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## Differential equations

# Introduction to the convolution

Introduction to the Convolution. Created by Sal Khan.

## Want to join the conversation?

- Nice to see an example.. but what's the use of convolution in the real world ?(54 votes)
- We use it in electrical engineering for circuit analysis when it is difficult or impossible to find a laplace transform for our circuit equations.(92 votes)

- Why are you only integrating from 0 to t instead of - to + infinity? Is it because your leading into Laplace, which only works from 0 to infinity? I thought the convolution operation covered -infinity < t < +infinity. Along those lines, what is the physical significance or intuition of convolution? In the case of the sigma function (width and height of one, centered @ origin) being convolved with any other function I deduce that it is the intersecting areas under the curves of the two functi(21 votes)
- Convolution is a complicated topic, and is studied in more depth in classes after Diff EQ (such as in engineering classes, as SAL said...). This is just to show how to calculate an example.

However there are a few things he glossed over a bit that would clear things up...

first of all convolution is in fact defined as integrating from -infinity to infinity. The reason he integrated from 0 to t is that the functions he is considering sin(t) and cos(t) starting at t = 0. So more specifically, the functions SAL is REALLY USING are:

f(t) = sin(t) for t >=0, 0 for t<0;

g(t) = cos(t) for t >=0, 0 for t<0;

Knowing this, the convolution integral will be 0 for values outside of the interval from 0 to t, and there is no reason to integrate from -infinity to infinity. The integration thus simplifies to limits of 0 to t.(25 votes)

- why at 10.10, can Sal change the integral with respect to 'tau' to an integral with respect to 'u', but not change his limits?(11 votes)
- Because the substitution was only temporary. He switched back from u to tau at12:25after the integral was done, and then evaluated them with tau-related limits ;)(21 votes)

- i would like to know more about the convolution on discrete data rather than the continuous functions(10 votes)
- In the video f(t) = sin(t) and the convolution of f(t) is Integral[sin(t-tau)].

My question is this: If I have f(t) = sin3(t) would the the integral of then be Integral[sin3(t-tau)](5 votes)- Yes, You introduce (t-tau) where t is:

Right: cos (3t+2) ->>cos (3*(t-tau)+2)

Wrong: cos (3t+2-tau) Don't do this!(3 votes)

- So, convolution is commutative? It'd would be helpful if you would go through the properties of the operation :)(5 votes)
- Yes

Proof: (note - I'm using S(x=A,x=B) f(x) dx to represent the integral of f(x) from A to B.)

let u=t-τ, so τ=t-u and du=-τ, then

(f*g)(t) = S(τ=-∞,τ=∞) f(t-τ)g(τ) dτ

= S(t-u=-∞,t-u=∞) f(u)g(t-u) -du

= -S(u=∞,u=-∞) g(t-u)f(u) du

= S(u=-∞,u=∞) g(t-u)f(u) du

= (g*f)(t)(1 vote)

- Are tau and t both a variable? I think Sal need to define this the very first time he introduce the formula so that we will not be confuse the denotations used in the formula.(4 votes)
- They both are variables but we are interested in taking the integral only for tau so we assume that t is a constant. In fact, t is our independent variable. It should be obvious to see which are variables. I think Sal is clear about it.(2 votes)

- I have a question about the definition of convolution. Why would that integral be chosen as the definition of convolution? What's so special about that integral? I can follow the algebraic computation, but it's like someone tells me that a piece of paper falls from the sky and the definition of convolution was written on the paper; therefore, we need to just accept it. I dislike learning something without understanding the reason behind it. Thanks in advance.(4 votes)
- Computationally fantastic....but visually....what is a convolution?

Does this have anything to do with the fact that tau = 2pi = a full circle?(2 votes)- Very unlikely. The use of the symbol tau for 2pi, I believe, is only a relatively recent idea; in 1958, Albert Eagle suggested tau=1/2 pi , and later the concept of tau=2pi became popular. The convolution has existed since long before that. Also, in your journey in mathematics, you will see that greek letters are used to denote entirely different things, so its more likely a coincidence.(5 votes)

- On Wikipedia (and in my textbook), the convolution integral is defined somewhat differently - it has minus infinity and plus infinity as integration limits. Of course, if the integrand is zero when tao is not in [0, t] the integration limits are reduced to 0 and t. But here, this is not the case so why does Sal define convolution in this way?(2 votes)
- Sal said that f(t) = sin(t) and g(t) = cos(t). However, this wasn't quite right.

