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

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    $\begingroup$ Did you find the answers to your previous questions helpful? If so, please accept and/or upvote them, or leave a comment explaining why they were not helpful, so they can be improved. Otherwise people might get discouraged answering any of your newer questions. $\endgroup$
    – Matt L.
    May 9 '19 at 20:11
  • $\begingroup$ Apologies, I hadn't realised this was the case! $\endgroup$
    – Janitt
    May 9 '19 at 20:17
  • $\begingroup$ it appears that Janitt check-marked the answers as succeeding to answer the question. must she also upvote them, @MattL. ? $\endgroup$ May 9 '19 at 20:18
  • $\begingroup$ @robertbristow-johnson: Of course not, but the answers hadn't been accepted till a few minutes ago :) $\endgroup$
    – 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:

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

  • $\begingroup$ i'll return to this later and put in some definitions. $\endgroup$ May 9 '19 at 21:33

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