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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.

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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 $$ Y(z)=(a+ b z^{-1}) X(z)$$

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  • $\begingroup$ 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? $\endgroup$ – Sylvain Leroux Dec 4 '19 at 7:22
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    $\begingroup$ @SylvainLeroux yeah, pretty much. See: the z-Transform :D $\endgroup$ – Marcus Müller Dec 4 '19 at 8:34
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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.

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

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  • $\begingroup$ 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? $\endgroup$ – Sylvain Leroux Dec 3 '19 at 22:45

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