# Using Z-transform with continous time signals

Started by 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

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&#4294967295;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&#4294967295;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.
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```
```>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,

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

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.

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
```