Well hello everyone and welcome to our fifth class on Laplace Transforms.
In our last class we introduced the unit step function,
a function that can be used to model a signal that suddenly switches on at a specific time and stays on indefinitely.
Today we're going to look at how to take the Laplace Transform of a unit step function
and this will be extremely useful when we come to solving certain initial value problems.
We'll then move on and introduce the second shifting theorem
and this will enable us to take Laplace Transforms and the inverse Laplace Transforms of functions that involve step functions
and again it's particularly useful for solving ODE's and initial value problems.
Okay so just a quick recap on the unit step function.
So we discussed how in real-world applications we often come across functions whose values change suddenly at specific values of time
and an example of that would be a voltage that's switched on or off in an electrical circuit at a given time
and this switching process can be described mathematically by a unit step function.
And solving differential equations involving step functions is made considerably easier using Laplace Transforms.
So we therefore need to know how to take the Laplace Transform of a step function and related functions.
It's functions that involve step functions just like we saw last time when we saw various products
and sums and differences of unit step functions.
And these will typically appear as floating terms in initial value problems.
So the unit step function as we know suddenly switches on at a given value and stays on forever.
So in this example my unit step function is zero all the way up to time t equals zero
and then it suddenly switches on, takes the value one and holds that value forever and ever.
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
and is equal to one for t greater than zero.
Now as we saw that unit step function switched on at time t equals zero
it doesn't necessarily have to switch on at that value of time, it can switch on for any positive value of time.
And this is an example of we might call the delayed or shifted unit step function.
So it switches on at a time a little later than t equals zero.
So in this case our unit step function has switched on at time t equals capital T
and we call this the critical value of the step function.
For the previous case the critical value was capital T is equal to zero.
In this case the unit step function switches on at t is equal to capital T
and mathematically we describe it like so u of t minus t that's our critical value given by the capital T.
So it's zero for t less than capital T and it's one for t greater than capital T.
And you can see that in the diagram here and we saw that of course in our last lecture.
So we would now like to know how to take the Laplace transform of a unit step function
if we're going to use these types of functions in differential equations to model voltages, forces and so on.
So let's have a look at how we do that.
Now we're going to do it here from the definition.
You will not be asked to reproduce this but it's still useful to know where this comes from.
So if you remember the definition for the Laplace transform,
we calculate it from first principles, we've got Laplace transform of function f of t is the integral from zero to infinity.
It's an improper integral because infinity is one of the limits.
The integral from zero to infinity of f of t times e to the minus st dt.
In this case our function f of t is given by our step function u of t minus capital T.
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.
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.
So we're going to split our integral into two separate intervals.
We're going to calculate it over two separate intervals.
The first being from zero to the critical value zero to capital T and then from capital T out to infinity.
Representing where the unit step has now switched on.
So in the first interval between zero and capital T we can see that our unit step function has the value zero.
So that's what I've done here.
So my integral from zero to capital T, the step function has the value zero times e to the minus st dt.
Now we know that this just returns a value of zero.
So when you integrate zero you get zero.
The second integral here takes into consideration the interval that goes from capital T out to infinity.
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.
Now the value of the unit step function over this interval as you can see is just one.
So my f of t here which is the unit step function is equal to one between t out to infinity.
There it is. You can see that from the diagram times e to the minus st dt.
Now I can integrate this exponential and I get minus one over s e to the minus st.
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.
It's a negative exponential so we'll have exponential decay.
So this will all go off to zero.
At the lower limit we substitute in t and this will evaluate to zero minus one over s e to the minus st dt.
And that will work out to give me one over s e to the minus s times capital T.
And that is the Laplace transform of the unit step function.
Okay whose critical value is located at t.
We've also called this as I said earlier the shifted or delayed unit step function.
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.
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.
In other words one over s e to the zero e to the zero is one so just one over s.
And that makes sense because this function here for t greater than zero is just the constant one.
It's just equal to one and we know that the Laplace transform of one is just one over s.
It's the first transform that's given in our tables if you wanted to check that.
