.000
 Okay, so just a couple of words here about this shift and truncated function.

198
00:25:05.000 --> 00:25:09.000
 So we start off with our original function.

199
00:25:09.000 --> 00:25:13.000
 We start off with f of t is equal to t.

200
00:25:13.000 --> 00:25:17.000
 So f of t is equal to t, the 45 degree line,

201
00:25:17.000 --> 00:25:24.000
 t minus 4, well. we shift this function, 4 units to the right,

202
00:25:24.000 --> 00:25:28.000
 So if we cut the horizontal axis, the t axis at t equals 4.

203
00:25:28.000 --> 00:25:39.000
 So this is shown in the green dotted line here, that's just the graph of f of t is equal to t, shifted 4 units to the right.

204
00:25:39.000 --> 00:25:42.000
 There we are, that's my green function.

205
00:25:42.000 --> 00:25:50.000
 And then I'm going to truncate this function and to do that, I multiply it by my step function whose critical value is 4.

206
00:25:50.000 --> 00:25:59.000
 And we know that has the effect of setting f into 0 all the way up to the critical value of 4.

207
00:25:59.000 --> 00:26:07.000
 And then it switches on the function because the step function takes the value 1 from then onwards.

208
00:26:07.000 --> 00:26:12.000
 So basically from 4 onwards, you just multiply t minus 4 by 1.

209
00:26:12.000 --> 00:26:20.000
 So you've just got the graph of t minus 4 from 4 onwards.

210
00:26:20.000 --> 00:26:25.000
 So that's what this function, the shift and truncate function is plotted in blue.

211
00:26:25.000 --> 00:26:32.000
 0 all the way up to 4 because the units step function hasn't been switched on.

212
00:26:32.000 --> 00:26:35.000
 So it's therefore the value of 0 all the way up to 4.

213
00:26:35.000 --> 00:26:40.000
 So anything that multiplies equal to 0, so the whole thing is equal to 0 up to t equals 4.

214
00:26:40.000 --> 00:26:45.000
 And then the step function switches on the t equals 4, takes the value 1.

215
00:26:45.000 --> 00:26:55.000
 So my g of t function will just be equal to t minus 4 from t equals 4 onwards forever and ever.

216
00:26:55.000 --> 00:26:58.000
 And it will just keep growing like that.

217
00:26:58.000 --> 00:27:01.000
 Okay, so let's do another example now.

218
00:27:01.000 --> 00:27:10.000
 So determine the Laplace transform of g of t is equal to t minus 4 all squared times the step function ut minus 4.

219
00:27:10.000 --> 00:27:18.000
 So this is along the same lines as the previous one except now that my function,

220
00:27:18.000 --> 00:27:24.000
 multiplying the step function, my shifted function is t minus 4 all squared.

221
00:27:24.000 --> 00:27:27.000
 It was just t minus 4 in the previous example.

222
00:27:27.000 --> 00:27:36.000
 So again, we're going to refer to a diagram, but first let's have a look and see what the plot of this function looks like.

223
00:27:36.000 --> 00:27:41.000
 I've got it in a bit more, this is it here, but I've got it in a bit more detail here.

224
00:27:41.000 --> 00:27:47.000
 So we want to ignore the step function and we want to ignore the shift.

225
00:27:47.000 --> 00:27:56.000
 That's what we do, ignore the step function, ignore the shift and that just leaves me that my function,

226
00:27:56.000 --> 00:28:01.000
 unshift function f of t will equal to t squared.

227
00:28:01.000 --> 00:28:03.000
 And that's what I plotted here in red.

228
00:28:03.000 --> 00:28:08.000
 It's your smiley face graph, f of t is equal to t squared.

229
00:28:08.000 --> 00:28:14.000
 Now I'm going to shift this function for units to the right.

230
00:28:14.000 --> 00:28:22.000
 Okay, so then that becomes t minus 4 all squared.

231
00:28:22.000 --> 00:28:25.000
 And that's what you can see in blue here.

232
00:28:25.000 --> 00:28:36.000
 So this is the function t squared, shifted 4 units to the right, gives me the function t minus 4 all squared.

233
00:28:36.000 --> 00:28:43.000
 Just be careful here, the temptation might be to write t squared minus 4,

234
00:28:43.000 --> 00:28:51.000
 but that will just shift the function up and down the vertical axis by 4 units.

235
00:28:51.000 --> 00:28:58.000
 To shift it along the horizontal axis, we write it as t minus 4 all squared.

236
00:28:58.000 --> 00:29:06.000
 Then the final step is to multiply by the unit step function whose critical value is 4

237
00:29:06.000 --> 00:29:13.000
 and that will have the effect of switching this function off all the way up to t equals 4.

238
00:29:13.000 --> 00:29:17.000
 And that is shown in black in the diagram down here.

239
00:29:17.000 --> 00:29:20.000
 It doesn't get switched on until the value t equals 4.

240
00:29:20.000 --> 00:29:29.000
 And that's because this step function, this unit step function, takes the value 0 all the way up to the critical value 4.

241
00:29:29.000 --> 00:29:35.000
 So all the way up to 4, you're multiplying this quadratic function by 0.

242
00:29:35.000 --> 00:29:40.000
 So my function g of t will be 0 all the way up to the critical value.

243
00:29:40.000 --> 00:29:47.000
 Then suddenly this unit step function switches on at t equals 4 and takes the value 1.

244
00:29:47.000 --> 00:29:59.000
 So from t equals 4, the critical value onwards, my function g of t will be t minus 4 all squared times 1, or just t minus 4 all squared.

