Normalizing Two-Pole Filter Gain at Resonance

The question we now pose is how to best compensate the tunable two-pole resonator of §10.1.3 so that its peak gain is the same for all tunings. Looking at Fig.10.18, and remembering the graphical method for determining the amplitude response,^11.5 it is intuitively clear that we can help matters by adding two zeros to the filter, one near dc and the other near $f_s/2$. A zero exactly at dc is provided by the term $(1-z^{-1})$ in the transfer function numerator. Similarly, a zero at half the sampling rate is provided by the term $(1+z^{-1})$ in the numerator. The series combination of both zeros gives the numerator $B(z)=(1-z^{-1})(1+z^{-1})=1-z^{-2}$. The complete second-order transfer function then becomes

$$H(z) = \frac{B(z)}{A(z)} = \frac{1 - z^{-2}}{1-2R\cos(\theta_c)z^{-1}+ R^2z^{-2}}$$

corresponding to the difference equation

$$y(n) = x(n) - x(n-2) + [2R\cos(\theta_c)] y(n-1) - R^2 y(n-2). \protect$$ (11.13)

Checking the gain for the case $\theta_c=\pi/2$, we have

$$\begin{eqnarray*} H(\pm1) &=& 0\\ \left.H(e^{j\theta_c})\right\vert _{\theta_c=\pi/2} &=& \frac{2}{1-R^2} \end{eqnarray*}$$

which is better behaved, but now the response falls to zero at dc and $f_s/2$ rather than being heavily boosted, as we found in Eq. (10.12).