Converting a simple analog Butterworth filter into a digital filter

Question

I was just studying an old circuit analysis textbook that was describing how to design a Butterworth filter, and that seemed easy enough.. then, I started to wonder if I can take this analog filter and convert it into a digital filter. Its not really an exercise in the textbook, i was just curious how to convert the analog filter into a digital filter just for fun without the heavy DSP theory.

So I was tried to taking a toy Butterworth filter to do just that. For example, Let's suppose I had an analog filter:

$$H_a(j\Omega) = \frac{1}{(1+j\Omega)(2+j\Omega)}$$

and i wanted to convert this into a digital filter with say a sampling period $T=200\pi$ rad/sec, and neglecting the effects of aliasing, using this formula:

$$H(e^{j\omega}) = H_a(j\Omega)\Big|_{\Omega = \omega/T}$$

What would $H(z)$ and $h[n]$ look like for the digital filter?

Answer 1

Converting the analog filter $H_a(s)$ into a digital filter $H_d(z)$ using the bilinear transform where T is the sampling period:

$\Large H_d(z) = H_a(s)\bigg|_{s=\frac{2}{T}\frac{z-1}{z+1}}$

Example:

Given a first order Butterworth filter

$H_a(s) = \frac{1}{1+RCs}$

$H_d(z) = H_a\bigg( \frac{2}{T} \frac{z-1}{z+1} \bigg)$

$H_d(z) = \frac{1}{1+RC\Big(\frac{2}{T}\frac{z-1}{z+1} \Big)}$

$H_d(z) = \frac{1}{1+RC(\frac{2}{T}\frac{z-1}{z+1})}$

$H_d(z) = \frac{1+z}{(1-2RC/T)+(1+2RC/T)z}$

$H_d(z) = \frac{1+z^{-1}}{(1+2RC/T)+(1-RC/T)z^{-1}}$

The coefficients of the denominator are the 'feed-backward' coefficients and the coefficients of the numerator are the 'feed-forward' coefficients used to implement a real-time digital filter.