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Convolution Theorem

The convolution theorem for Fourier transforms states that convolution in the time domain equals multiplication in the frequency domain. The continuous-time convolution of two signals $ x(t)$ and $ y(t)$ is defined by

$\displaystyle (x\ast y)(t) \isdef \ensuremath{\int_{-\infty}^{\infty}}x(\tau)y(t-\tau)d\tau.$ (B.15)

The Fourier transform is then

\begin{eqnarray*}
\hbox{\sc FT}_\omega(x\ast y) &\isdef &
\int_{-\infty}^\infty
\left[\ensuremath{\int_{-\infty}^{\infty}}x(\tau)y(t-\tau)d\tau\right]
e^{-j\omega t}dt\\
&=&
\int_{-\infty}^\infty d\tau\, x(\tau)
\ensuremath{\int_{-\infty}^{\infty}}dt\, y(t-\tau)e^{-j\omega t}\\
&=&
\int_{-\infty}^\infty d\tau\, x(\tau) e^{-j\omega\tau}Y(\omega)
\quad\mbox{(by the \emph{shift theorem})}\\
&=& X(\omega)Y(\omega),
\end{eqnarray*}

or,

$\displaystyle \zbox {x\ast y \;\longleftrightarrow\;X\cdot Y.}$ (B.16)

Exercise: Show that

$\displaystyle \zbox {x\cdot y \;\longleftrightarrow\;\frac{1}{2\pi}X\ast Y}$ (B.17)

when frequency-domain convolution is defined by

$\displaystyle (X\ast Y)(\omega) \isdef \ensuremath{\int_{-\infty}^{\infty}}X(\nu) Y(\omega-\nu) d\nu,$ (B.18)

where $ \nu$ is in radians per second, and that

$\displaystyle \zbox {x\cdot y \;\longleftrightarrow\;X\ast Y}$ (B.19)

when frequency-domain convolution is defined by

$\displaystyle (X\ast Y)(f) \isdef \ensuremath{\int_{-\infty}^{\infty}}X(\nu) Y(f-\nu) d\nu,$ (B.20)

with $ \nu$ in Hertz.


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``Spectral Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2011, ISBN 978-0-9745607-3-1.
Copyright © 2016-07-18 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA