| In the context of linear time-invariant filter analysis, a zero is a point in the complex s or z plane at which the transfer function goes to zero. A pole is a point in the s or z plane at which the transfer function goes to infinity. The s plane is used to analyze analog filters while the z plane is used for digital filters. — Click for https://ccrma.stanford.edu/~jos/filters/Pole_Zero_Analysis_I.html |
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and 
. 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*}](img699.png)
