# Z-transform generalization

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1. Aug 11, 2015

### MathematicalPhysicist

Has there been any applications in engineering to the following generelization to Z-transform?

$X(z) = \sum_{n=-\infty}^\infty x[n] e^{-zn}$

We get the $z$-trasnform by using the substitution $z \mapsto Log(z)$.

2. Aug 11, 2015

### FactChecker

It's looks so closely related to the Laplace transform (and maybe the Fourier transform) that any any application of it would probably be satisfied by one of those.

3. Aug 12, 2015

### jasonRF

Yes. The standard Z transform:
$$X(z) = \sum_{n=-\infty}^\infty x[n] z^{-n}$$
can be evaluated on the unit circle, $z=e^{i \Omega}$ and you get the discrete time Fourier transform. If you have a finite number of elements in your sequence $x[n]$ this is essentially the Discrete Fourier Transform(DFT) if you chose $\Omega$ properly. If you move off the unit circle (or don't have the nice spacing of the DFT) you can get something sometimes called the chirp-z transform (and may have other names), which in general samples along spirals int he complex plane. It can be used to try and detect poles off the unit circle, or sometimes is used to sample only the part of the unit circle that you want.

jason

4. Aug 13, 2015

### MathematicalPhysicist

@jasonRF but in your case $\Omega$ is real, in my case, the $z$ in the exponent is complex. As in $e^{x+iy}$.
So it maybe called the discrete laplace transform.

5. Aug 13, 2015

### FactChecker

Right. But also the summation goes negative, where the Laplace integral does not. I can't see what the relationship would be. A sampled signal mixed with a Laplace?

6. Aug 13, 2015

### jasonRF

Sorry I wasn't more clear; that is what I meant by "off the unit circle". More explicitly, for a finite time sequence people have used,
$$X[k] = \sum_{n=0}^{N-1} x[n] e^{-\sigma n k - i 2 \pi k n /N}$$
with $k$ varying from 0 to N-1. This is like a DFT, except on a spiral instead of the unit circle. You can think of this as
$$X(z_k) = \sum_{n=0}^{N-1} x[n] z_k^{-n}$$
with
$$z_k = e^{\sigma k - i 2 \pi k /N}$$
just as you were asking.

If you are thinking an infinite sequence, then your example is conceptually no different than the Z transform.

A classic reference on this (I must admit that I have not read the paper) is:
http://www.researchgate.net/publication/3205060_Enhancement_of_Poles_in_Spectral_Analysis [Broken]

jason

Last edited by a moderator: May 7, 2017
7. Aug 14, 2015

### MathematicalPhysicist

There's one sided Laplace transform and there's the two sided Laplace transform, i.e from -\infty to \infty.

8. Aug 31, 2015

### donpacino

the z-transform is used a lot in DSP and controls to name a few