Talk:Z-transform/Archive 1

Archive 1

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)

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)

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)

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)

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)

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

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)

"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)

Influence of sampling time T

The differentiation property in the table with transform properties seems to assume that the sampling time T is equal to 1 second. The same holds for some of given Z-transform pairs in the next section. This table seems to be a bit academic because a transform pair for a first order or exponential function is missing which is used very much in practical cases. —Preceding unsigned comment added by 141.252.42.10 (talk) 08:05, 17 September 2008 (UTC)

Unilateral and bilateral transforms, transform tables, and discovery

"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)

Yes, it was published by Ragazzini and Zadeh in Transactions of the American Institute of Electrical Engineers in 1952, and this fact is noted in Jury's book. I have no idea why Jury is credited here with inventing the z-transform, he would be horrified at such a misattribution, I have no doubt. I will change the article to reflect the literature. Safulop (talk) 07:02, 18 April 2009 (UTC)

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)

If I understand the original poster right, what he means is that he misses some sort of link to Laurent series at the bottom of the page. User:ehasl —Preceding unsigned comment added by 77.40.128.194 (talk) 20:55, 7 December 2009 (UTC)

Labels for tables

Hello. Looks like the table for properties of the Z transform and the table of some transform pairs are both assuming the two-sided definition. They should be labeled as such, agreed? I'll do so if nobody complains. On a slightly broader note, all of the transform pairs listed have u[n] in them which makes them essentially one-sided. How about if we cut out the u[n] and just label the table as a table of one-sided transforms? FWIW & HAND. 4.227.238.139 (talk) 21:37, 28 December 2008 (UTC)

if we have Z domian then we try to transfer to time domian please tell me hou can we do it —Preceding unsigned comment added by 118.101.211.68 (talk) 15:27, 16 July 2010 (UTC)

Numerical evaluation of (inverse) Z-transform

Would this be an appropriate section to add here? I'm only aware of one method (which I used myself) and that is (assuming the transform is stable) to do a numerical approximation of the inverse contour integral theorem. Interesting enough if one uses a basic trapetzoid approximation the inverse z-transform can easily be formulated as a IDFT problem and hence leading itself to reasonable efficient numerical calculation.

However, I'm not sure about the numerical stability and error analysis for this method (or if this is indeed a commonly used method - I have not seen it in the literature as such in conjunction with the Z-transform) — Preceding unsigned comment added by Johnbanjo (talk • contribs) 11:23, 6 February 2011 (UTC)

If it's not in the literature, it doesn't belong in wikipedia. Speaking of which, can you tell us your sources for the history update you did? Thanks. I found a few relevant sources, but more are needed for the history bits. Dicklyon (talk) 01:28, 7 February 2011 (UTC)

Actually reading this article more closely this idea of numerical inversion is at least indirectly mentioned in the subsection on inversion (although the idea is not clearly expanded there - It just states that the inversion integral can be evaluated efficiently with the IDFT),

Regarding the reference for the naming. This was passed on to me by one of my prof. in grad school and of course this is not a proper reference. I'm going to dig a bit in my old archives (which are paper based) but if memory doesn't fail me (grad school was a long time back). My prof. got this information from Zadeh with whom he he met briefly during a sabbatical. If I cannot find a proper reference for this by end of next week I'll edit it out. My old prof. is dead now so asking him is no longer an option. — Preceding unsigned comment added by Johnbanjo (talk • contribs) 22:23, 8 February 2011 (UTC)

"Matched" Z-transform

The Butterworth filter article links to this one while referring to a "matched" Z-transform, perhaps this article should say what that means, somewhere... A5 (talk) 16:19, 13 October 2008 (UTC)

I fixed that. It's not about a type of Z-transform, but rather an application, the "matched Z-transform method". It needs an article, like bilinear transform and impulse invariance, which are alternative methods. Here is a source. I think it's also called the "pole–zero mapping method" like here. Indeed, this book says they're the same. I'll stub in Matched Z-transform method and Pole–zero mapping method and such. Dicklyon (talk) 18:39, 7 February 2011 (UTC)

See the new article I made on Matched Z-transform method. Dicklyon (talk) 17:01, 11 February 2011 (UTC)

Other meanings

Coming from an informatics/maths background I was confused by psychology for a while, as everyone was talking about z-transforms in contexts where it made no sense. Turns out they were talking about z-scores (i.e. subtracting mean and dividing by s.d.). Maybe a link to the 'z score' page, at the top, to redirect confused people? Lionfish0 (talk) 15:53, 13 October 2011 (UTC)

It might also refer to the http://en.wikipedia.org/wiki/Fisher_transformation, which some people seem to call the Fisher z-transformation... Lionfish0 (talk) 15:53, 13 October 2011 (UTC)

(I've been trying to figure out what this paper meant for a while!) Eventually found it! People refer to the inverse-normal cumulative distribution as the 'z-transform'. See http://en.wikipedia.org/wiki/D%27. Lionfish0 (talk) 16:00, 13 October 2011 (UTC)

Maybe any or all of these could be linked from the top. Maybe a disambiguation page?! Lionfish0 (talk) 16:00, 13 October 2011 (UTC)

Letter z

Boodlepounce suspects that the unreferenced speculation about the origin of the letter z in the name is nonsense and has removed it. Boodlepounce (talk) 05:22, 1 November 2011 (UTC)

Relation to Laplace Transform

The relation between Z_Transform and Laplace_transform via the bi-linear transform is only a design tool (as far as I know) for mapping stable systems in continuous domain to stable systems in discrete domain.

