Convolution Theorem

Prerequisites

• Fourier Transforms (link?)
• Integral Calculus (link?)

Short Topical Videos

• Introduction to the Convolution by Khan Academy

Convolution Theorem

Fourier Transform

Here are the definitions we will use for the forward (${\mathcal {F}}$) and inverse (${\mathcal {F}}^{-1}$) Fourier transforms:

${\begin{aligned}f(t)&={\mathcal {F}}({\hat {f}}(\omega ))={\frac {1}{2\pi }}\int {{\hat {f}}(\omega )e^{i\omega t}d\omega }\\{\hat {f}}(\omega )&={\mathcal {F}}^{-1}(f(t))=\int {f(t)e^{-i\omega t}dt}\end{aligned}}\,\!$

where $\omega \equiv 2\pi \nu$ is the angular frequency coordinate that is the Fourier complement of time $t$, and a top-hat is generally used to denote Fourier-domain quantities.

Convolution Theorem

The convolution is a useful operation with applications ranging from photo editing (blurring) to crystallography to astronomy.

${\begin{aligned}[f*g](\tau )&\equiv \int {f(t)g(\tau -t)dt}\\&={\frac {1}{(2\pi )^{2}}}\int \!\!\!\int {{\hat {f}}(\omega _{1})e^{i\omega _{1}t}d\omega _{1}\,{\hat {g}}(\omega _{2})e^{i\omega _{2}(\tau -t)}d\omega _{2}\,dt}\\&={\frac {1}{(2\pi )^{2}}}\int \!\!\!\int {{\hat {f}}(\omega _{1}){\hat {g}}(\omega _{2})e^{i(\omega _{1}-\omega _{2})t}e^{i\omega _{2}\tau }d\omega _{1}\,d\omega _{2}\,dt}\\&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega ){\hat {g}}(\omega )e^{i\omega \tau }d\omega }.\end{aligned}}\,\!$

Renaming $\tau$ to be $t$ (which we are totally free to do), we get a statement of the convolution theorem:

$f(t)*g(t)=\int {{\hat {f}}(\omega ){\hat {g}}(\omega )e^{i\omega t}d\omega }={\mathcal {F}}^{-1}\!\!\left({\mathcal {F}}(f)\cdot {\mathcal {F}}(g)\right).\,\!$

Convolution vs. Correlation

Correlation is very similar to convolution, and it is best defined through its equivalent “correlation theorem”:

$f(t)\star g(t)=\int {{\hat {f}}(\omega ){\hat {g}}^{*}(\omega )e^{i\omega t}d\omega }.\,\!$

The difference between correlation and convolution is that that when correlating two signals, the Fourier transform of the second function (${\hat {g}}(\omega )$ in equation None) is conjugated before multiplying and integrating. Using that

${\begin{aligned}g^{*}(-t)&={\frac {1}{2\pi }}\int {{\hat {g}}^{*}(\omega )e^{-i\omega (-t)}d\omega }\\&={\frac {1}{2\pi }}\int {{\hat {g}}^{*}(\omega )e^{i\omega t}d\omega },\end{aligned}}\,\!$

we can show that correlating $f(t)$ and $g(t)$ is equivalent to convolving $f(t)$ with a conjugated, time-reversed version of $g(t)$:

$f(t)*g^{*}(-t)=f(t)\star g(t).\,\!$

Although this relation between convolution and correlation is often mentioned in the literature, I don’t personally find it very intuitively illuminating. I much prefer the “correlation theorem” in equation (None), because when it is combined with the expression of a time-shifted signal in Fourier domain:

${\begin{aligned}f(t-\tau )&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega )e^{i\omega (t-\tau )}d\omega }\\&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega )e^{i\omega t}e^{-i\omega \tau }d\omega },\end{aligned}}\,\!$

it shows that correlating a flat-spectrum signal with a time-shifted version of itself yields a measure of the power of the signal at the delay corresponding to the time shift:

${\begin{aligned}f(t)\star f(t-\tau )&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega ){\hat {f}}^{*}(\omega )e^{-i\omega \tau }e^{i\omega t}d\omega }\\&={\frac {1}{2\pi }}\int {|f|^{2}e^{i\omega (t-\tau )}d\omega }\\&=|f|^{2}\cdot \delta (t-\tau ).\end{aligned}}\,\!$

Related Subjects

• Interferometric Imaging