245
00:29:59.000 --> 00:30:03.000
 So that's where this graph here comes from.

246
00:30:03.000 --> 00:30:06.000
 And that's what's shown here in blue.

247
00:30:06.000 --> 00:30:11.000
 So we want to find its Laplace transform.

248
00:30:11.000 --> 00:30:21.000
 So again, we're going to use our diagram starting the bottom left hand corner and move around the diagram in a clockwise direction.

249
00:30:21.000 --> 00:30:26.000
 And that will move us from the t domain to the s domain.

250
00:30:26.000 --> 00:30:33.000
 Okay, so step one, ignore the step function and ignore the shift.

251
00:30:33.000 --> 00:30:40.000
 So if I do that, that leaves me the function f of t equals t squared.

252
00:30:40.000 --> 00:30:47.000
 I then want to find f of s, capital F of s, which is just the Laplace transform of f of t.

253
00:30:47.000 --> 00:30:58.000
 And that means taking the Laplace transform of t squared, which I can do from memory or either that, or just use number three, I think, in the table.

254
00:30:58.000 --> 00:31:06.000
 And it'll tell me that f of s is equal to 2 over s cubed, number three in the table.

255
00:31:06.000 --> 00:31:21.000
 Then finally, my final step here is to multiply f of s that I just found at step two, the Laplace transform of f of t, my unshifted function f of t, it's Laplace transform f of s.

256
00:31:21.000 --> 00:31:31.000
 Multiply that by e to the minus s t, where t is the critical value of the step function, which equals the shift.

257
00:31:31.000 --> 00:31:39.000
 And that's four. So I end up with 2 over s cubed, that's my f of s, times e to the minus 4s.

258
00:31:39.000 --> 00:31:51.000
 So that's the Laplace transform of this shifted untrunkated function, shown in blue in the diagram here.

259
00:31:51.000 --> 00:32:02.000
 Okay, moving on, here's another function, just maybe slightly more complicated perhaps.

260
00:32:02.000 --> 00:32:09.000
 We've got our function g of t is equal to sine t minus 2 pi times the step function, whose critical value is 2 pi.

261
00:32:09.000 --> 00:32:13.000
 And we want the Laplace transform of this function.

262
00:32:13.000 --> 00:32:32.000
 So once again, we'll start in the bottom left in the time domain, move around the diagram in a clockwise direction to get our s domain function, which is the Laplace transform of our shifted untrunkated function given here.

263
00:32:32.000 --> 00:32:44.000
 So the first thing we do is we ignore the step function and we ignore the shift and that just leaves me my function f of t is equal to sine t.

264
00:32:44.000 --> 00:32:51.000
 There it is. So I'm up here in the diagram. Now I want the Laplace transform of f of t.

265
00:32:51.000 --> 00:32:58.000
 Now that's just number 8 in the table and it's sine t, so it's sine 1t, so omega is equal to 1.

266
00:32:58.000 --> 00:33:10.000
 So the Laplace transform of sine t would just be 1 over s squared plus 1, using this here, number 8 with omega equal to 1.

267
00:33:10.000 --> 00:33:28.000
 I then need to identify the value of capital T and that's just the critical value of my step function, which equals the shift, which in this case is 2 pi.

268
00:33:28.000 --> 00:33:36.000
 And my table tells me that I should multiply f of s by e to the minus s times 2 pi.

269
00:33:36.000 --> 00:33:52.000
 So when I do that, I obtain that the Laplace transform of this shifted untrunkated function, that's the Laplace transform of sine t minus 2 pi times the unit step function t minus 2 pi is simply 1 over s squared plus 1,

270
00:33:52.000 --> 00:34:09.000
 and that's the Laplace transform of the unshifted function f of t equals sine t, that's the Laplace transform of that, which we call f of s, and we multiply that by e to the minus 2 pi s, where 2 pi was the critical value of the step function,

271
00:34:09.000 --> 00:34:14.000
 which was equal to the shift, which is also called the delay.

272
00:34:14.000 --> 00:34:24.000
 So you can see the diagram here, the plot of my function sine of 2t minus 2 pi times the unit step t minus 2 pi.

273
00:34:24.000 --> 00:34:27.000
 So first of all, let's see what happens.

274
00:34:27.000 --> 00:34:30.000
 So we've got a function sine of t minus 2 pi.

275
00:34:30.000 --> 00:34:37.000
 Now that is just the graph of sine t shifted 2 pi units to the right.

276
00:34:37.000 --> 00:34:50.000
 And then we remember that sine is 2 pi periodic, it repeats the shape after every interval of length 2 pi s.

277
00:34:50.000 --> 00:35:08.000
 And we multiply this by our step function, which sets everything to zero up to the critical value 2 pi, and then step function is equal to 1 from there onwards, and it just multiplies this to give us this graph here.

278
00:35:08.000 --> 00:35:28.000
 We remember of course that because sine of t is 2 pi periodic, in here really it repeats after every interval with 2 pi, so in here we'd have a graph that could go up, down, and up, basically we'd have this graph again, but the step function has set it to zero.

279
00:35:28.000 --> 00:35:41.000
 So in effect we've really just got the graph of sine t carrying on forever and ever from 2 pi onwards, because it's a periodic function.

280
00:35:41.000 --> 00:35:48.000
 Okay, right, so what do we got now?

281
00:35:48.000 --> 00:35:59.000
 We've got another function here, g of t is equal to t minus 2 old cubed times the unit step function t minus 2.

