# Usefulness of Matched $z$ transform Method

I'm aware that the matched $$z$$ transform method maps between the continuous $$s$$ plane and the discrete/digital $$z$$ plane but my question is - when would this be necessary? Why would we need to convert between the two? Thanks

• Apologies, I hadn't realised this was the case! – Janitt May 9 '19 at 20:17
• it appears that Janitt check-marked the answers as succeeding to answer the question. must she also upvote them, @MattL. ? – robert bristow-johnson May 9 '19 at 20:18
• @robertbristow-johnson: Of course not, but the answers hadn't been accepted till a few minutes ago :) – Matt L. May 9 '19 at 20:21

the Matched Z method is the simplest method to convert an analog filter design with transfer function $$H_\mathrm{a}(s)$$ to a digital filter design $$H(z)$$. It does it by mapping every pole and every zero from the $$s$$-plane to the $$z$$-plane using:

$$z \leftarrow e^{sT}$$

where $$T \triangleq \frac{1}{f_\mathrm{s}}$$ is the sampling period and the reciprocal of the sample rate, $$f_\mathrm{s}$$. Stable analog filters get mapped to stable digital filters. That's all that it is.

The other two common methods are the Bilinear Transform, which approximates the above mapping:

$$z \leftarrow \frac{1+\frac{sT}{2}}{1-\frac{sT}{2}} \approx \frac{e^{\frac{sT}{2}}}{e^{\frac{-sT}{2}}} = e^{sT}$$

and the Impulse Invariant method which is a time-domain way of looking at things:

$$h[n] \leftarrow h_\mathrm{a}(nT)$$

• i'll return to this later and put in some definitions. – robert bristow-johnson May 9 '19 at 21:33