Reverb Is Convolution

Clap once in a hall and you hear the clap fade away: that's its impulse response. Any sound is a stream of tiny claps, each fading the same way. Add up all the fading copies and you get the sound with reverb. That sum is a convolution.

How it works

Convolution integrals

In one sentence

The convolution of $f$ and $g$ is $(f * g)(t) = \int_0^t f(t - \tau)g(\tau)\,d\tau$ , and $\mathcal{L}\{f * g\} = F(s)G(s)$ ; a linear system starting at rest responds with its impulse response convolved with its input.

General formula

{\htmlData{sym=0}{y(t)}} = ({\htmlData{sym=1}{h}} * x)(t) = \int_0^t {\htmlData{sym=1}{h}}(t - {\htmlData{sym=3}{\tau}})\,{\htmlData{sym=2}{x(\tau)}}\,d{\htmlData{sym=3}{\tau}}, \qquad {\htmlData{sym=1}{h}}(t) = \frac{1}{\htmlData{sym=4}{{T}}}e^{-t/{\htmlData{sym=4}{T}}}

At each moment, look back over the input, weight each past instant by how much it still echoes ( $h$ of how long ago), and add.

What the symbols mean

$y(t)$
Output
The sound you hear, with reverb.
$h$
Impulse response
How the room answers one instant clap: a fading echo.
$x(\tau)$
Input
The original dry sound at the earlier time τ.
$\tau$
Past time
The moment each tiny clap was made.
$T$
Decay time
How long the echo lasts.

The impulse response $h(t)$ is how the room answers a single instant clap.
An input sound $x$ is a stream of tiny claps; the one at time $\tau$ produces a fading copy $h(t - \tau)x(\tau)\,d\tau$ .
Adding all the copies is the convolution integral: the output at time $t$ is a weighted memory of all past input.
In the Laplace domain convolution becomes multiplication: $Y(s) = H(s)X(s)$ .

Learn the math

Variation of ParametersDiff Eq · Lesson 20, Second Order Differential Equations Convolution IntegralsDiff Eq · Lesson 29, Laplace Transforms