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# The Convolution and the Laplace Transform : Understanding how the product of the Transforms of two functions relates to their convolution.

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- Now that you've had a little bit of exposure to what a
- convolution is, I can introduce you to the
- convolution theorem, or at least in the context of--
- there may be other convolution theorems-- but we're talking
- about differential equations and Laplace transforms. So
- this is the convolution theorem as applies to Laplace
- transforms. And it tells us that if I have a function f of
- t-- and I can define its Laplace transform as, let's
- see, the Laplace transform of f of t is capital F of s.
- We've done that before.
- And if I have another function, g of t, and I take
- its Laplace transform, that of course is capital G of s.
- Then if we were to convolute these two functions, so if I
- were to take f and I were to convolute it with g, which is
- going to be another function of t-- and
- we already saw this.
- We saw that in the last video.
- I convoluted sine and cosine.
- So this is going to be a function of t.
- That the Laplace transform of this thing, and this the crux
- of the theorem, the Laplace transform of the convolution
- of these two functions is equal to the products of their
- Laplace transforms. It equals F of s, big capital F of s,
- times big capital G of s.
- Now, this might seem very abstract and very, you know,
- hard to kind of handle for you right now.
- So let's do an actual example.
- And actually, even better, let's do an inverse Laplace
- transform with an example.
- And actually, let me write one more thing.
- If this is true, then we could also do it the other way.
- We could also say that f-- and I'll just do it all in yellow;
- it takes me too much time to keep switching colors-- that
- the convolution of f and g, this is just a function of t,
- I can just say it's the inverse Laplace transform.
- It's just the inverse Laplace transform of F of
- s times G of s.
- Although I couldn't resist it.
- Let me switch colors.
- There you go.
- Now, what good does all of this do?
- Well, we can take inverse Laplace transforms. Let's just
- say that I had-- let me write it down here-- let's say I
- told you that the following expression or function, let's
- say H of s-- let me write it this way-- H of s is equal to
- 2s over s squared plus 1.
- Now, we did this long differential equations at the
- end, we end up with this thing and we have to take the
- inverse Laplace transform of it.
- So we want to figure out the inverse Laplace transform of H
- of s, or the inverse Laplace transform of
- this thing right there.
- So we want to figure out the inverse Laplace transform of
- this expression right here, 2s over s squared plus 1 squared.
- I don't want to lose that.
- Right there.
- Now, can we write this as the product of two Laplace
- transforms that we do know?
- Let's try to do it.
- So we can rewrite this.
- And so this is the inverse Laplace transform.
- So let me rewrite this expression down here.
- So I can rewrite 2s over s squared plus 1 squared.
- This is the same thing as-- let me write it this way-- 2
- times 1 over s squared plus 1, times s over s squared plus 1.
- I just kind of broke it up.
- If you multiply the numerators here, you get 2 times
- 1, times s, or 2s.
- If you multiply the denominators here, s squared
- plus 1, times s squared plus 1, well, that's just s squared
- plus 1 squared.
- So this is the same thing.
- So if we want to take the inverse Laplace transform of
- this, it's the same thing as taking the inverse Laplace
- transform of this right here.
- Now, something should hopefully start
- popping out at you.
- If these were separate transforms, if they were on
- their own, we know what this is.
- If we call this F of s, if we said this is the Laplace
- transform of some function, we know what that function is.
- This is this piece right here.
- I'm just doing a little dotted line around it.
- This is the Laplace transform of sine of t.
- And then if we draw a little box around this one right
- here, this is the Laplace transform of
- cosine of t, G of s.
- So this is the Laplace transform of sine of t, or we
- could write that this implies that f of t is
- equal to sine of t.
- You should recognize that one by now.
- And this implies that g of t, if we define this as the
- Laplace transform of g, this means that g of t is equal to
- cosine of t.
- And, of course, when you take the inverse Laplace
- transforms, you could take the 2's out.
- So now what can we say?
- We can now say that the-- let me write it this way-- the
- inverse-- so actually, let me write it this way.
- Or, actually, a better thing to do, instead of taking the 2
- out, so I can leave it nice and clean, we could, if we
- were to draw a box around this whole thing, and define this
- whole thing as F of s, then F of s is the Laplace transform
- of 2 sine of t.
- I just wanted to include that 2.
- I didn't want to leave that out and confuse the issue.
- I wanted a very pure F of s times G of s.
- So this expression right here is the product of the Laplace
- transform of 2 sine of t, and the Laplace transform of
- cosine of t.
- Now, our convolution theorem told us this right here.
- That if we want to take the inverse Laplace transform of
- the Laplace transforms of two functions-- I know that sounds
- very confusing --but you just kind of pattern match.
- You say, OK look, this thing that I had here, I could
- rewrite it as a product of two Laplace
- transforms I can recognize.
- This right here is the Laplace transform of 2 sine of t.
- This is the Laplace transform of cosine of t.
- And we just wrote that as G of s, and F of s.