282
00:35:59.000 --> 00:36:07.000
 Is this in the form of that we need to use the second shift in theorem?

283
00:36:07.000 --> 00:36:10.000
 Well the answer is yes it is.

284
00:36:10.000 --> 00:36:23.000
 This is a function f of t minus 2, and it's multiplied by a step function ut minus 2, and the critical value of the step function matches the shift.

285
00:36:23.000 --> 00:36:28.000
 So we can use the second shift in theorem here.

286
00:36:28.000 --> 00:36:31.000
 So maybe let's just have a look at what this looks like.

287
00:36:31.000 --> 00:36:50.000
 Here's the graph of t cubed, there it is passing through the origin, the point of inflection there, passes through the origin, goes off to infinity in that, in the positive t direction, and goes to minus infinity moving the negative t direction.

288
00:36:50.000 --> 00:37:01.000
 So we're going to shift it two units to the right, so it's just basically the red graph, shift it two units to the right, so when it passes through the origin here it now passes through two here.

289
00:37:01.000 --> 00:37:04.000
 So that's what it looks like.

290
00:37:04.000 --> 00:37:16.000
 And then we multiplied it by a unit step function, this critical value was 2, so the effect was to the left of 2 everything is set to zero,

291
00:37:16.000 --> 00:37:31.000
 and then we've just basically got this graph from t equals 2 onwards, and that is exactly what we can see here, set to zero, all the way up to t equals 2, and then this function comes alive then.

292
00:37:31.000 --> 00:37:43.000
 So again we'll use our table, so we're going to ignore the step function, ignore the shift, and that leaves me that my function f of t is equal to t cubed.

293
00:37:43.000 --> 00:37:54.000
 Okay, I ignore the step function, I ignore the shift, it gives me f of t, which in this case, ignore that, ignore the shift, it gives me that f of t is equal to t cubed.

294
00:37:54.000 --> 00:38:12.000
 I want the Laplace transform of t cubed, that'll be number four in the table, we have n factorial over s to the n plus one, n is equal to three, so I've got three factorial over s to the three plus one, or s to the four in other words.

295
00:38:12.000 --> 00:38:25.000
 Three factorial is three times two times one, so my Laplace transform of f of t, f of s in other words, is equal to six over s to the power four.

296
00:38:25.000 --> 00:38:48.000
 Then my final step, I've got to multiply this f of s by e to the minus s t, that's capital T, where capital T is the critical value of the step function, and that's equal to the shift, and in this case that's easily identified as capital T is equal to two.

297
00:38:48.000 --> 00:39:08.000
 So I multiply f of s from the previous step by e to the minus two s, and there we are, that is the Laplace transform of my shifted and truncated function, six over s to the power four e to the minus two s.

298
00:39:08.000 --> 00:39:26.000
 Okay, so here's one that's slightly different, we want to find the Laplace transform of t u t minus eight, so it's a function t, multiply by the step function, its critical value is eight.

299
00:39:26.000 --> 00:39:35.000
 Now the question we need to ask ourselves is can we use the second shifting theorem directly on that function.

300
00:39:35.000 --> 00:39:44.000
 So we'll compare this function with a function the bottom left in our time domain of our diagram shown here.

301
00:39:44.000 --> 00:40:02.000
 Now one thing we do know is that to use for the second shifting theorem to apply the critical value of the step function must equal the shift in our function, by the amount that our function f of t has been shifted,

302
00:40:02.000 --> 00:40:18.000
 and we can see this just go back here, in our previous case in this one we looked and we saw that our critical value of the step function was t is equal to two, and we saw that our function f of t was equal to t cubed was shifted horizontally to units,

303
00:40:18.000 --> 00:40:25.000
 and we got t minus two all cubed, so that the critical value of the step function matched the shift.

304
00:40:25.000 --> 00:40:36.000
 The problem here is that the critical value of the step function is eight, the shift here if you like is zero, you know you could write this as t minus zero.

305
00:40:36.000 --> 00:40:57.000
 So we've got to perform some algebraic manipulation before we can get this function to conform to the form that's given in the bottom left of the table here, and that would allow us to use the second shifting theorem.

306
00:40:57.000 --> 00:41:24.000
 Now we have to rewrite this function f of t equals t so that it includes a minus capital T, because if you look this is a step function u t minus capital T, our multiplying function must have the same shift equaling the critical value of the step function.

307
00:41:25.000 --> 00:41:39.000
 So we must therefore, we must include t minus eight somehow, but we're not allowed to change what we start out with, f of t was equal to t.

308
00:41:39.000 --> 00:41:58.000
 So the trick we're going to use is we're going to write down t minus eight, that's what we've got to have, we've got to have a function t minus eight, but to overcome the fact that we've changed it by subtracting eight, we can add eight back in.

309
00:41:58.000 --> 00:42:07.000
 So if you simplify this term in square brackets t minus eight plus eight, it's just t and that's the function that we started out with here.

310
00:42:07.000 --> 00:42:22.000
 So we haven't changed anything, but we've rewritten this in the form of f of t minus eight, you can think of this as a function of t minus eight.

311
00:42:23.000 --> 00:42:40.000
 But we haven't changed anything, so g of t from the previous slide, there it is, g of t is equal to t times u t minus eight, I've written it as t minus eight plus eight times our step function.

312
00:42:40.000 --> 00:42:47.000
 And this thing here, what I've underlined in blue here, is a function of t minus eight.

313
00:42:47.000 --> 00:43:09.000
 Now if you need more convincing of that, think of it this way, if my function f of t is t plus eight, then if I replace t with t minus eight, I'll have that my function f of t minus eight will just be t minus eight plus eight.

