# Convolution Theorem

## Fourier Transform

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

{\displaystyle {\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 ${\displaystyle \omega \equiv 2\pi \nu }$ is the angular frequency coordinate that is the Fourier complement of time ${\displaystyle t}$, 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 to crystallography to astronomy. In words, the convolution of two functions ${\displaystyle f,g}$ is what you get when you smooth one function (${\displaystyle f}$) by another (${\displaystyle g}$). Note that the order of ${\displaystyle f}$ and ${\displaystyle g}$ does not matter, though people often call the latter the “kernel”. Smoothing ${\displaystyle f}$ by ${\displaystyle g}$ means that you slide ${\displaystyle g}$ along ${\displaystyle f}$, and at each step along the way, you sum up all of the parts of ${\displaystyle f}$ with weights drawn from the value of ${\displaystyle g}$ at the point you slid it to. In essence, you are blurring ${\displaystyle f}$ by ${\displaystyle g}$.

Mathematically, this is described as:

{\displaystyle {\begin{aligned}\left[f*g\right](\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 ${\displaystyle \tau }$ to be ${\displaystyle t}$ (which we are totally free to do), we get a statement of the convolution theorem:

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

### Convolution vs. Correlation

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

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

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

{\displaystyle {\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 ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ is equivalent to convolving ${\displaystyle f(t)}$ with a conjugated, time-reversed version of ${\displaystyle g(t)}$:

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

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:

{\displaystyle {\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:

{\displaystyle {\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}}\,\!}