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While searching resources for generating pink noise (and with your help in the comments and answers of other questions), I came to such kind of formula:

$$ H(z) = { .041 - .096z^{-1} + .051z^{-2} - .004z^{-3} \over{} -2.495z^{-1} + 2.017z^{-2} - .522z^{-3}} $$

(from https://dsp.stackexchange.com/a/376/46389)

What is $z$?

I know it should be obvious for anyone familiar with digital signal processing but for people coming from a different field, it is not obvious to know what this represents and how it relates to the signal amplitude or frequency or any other "tangible" parameter.

Edit: amusingly enough, the site suggested me the tag for this question. I suspect I should consider that as a clue.

Sylvain Leroux
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    Please see this post as it goes into detail on what z is (z is any complex number): https://dsp.stackexchange.com/questions/31830/how-why-are-the-mathcal-z-transform-and-unit-delays-related/31841#31841 – Dan Boschen Dec 04 '19 at 00:32
  • Thanks for having pointed me to that great answer @Dan! – Sylvain Leroux Dec 04 '19 at 07:04

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$z$ is interpreted as the time advance operator. $z^{-1}$ is the time delay operator.

So for a difference equation like $$ y[n]=a x[n]+b x[n-1] $$ in the $z$ domain $$ Y(z)=(a+ b z^{-1}) X(z)$$

  • Thanks, @Stanley. It's kind of a revelation. Does... that... mean... I can simply replace all $z^{-k}$ terms by $s[n-k]$ in the filter mentioned in the question above to have its expression in the time domain? Or do I need to plug some extra magic transformation somewhere? – Sylvain Leroux Dec 04 '19 at 07:22
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    @SylvainLeroux yeah, pretty much. See: the z-Transform :D – Marcus Müller Dec 04 '19 at 08:34
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I don't understand much of this myself. Probably less than you. :) But I think something you're looking for is this:

https://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

It provides basic substitution functions (if I am understanding correctly) for converting between the time, z, and laplace domains.

mike
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$z = e^{j\omega}$ is a convenient substitution for a complex valued function.

Samuel
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  • Thanks, @Samuel. I know complex numbers and I have (had?) some familiarities with Euler's complex notation. But starting from plain time-domain samples, how do I obtain that $z$? Or do I have just to consider my samples are the real part of that imaginary number? I doubt this would be so simple. Or it is? – Sylvain Leroux Dec 03 '19 at 22:45
  • Can someone please clarify why I got downvoted on this? I answered the question "what is z" I don't see how this deserves a downvote. – Samuel May 28 '20 at 22:42
  • Sylvain, I know it might be late, but starting with time domain samples, as we can see $z = e^{j\omega} = cos(\omega t) + jsin(\omega t)$... so in short you can convert your time domain samples to a complex number "z" by choosing some $\omega$ and using the relationship I described. – Samuel May 28 '20 at 22:44