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Commutativity of Convolution

Convolution (cyclic or acyclic) is commutative, i.e.,

$\displaystyle \zbox {x\circledast y = y\circledast x .}
$



Proof:

\begin{eqnarray*}
(x\circledast y)_n &\isdef & \sum_{m=0}^{N-1}x(m) y(n-m) =
\sum_{l=n}^{n-(N-1)} x(n-l) y(l)\\
&=& \sum_{l=0}^{N-1}y(l) x(n-l) \\
&\isdef & (y \circledast x)_n
\end{eqnarray*}

In the first step we made the change of summation variable $ l\isdeftext n-m$ , and in the second step, we made use of the fact that any sum over all $ N$ terms is equivalent to a sum from 0 to $ N-1$ .


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2014-04-21 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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