Forced Solution

The output of our digital filter is $y(n)$. We have not so far provided a mechanism to input a series of samples of a time function into the filter. An input is necessary if we wish to use the filter in a traditional way--that is, to ``filter'' samples of a sound wave. A number of possibilities exist for adding input samples to our difference equation. We have chosen to add input samples to $x(n)$. With this modification, Equations (2-3) become

$$x(n+1) = x_1 x(n) - y_1y(n)+u(n) \protect$$ (12)

$$y(n+1) = y_1x(n) + x_1y(n) \protect$$ (13)

where $u(n)$ are samples of the input to the filter. The transfer function of the filter is then

$$H(z) \mathrel{\stackrel{\mathrm{\Delta}}{=}} \frac{Y(z)}{U(z)}=\frac{y_1 z^{-2}}{1-2 x_1z^{-1}+ (x_1^2+y_1^2)z^{-2}}$$ (14)

$$= \frac{r_1\sin(\theta_1) z^{-2}}{1-2r_1\cos(\theta_1)z^{-1}+ r_1^2z^{-2}}$$ (15)

This is a so-called ``all pole'' filter, since its two zeros are at $z=\infty$ and therefore do not affect the magnitude response. This choice of input can be interpreted as driving the poles in series, since the transfer functions of series (cascade) filter combinations simply multiply.

We have compared both the sound and the spectrum of our filter output with the outputs of other filter difference equations realizations and we have detected no differences.

We may alternatively add the input samples $u(n)$ on the right-hand side of Eq.$\,$(13) to obtain the transfer function

$$H(z) = \frac{z^{-1}-x_1 z^{-2}}{1-2 x_1z^{-1}+ (x_1^2+y_1^2)z^{-2}}$$ (16)

$$= \frac{z^{-1}-r_1\cos(\theta_1) z^{-2}}{1-2r_1\cos(\theta_1)z^{-1}+ r_1^2z^{-2}}.$$ (17)

This choice of input introduces a finite zero at $z=x_1=r_1\cos(\theta_1)$, which is the real part of the location of both poles. This can be interpreted as driving the poles in parallel, since denoting the complex conjugate of $z = x+jy$ by $\overline{z}= x-jy$, we have

$$z_1^n+\overline{z}_1^n \leftrightarrow \frac{1}{1-z_1z^{-1}} + \frac{1}{1-\overline{z}_1z^{-1}}$$ (18)

$$= \frac{2-2\mbox{re}\left\{z_1\right\}z^{-1}}{1-2\mbox{re}\left\{z_1\right\}z^{-1}+\vert z_1\vert^2z^{-2}}$$ (19)

$$= 2\frac{1-x_1z^{-1}}{1-2x_1z^{-1}+r_1^2z^{-2}} = 2zH(z).$$ (20)