Okay so just remind yourselves some other step functions.
Here we've got a step function whose critical value is t equals four and it's got amplitude six.
So we just basically multiply the unit step function u t minus four by six to represent this function.
And we can take the Laplace transform of that quite easily.
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.
It's number 26 in our tables with capital T equal to four.
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.
Now here's our step function, our unit step function whose critical value is at the origin.
It switches on at time t equals zero.
We can either write that as u of t or we can write as u of t minus zero.
This represents the capital T value.
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.
Or one over s e to the zero which is just one over s.
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.
Okay so let's now move on and look at the second shifting theorem.
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,
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.
So this is a multiplication by an exponential in the time domain.
Multiplication of some function f of t by an exponential in the time domain corresponds to a shift by alpha in the s domain.
So basically we have the Laplace transform of f of t, our function here, f of s into the Laplace transform of this function.
And that's shifted by an amount alpha in the estimate where the alpha is given here.
So that was the first shifting theorem. Today we're going to look at the second shifting theorem.
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.
I mean you saw the idea of a delayed function earlier on in these for example.
So this is a delayed unit step function. It's basically switched off up until the value t equals 4.
So that's kind of an example of a delayed or shifted function.
Okay so suppose now we've got a function f of t, we can write it as f of t times u of t.
Basically it switches on the time t equals 0. Remember that's what the step function can be used for.
Use can be used to switch on and off other functions.
We saw that in the last lecture when we looked at products of functions and the step function.
We looked at the graphs of these functions.
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.
The corresponding delayed function would be given like so.
f of t minus t times the step function is critical value is capital t.
So this function because of the minus it gets shifted to the right.
So we'll discuss that a little more later on but be careful here.
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.
And you'll see that we'll be calling these shifted and truncated functions.
Truncated in the sense that it gets killed off or set to 0 all the way up to the critical value t.
So the second shifting theorem states that if we take the Laplace transform of this function here.
Shifted and truncated function if you like.
Then the Laplace transform of this time delayed function is a Laplace transform of the function without the delay.
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.
So that's what I say down here. This function f of t minus capital t times its f function with critical value t.
It's often referred to as shifted and truncated function.
And I'll just go once more over that.
So the Laplace transform of a shifted and truncated function is simply the Laplace transform of the unshifted function.
That's f of s because the unshifted function is simply f of t Laplace transform of f of t is f of s.
So we take the Laplace transform f of s of the unshifted function.
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.
So here's just another example of a shift and truncated function.
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.
The original function f of t times u of t.
There it is.
Just basically f of t for positive values of t because the effect of this step function whose critical value is zero.
It's basically to switch everything off right up to t equals zero.
So it's just the function f of t for positive values of t.
So if we want to shift and truncate this function, we can shift f function by capital t units to the left.
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.
So my function here has been shifted by capital t units to the right and it's been truncated.
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.
Okay, so here's just another example of a shift and truncated function.
My original function here is f of t is equal to t just the 45 degree line.
I'm going to shift this function two units to the right and I get my blue function which I call g of t.
So g of t is t minus two because my shift is to the right.
And you can see that this makes sense because if you set y, let's call y is equal to t minus two.
So where is y equal to zero? Well, it's when t is equal to two.
So the minus is the correct sign here.
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.
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.
And then I'm going to truncate this function.
I'm going to set it to zero all the way up to the value t equals two.
And to do that, I multiply by the unit step function, which remember whose critical value is equal to two,
which as we know will set everything equal to zero all the way up to the critical value, which is t equals two.
So this is what we get.
Our function h of t, I've called it, is zero all the way up to value two.
And then it's simply the graph of t minus two from there onwards.
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.
Okay, so there we are.
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,
if that's equal to capital F of s, then the Laplace transform of the shifted and truncated function, as I said earlier on,
is simply the product of the Laplace transform of the unshifted function.
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,
where t is the shift or the delay if you prefer, and it's also the critical value that appears in our step function.