What he meant to say is

f(t) = sin(t) for t>=0 and 0 for t<0

g(t) = cos(t) for t>=0 and 0 for t<0

Thus, the integrand is zero when 0>τ>t and the limits of integration reduce to 0 and t.(1 vote)

## Video transcript

In this video, I'm going to
introduce you to the concept of the convolution, one of the
first times a mathematician's actually named something
similar to what it's actually doing. You're actually convoluting
the functions. And in this video, I'm not
going to dive into the intuition of the convolution,
because there's a lot of different ways you
can look at it. It has a lot of different
applications, and if you become an engineer really of any
kind, you're going to see the convolution in kind of a
discrete form and a continuous form, and a bunch of
different ways. But in this video I just want
to make you comfortable with the idea of a convolution,
especially in the context of taking Laplace transforms. So the convolution theorem--
well, actually, before I even go to the convolution theorem,
let me define what a convolution is. So let's say that I have
some function f of t. So if I convolute f with g-- so
this means that I'm going to take the convolution of f and
g, and this is going to be a function of t. And so far, nothing I've written
should make any sense to you, because I haven't
defined what this means. This is like those SAT problems
where they say, like, you know, a triangle b means a
plus b over 3, while you're standing on one leg or
something like that. So I need to define this
in some similar way. So let me undo this silliness
that I just wrote there. And the definition of a
convolution, we're going to do it over a-- well, there's
several definitions you'll see, but the definition we're
going to use in this, context there's actually one other
definition you'll see in the continuous case, is the integral
from 0 to t of f of t minus tau, times g of t-- let me
just write it-- sorry, it's times g of tau d tau. Now, this might seem like a very
bizarro thing to do, and you're like, Sal, how
do I even compute one of these things? And to kind of give you that
comfort, let's actually compute a convolution. Actually, it was hard to find
some functions that are very easy to analytically compute,
and you're going to find that we're going to go into a lot of
trig identities to actually compute this. But if I say that f of t, if I
define f of t to be equal to the sine of t, and I define
cosine of t-- let me do it in orange-- or I define g of t to
be equal to the cosine of t. Now let's convolute
the two functions. So the convolution of f with g,
and this is going to be a function of t, it equals this. I'm just going to show you how
to apply this integral. So it equals the integral--
I'll do it in purple-- the integral from 0 to t of
f of t minus tau. This is my f of t. So it's is going to be sine of
t minus tau times g of tau. Well, this is my g of t, so
g of tau is cosine of tau, cosine of tau d tau. So that's the integral, and
now to evaluate it, we're going to have to break out
some trigonometry. So let's do that. This almost is just a very
good trigonometry and integration review. So let's evaluate this. But I wanted to evaluate this
in this video because I want to show you that this isn't some
abstract thing, that you can actually evaluate
these functions. So the first thing I want to
do-- I mean, I don't know what the antiderivative of this is. It's tempting, you see a sine
and a cosine, maybe they're the derivatives of each
other, but this is the sine of t minus tau. So let me rewrite that sine of
t minus tau, and we'll just use the trig identity, that
the sine of t minus tau is just equal to the sine of t
times the cosine of tau minus the sine of tau times
the cosine of t. And actually, I just made a
video where I go through all of these trig identities really
just to review them for myself and actually to make a
video in better quality on them as well. So if we make this subsitution,
this you'll find on the inside cover of any
trigonometry or calculus book, you get the convolution of f and
g is equal to-- I'll just write that f-star g; I'll just
write it with that-- is equal to the integral from 0 to t of,
instead of sine of t minus tau, I'm going to write this
thing right there. So I'm going to write the sine
of t times the cosine of tau minus the sine of tau times the
cosine of t, and then all of that's times the
cosine of tau. I have to be careful with my
taus and t's, and let's see, t and tau, tau and t. Everything's working so far. So let's see, so
