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

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
• 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.
• Did they invent convolution when solving DE using laplace transform? Or convolution appears in a totally irrelevant situation?
• 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.
• 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.
• 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.
• Does this also mean that the Laplace transform of the function tsint is 2s/(s^2+1)^2...
(1 vote)
• can we have the proof of convolution theorem?
• 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)?
• What IS a Laplace transform? Is it the system function as opposed to the impulse or input function or something like that?
• 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^^