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Normalizing Two-Pole Filter Gain at Resonance

The question we now pose is how to best compensate the tunable two-pole resonator of §B.1.3 so that its peak gain is the same for all tunings. Looking at Fig.B.17, and remembering the graphical method for determining the amplitude response,B.6 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

$\displaystyle 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

$\displaystyle y(n) = x(n) - x(n-2) + [2R\cos(\theta_c)] y(n-1) - R^2 y(n-2). \protect$ (B.15)

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

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

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.(B.14).

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition)
Copyright © 2024-05-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University