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 $f(t)$ and $g(t)$ is defined in one dimension, as

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

Fourier domain equivalent

Replacing $f(\tau )$ and $g(t-\tau )$ by their Fourier domain representations

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

and

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

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

Substituting these representations into the original integral representation of convolution yields

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

We may rearrange the order of integrations

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

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

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

permits us to write the equivalent expression

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

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

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

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.

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

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

where the constant $K$ remains to be determined. As above, we substitute the Fourier representations of $F(\omega )$ and $G(\omega )$

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

and

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

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

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

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

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

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

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

Hence, the factor of $K$ is determined to be $1/2\pi .$ Thus the the correct representation of convolution in the frequency domain, with these Fourier transform definitions is

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