Homogeneous Solution

Equations (1-3) can be modified to compute the response of a finite $Q$ filter simply by changing $r_1$ to some value different from unity. From Eq.$\,$(1), $\vert z(n)\vert$ will be $r_1$ raised to the $n$th power, and $y(n)$ will be

$$y(n)=r_1^n\sin(n\theta_1) \protect$$ (4)

If $r_1$ is less than one, Eq.$\,$(4) shows the homogeneous solution to Eq.$\,$(1) is a sinusoid times a decaying exponential. This solution is the response of a two pole filter.

Equations (2-3) are unchanged except that the coefficients $x_1$ and $y_1$ become

$$x_1 = r_1 \cos(\theta_1) \protect$$ (5)

$$y_1 = r_1 \sin(\theta_1) \protect$$ (6)

If $y(n)$ is the input to a digital-to-analog converter operating at a sampling rate of $f_s$ samples per second, then $x_1$ and $y_1$ can be written

$$x_1 = e^{-\frac{1}{\tau f_s}} \cos\left(2\pi \frac{f}{f_s}\right) \protect$$ (7)

$$y_1 = e^{-\frac{1}{\tau f_s}} \sin\left(2\pi \frac{f}{f_s}\right) \protect$$ (8)

where

$$\tau \mathrel{\stackrel{\mathrm{\Delta}}{=}} \hbox{time in seconds for filter to decay to $1/e$} \protect$$ (9)

$$f \mathrel{\stackrel{\mathrm{\Delta}}{=}} \hbox{resonant frequency of filter in Hz} \protect$$ (10)

$$f_s \mathrel{\stackrel{\mathrm{\Delta}}{=}} \hbox{sampling rate in samples per second} \protect$$ (11)

Equations (7-10) are valid for $f$ less than $f_s/2$. For $f$ between $f_s/2$ and $f_s$, the generated frequency will be reflected about $f_s/2$.

Negative values of $\tau$ will generate exponentially increasing sinusoids which can be musically useful.