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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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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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