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Introduction to Digital Filters
    Transfer Function Analysis
       Partial Fraction Expansion
          Example

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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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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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