Sampled Data System and Discrete-Time System books (like B.C.Kuo and K.Ogata) give a different and more physical relationship arising from sampling a continuous signal. When an amplitude modulated train of impulses is fed to a continuous time LTI system, the relation between Laplace_Transform, Starred transform and Z_transform can be seen to be the variable substitution z = exp(sT). this also may be added to the article if it is appropriate. Cplusplusboy (talk) 17:11, 12 June 2012 (UTC)

Yes, the point of the bilateral transform is very poorly explained, and the sampling relationship that you point out is the more fundamental one. Please do work on, based on your sources, and cite them. Dicklyon (talk) 18:30, 12 June 2012 (UTC)

I have added stuff based on sampled data systems books by Ogata, Kuo.B.C etc. But it looks like the derivation like presentation is not suitable and only the relationship need to be given. Or may be just linking to Star_transform may be sufficient. Cplusplusboy (talk) 17:03, 19 June 2012 (UTC)

I did some basic copy edits for case and formatting. See if I messed anything up. I haven't checked the details yet. Still need to work on that bilateral mess. Dicklyon (talk) 17:26, 19 June 2012 (UTC)

Property table and Parseval

The property table lists time-functions and their Z-transforms. The last entry "Parseval" however is an identity (between 1^st and 2^nd column), not a Z-transform. It should be taken out of the table, and added to the two relations immediately following the table. Thanks for the good work. Herbmuell (talk) 07:56, 31 May 2013 (UTC)

Done. Glrx (talk) 16:36, 1 June 2013 (UTC)

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 ($j^{2}=-1$)

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)

z in the z-transform is simply a complex number. The unit circle on the complex plane is of special importance for the z-transform because when the z-transform is evaluated on the unit circle it is equivalent to the Discrete Fourier Transform. However, the location of any poles and zeros can be anywhere on the complex plane. — Preceding unsigned comment added by 66.18.199.236 (talk) 21:15, 23 July 2013 (UTC)

TeX equations need to be displayed

Urgent! The TeX equations need to be displayed properly. Please review them. -210.212.162.140 (talk) 05:34, 10 October 2013 (UTC)

Answer

As in Transform Theory, "L" represent Laplace Transform, named after "Laplace,Pierre-Simon" same with Fourier Transform, we use "F" in honour of its inventor "Fourier, Joseph ". So, similarly, "Z" in Z-Transform holds the name of an american scientist linked to this transform, named "Zadeh, Lotfali Askar" — Preceding unsigned comment added by 117.55.241.3 (talk) 14:21, 23 November 2013 (UTC)

good faith edits (‎by Kameenakaka) introduce a contradiction

The article now states:

Convergengence of z-transform depends only on the magnitude of z, i.e:r=|z|, not on the angle of z.

and

The region of convergence (ROC) is the set of points in the complex plane for which the Z-transform summation converges.

$ROC=\left\{z:\left|\sum _{n=-\infty }^{\infty }x[n]z^{-n}\right|<\infty \right\}$

The latter means that the ROC for sequence $e^{i\theta n}$ includes $z=e^{i\omega },$ for almost all values of $\omega .$ And of course it excludes the point $\omega =\theta$.

The article also now states

The Z-transform evaluated on unit circle corresponds the fourier transform.

But the Fourier transform is not defined for discrete data. That whole subject is covered in section Z-transform#Relationship_to_Fourier_series_and_Fourier_transform.

--Bob K (talk) 16:37, 28 September 2014 (UTC)

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 $z\neq 0.5$. Examples 2 and 3 have different series, which have different (in this case, complementary) ROCs. Oli Filth^(talk) 12:39, 20 January 2008 (UTC)

The term $1/(1-0.5z^{-1})$ at the end of example 3 should actually be $0.5z^{-1}/(1-0.5z^{-1})$. Shouldn't it be? Aravind V R (talk) 13:27, 11 May 2013 (UTC)

It should be end with $0.5z^{-1}/(1-0.5z^{-1})$. Therefore, both the result and the comment made for that result are wrong. ROC is just a condition for simplifying the series. I am not sure that it is possible, but you need to put a different example for showing that two different transform can converge to a same thing at the same time having different ROCs. Mnemonicker 12:10, 5 January 2015 (UTC)

In example 1, is not {0.5} a ROC? —Preceding unsigned comment added by 77.40.128.194 (talk) 20:57, 7 December 2009 (UTC)

Inverse Z-transform formula: formula name or reference

Is there some reference to this formula, as in name, history, proof, etc? — Preceding unsigned comment added by 5.12.77.107 (talk) 04:45, 3 November 2014 (UTC)

A signal is generated by sampling e*-1 every T seconds beginning at t=0 (a) write as a series the Z-Transform of the signal — Preceding unsigned comment added by 41.66.208.151 (talk) 09:06, 1 April 2015 (UTC)

2.146.131.215 (talk) 17:05, 9 December 2015 (UTC)