If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Differential equations

### Course: Differential equations>Unit 3

Lesson 4: The convolution integral

# 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 ?
• We use it in electrical engineering for circuit analysis when it is difficult or impossible to find a laplace transform for our circuit equations.
• 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
• 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.
• 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?
• Because the substitution was only temporary. He switched back from u to tau at after the integral was done, and then evaluated them with tau-related limits ;)
• i would like to know more about the convolution on discrete data rather than the continuous functions
• 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)]
• 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!
• So, convolution is commutative? It'd would be helpful if you would go through the properties of the operation :)
• 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.
• 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.
• 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.
• Computationally fantastic....but visually....what is a convolution?

Does this have anything to do with the fact that tau = 2pi = a full circle?
• 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.