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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

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\,.

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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