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:

$$H(z) = \frac{g}{(1 - p z^{-1}) (1 - \overline{p} z^{-1})} = \frac{g_1}{1-pz^{-1}} + \frac{g_2}{1-\overline{p}z^{-1}} \protect$$ (11.7)

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.10.10. 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 10.10: Frequency response (solid lines) of the two-pole resonator

To show Eq. (10.7) 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

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

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. (10.7)) is clearly

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

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.