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Frequency Response as a Ratio of DTFTs

From Eq.$ \,$ (6.5), we have $ H(z)=B(z)/A(z)$ , so that the frequency response is

$\displaystyle H(e^{j\omega T})=\frac{B(e^{j\omega T})}{A(e^{j\omega T})},
$

and

\begin{eqnarray*}
B(e^{j\omega T}) &=& \mbox{{\sc DTFT}}_{\omega T}({\underline{b}})\\
A(e^{j\omega T}) &=& \mbox{{\sc DTFT}}_{\omega T}({\underline{a}}),
\end{eqnarray*}

where

\begin{eqnarray*}
{\underline{b}}&\isdef & [b_0,b_1,\ldots,b_M,0,\ldots]\\
{\underline{a}}&\isdef & [1,a_1,\ldots,a_N,0,\ldots],
\end{eqnarray*}

and the DTFT is as defined in Eq.$ \,$ (7.1).

From the above relations, we may express the frequency response of any IIR filter as a ratio of two finite DTFTs:

$\displaystyle H(e^{j\omega T}) \eqsp \frac{\mbox{{\sc DTFT}}_{\omega T}({\underline{b}})}{\mbox{{\sc DTFT}}_{\omega T}({\underline{a}})} \isdefs \frac{\displaystyle\sum_{m=0}^M b_m e^{-j\omega mT}}{\displaystyle\sum_{n=0}^N a_n e^{-j\omega nT}} \protect$ (8.5)

This expression provides a convenient basis for the computation of an IIR frequency response in software, as we pursue further in the next section.



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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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