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

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

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:

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