then that's dt. Oh, sorry, d tau. Let me be very careful here. Now let's distribute this
cosine of tau out, and what do we get? We get this is equal to-- so f
convoluted with g, I guess we call it f-star g, is equal to
the integral from 0 to t of sine of t times cosine of
tau times cosine of tau. I'm just distributing
this cosine of tau. So it's cosine squared of tau,
and then minus-- let's rewrite the cosine of t first, and I'm
doing that because we're integrating with
respect to tau. So I'm just going to write my
cosine of t first. So cosine of t times sine of tau times
the cosine of tau d tau. And now, since we're taking
the integral of really two things subtracting from each
other, let's just turn this into two separate integrals. So this is equal to the integral
from 0 to t, of sine of t, times the cosine squared
of tau d tau minus the integral from 0 to t of cosine
of t times sine of tau cosine of tau d tau. Now, what can we do? Well, to simplify it more,
remember, we're integrating with respect to-- let
me be careful here. We're integrating with
respect to tau. I wrote a t there. We're integrating with
respect to tau. So all of these, this
cosine of t right here, that's a constant. The sine of t is a constant. For all I know, t could
be equal to 5. It doesn't matter that one
of the boundaries of our integration is also a t. That t would be a 5,
in which case these are all just constants. We're integrating only with
respect to the tau, so if cosine of 5, that's a constant,
we can take it out of the integral. So this is equal to sine of t
times the integral from 0 to t of cosine squared of tau d tau
and then minus cosine of t-- that's just a constant; I'm
bringing it out-- times the integral from 0 to t of sine
of tau cosine of tau d tau. Now, this antiderivative is
pretty straightforward. You could do u substitution. Let me do it here, instead
of doing it in our heads. This is a complicated
problem, so we don't want to skip steps. If we said u is equal to sine of
tau, then du d tau is equal to the cosine of tau, just
the derivative of sine. Or we could write that
du is equal to the cosine of tau d tau. We'll undo the substitution
before we evaluate the endpoints here. But this was a little bit
more of a conundrum. I don't know how to take the
antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going
to break out some more trigonometric identities. And in a video I just recorded,
it might not be the last video in the playlist, I
showed that the cosine squared of tau-- I'm just using tau as
an example-- is equal to 1/2 times 1 plus the cosine
of 2 tau. And once again, this is just a
trig identity that you'll find really in the inside cover of
probably your calculus book. So we can make this substitution
here, make this substitution right there, and
then let's see what our integrals become. So the first one over here,
let me just write it here. We get sine of t times the
integral from 0 to t of this thing here. Let me just take the 1/2 out,
to keep things simple. So I'll put the 1/2 out here. That's this 1/2. So 1 plus cosine of 2 tau and
all of that is d tau. That's this integral
right there. And then we have this integral
right here, minus cosine of t times the integral from--
let me be very clear. This is tau is equal to 0
to tau is equal to t. And then this thing right
here, I did some u subsitution. If u is equal to sine of
t, then this becomes u. And we showed that du is equal
to cosine-- sorry, u is equal to sine of tau. And then we showed that du is
equal to cosine tau d tau, so this thing right here
is equal to du. So it's u du, and let's see if
we can do anything useful now. So this integral right here, the
antiderivative of this is pretty straightforward, so
what are we going to get? Let me write this
outside part. So we have 1/2 times
the sine of t. And now let me take the
antiderivative of this. This is going to be tau plus
the antiderivative of this. It's going to be 1/2
sine of 2 tau. I mean, we could have done
the u substitution. we could have said u is equal to
2 tau and all of that, but I think you could do that from
recognition, and if you don't believe me, you just have to
take the derivative of this. 1/2 sine of 2 tau is the
derivative of this. You multiply, you take the
derivative of the inside, so that's 2, so the 2 and the 1/2
cancel out, and the derivative of the outside, so
cosine of 2 tau. And you're going to evaluate
that from 0 to t. And then we have minus
cosine of t. When we take the antiderivative
of this-- let me do this on the side. So the integral of u du,
