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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}(B)\\
A(e^{j\omega T}) &=& \mbox{{\sc DTFT}}_{\omega T}(A).
\end{eqnarray*}

As before,

   DTFT$\displaystyle _{\omega T}(x) \isdef \sum_{n=-\infty}^\infty x(n) e^{-j\omega nT}
$

denotes the (bilateral) discrete time Fourier transform (DTFT).

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

$\displaystyle H(e^{j\omega T}) = \frac{\mbox{{\sc DTFT}}_{\omega T}(B)}{\mbox{{...
...^M b_m e^{-j\omega mT}}{\displaystyle\sum_{n=0}^N a_n e^{-j\omega nT}} \protect$ (8.4)

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, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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