Example: Digital Lowpass Filter

Suppose we want to construct a digital filter with cutoff \(f_c\) from a 2-pole Butterworth lowpass filter. Then start with its transfer function

\[ H(s) = \frac{1}{s^2 + \sqrt{2}s + 1} \]

which has \(1\) as the cutoff freqency.

In order for the transformation to work we need the cutoff to be \(\tan(a/2)\) instead of \(1\), where \(a = 2 \pi f_c / f_s\). So we compute \(\omega = tan(\pi f_c/f_s)\) and replace \(s\) with \(s/\omega\) so that the cutoff is at \(i\omega\) instead of \(i\). Then we perform a bilinear transform by replacing \(s\) with \(\frac{z-1}{z+1}\). We end up with

\[H(z) = \frac{1}{\frac{1}{\omega^2} (\frac{z-1}{z+1})^2 + \sqrt{2} \frac{1}{\omega}\frac{z-1}{z+1} + 1} = \frac{z^2 + 2z + 1}{(\frac{1}{\omega^2} + \frac{\sqrt{2}}{\omega} + 1)z^2 + (2 - \frac{2}{\omega^2})z + (\frac{1}{\omega^2} - \frac{\sqrt{2}}{\omega} + 1)}\]

So letting \(d = \frac{1}{\omega}\), \(c = 1 /(d^2 + \sqrt{2}d + 1)\), we have the filter coefficients:

\[a_0 = c, a_1 = 2c, a_2 = c, b_1 = (2 - 2d^2)c, b_2 = (d^2 + 1 - \sqrt{2}d)c \]

Ben Lynn blynn@cs.stanford.edu