314
00:43:10.000 --> 00:43:19.000
 So that I've replaced the t here with t minus eight and that hopefully will convince you that this is a function of t minus eight.

315
00:43:19.000 --> 00:43:35.000
 And we've now got it in the form that conforms with the bottom left hand corner of our diagram and that allows us to use the second shifted theorem to transform our function g of t.

316
00:43:36.000 --> 00:43:40.000
 Okay, so I set up my problem again.

317
00:43:40.000 --> 00:43:49.000
 This is the way I rewrote the function, but we know that it's actually equal to the original function which we were given.

318
00:43:49.000 --> 00:43:54.000
 Again, we start the bottom left and travel round in a clockwise direction.

319
00:43:54.000 --> 00:43:58.000
 We ignore the step function, we ignore the shift.

320
00:43:58.000 --> 00:44:02.000
 So just been very careful here.

321
00:44:02.000 --> 00:44:04.000
 Ignore the step function, well that's easy.

322
00:44:04.000 --> 00:44:07.000
 That's easy, that's just the u t minus eight.

323
00:44:07.000 --> 00:44:11.000
 So let me just get the pen.

324
00:44:11.000 --> 00:44:16.000
 So I'm going to ignore my step function.

325
00:44:16.000 --> 00:44:18.000
 Now ignore the shift.

326
00:44:18.000 --> 00:44:24.000
 Now the shift is the quantity that subtracted from the t.

327
00:44:24.000 --> 00:44:27.000
 It's that one.

328
00:44:27.000 --> 00:44:32.000
 Okay, so we ignore that and that leaves us t plus eight.

329
00:44:32.000 --> 00:44:42.000
 I'm making a point of this because later on when we need to do some other form manipulation, see that these values will not necessarily be equal.

330
00:44:42.000 --> 00:44:49.000
 So it's important that you know at this stage which one is the shift and that's the one to ignore.

331
00:44:49.000 --> 00:44:53.000
 So the shift is the one that you're subtracting from the t.

332
00:44:53.000 --> 00:44:57.000
 I know in this case they are equal but they won't always be.

333
00:44:57.000 --> 00:45:00.000
 So we ignore the eight here.

334
00:45:00.000 --> 00:45:02.000
 What does that leave me?

335
00:45:02.000 --> 00:45:06.000
 Well that will leave me t plus eight.

336
00:45:06.000 --> 00:45:14.000
 Okay, I'm ignoring that minus eight so that will leave me that f of t is equal to t plus eight.

337
00:45:14.000 --> 00:45:20.000
 So you know just pointing out that that's a mistake that is quite often made.

338
00:45:20.000 --> 00:45:30.000
 Even when the values are equal, sometimes people ignore this here, the plus eight, and they write out that the function is t minus eight.

339
00:45:30.000 --> 00:45:31.000
 It's not.

340
00:45:31.000 --> 00:45:34.000
 The shift is what's been subtracted from the t.

341
00:45:34.000 --> 00:45:41.000
 So if you ignore that, that leaves you that your function f of t is t plus eight.

342
00:45:41.000 --> 00:45:46.000
 So we're good then to go on to step two and take a little plus transform of that function.

343
00:45:46.000 --> 00:45:47.000
 That's pretty easy.

344
00:45:47.000 --> 00:45:56.000
 It's just number two and number one in the table to give me one over s squared plus eight over s.

345
00:45:56.000 --> 00:46:03.000
 And my final step, that's f of s, my final step is to multiply that f of s by e to the minus st.

346
00:46:03.000 --> 00:46:11.000
 Capital t is the critical value of the step function which actually equals, as always equals the shift.

347
00:46:11.000 --> 00:46:13.000
 And in this case, that's equal to eight.

348
00:46:13.000 --> 00:46:19.000
 So I've got to multiply f of s from step two by e to the minus eight s.

349
00:46:19.000 --> 00:46:26.000
 And that gives me the Laplace transform of my original function.

350
00:46:26.000 --> 00:46:35.000
 So just very quickly recap that we were asked to find the Laplace transform g of t is equal to t times ut minus eight.

351
00:46:35.000 --> 00:46:42.000
 We looked at this and it certainly didn't appear to be in a form that we could use the second shift in theorem.

352
00:46:42.000 --> 00:46:47.000
 But with a little bit of algebraic manipulation, we can.

353
00:46:47.000 --> 00:46:55.000
 And so we had to rewrite this function, this multiplying function by a unshifted function at the moment.

354
00:46:55.000 --> 00:46:56.000
 It's just t.

355
00:46:56.000 --> 00:47:02.000
 We had to rewrite that with a shift to match the critical value of the step function.

356
00:47:02.000 --> 00:47:07.000
 So we had to write, we had to have a t minus eight in there.

357
00:47:07.000 --> 00:47:16.000
 But that would have changed the function and to make up for the subtracting eight, we simply add eight.

358
00:47:16.000 --> 00:47:20.000
 And that now gives me a function of t minus eight.

359
00:47:20.000 --> 00:47:28.000
 So we're in the form of f of t minus capital t times the step function ut minus capital t.

360
00:47:29.000 --> 00:47:38.000
 And then we were able just to follow the diagram round and obtain the Laplace transform of my shifted and truncated function.

361
00:47:38.000 --> 00:47:42.000
 Okay.

362
00:47:42.000 --> 00:47:45.000
 Moving on now.

363
00:47:45.000 --> 00:47:49.000
 This is slightly different.