So that's what the second shifting theorem says, and it allows us to take Laplace transforms and inverse Laplace transforms of step functions
and other functions that will involve possibly products of step functions with other functions.
Now, we're going to now show how we can apply the second shifting theorem to calculate the Laplace transforms of certain functions,
and we're going to do that using this diagram here.
So we're going to use these shifted and truncated functions in our examples.
So we would like to be able to go directly over here, but we can't do it right away like this.
So what we do is we use the diagram.
First of all, we ignore the step function, we ignore the shift, and that leaves us f of t.
We take the Laplace transform of f of t to give us f of s, and our final step, as we talked about before,
is we multiply f of s, the Laplace transform of the unshifted function.
We multiply that by e to the minus s t, where t represents the shift or the delay if you prefer,
and it also represents the critical value of the step function.
So we have one, two, three steps to go from the time domain where we have a shifted and truncated function
to go from the time domain to the s domain.
In other words, to get the Laplace transform of this shifted and truncated function.
Okay, so here we are then, we've got a function g of t is equal to t minus 4 ut minus 4.
Now, does this match the type of function that we need for applying the second shifted theorem?
So basically, I've got to have some function of t minus capital T,
multiplied by a step function whose critical value is the same as the amount of shift in our function.
Is that the case? Well, yes it is. The critical value of the step function is 4.
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.
I just get my tablet, so just say a couple of words about that.
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?
Well, I get that simply by replacing t minus 4 wherever I see t in the definition of the function.
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.
So, we can move on now and look at the calculating the last transform of this shifted and truncated function.
So we start off on the bottom left hand corner of the diagram in the time domain.
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.
So that's what I say here. Start bottom left and move around clockwise.
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.
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.
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.
So f of s, the Laplace transform of f of t is simply 1 over s squared, so there it is.
And my final step says multiply f of s, we just found at step two.
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.
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.
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.
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.
Okay, so just a couple of words here about this shift and truncated function.
So we start off with our original function.
We start off with f of t is equal to t.
So f of t is equal to t, the 45 degree line,
t minus 4, well. we shift this function, 4 units to the right,
So if we cut the horizontal axis, the t axis at t equals 4.
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.
There we are, that's my green function.
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.
And we know that has the effect of setting f into 0 all the way up to the critical value of 4.
And then it switches on the function because the step function takes the value 1 from then onwards.
So basically from 4 onwards, you just multiply t minus 4 by 1.
So you've just got the graph of t minus 4 from 4 onwards.
So that's what this function, the shift and truncate function is plotted in blue.
0 all the way up to 4 because the units step function hasn't been switched on.
So it's therefore the value of 0 all the way up to 4.
So anything that multiplies equal to 0, so the whole thing is equal to 0 up to t equals 4.
And then the step function switches on the t equals 4, takes the value 1.
So my g of t function will just be equal to t minus 4 from t equals 4 onwards forever and ever.
And it will just keep growing like that.
Okay, so let's do another example now.
So determine the Laplace transform of g of t is equal to t minus 4 all squared times the step function ut minus 4.
So this is along the same lines as the previous one except now that my function,
multiplying the step function, my shifted function is t minus 4 all squared.
It was just t minus 4 in the previous example.
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.
I've got it in a bit more, this is it here, but I've got it in a bit more detail here.
So we want to ignore the step function and we want to ignore the shift.
That's what we do, ignore the step function, ignore the shift and that just leaves me that my function,
unshift function f of t will equal to t squared.
And that's what I plotted here in red.
It's your smiley face graph, f of t is equal to t squared.
Now I'm going to shift this function for units to the right.
Okay, so then that becomes t minus 4 all squared.
And that's what you can see in blue here.
So this is the function t squared, shifted 4 units to the right, gives me the function t minus 4 all squared.
Just be careful here, the temptation might be to write t squared minus 4,
but that will just shift the function up and down the vertical axis by 4 units.
To shift it along the horizontal axis, we write it as t minus 4 all squared.
Then the final step is to multiply by the unit step function whose critical value is 4
and that will have the effect of switching this function off all the way up to t equals 4.
