# Dictionary:Convolution theorem

The Fourier transform of the convolution of two functions is equal to the product of their individual transforms (or multiplying their amplitude spectra and summing their phase spectra). See Figures F-20 and F-22.

### Integral definition

The process of convolution of two functions ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ is defined in one dimension, as

${\displaystyle (f\star g)(t)=\int _{-\infty }^{\infty }f(\tau )g(t-\tau )\;d\tau .}$

### Fourier domain equivalent

We may replace ${\displaystyle f(\tau )}$ and ${\displaystyle g(t-\tau )}$ by their Fourier domain representations

${\displaystyle f(\tau )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }F(\omega )e^{-i\omega \tau }\;d\omega }$

and

${\displaystyle g(t-\tau )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }G(\Omega )e^{-i\Omega (t-\tau )}\;d\Omega }$

where ${\displaystyle F(\omega )}$ and ${\displaystyle G(\omega )}$ are the Fourier transforms of ${\displaystyle f(t)}$ and ${\displaystyle g(t),}$ respectively.

Substituting these representations into the original integral representation of convolution yields

${\displaystyle (f\star g)(t)=\int _{-\infty }^{\infty }\left({\frac {1}{2\pi }}\int _{-\infty }^{\infty }F(\omega )e^{-i\omega \tau }\;d\omega \right)\left({\frac {1}{2\pi }}\int _{-\infty }^{\infty }G(\Omega )e^{-i\Omega (t-\tau )}\;d\Omega \right)\;d\tau .}$

We may rearrange the order of integrations

${\displaystyle (f\star g)(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\;d\omega \int _{-\infty }^{\infty }\;d\Omega \;F(\omega )G(\Omega )e^{-i\Omega t}\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{-i(\omega -\Omega )\tau }\;d\tau \right].}$

Recognizing the factor in ${\displaystyle [...]}$ as the frequency domain representation of the Dirac delta function,

${\displaystyle \delta (\omega -\Omega )=\delta (\Omega -\omega )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{-i(\omega -\Omega )\tau }\;d\tau .}$

permits us to write the equivalent expression

${\displaystyle (f\star g)(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\;d\omega \int _{-\infty }^{\infty }\;d\Omega \;F(\omega )G(\Omega )e^{-i\Omega t}\delta (\Omega -\omega ).}$

The ${\displaystyle \Omega }$ integral may be performed, exploiting the sifting property of the delta function to convert the ${\displaystyle \Omega }$ to ${\displaystyle \omega }$ yields the equivalence of multiplication in the frequency domain to convolution in the time domain

${\displaystyle (f\star g)(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }F(\omega )G(\omega )e^{-i\Omega t}\;d\omega .}$

### Convolution in the Frequency domain

Similarly, multiplication in the time domain may be interpreted as convolution in the frequency domain, to within a constant factor. Such a frequency domain convolution representation is useful, for example, if we were interested in finding the Fourier transform of the product of functions of known Fourier transform.

Paralleling the derivation above, we write the convolution in the frequency domain

${\displaystyle (F\star G)(\omega )=\int _{-\infty }^{\infty }F(\Omega )G(\omega -\Omega )\;d\Omega }$

As above, we substitute the Fourier representations of ${\displaystyle F(\omega )}$ and ${\displaystyle G(\omega )}$

${\displaystyle F(\Omega )=\int _{-\infty }^{\infty }f(t)e^{i\Omega t}\;dt}$

and

${\displaystyle G(\omega -\Omega )=\int _{-\infty }^{\infty }g(\tau )e^{i(\omega -\Omega )\tau }\;d\tau .}$

As in the derivation above, we substitute the Fourier representations of ${\displaystyle F(\omega )}$ and ${\displaystyle G(\Omega )}$ and rearrange the terms to yield

${\displaystyle (F\star G)(\omega )=K\int _{-\infty }^{\infty }\;dt\int _{-\infty }^{\infty }\;d\tau \;f(t)g(\tau )\;\left[\int _{-\infty }^{\infty }e^{i\Omega (t-\tau )}\;d\Omega \right].}$

We recognize the term in ${\displaystyle [...]}$ as the Fourier form of the Dirac delta function

${\displaystyle 2\pi \delta (t-\tau )=2\pi \delta (\tau -t)=\int _{-\infty }^{\infty }e^{i\Omega (t-\tau )}\;d\Omega .}$

As before, we apply the sifting property of the delta function, in this case to perform the ${\displaystyle \tau }$ integration to yield

${\displaystyle (F\star G)(\omega )=(2\pi )\int _{-\infty }^{\infty }f(t)g(t)e^{i\omega t}\;dt}$

There is an extra factor of ${\displaystyle 2\pi .}$ Thus, if we were representing the Fourier transform of the product ${\displaystyle f(t)g(t)}$ as the Frequency domain convolution of their respective Fourier transforms, we would need to include a factor of ${\displaystyle 1/2\pi }$ in the convolution

${\displaystyle {\frac {1}{2\pi }}(F\star G)(\omega )=\int _{-\infty }^{\infty }f(t)g(t)e^{i\omega t}\;dt.}$

To come full circle, if we were to perform the inverse Fourier transform of ${\displaystyle F(\omega )}$ we would need to treat this as a contour integral, and choose the contour along the real axis to pass over any poles that would be on the real axis, to obtain the causal function ${\displaystyle f(t).}$