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Introduction to Digital Filters
    Elementary Audio Digital Filters
       Elementary Filter Sections
          Complex Resonator
             Two-Pole Partial Fraction Expansion

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Two-Pole Partial Fraction Expansion

Note that every real two-pole resonator can be broken up into a sum of two complex one-pole resonators:

$\displaystyle H(z) = \frac{g}{(1 - p z^{-1}) (1 - \overline{p} z^{-1})} = \frac{g_1}{1-pz^{-1}} + \frac{g_2}{1-\overline{p}z^{-1}} \protect$ (B.7)

where $ g_1$ and $ g_2$ are constants (generally complex). In this ``parallel one-pole'' form, it can be seen that the peak gain is no longer equal to the resonance gain, since each one-pole frequency response is ``tilted'' near resonance by being summed with the ``skirt'' of the other one-pole resonator, as illustrated in Fig.B.9. This interaction between the positive- and negative-frequency poles is minimized by making the resonance sharper ( $ \left\vert p\right\vert\to1$), and by separating the pole frequencies $ 0\ll\angle p \ll \pi$. The greatest separation occurs when the resonance frequency is at one-fourth the sampling rate ( $ \angle p =\pi/2$). However, low-frequency resonances, which are by far the most common in audio work, suffer from significant overlapping of the positive- and negative-frequency poles.

Figure B.9: Frequency response (solid lines) of the two-pole resonator
$ H(z)=1/(1-2R\cos (\theta _c)z^{-1}+ R^2z^{-2})$,
for $ R=0.8$ and $ \theta _c = \pi /8$, overlaid with the frequency responses (dashed lines) of its positive- and negative-frequency complex one-pole components. Also marked (dashed lines) are the two resonance frequencies; the peak frequencies can be seen to lie slightly outside the resonance frequencies.
\includegraphics[width=\twidth ]{eps/tppfe}

To show Eq.$ \,$(B.7) is always true, let's solve in general for $ g_1$ and $ g_2$ given $ g$ and $ p$. Recombining the right-hand side over a common denominator and equating numerators gives

$\displaystyle g = g_1 - g_1 \overline{p}z^{-1}+ g_2 - g_2 pz^{-1} $

which implies

\begin{eqnarray*} g_1+g_2 &=& g\\ g_1 \overline{p} + g_2 p &=& 0. \end{eqnarray*}

The solution is easily found to be

\begin{eqnarray*} g_1 &=& g \frac{p}{2\mbox{im}\left\{p\right\}}\\ g_2 &=& -g \frac{\overline{p}}{2\mbox{im}\left\{p\right\}} \end{eqnarray*}

where we have assumed im$ \left\{p\right\}\neq 0$, as necessary to have a resonator in the first place.

Breaking up the two-pole real resonator into a parallel sum of two complex one-pole resonators is a simple example of a partial fraction expansion (PFE) (discussed more fully in §6.8).

Note that the inverse z transform of a sum of one-pole transfer functions can be easily written down by inspection. In particular, the impulse response of the PFE of the two-pole resonator (see Eq.$ \,$(B.7)) is clearly

$\displaystyle h(n) = g_1 p^n + g_2 \overline{p}^n,\qquad n=0,1,2,\ldots $

Since $ h(n)$ is real, we must have $ g_2=\overline{g}_1$, as we found above without assuming it. If $ \left\vert p\right\vert=1$, then $ h(n)$ is a real sinusoid created by the sum of two complex sinusoids spinning in opposite directions on the unit circle.

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Previous: Complex Resonator
Next: The BiQuad Section
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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