# I often see here formula expressed in term of $z$. But what is $z$?

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:

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

## 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.

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

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.

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

• 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 3 '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 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 at 22:44