364
00:47:49.000 --> 00:47:53.000
 Same idea really as the one we've just seen.

365
00:47:53.000 --> 00:48:02.000
 We don't have the shift or the delay value matching the critical value of the step function.

366
00:48:02.000 --> 00:48:11.000
 So as it is, this is not in a form where we can directly use the second shift in theorem.

367
00:48:11.000 --> 00:48:22.000
 For the simple reason is that we do not have the shift capital t equal the critical value of the step function capital t.

368
00:48:22.000 --> 00:48:26.000
 The shift is two and a critical value is nine.

369
00:48:26.000 --> 00:48:36.000
 But we can manipulate this in a similar way to the way we did with the last example.

370
00:48:36.000 --> 00:48:45.000
 So I've got to start with t minus two and that's multiplying the step function ut minus nine.

371
00:48:45.000 --> 00:48:51.000
 Now one thing we know is we must have the shift matching the critical value.

372
00:48:51.000 --> 00:48:56.000
 In here, I must have a t minus nine.

373
00:48:56.000 --> 00:49:07.000
 I've got to have a t minus nine and that's going to multiply my step function ut minus nine.

374
00:49:07.000 --> 00:49:11.000
 But the thing here is I've now changed this.

375
00:49:11.000 --> 00:49:14.000
 I start with t minus two.

376
00:49:14.000 --> 00:49:16.000
 I've got a t minus nine here.

377
00:49:16.000 --> 00:49:25.000
 But hopefully it's obvious that to get back, retrieve my original function, I just add seven.

378
00:49:25.000 --> 00:49:31.000
 So if you work that out, t minus nine plus seven, that's simply t minus two.

379
00:49:31.000 --> 00:49:35.000
 And that perfectly matches what I was given in the question.

380
00:49:35.000 --> 00:49:45.000
 And now this thing here is actually just a function of t minus nine.

381
00:49:45.000 --> 00:49:55.000
 So that's a function of t minus nine times a step function whose critical value is nine.

382
00:49:55.000 --> 00:50:05.000
 So I'm now in the form of f of t minus capital T times the unit step function whose critical value is capital T.

383
00:50:05.000 --> 00:50:08.000
 So I can use the second shifting theorem.

384
00:50:08.000 --> 00:50:11.000
 So that's exactly what I'm going to do now.

385
00:50:11.000 --> 00:50:19.000
 I'm going to take the diagram and use my function g of t as given here.

386
00:50:19.000 --> 00:50:22.000
 So the first thing I do is I ignore the step function.

387
00:50:22.000 --> 00:50:25.000
 U of t minus nine gets ignored.

388
00:50:25.000 --> 00:50:28.000
 I ignore the shift and again be careful.

389
00:50:28.000 --> 00:50:33.000
 The shift is the quantity that's subtracted from the t.

390
00:50:33.000 --> 00:50:43.000
 So I ignore the nine and that leaves me my function f of t is equal to t plus seven.

391
00:50:43.000 --> 00:50:48.000
 So that's what I say here at step one.

392
00:50:48.000 --> 00:50:51.000
 So the shift of t is equal to t plus seven.

393
00:50:51.000 --> 00:50:59.000
 I ignore the shift which is equal to nine because it matches what's the critical value of the step function.

394
00:50:59.000 --> 00:51:02.000
 It leaves me f of t plus t plus seven.

395
00:51:02.000 --> 00:51:07.000
 Again a simple matter now of calculating the Laplace transform of this function.

396
00:51:07.000 --> 00:51:11.000
 It will just be one over s squared plus seven over s.

397
00:51:11.000 --> 00:51:13.000
 That's my big f of s.

398
00:51:13.000 --> 00:51:16.000
 I'm up here in the top right hand corner of my diagram.

399
00:51:16.000 --> 00:51:18.000
 There's my f of s.

400
00:51:18.000 --> 00:51:24.000
 The final step now is to multiply f of s by e to the minus s t.

401
00:51:24.000 --> 00:51:26.000
 That's capital T.

402
00:51:26.000 --> 00:51:32.000
 Where capital T is the critical value of the step function and it equals the shift.

403
00:51:32.000 --> 00:51:35.000
 And in this case that's equal to nine.

404
00:51:35.000 --> 00:51:39.000
 So I multiply f of s by e to the minus nine s.

405
00:51:39.000 --> 00:51:43.000
 That's given in the table here, my diagram here.

406
00:51:43.000 --> 00:51:52.000
 I end up with the Laplace transform of t minus two times the unit step t minus nine.

407
00:51:52.000 --> 00:52:00.000
 That's equal to one over s squared plus seven s, seven over s, all multiplied by e to the minus nine s.

408
00:52:00.000 --> 00:52:06.000
 So that's another example and that's what our function looks like.

409
00:52:06.000 --> 00:52:13.000
 Basically we've got the graph of t minus two.

410
00:52:13.000 --> 00:52:14.000
 That would be it there.

411
00:52:14.000 --> 00:52:17.000
 If I drew the graph of t minus two, it would carry on down like this.

412
00:52:17.000 --> 00:52:21.000
 It would cut the horizontal t axis at t equals two.

413
00:52:21.000 --> 00:52:27.000
 That makes sense because when t equals two, this function would be equal to zero and that's perfectly correct.

414
00:52:27.000 --> 00:52:35.000
 That's it there and the graph would carry on like this all the way to the right and all the way to the left like so.

415
00:52:35.000 --> 00:52:44.000
 But we're going to switch it off by multiplying by a critical value, by a step function whose critical value is equal to nine.

