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Using Z-transform with continous time signals

Started by Communications_engineer January 27, 2009
Hello, can we use Z-transform on continuous time signals to find their
z-domain analysis. (what do we really realize in z-domain and how is
it different from frequency domain analysis) and also can we use
Laplace for discrete time signals.

Also what kind of analysis/information can we get from

i) Laplace Transform
ii) Z Transform
iii) Fourier Transform
iv) Hilbert Transform


(PS. no arrogance)
On Tue, 27 Jan 2009 06:24:54 -0800 (PST), Communications_engineer
<communications_engineer@yahoo.com> wrote:

>Hello, can we use Z-transform on continuous time signals to find their >z-domain analysis. (what do we really realize in z-domain and how is >it different from frequency domain analysis) and also can we use >Laplace for discrete time signals. > >Also what kind of analysis/information can we get from > >i) Laplace Transform >ii) Z Transform >iii) Fourier Transform >iv) Hilbert Transform
Well, to find those answers I'd probably ask a Communications Engineer. Oops, wait ...
>(PS. no arrogance)
But snark, that's okay. -- Rich Webb Norfolk, VA
Communications_engineer wrote:
> Hello, can we use Z-transform on continuous time signals to find their > z-domain analysis. (what do we really realize in z-domain and how is > it different from frequency domain analysis) and also can we use > Laplace for discrete time signals. > > Also what kind of analysis/information can we get from > > i) Laplace Transform > ii) Z Transform > iii) Fourier Transform > iv) Hilbert Transform > > > (PS. no arrogance)
As you may know, the "z-domain" is for sampled data / systems much as the s-domain is for continuous data / systems. Sounds like homework to me. Good luck. Fred
Communications_engineer <communications_engineer@yahoo.com> writes:

> Hello, can we use Z-transform on continuous time signals to find their > z-domain analysis. (what do we really realize in z-domain and how is > it different from frequency domain analysis) and also can we use > Laplace for discrete time signals. > > Also what kind of analysis/information can we get from > > i) Laplace Transform > ii) Z Transform > iii) Fourier Transform > iv) Hilbert Transform > > > (PS. no arrogance)
The answers to these questions are covered (directly or indirectly) in [signalsandsystems]. --Randy @BOOK{signalsandsystems, title = "{Signals and Systems}", author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}", publisher = "Prentice Hall", year = "1983"} -- % Randy Yates % "How's life on earth? %% Fuquay-Varina, NC % ... What is it worth?" %%% 919-577-9882 % 'Mission (A World Record)', %%%% <yates@ieee.org> % *A New World Record*, ELO http://www.digitalsignallabs.com
On Jan 28, 3:24&#2013266080;am, Communications_engineer
<communications_engin...@yahoo.com> wrote:
> Hello, can we use Z-transform on continuous time signals to find their > z-domain analysis. (what do we really realize in z-domain and how is > it different from frequency domain analysis) and also can we use > Laplace for discrete time signals. > > Also what kind of analysis/information can we get from > > i) Laplace Transform > ii) Z Transform > iii) Fourier Transform > iv) Hilbert Transform > > (PS. no arrogance)
Definately not!
On Jan 27, 8:24&#2013266080;am, Communications_engineer
<communications_engin...@yahoo.com> wrote:
.........
> > (PS. no arrogance)
When you say "(PS. no arrogance)" are you claiming that your request exhibits no arrogance, or are you demanding that the responses show no arrogance but proffer themselves for your delectation as humbly as possible?
Communications_engineer wrote:
> Hello, can we use Z-transform on continuous time signals to find their > z-domain analysis. (what do we really realize in z-domain and how is > it different from frequency domain analysis) and also can we use > Laplace for discrete time signals. > > Also what kind of analysis/information can we get from > > i) Laplace Transform > ii) Z Transform > iii) Fourier Transform > iv) Hilbert Transform > > > (PS. no arrogance)
Clue me in here. Are you really a communications_engineer, as your handle claims? Look up the terms that interest you using, say, Google. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
>Hello, can we use Z-transform on continuous time signals to find their >z-domain analysis. (what do we really realize in z-domain and how is >it different from frequency domain analysis) and also can we use >Laplace for discrete time signals. > >Also what kind of analysis/information can we get from > >i) Laplace Transform >ii) Z Transform >iii) Fourier Transform >iv) Hilbert Transform > > >(PS. no arrogance)
Dear (aspiring) Communications Engineer, It is admirable that you are curious to learn the tools of your trade. But you should really start by reading these topics in your textbooks and/or Wikipedia first. This will give you the most information in the shortest time. (Plus you can choose not to read authors that you find arrogant, if any.) People here are usually helpful, even more so when you show at least some effort. You should do your homework first and not ask for what is easily and widely accessible. Then, you should be more specific in your questions to get a satisfactory answer. Having said that, your first question is the closest to having these features. The answer, as other posters pointed, is "NO" (and a big one, too.) Z-transform is a mapping *defined* for discrete signals. That means, it doesn't apply to continuous signals. Period. If you had asked "What happens when you increase your sampling rate indefinitely? What does the transform approach in the limit?" that would be a more interesting question. (I'm not claiming that this questions makes sense, but it certainly requires a bit more thought. See what I mean?) Hope this helps. Emre
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message 
news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net...
> > As you may know, the "z-domain" is for sampled data / systems much as the > s-domain is for continuous data / systems.
The z transform is for delayed signals, irrespective of whether or not they are sampled. z = e^(sT) is an analogue transformation.
"Peter \"Timmy\" Timmins" <no.spam.thank.you@invalid.invalid> writes:

> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... >> >> As you may know, the "z-domain" is for sampled data / systems much as the >> s-domain is for continuous data / systems. > > The z transform is for delayed signals, irrespective of whether or not > they are sampled.
Excuse me? Please show me a "t" in the z-transform.
> z = e^(sT) is an analogue transformation.
This is a mapping from C to C (C = the complex plane) - it is not a transform. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://www.digitalsignallabs.com