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.