416
00:52:44.000 --> 00:52:52.000
 So it will set this function equal to zero all the way up to nine and then it suddenly kicks in.

417
00:52:52.000 --> 00:52:55.000
 It switches on a t equals nine.

418
00:52:55.000 --> 00:53:05.000
 So what you have is the graph of the function t minus two all the way from nine onwards forever and ever.

419
00:53:05.000 --> 00:53:14.000
 And again that's because the unit step function is equal to one from nine onwards.

420
00:53:14.000 --> 00:53:22.000
 So you just multiply t minus two by one and just get t minus two back in return.

421
00:53:22.000 --> 00:53:26.000
 And that's from nine onwards up to t equals nine.

422
00:53:26.000 --> 00:53:29.000
 The step function takes the value zero.

423
00:53:29.000 --> 00:53:35.000
 So of course you multiply anything by zero, you get zero and that's as you can see in the diagram there.

424
00:53:35.000 --> 00:53:41.000
 Okay, so here's another example.

425
00:53:41.000 --> 00:53:51.000
 We've been given another function g of t which once again in this case the critical value of the step function is two.

426
00:53:51.000 --> 00:53:53.000
 That certainly doesn't match the shift here.

427
00:53:53.000 --> 00:53:59.000
 That certainly does not match our shift.

428
00:53:59.000 --> 00:54:02.000
 So how do we go about this?

429
00:54:02.000 --> 00:54:04.000
 Well it's a similar idea really.

430
00:54:04.000 --> 00:54:10.000
 Once again, if I just take the tablet once again, we've got our step function.

431
00:54:10.000 --> 00:54:12.000
 We can't change that.

432
00:54:12.000 --> 00:54:16.000
 Our step function decides when things are switched on and switched off.

433
00:54:16.000 --> 00:54:22.000
 So what we do is we've got the function t plus four.

434
00:54:22.000 --> 00:54:31.000
 Now what we know is we've got to have a shift that matches the critical value of the step function.

435
00:54:31.000 --> 00:54:35.000
 So somewhere here we must have a t minus two.

436
00:54:35.000 --> 00:54:38.000
 But that will change.

437
00:54:38.000 --> 00:54:43.000
 If we go t minus two times ut minus two, that's certainly not what we started out with.

438
00:54:43.000 --> 00:54:51.000
 We've got to try to get this function here that's multiplying the step function to match the original function.

439
00:54:51.000 --> 00:54:54.000
 So we start with t plus four.

440
00:54:54.000 --> 00:54:56.000
 We've got t minus two here now.

441
00:54:56.000 --> 00:55:00.000
 So I think to go from here to here we just add six.

442
00:55:00.000 --> 00:55:05.000
 This t minus two plus six is equal to t plus four.

443
00:55:05.000 --> 00:55:16.000
 We're definitely in the correct form now because this function here is f of t minus two.

444
00:55:16.000 --> 00:55:19.000
 And that function is ut minus two.

445
00:55:19.000 --> 00:55:31.000
 So that's absolutely in the form given in the bottom left of our diagram where capital T, the critical value of this step function, is equal to the shift and it's equal to two.

446
00:55:31.000 --> 00:55:39.000
 So we can now use our diagram directly to calculate the Laplace transform of our given function g of t.

447
00:55:39.000 --> 00:55:42.000
 So let's just do that then.

448
00:55:42.000 --> 00:55:46.000
 So g of t is given by this expression here.

449
00:55:46.000 --> 00:55:53.000
 So as always I must ignore the step function and I ignore the shift.

450
00:55:53.000 --> 00:55:56.000
 So ignore the step function that's ut minus two.

451
00:55:56.000 --> 00:55:57.000
 That's easy.

452
00:55:57.000 --> 00:55:58.000
 That's gone.

453
00:55:58.000 --> 00:56:03.000
 Now remember the shift is the thing that you're subtracting from the t.

454
00:56:03.000 --> 00:56:05.000
 So you're ignoring the shift here.

455
00:56:05.000 --> 00:56:07.000
 So that's ignored the two.

456
00:56:07.000 --> 00:56:15.000
 Basically if you ignore the quantity that matches the critical value of the step function, critical value of the step function is two.

457
00:56:15.000 --> 00:56:17.000
 So we ignore the two here.

458
00:56:17.000 --> 00:56:23.000
 And that leaves us a function that f of t is equal to t plus six.

459
00:56:23.000 --> 00:56:27.000
 Again that's a very easy Laplace transform to calculate.

460
00:56:27.000 --> 00:56:36.000
 So I'm going to do number two for this one and number one for this one to give me that f of s is equal to one over s squared plus six over s.

461
00:56:36.000 --> 00:56:38.000
 So I'm up here now.

462
00:56:38.000 --> 00:56:40.000
 I'm up in the top right of my diagram.

463
00:56:40.000 --> 00:56:47.000
 I've got f of s and the last step is to multiply f of s that I've gotten step two, this quantity here.

464
00:56:47.000 --> 00:56:56.000
 Multiply that by e to the minus s t, which is capital T is equal to the critical value of the step function and also equal to the shift.

465
00:56:56.000 --> 00:56:58.000
 So that value was two.

466
00:56:58.000 --> 00:57:02.000
 So I multiply f of s, that's one over s squared plus six over s.

467
00:57:02.000 --> 00:57:05.000
 I multiply that by e to the minus two s.

468
00:57:05.000 --> 00:57:15.000
 And that gives me the Laplace transform of the function g of t in the question.

