---
title: "Convolution theorem"
date: 2021-02-22
categories:
- Mathematics
layout: "concept"
---

The **convolution theorem** states that a convolution in the direct domain
is equal to a product in the frequency domain. This is especially useful
for computation, replacing an $\mathcal{O}(n^2)$ convolution with an
$\mathcal{O}(n \log(n))$ transform and product.

## Fourier transform

The convolution theorem is usually expressed as follows, where
$\hat{\mathcal{F}}$ is the [Fourier transform](/know/concept/fourier-transform/),
and $A$ and $B$ are constants from its definition:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
            B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
        \end{aligned}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-fourier"/>
<label for="proof-fourier">Proof</label>
<div class="hidden">
<label for="proof-fourier">Proof.</label>
We expand the right-hand side of the theorem and
rearrange the integrals:

$$\begin{aligned}
    \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k}
    \\
    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'}
    \\
    &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
    = A \cdot (f * g)(x)
\end{aligned}$$

Then we do the same again,
this time starting from a product in the $x$-domain:

$$\begin{aligned}
    \hat{\mathcal{F}}\{f(x) \: g(x)\}
    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
    \\
    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
    \\
    &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
    = B \cdot (\tilde{f} * \tilde{g})(k)
\end{aligned}$$
</div>
</div>


## Laplace transform

For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$,
the convolution theorem can also be stated using
the [Laplace transform](/know/concept/laplace-transform/):

$$\begin{aligned}
    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
\end{aligned}$$

Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
unpleasant, the theorem is often stated using the forward transform
instead:

$$\begin{aligned}
    \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-laplace"/>
<label for="proof-laplace">Proof</label>
<div class="hidden">
<label for="proof-laplace">Proof.</label>
We expand the left-hand side.
Note that the lower integration limit is 0 instead of $-\infty$,
because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:

$$\begin{aligned}
    \hat{\mathcal{L}}\{(f * g)(t)\}
    &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t}
    \\
    &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
\end{aligned}$$

Then we define a new integration variable $\tau = t - t'$, yielding:

$$\begin{aligned}
    \hat{\mathcal{L}}\{(f * g)(t)\}
    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
    \\
    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
    \\
    &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
    = \tilde{f}(s) \: \tilde{g}(s)
\end{aligned}$$
</div>
</div>



## References
1.  O. Bang,
    *Applied mathematics for physicists: lecture notes*, 2019,
    unpublished.