that's trivially easy. That's 1/2 u squared. Now, that's 1/2 u squared, but
what was u to begin with? It was sine of tau. So the antiderivative of this
thing right here is 1/2 u squared, but u is sine of tau. So it's going to be 1/2u, which
is sine of tau squared. And we're going to evaluate
that from 0 to t. And we didn't even have to do
all this u substitution. The way I often do it in my
head, I see the sine of tau, cosine of tau. if I have a function and I have
its derivative, I can treat that function just like
as if I had an x there, so it'd be sine squared of tau over
2, which is exactly what we have there. So it looks like we're
in the home stretch. We're taking the convolution of
sine of t with cosine of t. And so we get 1/2 sine of t. Now, if I evaluate this thing
at t, what do I get? I get t plus 1/2 sine
of 2t, that's when I evaluated it at t. And then from that I need to
subtract it evaluated at 0, so minus 0 minus 1/2 sine of
2 times 0, which is just sine of 0. So this part right here, this
whole thing right there, what does that simplify to? Well this is 0, sine of 0
is 0, so this is all 0. So this first integral right
there simplifies to 1/2 sine of t times t plus
1/2 sine of 2t. All right, now what does this
one simplify to over here? Well, this one over here, you
have minus cosine of t. And we're going to evaluate this
whole thing at t, so you get 1/2 sine squared of t
minus 1/2 the sine of 0 squared, which is just 0,
so that's just minus 0. So far, everything that we have
written simplifies to-- let me multiply it all out. So I have 1/2-- let me just pick
a good color-- 1/2t sine of t-- I'm just multiplying
those out-- plus 1/4 sine of t sine of 2t. And then over here I have minus
1/2 sine squared t times cosine of t. I just took the minus cosine t
and multiplied it through here and I got that. Now, this is a valid answer,
but I suspect that we can simplify this more, maybe using
some more trigonometric identities. And this guy right there
looks ripe to simplify. And we know that the sine of
2t-- another trig identity you'll find in the inside cover
of any of your books-- is 2 times the sine of t
times the cosine of t. So if you substitute that there,
what does our whole expression equal? You get this first term. Let me scroll down
a little bit. You get 1/2t times the sine of
t plus 1/4 sine of t times this thing in here, so times
2 sine of t cosine of t. Just a trig identity, nothing
more than that. And then finally I have this
minus 1/2 sine squared t cosine of t. No one ever said this was
going to be easy, but hopefully it's instructive
on some level. At least it shows you that you
didn't memorize your trig identities for nothing. So let me rewrite the whole
thing, or let me just rewrite this part. So this is equal to 1/4. Now, I have-- well let
me see, 1/4 times 2. 1/4 times 2 is 1/2. And then sine squared
of t, right? This sine times this sine is
sine squared of t cosine of t. And then this one over here is
minus 1/2 sine squared of t cosine of t. And luckily for us, or lucky
for us, these cancel out. And, of course, we had this
guy out in the front. We had this 1/2t sine
t out in front. Now, this guy cancels with this
guy, and all we're left with, through this whole hairy
problem, and this is pretty satisfying, is 1/2t sine of t. So we just showed you that the
convolution-- if I define-- let me write our result. I feel like writing this
in stone because this was so much work. But if we write that f of t is
equal to sine of t, and g of t is equal to cosine of t, I
just showed you that the convolution of f with g, which
is a function of t, which is defined as the integral from 0
to t of f of t minus tau times g of tau d tau, which was equal
to-- and I'll switch colors here-- which was equal to
the integral from 0 to t of sine of t minus tau times g of
tau d tau, that all of this mess, all of this convolution,
it all equals-- and this is pretty satisfying-- it all
equals 1/2t sine of t. And the whole reason why I went
through all of this mess and kind of bringing out the
neurons that had the trig identities memorized or having
to reproof them or whatever else is to just show you that
this convolution, it is convoluted and it seems a little
bit bizarre, but you really can take the convolutions
of actual functions and get an
actual answer. So the convolution of sine
of t with cosine of t is 1/2t sine of t. So, hopefully, you have a little
of intuition of-- well, not intuition, but you at least
have a little bit of hands-on understanding of how
the convolution can be calculated.