469
00:57:15.000 --> 00:57:25.000
 Okay, so we've got another one here, which is slightly tricky perhaps.

470
00:57:25.000 --> 00:57:31.000
 So we've got a function g of t is equal to t squared times the step function is critical value is 10.

471
00:57:31.000 --> 00:57:46.000
 Well, clearly the shift here is zero and that definitely does not match the shift in, sorry, does not match the critical value of my step function.

472
00:57:46.000 --> 00:57:49.000
 So I've got to do some algebraic manipulation.

473
00:57:49.000 --> 00:57:52.000
 So let's have a look at this.

474
00:57:52.000 --> 00:57:57.000
 So I'll write down my step function u t minus 10.

475
00:57:57.000 --> 00:58:05.000
 Now I know that in order to use the second shift in theorem, my shift must match the critical value.

476
00:58:05.000 --> 00:58:13.000
 So in here somewhere in my square brackets, I must have a t minus 10, must have a t minus 10.

477
00:58:13.000 --> 00:58:15.000
 So I write that in.

478
00:58:15.000 --> 00:58:20.000
 Then to compensate for introducing the minus 10, I add 10.

479
00:58:20.000 --> 00:58:27.000
 So that when you simplify this square bracket term, t minus 10 plus 10 is just t.

480
00:58:27.000 --> 00:58:32.000
 So you could t squared and that's exactly what you started out with.

481
00:58:32.000 --> 00:58:40.000
 So basically I've written this in a different form, but an equal form.

482
00:58:40.000 --> 00:58:43.000
 I haven't changed anything.

483
00:58:43.000 --> 00:58:49.000
 But by writing this form, it allows me to use my second shift in theorem.

484
00:58:49.000 --> 00:59:00.000
 So this function here in the square brackets is f of t minus 10 and that's multiplying a step function u t minus 10.

485
00:59:00.000 --> 00:59:08.000
 And the critical value of that step function matches the shift in the multiplying function.

486
00:59:08.000 --> 00:59:13.000
 So let's have a look now.

487
00:59:13.000 --> 00:59:14.000
 We're not quite ready yet.

488
00:59:14.000 --> 00:59:18.000
 We need to do a little bit more work on this function.

489
00:59:18.000 --> 00:59:22.000
 So we'll go to the next slide and have a look at that.

490
00:59:22.000 --> 00:59:25.000
 So let's expand this bracket.

491
00:59:25.000 --> 00:59:33.000
 So t squared as we saw was equal to this function t minus 10 plus 10 certainly equal to t squared.

492
00:59:33.000 --> 00:59:38.000
 So I'm going to expand this.

493
00:59:38.000 --> 00:59:40.000
 So it's just it squared.

494
00:59:40.000 --> 00:59:41.000
 So there it is.

495
00:59:41.000 --> 00:59:45.000
 It's t minus 10 plus 10 times itself.

496
00:59:45.000 --> 00:59:46.000
 That's what it is.

497
00:59:46.000 --> 00:59:50.000
 I'm going to consider the t minus 10 as a single unit if you like.

498
00:59:50.000 --> 00:59:56.000
 So I'll have t minus 10 times t minus 10 to give me t minus 10 squared.

499
00:59:56.000 --> 01:00:10.000
 I'll have t minus 10 times 10 and also I've got t minus 10 times 10.

500
01:00:10.000 --> 01:00:15.000
 So that's that one times that one.

501
01:00:15.000 --> 01:00:21.000
 And then I'll have 10 times t minus 10 plus 10 times t minus 10.

502
01:00:21.000 --> 01:00:25.000
 And you can see that's just going to give me 20 lots of t minus 10.

503
01:00:25.000 --> 01:00:29.000
 So that's where that second term comes from that term there.

504
01:00:29.000 --> 01:00:33.000
 And then I'm going to have 10 times 10 and that gives me the 100.

505
01:00:33.000 --> 01:00:42.000
 So and of course we just saw at the very beginning t minus 10 times t minus 10 gives me t minus 10 all squared.

506
01:00:42.000 --> 01:00:49.000
 So this now is clearly a function of t minus 10.

507
01:00:49.000 --> 01:00:51.000
 I just clear these away.

508
01:00:51.000 --> 01:00:59.000
 So this function here underlined in blue is clearly a function of t minus 10.

509
01:00:59.000 --> 01:01:07.000
 You know if you need a little bit more convincing suppose we've got a function f of t is equal to t squared plus 20t plus 100.

510
01:01:07.000 --> 01:01:16.000
 Then a function of t minus 10 which is involved with replacing every occurrence of t by t minus 10 to give us this expression here.

511
01:01:16.000 --> 01:01:18.000
 And that's exactly what I've got up here.

512
01:01:18.000 --> 01:01:23.000
 So that is a function of t minus 10.

513
01:01:23.000 --> 01:01:36.000
 So I can now write my original function g of t which was t squared times the step function critical value 10 as this thing I've just calculated f of t minus 10 times t minus 10.

514
01:01:36.000 --> 01:01:39.000
 Times the same step function.

515
01:01:39.000 --> 01:01:47.000
 So I am definitely now in a form that I can use the second shifting theorem and the diagram up top here.

516
01:01:47.000 --> 01:02:04.000
 So I've got f of t minus 10 which is f of t minus capital T times the step function ut minus 10 ut minus capital T and the critical value step function matches the shift.

517
01:02:04.000 --> 01:02:06.000
 As you can see.

518
01:02:06.000 --> 01:02:11.000
 So the shift is 10 the critical value of the step function is 10.

