Example

Consider the two-pole filter

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

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

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

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

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

as expected.