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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.\f... ...\ &=& \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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Complex Example
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Parallel Case