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## Differential equations

### Course: Differential equations > Unit 3

Lesson 4: The convolution integral# The convolution and the Laplace transform

Understanding how the product of the Transforms of two functions relates to their convolution. Created by Sal Khan.

## Want to join the conversation?

- Does this mean that convolutions are commutative? That is f * g = g * f(24 votes)
- Here are some properties of convolution.

From my DE book:

Let f(t), g(t), and h(t) be piecewise continuous on [0, infinity), then:

1: f*g=g*f,

2: f*(g+h)=(f*g)+(f*h),

3: (f*g)**g=f**(g*h),

4: f*0=0.(53 votes)

- Did they invent convolution when solving DE using laplace transform? Or convolution appears in a totally irrelevant situation?(8 votes)
- Both convolution and Laplace transform have uses of their own, and were developed around the same time, around mid 18th century, but absolutely independently. As a matter of fact the convolution appeared in math literature before Laplace work, though Euler investigated similar integrals several years earlier. The connection between the two was discovered long after their works.(7 votes)

- But the RHS of the theorem doesn't match with LHS. Because convolution of 2sint and cost is equal to tsint, whereas Inverse Laplace of the given F(s)G(s) is equal to (2sint)(cost). The proof needs more elaboration. I mean Inverse Laplace of H(s) should result directly into tsint according to statement of the theorem.(3 votes)
- Convolution theorem states that if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms.

Convolution theorem gives us the ability to break up a given Laplace transform, H(s), and then find the inverse Laplace of the broken pieces individually to get the two functions we need [instead of taking the inverse Laplace of the whole thing, i.e. 2s/(s^2+1)^2; which is more difficult].

I hope it clears the confusion.(3 votes)

- Does this also mean that the Laplace transform of the function tsint is 2s/(s^2+1)^2...(1 vote)
- Yes, in (http://www.stanford.edu/~boyd/ee102/laplace-table.pdf) there is a rule stating that:

L( t*f(t) ) = -dF(s)/ds.

Applying this to our problem gives us:

L( t*sint ) = -d/ds * 1/(s^2+1) = ... = 2s/(s^2+1)^2.

You can always double check with WolframAlpha:

http://www.wolframalpha.com/input/?i=Laplace+transform%28tsint%29(4 votes)

- can we have the proof of convolution theorem?(2 votes)
- You can find in djairo figueiredo book the proof for Fourier transforms, : "Fourier analysis and differential partial equations ", and Simmons " diferential equations with aplications and historical notes" should have it too,but it shouldn't be too hard to prove it yourself with a little of calculus 3 .(1 vote)

- Is there some form of "inverse" convolution theorem giving L{f(t)g(t)} as some convolution of F(s) and G(s)?(2 votes)
- What IS a Laplace transform? Is it the system function as opposed to the impulse or input function or something like that?(2 votes)
- It's literally just the integral, i.e. the integrand is f(t)*e^-st dt from whatever bounds you're using.(1 vote)

- Somewhere in the video you write L(inverse){F(s).G(s)}= [L(inv){F(s)}].[L(inv){G(s)}]. Actually as far as i understand it that may not be right. The theorem says L{f*g}=[L{f(t)}].[L{g(t)}], therefore giving f*g=L(inv){L(inv){F(s)}.L(inv){G(s)}}

And by that definition it must be L(inverse){F(s).G(s)}= L(inv){[L(inv){F(s)}].[L(inv){G(s)}]}

P.S Sorry for the symbolic jumble up there^^(2 votes) - Why do you need to do the convolution of the two functions, let alone even find the two functions if you have the laplace formula of tsin(at)=2as/(s^2+1)^2. Is it just to show how convolution works? maybe a problem where there really isn't a formula for determining the laplace transform. Also, is convolution the method for determining all of these formulas, cause if so it totally makes sense.(1 vote)
- The convolution theorem allows us to use the convolution to solve inverse Laplace transforms where partial fractions would otherwise be necessary. There is also products that we will not be able to solve normally.(2 votes)

- would we arrive at the same answer if we did Partial fraction expansion and solved the LT of those?(1 vote)

## Video transcript

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.