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Z Transform of Convolution

From Eq.$ \,$ (5.5), we have that the output $ y$ from a linear time-invariant filter with input $ x$ and impulse response $ h$ is given by the convolution of $ h$ and $ x$ , i.e.,

$\displaystyle y(n) \eqsp (h \ast x)(n) \protect$ (7.3)

where ``$ \ast $ '' means convolution as before. Taking the z transform of both sides of Eq.$ \,$ (6.3) and applying the convolution theorem from the preceding section gives

$\displaystyle Y(z) \eqsp H(z)X(z) \protect$ (7.4)

where H(z) is the z transform of the filter impulse response. We may divide Eq.$ \,$ (6.4) by $ X(z)$ to obtain

$\displaystyle H(z) \eqsp \frac{Y(z)}{X(z)} \;\isdef \; \hbox{transfer function}.
$

This shows that, as a direct result of the convolution theorem, the z transform of an impulse response $ h(n)$ is equal to the transfer function $ H(z)=Y(z)/X(z)$ of the filter, provided the filter is linear and time invariant.


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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