And that is shown in black in the diagram down here.
It doesn't get switched on until the value t equals 4.
And that's because this step function, this unit step function, takes the value 0 all the way up to the critical value 4.
So all the way up to 4, you're multiplying this quadratic function by 0.
So my function g of t will be 0 all the way up to the critical value.
Then suddenly this unit step function switches on at t equals 4 and takes the value 1.
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.
So that's where this graph here comes from.
And that's what's shown here in blue.
So we want to find its Laplace transform.
So again, we're going to use our diagram starting the bottom left hand corner and move around the diagram in a clockwise direction.
And that will move us from the t domain to the s domain.
Okay, so step one, ignore the step function and ignore the shift.
So if I do that, that leaves me the function f of t equals t squared.
I then want to find f of s, capital F of s, which is just the Laplace transform of f of t.
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.
And it'll tell me that f of s is equal to 2 over s cubed, number three in the table.
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.
Multiply that by e to the minus s t, where t is the critical value of the step function, which equals the shift.
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.
So that's the Laplace transform of this shifted untrunkated function, shown in blue in the diagram here.
Okay, moving on, here's another function, just maybe slightly more complicated perhaps.
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.
And we want the Laplace transform of this function.
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.
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.
There it is. So I'm up here in the diagram. Now I want the Laplace transform of f of t.
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.
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.
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.
And my table tells me that I should multiply f of s by e to the minus s times 2 pi.
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,
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,
which was equal to the shift, which is also called the delay.
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.
So first of all, let's see what happens.
So we've got a function sine of t minus 2 pi.
Now that is just the graph of sine t shifted 2 pi units to the right.
And then we remember that sine is 2 pi periodic, it repeats the shape after every interval of length 2 pi s.
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.
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.
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.
Okay, right, so what do we got now?
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.
Is this in the form of that we need to use the second shift in theorem?
Well the answer is yes it is.
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.
So we can use the second shift in theorem here.
So maybe let's just have a look at what this looks like.
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.
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.
So that's what it looks like.
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,
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.
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.
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.
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.
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.
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.
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.
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.
Now the question we need to ask ourselves is can we use the second shifting theorem directly on that function.
So we'll compare this function with a function the bottom left in our time domain of our diagram shown here.
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,
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,
and we got t minus two all cubed, so that the critical value of the step function matched the shift.
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.
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.
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.
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.
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.
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.
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.
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.
And this thing here, what I've underlined in blue here, is a function of t minus eight.
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.
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.
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.
Okay, so I set up my problem again.
This is the way I rewrote the function, but we know that it's actually equal to the original function which we were given.
Again, we start the bottom left and travel round in a clockwise direction.
We ignore the step function, we ignore the shift.
So just been very careful here.
Ignore the step function, well that's easy.
That's easy, that's just the u t minus eight.
So let me just get the pen.
So I'm going to ignore my step function.
Now ignore the shift.
Now the shift is the quantity that subtracted from the t.
It's that one.
Okay, so we ignore that and that leaves us t plus eight.
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.
So it's important that you know at this stage which one is the shift and that's the one to ignore.
So the shift is the one that you're subtracting from the t.
I know in this case they are equal but they won't always be.
So we ignore the eight here.
What does that leave me?
Well that will leave me t plus eight.
Okay, I'm ignoring that minus eight so that will leave me that f of t is equal to t plus eight.
So you know just pointing out that that's a mistake that is quite often made.
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.
It's not.
The shift is what's been subtracted from the t.
So if you ignore that, that leaves you that your function f of t is t plus eight.
So we're good then to go on to step two and take a little plus transform of that function.
That's pretty easy.
It's just number two and number one in the table to give me one over s squared plus eight over s.
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.
Capital t is the critical value of the step function which actually equals, as always equals the shift.
And in this case, that's equal to eight.
So I've got to multiply f of s from step two by e to the minus eight s.
And that gives me the Laplace transform of my original function.
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.
We looked at this and it certainly didn't appear to be in a form that we could use the second shift in theorem.
