Convolution · Differential Equations

Intuition

A linear system answers an impulse with its impulse response, and any forcing can be thought of as a stream of impulses, one at every instant, each as large as the forcing is then. The response to all of them is the sum of the impulse responses they set off, each delayed to its own instant. That sum is an integral, the convolution of the forcing with the impulse response, and its transform is simply the product of the two transforms.

The echo in a large hall. Every sound adds its own fading echo, and what you hear at any moment is the sum of the echoes of everything said before, each faded by how long ago it was said.

$y'+y=f$ from rest, with $f$ a pulse from $1$ to $3$ . The response is the convolution $y(t)=\int_{0}^{t}f(\tau)e^{-(t-\tau)}\,d\tau$ : each instant of the pulse sets off an impulse response $e^{-t}$ , delayed to that instant, and their sum rises while the pulse lasts and fades after it.

The response as an integral

The convolution of $f$ and $g$ is the function $f*g$ below. It is symmetric, $f*g=g*f$ , and its transform is the product . So the solution from rest of is , where is the impulse response: each value is an impulse at , answered by , and the integral adds them up.

(f*g)(t)=\int_{0}^{t}f(\tau)\,g(t-\tau)\,d\tau

Using convolution

$1*1=\int_{0}^{t}1\,d\tau=t$ , and indeed .

The convolution theorem

\mathcal{L}\{f*g\}=F(s)\,G(s)

Write the transform of the convolution as a double integral over the region where tau runs from zero to t. Exchange the order: for each tau, t runs from tau to infinity. Shift the inner variable so that it starts at zero; it produces e to the minus s tau times the transform of g. The outer integral is then the transform of f, times the transform of g.

Proof steps

\mathcal{L}\{f*g\}=\int_{0}^{\infty}\!e^{-st}\!\int_{0}^{t}f(\tau)g(t-\tau)\,d\tau\,dt

Write out the transform of the convolution.

=\int_{0}^{\infty}\!f(\tau)\!\int_{\tau}^{\infty}e^{-st}g(t-\tau)\,dt\,d\tau

Exchange the order of integration over the region where tau lies between 0 and t.

\int_{\tau}^{\infty}e^{-st}g(t-\tau)\,dt=e^{-s\tau}G(s)

Shift the inner variable to start at zero.

=\int_{0}^{\infty}f(\tau)e^{-s\tau}\,d\tau\;G(s)=F(s)\,G(s)

What remains outside is the transform of f.

Applications

Practice

Products of Transforms

The transform of a convolution is the product of the transforms, so a product of transforms inverts to a convolution.

\mathcal{L}\{f*g\}=F(s)\,G(s)

Try it

Which function has the transform $\dfrac1s\cdot\dfrac{1}{s^{2}+1}$ ?