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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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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2015-04-22 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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