But with a little bit of algebraic manipulation, we can.
And so we had to rewrite this function, this multiplying function by a unshifted function at the moment.
It's just t.
We had to rewrite that with a shift to match the critical value of the step function.
So we had to write, we had to have a t minus eight in there.
But that would have changed the function and to make up for the subtracting eight, we simply add eight.
And that now gives me a function of t minus eight.
So we're in the form of f of t minus capital t times the step function ut minus capital t.
And then we were able just to follow the diagram round and obtain the Laplace transform of my shifted and truncated function.
Okay.
Moving on now.
This is slightly different.
Same idea really as the one we've just seen.
We don't have the shift or the delay value matching the critical value of the step function.
So as it is, this is not in a form where we can directly use the second shift in theorem.
For the simple reason is that we do not have the shift capital t equal the critical value of the step function capital t.
The shift is two and a critical value is nine.
But we can manipulate this in a similar way to the way we did with the last example.
So I've got to start with t minus two and that's multiplying the step function ut minus nine.
Now one thing we know is we must have the shift matching the critical value.
In here, I must have a t minus nine.
I've got to have a t minus nine and that's going to multiply my step function ut minus nine.
But the thing here is I've now changed this.
I start with t minus two.
I've got a t minus nine here.
But hopefully it's obvious that to get back, retrieve my original function, I just add seven.
So if you work that out, t minus nine plus seven, that's simply t minus two.
And that perfectly matches what I was given in the question.
And now this thing here is actually just a function of t minus nine.
So that's a function of t minus nine times a step function whose critical value is nine.
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.
So I can use the second shifting theorem.
So that's exactly what I'm going to do now.
I'm going to take the diagram and use my function g of t as given here.
So the first thing I do is I ignore the step function.
U of t minus nine gets ignored.
I ignore the shift and again be careful.
The shift is the quantity that's subtracted from the t.
So I ignore the nine and that leaves me my function f of t is equal to t plus seven.
So that's what I say here at step one.
So the shift of t is equal to t plus seven.
I ignore the shift which is equal to nine because it matches what's the critical value of the step function.
It leaves me f of t plus t plus seven.
Again a simple matter now of calculating the Laplace transform of this function.
It will just be one over s squared plus seven over s.
That's my big f of s.
I'm up here in the top right hand corner of my diagram.
There's my f of s.
The final step now is to multiply 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 it equals the shift.
And in this case that's equal to nine.
So I multiply f of s by e to the minus nine s.
That's given in the table here, my diagram here.
I end up with the Laplace transform of t minus two times the unit step t minus nine.
That's equal to one over s squared plus seven s, seven over s, all multiplied by e to the minus nine s.
So that's another example and that's what our function looks like.
Basically we've got the graph of t minus two.
That would be it there.
If I drew the graph of t minus two, it would carry on down like this.
It would cut the horizontal t axis at t equals two.
That makes sense because when t equals two, this function would be equal to zero and that's perfectly correct.
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.
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.
So it will set this function equal to zero all the way up to nine and then it suddenly kicks in.
It switches on a t equals nine.
So what you have is the graph of the function t minus two all the way from nine onwards forever and ever.
And again that's because the unit step function is equal to one from nine onwards.
So you just multiply t minus two by one and just get t minus two back in return.
And that's from nine onwards up to t equals nine.
The step function takes the value zero.
So of course you multiply anything by zero, you get zero and that's as you can see in the diagram there.
Okay, so here's another example.
We've been given another function g of t which once again in this case the critical value of the step function is two.
That certainly doesn't match the shift here.
That certainly does not match our shift.
So how do we go about this?
Well it's a similar idea really.
Once again, if I just take the tablet once again, we've got our step function.
We can't change that.
Our step function decides when things are switched on and switched off.
So what we do is we've got the function t plus four.
Now what we know is we've got to have a shift that matches the critical value of the step function.
So somewhere here we must have a t minus two.
But that will change.
If we go t minus two times ut minus two, that's certainly not what we started out with.
