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

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

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

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:

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