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Example

Consider the two-pole filter

$\displaystyle H(z) \eqsp \frac{1}{(1-z^{-1})(1-0.5z^{-1})}.
$

The poles are $ p_1=1$ and $ p_2=0.5$ . The corresponding residues are then

\begin{eqnarray*}
r_1 &=& \left.(1-z^{-1})H(z)\right\vert _{z=1}
\eqsp \left.\frac{1-z^{-1}}{(1-z^{-1})(1-0.5z^{-1})}\right\vert _{z=1} \\
&=& \left.\frac{1}{1-0.5z^{-1}}\right\vert _{z=1}
\eqsp 2\,,\mbox{ and}\\ [10pt]
r_2 &=& \left.(1-0.5z^{-1})H(z)\right\vert _{z=0.5}
\eqsp \left.\frac{1-0.5z^{-1}}{(1-z^{-1})(1-0.5z^{-1})}\right\vert _{z=0.5} \\
&=& \left.\frac{1}{1-z^{-1}}\right\vert _{z=0.5}
\eqsp -1\,.
\end{eqnarray*}

We thus conclude that

$\displaystyle H(z) \eqsp \frac{2}{1-z^{-1}} - \frac{1}{1-0.5z^{-1}}.
$

As a check, we can add the two one-pole terms above to get

$\displaystyle \frac{2}{1-z^{-1}} - \frac{1}{1-0.5z^{-1}} \eqsp \frac{2-z^{-1}- 1 + z^{-1}}{(1-z^{-1})(1-0.5z^{-1})} \eqsp \frac{1}{(1-z^{-1})(1-0.5z^{-1})}
$

as expected.


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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
CCRMA