We've got to try to get this function here that's multiplying the step function to match the original function.
So we start with t plus four.
We've got t minus two here now.
So I think to go from here to here we just add six.
This t minus two plus six is equal to t plus four.
We're definitely in the correct form now because this function here is f of t minus two.
And that function is ut minus two.
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.
So we can now use our diagram directly to calculate the Laplace transform of our given function g of t.
So let's just do that then.
So g of t is given by this expression here.
So as always I must ignore the step function and I ignore the shift.
So ignore the step function that's ut minus two.
That's easy.
That's gone.
Now remember the shift is the thing that you're subtracting from the t.
So you're ignoring the shift here.
So that's ignored the two.
Basically if you ignore the quantity that matches the critical value of the step function, critical value of the step function is two.
So we ignore the two here.
And that leaves us a function that f of t is equal to t plus six.
Again that's a very easy Laplace transform to calculate.
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.
So I'm up here now.
I'm up in the top right of my diagram.
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.
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.
So that value was two.
So I multiply f of s, that's one over s squared plus six over s.
I multiply that by e to the minus two s.
And that gives me the Laplace transform of the function g of t in the question.
Okay, so we've got another one here, which is slightly tricky perhaps.
So we've got a function g of t is equal to t squared times the step function is critical value is 10.
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.
So I've got to do some algebraic manipulation.
So let's have a look at this.
So I'll write down my step function u t minus 10.
Now I know that in order to use the second shift in theorem, my shift must match the critical value.
So in here somewhere in my square brackets, I must have a t minus 10, must have a t minus 10.
So I write that in.
Then to compensate for introducing the minus 10, I add 10.
So that when you simplify this square bracket term, t minus 10 plus 10 is just t.
So you could t squared and that's exactly what you started out with.
So basically I've written this in a different form, but an equal form.
I haven't changed anything.
But by writing this form, it allows me to use my second shift in theorem.
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.
And the critical value of that step function matches the shift in the multiplying function.
So let's have a look now.
We're not quite ready yet.
We need to do a little bit more work on this function.
So we'll go to the next slide and have a look at that.
So let's expand this bracket.
So t squared as we saw was equal to this function t minus 10 plus 10 certainly equal to t squared.
So I'm going to expand this.
So it's just it squared.
So there it is.
It's t minus 10 plus 10 times itself.
That's what it is.
I'm going to consider the t minus 10 as a single unit if you like.
So I'll have t minus 10 times t minus 10 to give me t minus 10 squared.
I'll have t minus 10 times 10 and also I've got t minus 10 times 10.
So that's that one times that one.
And then I'll have 10 times t minus 10 plus 10 times t minus 10.
And you can see that's just going to give me 20 lots of t minus 10.
So that's where that second term comes from that term there.
And then I'm going to have 10 times 10 and that gives me the 100.
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.
So this now is clearly a function of t minus 10.
I just clear these away.
So this function here underlined in blue is clearly a function of t minus 10.
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.
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.
And that's exactly what I've got up here.
So that is a function of t minus 10.
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.
Times the same step function.
So I am definitely now in a form that I can use the second shifting theorem and the diagram up top here.
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.
As you can see.
So the shift is 10 the critical value of the step function is 10.
So let's have a look now.
So here's my function g of t.
So I want to ignore the step function and ignore the shift.
Now just bear in mind when you're ignoring the shift you're ignoring the quantity that's subtracted from the variable t.
So first of all ignore the step function so that goes out.
Now ignore the shift so I ignore anything that's subtracted from the t.
So I ignore the minus 10 there.
Ignore the minus 10 there so and that's everything.
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.
Ignoring the shift will be t squared therefore that time plus 20 t plus 100 and that's f of t.
That's what I've written down here.
So I can now take the Laplace transform of that function f of t and that's fairly straightforward really.
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.
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.
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.
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.
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.
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.
We now know how to take Laplace transform of that type of function.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
So we're going to solve but Laplace transforms allow us to solve these equations much quicker and much easier.
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.
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.