Talk:Z-transform

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"The former is sometimes called a unilateral Z-transform and the latter a bilateral or doubly infinite Z-transform."

Are "former" and "latter" the wrong way around here?
Yep. Thanks. Fixed.


I think a section on common Z-Transform pairs should be added. Can someone please do it? I don't know to use the Math functions.

added some Z-Transform pairs SKvalen 20:43, 7 December 2005 (UTC)

Does anyone know who discovered the z-transform and its usefulness? Maybe we could put a short blurb about it?

IIRC, my CRC math handbook says something about it. i'll dig it up.
Didn't Lotfi Zadeh publish it for the first time under than name z-transform in 1952? (with Raggazzini)

Contents

[edit] circular coordinate representation of "z"

Although for some it may be obvious, I still think that, for clarity, this should be added to the definition of the Z transform: ... where e is the base of the natural logarithm and j is the imaginary unit (j2 = − 1)

IMO, i see no reason to say that z is a "circular" complex number. z is just a complex variable and there is no reason to say it has to be in polar form. maybe we should take that whole thing out. r b-j 02:53, 7 May 2005 (UTC)
Most, if not all, texts I have seen use circular and I see no reason to drop it from the WP article... Cburnett 03:45, May 7, 2005 (UTC)
none of mine do until they start talking about frequency response or maybe ROC. computing those summations in z for various transforms and theorems does not require it. i'm happy to leave it. but then maybe we should respond to the suggestion that we explicitly define "e" and "j". r b-j 05:09, 7 May 2005 (UTC)
Well, a Z-transform without the ROC is meaningless and I can't recall seeing a ROC that was non-cicular. A non-circular/cartesian Z-transform would be more like a Laplace transform... Cburnett 06:33, May 7, 2005 (UTC)

[edit] Can anyone figure out why row 3 and row 9 will not render in tex?

in the table of Z transforms, all's i did was change some pareths into brackets for discrete defined functions such as the discrete impulse and discrete unit step. but two rows will not render correctly and i can see no rhyme nor reason for it. r b-j 23:43, 5 February 2006 (UTC)

i checked it out with another computer and different web browser and it looks okay. just chalk it up as another "feature" of Micro$hit IE. r b-j 02:01, 6 February 2006 (UTC)
Many times it helps if you insert one or two blank spaces in between different tokens or elements of the LaTeX code. Sometimes it seems that browser's (even those created by the Evil Empire) have trouble rendering the equation when they get very complex and you don't space things out a bit. This approach also has the added advantage that it makes the LaTeX code more readable for human editors as well. -- Metacomet 02:55, 6 February 2006 (UTC)

For example:

<math>{-b\pm\sqrt{b^2-4ac}\over2a}</math>

versus

<math> { -b \pm \sqrt{ b^2 - 4ac } \over 2a } </math>

The second version is less likely to create a problem for the browser during rendering. In addition, it is argualbly much easier for humans to read and modify than the first version

-- Metacomet 03:09, 6 February 2006 (UTC)

[edit] Applications of the Z transform

Can someone please find the time to write a section on the applications of the Z transform? Once you get a sampled sequence into the complex frequency domain using the Z transform, then what? What are the benefits of doing that? How is using the Z transform better/worse than using the DFT or rather the FFT? 193.251.135.126 23:00, 20 February 2006 (UTC)

[edit] Question

What do "z" and "n" stand for? All it says is that they are "complex number" and "integer" respectively. Sorry if I'm missing something obvious here. - FrancisTyers · 18:28, 30 July 2006 (UTC)

To continue in the sense of the 2 previous question, i am in ingenering, i would like to use Z-transform for signal processing but i don't understand how to do calculus with what is given.--82.247.82.41 20:48, 25 October 2006 (UTC)


[edit] Q

I don't think entry 15 is correct in the z-table transform. It should be (1+az^-1) in the numerator.

you mean, without the power ^3?? I calculated it, and it looked right, but if you look for it in Internet you can get the same result. Here is an example [1] Alessio Damato 17:19, 24 January 2007 (UTC)

[edit] Causality

I fixed a problem in the definition of a unilateral Z-transform wherein the unilateral z-transform is applied to "causal signals." Causality can only be applied to systems, not signals. Probably a typo.