- So if I have an expression written like this, I can take
- the inverse Laplace transform and it'll be equal to the
- convolution of the original functions.
- It'll be equal to the convolution of the inverse of
- g or the inverse of f.
- Let me write it this way.
- I could write it like this.
- We know that f of t is equal to the inverse Laplace
- transform of F of s.
- And we know that g-- I should have done it in a different
- color, but I'll do g in green-- we know that g of t is
- equal to the inverse Laplace transform of G of s.
- So we can rewrite the convolution theorem as the
- inverse-- and this might maybe confuse you more than help,
- but I'll give my best shot.
- The inverse Laplace transform of-- and I'll try to stay true
- to the colors-- of F of s times G of s is equal to-- I'm
- just restating this convolution
- theorem right here.
- This is equal to the convolution of the inverse
- Laplace transform of F of s.
- So it's equal to the convolution of the inverse
- Laplace transform of F of s with the inverse Laplace
- transform of G of s.
- With the inverse Laplace transform of
- capital G, of G of s.
- I'm not sure if that helps you or not, but if you go back to
- this example it might.
- This is F of s, this is F of s right here.
- 2 times-- I'll do it in the light blue-- this is 2 over s
- squared plus 1.
- That's F of s in our example.
- And the G of s was s over s squared plus 1.
- And all I got that from is I just broke this up into two
- things that I recognize.
- If I multiply this together, I get back to my original thing
- that I was trying to take the inverse Laplace transform of.
- And so the convolution theorem just says that, OK, well, the
- inverse Laplace transform of this is equal to the inverse
- Laplace transform of 2 over s squared plus 1, convoluted
- with the inverse Laplace transform of our G of s, of s
- over s squared plus 1.
- And we know what these things are.
- I already told them to you, but they should be somewhat
- second nature now.
- This is 2 times sine of t.
- You take the Laplace transform of sine of t, you get 1 over s
- squared plus 1, and then you multiply it by 2, you get the
- 2 up there.
- And you're going to have to convolute that with the
- inverse Laplace transform of this thing here.
- And we already went over this.
- This is cosine of t.
- So our result so far-- let me be very clear.
- It's always good to take a step back and just think about
- what we're doing, much less why we're doing it.
- But let's see, the inverse Laplace transform of this
- thing up in this top left corner, 2s over s squared plus
- 1 squared, which before we did what we're doing now was very
- hard to figure out-- actually, this would be a curly bracket
- right here, but you get the idea-- is equal to this.
- It's equal to 2 sine of t, convoluted with cosine of t.
- And you're like, Sal, throughout this whole process
- I've already forgotten what it means to convolute two
- functions, so let's convolute them.
- And I'll just write the definition, or the definition
- we're using of the convolution.
- That f convoluted with g-- it's going to be
- a function of g.
- I'll just write this short-hand-- is equal to the
- integral from 0 to t, of f of t minus tau,
- times g of tau, dtau.
- So 2 sine of t convoluted with cosine of t is equal to-- let
- me do a neutral color-- the integral from 0 to t, of 2
- sine of t, minus tau, times the cosine of tau, dtau.
- Now if you watched the very last video I made, I actually
- solved this, or I solved a very similar thing to this.
- If we take the 2 out we get 2, times the integral from 0 to
- t, of sine of t minus tau, times the cosine of tau.
- I actually solved this in the previous video.
- This right here, this is the convolution of sine of t and
- cosine of t.
- It's sine of t convoluted with cosine of t.
- And I show you in the previous video, just watch that video,
- where I introduce a convolution, that this thing
- right here is equal to 1/2t sine of t.
- Now, if this thing is equal to 1/2 t sine of t, and I have to
- multiply it by 2, then we get, our big result, that the
- inverse Laplace transform of 2s over s squared plus 1
- squared is equal to the convolution of 2 sine of t
- with cosine of t.
- Which is just 2 times this thing here, which is 2 times
- 1/2-- those cancel out-- so it equals t sine of t.
- And once you get the hang of it, you won't have to go
- through all of these steps.
- But the key is to recognize that this could be broken down
- as the products of two Laplace transforms that you recognize.
- This could be broken down as the product of two Laplace
- transforms we recognized.
- This is the Laplace transform of 2 sine of t.
- This was the Laplace transform of cosine of t.
- So the inverse Laplace transform of our original
- thing, or original expression, is just the convolution of
- that with that.
- And if you watched the previous video, you'd realize
- that actually calculating that convolution was no simple
- task, but it can be done.
- So you actually can get an integral form.
- Even if it can't be done, you can get your answer, at least,
- in terms of some integral.
- So I haven't proven the convolution
- theorem to you just yet.
- I'll do that in a future video.
- But hopefully, this gave you a little bit of a sense of how
- you can use it to actually take inverse Laplace
- transforms. And remember, the reason why we're learning to
- take inverse Laplace transforms, and we have all of
- these tools to do it, is because that's always that
- last step when you're solving these differential equations,
- using your Laplace transforms.

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