Bilinear Transform

transforming the analogue transfer function into a digital one

As a next step, we need to transform our analogue transfer functions $H(s$) to digital ones $H(z)$

This can be done with the Matched Z-Transform Method, however, the problem is that these map frequencies 1:1 between the analogue and digital domains, which isn’t right due to the Nyquist frequency, which means we cant deal with things such as infinite frequencies. The solution to this is essentially to map all analogue frequencies from $0..\infty$ to $\mathrm{0..0.5}$ in a non-linear way:

$$-\infty <\Omega <\infty ⟹-\pi \le \omega \le \pi$$

This is called the Bilinear Transform:

$$s=\frac{2}{T}\frac{z-1}{z+1}$$

This rule replaces all $s$ with $z$ in our analogue transfer function so that it’s now digital. However, the cutoff frequency ${\Omega }_c$ is still an analogue one but if we want to design a digital filter we want to specify a digital cutoff frequency. Plus remember that the frequency mapping is non-linear so that there is non-linear mapping also of the cutoff.

In the analogue domain the frequency is given as $s=j\Omega$ and in the sampled domain as $z=e^{j\omega }$. Therefore with this we can express:

Core Formula

$$\begin{array}{c}j\Omega =\frac{2}{T}\left[\frac{e^{j\omega }-1}{e^{j\omega }+1}\right]=\frac{2}{T}j\tan \frac{\omega }{2}\\ \Omega =\frac{2}{T}\tan \frac{\omega }{2}\\ \therefore {\Omega }_c=\frac{2}{T}\tan \frac{{\omega }_c}{2}\end{array}$$

Where $T$ is our sampling interval. This also has the benefit of meaning we can express in terms of normalised frequencies using ${\omega }_c=2\pi f_c$