A causal signal is defined as x[n] = 0 for all n < 0. Therefore the original definition of the unilateral Z-transform was correct. Oli Filth 20:12, 29 January 2007 (UTC)

[edit] Contradiction?

Currently, the article states that

The inverse Z-Transform is
x[n] = Z^{-1} \{X(z) \}= \frac{1}{2 \pi j} \oint_{C} X(z) z^{n-1} dz \
where C \ is a counterclockwise closed path encircling the origin and entirely in the region of convergence (ROC). The contour or path, C \ , must encircle all of the poles of X(z) \ .

What happens if the poles lie on C? --Abdull 17:37, 11 February 2007 (UTC)

The ROC cannot contain poles; therefore, if C \ is contained entirely within the ROC, there cannot be any poles lying on it. Oli Filth 22:16, 11 February 2007 (UTC)
The Z-transform is related to the Fourier transform in that the Fourier transform is the Z-transform evaluated around the unit circle. If a pole were to exist on the unit circle, the inverse time-domain signal would oscilate wildly between plus and minus infinity at the frequency corresponding to the location of the pole. —Preceding unsigned comment added by 68.145.40.189 (talk) 23:18, 13 March 2008 (UTC)

[edit] Example 3 math problem

Something goes wrong in the math in example 3, as the limit of the geometric series ends up with the same result as section 2.

BenJackson 08:17, 7 April 2007 (UTC)

That's the whole point. The two systems are different (one is causal, one is anticausal), but end up with the same Z-transform expression. Oli Filth 08:32, 7 April 2007 (UTC)
But if they end up with the same Z-transform expression (i.e. the same sum), then also the ROC would be the same? If I understand correctly, the ROC is derived from the infinity geometric series (1/(1-q)) constraint that |q| < 1. And since the series is the same, also the constraint would have to be the same? Nejko (talk) 10:55, 20 January 2008 (UTC)
The ROC is the area for which the original series (i.e. the things in the form \sum h_n z^{-n}) converges. It is only in this region that the series will equal the derived expression 1 / (1 − 0.5z − 1), as clearly this is defined at all z \ne 0.5. Examples 2 and 3 have different series, which have different (in this case, complementary) ROCs. Oli Filth(talk) 12:39, 20 January 2008 (UTC)

[edit] University of Penn's article available as PDF

I've just converted the University of Penn's tutorial on Z-transforms across to using LaTeX, as while I found it very useful, I found the formulae a little difficult to read.

It's available on my site (see the external links on the article page). I've also provided the LaTeX sources here in a 7z archive -- if people wish to make corrections to the document. I couldn't see any contact or author information on the university's page, so I figured I'd leave a note here. (And if the original author of that page is watching this -- feel free to mirror the PDF doc and LaTeX sources, or make updates.) Redhatter 13:22, 9 November 2007 (UTC)

The opening lines of the article say that the z-transform works on real numbers, but there is no reason it will not work for complex numbers as well. Indeed, it is often used for this purposes (e.g. signals acquired using a phase-sensitive detector).

82.36.124.238 (talk) 12:11, 19 January 2008 (UTC)gmr

[edit] Isn't Z-transform a special case of Laurent series?

One have nothing to learn about Z-transform if he studied complex analysis. Don't you think this article should refer to Laurent_series page? —Preceding unsigned comment added by 62.181.43.110 (talk • contribs) 10:01, 31 January 2008

Not really. The Z-transform is heavily used in engineering maths (specifically signal processing); it's one of the most important tools available. Almost no-one refers to the Laurent series (except in academic papers, journal articles, etc.). It would help no-one to redirect, and what would we do with all the information here?
It might be true that the Z-transform is a special case of the Laurent series, but by this logic, you might as well redirect Fourier Transform to Laplace Transform. Oli Filth(talk) 11:35, 31 January 2008 (UTC)

[edit] "Transform Realisation for Digital Controllers and Filters" section

This newly-added section discusses (in a somewhat confused manner) about a Direct-Form II implementation of a digital filter. I contend that this article is not the place to discuss implementations (that should be in the Digital filter article, or perhaps a new Direct form or Digital filter structures article); the scope of this article is the mathematical construct. What's more, the additions point out that there are many different filter structures; we can't possibly list them all here, so why list any? Oli Filth(talk) 11:40, 10 May 2008 (UTC)