519
01:02:11.000 --> 01:02:14.000
 So let's have a look now.

520
01:02:14.000 --> 01:02:16.000
 So here's my function g of t.

521
01:02:16.000 --> 01:02:21.000
 So I want to ignore the step function and ignore the shift.

522
01:02:21.000 --> 01:02:29.000
 Now just bear in mind when you're ignoring the shift you're ignoring the quantity that's subtracted from the variable t.

523
01:02:29.000 --> 01:02:34.000
 So first of all ignore the step function so that goes out.

524
01:02:34.000 --> 01:02:40.000
 Now ignore the shift so I ignore anything that's subtracted from the t.

525
01:02:40.000 --> 01:02:43.000
 So I ignore the minus 10 there.

526
01:02:43.000 --> 01:02:46.000
 Ignore the minus 10 there so and that's everything.

527
01:02:46.000 --> 01:02:57.000
 So that will leave me when I simplify this it leave me that my f of t the function I get by ignoring the step function.

528
01:02:57.000 --> 01:03:07.000
 Ignoring the shift will be t squared therefore that time plus 20 t plus 100 and that's f of t.

529
01:03:07.000 --> 01:03:10.000
 That's what I've written down here.

530
01:03:10.000 --> 01:03:24.000
 So I can now take the Laplace transform of that function f of t and that's fairly straightforward really.

531
01:03:24.000 --> 01:03:38.000
 Because it's just a matter of using number three, number two and number one in the table to give me that big f of s is 2 over s cubed plus 20 over s squared plus 100 over s.

532
01:03:38.000 --> 01:03:47.000
 Just using this one two three in the tables and abuse them there's number three on that one number two on that one number one on that one and there's big f of s.

533
01:03:47.000 --> 01:03:58.000
 And finally my last step is to multiply f of s that I found at step two by e to the minus s t where this t is the capital t.

534
01:03:58.000 --> 01:04:07.000
 That's critical value of the step function and it's also equal to the shift and in our case in this example that was equal to 10.

535
01:04:07.000 --> 01:04:23.000
 So I multiply big f of s this thing here by e to the minus 10 s and that gives me the Laplace transform of my function f of t equals t squared times u t minus 10.

536
01:04:23.000 --> 01:04:43.000
 So you know if this thing for example t squared u t minus 10 that function was for example there it is that was let's say the fortune term in a differential equation the fortune term is just the right hand side of the differential equation.

537
01:04:43.000 --> 01:04:50.000
 We now know how to take Laplace transform of that type of function.

538
01:04:50.000 --> 01:05:14.000
 So in summary then we started off today by calculating the Laplace transform of a step function we saw how to do it using the integral that's using the definition of the Laplace transform but we don't need to do that we can just use the tables I think it was number 26 in the tables allows to calculate the Laplace transform of a step function.

539
01:05:14.000 --> 01:05:39.000
 We then introduced the second shifting theorem and that enables us to transform more complicated time shifted or delayed functions functions that involves step functions and will include discontinuities but discontinuities I mean jumps like like let's see once like this one for example.

540
01:05:39.000 --> 01:05:55.000
 You know it's got the value zero all the way along here and then it jumps up to take the value of the function t plus four so that's a discontinuity where it jumps from zero here suddenly up to here in between we haven't really got a defined value.

541
01:05:55.000 --> 01:06:20.000
 Using this function was on the right hand side of a differential equation and using the time domain methods we've seen up to now it would be pretty complicated to solve the corresponding differential equation but using the plus transforms the second shift theorem this can be done with relative ease.

542
01:06:21.000 --> 01:06:30.000
 So that's what I say here so we need these results to solve initial value problems differential equations that involve step functions and time delays.

543
01:06:30.000 --> 01:06:49.000
 The next time we're going to look at using the second shifting theorem in reverse to obtain the inverse Laplace transforms of more complicated functions and again we'll need to know how to do this when we come to inverter solution to the algebra problems.

544
01:06:49.000 --> 01:07:12.000
 So let's again suppose we've got a differential equation with one of these shifting truncated functions on the right hand side when we solve when we take our Laplace transforms we've seen how to do that today we would solve the resulting estimate algebra problem but we have to get back to the time domain and we do that by taking in the Laplace transforms.

545
01:07:12.000 --> 01:07:29.000
 So the next time we're going to see how we're going to use the second shifting theorem in reverse to go from the estimate to the time domain and we're also going to look at solving differential equations that involve step functions on the right hand side.

546
01:07:30.000 --> 01:07:52.000
 These are what are called the fortune terms of on the differential equation the right hand side the differential equation and we're also going to look at impulse functions on the right hand side impulse functions are functions that allow us to model when huge forces are applied over a very short period of time.

547
01:07:52.000 --> 01:08:21.000
 For example a short circuit in an electrical circuit a lightning strike for example or perhaps more a mechanical example would be the hammer hitting a pipe or some object you got a very large force applied over a very short time interval time to be method for solving differential equations that model these situations time to be methods would just be very complicated very complicated.

548
01:08:22.000 --> 01:08:33.000
 So we're going to solve but Laplace transforms allow us to solve these equations much quicker and much easier.

549
01:08:33.000 --> 01:08:49.000
 So here's a list of questions that you'd now be able to do is to be able to attempt Mobius questions one to 14 and down here I've also listed some tutorial questions from the notes that you can try.

550
01:08:49.000 --> 01:08:58.000
 That's it for this class the next class will be our last on Laplace transforms so we'll leave it at that for now.