Using Z-transform with continous time signals

Started by January 27, 2009
On Jan 29, 6:55&#2013266080;pm, Communications_engineer
<communications_engin...@yahoo.com> wrote:
> On Jan 29, 12:31&#2013266080;am, Tim Wescott <t...@seemywebsite.com> wrote:
>
>
>
> > On Wed, 28 Jan 2009 07:36:39 -0500, Randy Yates wrote:
> > > "Peter \"Timmy\" Timmins" <no.spam.thank....@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.
>
> > He's right. &#2013266080;You _can_ use the z transform for analyzing systems that
> > contain delays that are all integer multiples of some characteristic
> > time. &#2013266080;Not that there are many like that, but should you stumble across a
> > system that's all done with equal-length bits of transmission line, you
> > have the mathematical tools to analyze it.
>
> > The 't' in the z-transform is the kT in z = e^(skT). &#2013266080;So if your system
> > is all in delays of T or greater, then your transfer function will all be
> > in terms of e^(skT).
>
> > >> z = e^(sT) is an analogue transformation.
>
> > > This is a mapping from C to C (C = the complex plane) - it is not a
> > > transform.
>
> > OK, it's an analog mapping. &#2013266080;It still shows how you can use the z
> > transform to analyze continuous-time systems.
>
> > --http://www.wescottdesign.com
>
> When Differential-PSK gives better result that coherent-PSK under
> observed this using scatter plots in Matlab using Rayleigh and Rician
> models and observed that when constellation was severely distortion
> the BER was much lower than what I &#2013266080;got with PSK) so why is it not a
> popular technique in modern cellular systems. Is there some cost
> problem or some technical issue? &#2013266080;Wouldn't 8-DPSK be better than pi/4
> shifted QPSK? So we can leave out the carrier recovery circuit.

Well from what I see, the "true" meaning of Z-transform and Laplace
Transform is not just ambiguous for me but for other people/experts
here as well. In fact this same ambiguity exists for Laplace as well.
Its not a transform since it doesn't transform from time-domain to
another. It just seems to me like a complimentary time-domain info. As
said by Tim Wescot &  Randy Yates its just a mapping in complex plane.
So can I infer that it is just complimentary time-domain mapping so
that our domain in complex and range is also complex so that we apply
complex input signals to our system and see if its stable check its
impulse response ..

Also any answer regarding Differential non-coherent modulation as a
better option than coherent modulation


This isn't what he is asking, but when I read 'Z-transform in
continuous time' I thought of the sliding FFT method with an
exponential window for probing points on the complex Z-plane 'in real
time'.

Essentially, you pass the signal through:

y[n] = exp(a+b*j)*y[n-1] + (1-exp(a))*x[n];

For all a,b of interest.

-Martin

Tim Wescott <tim@seemywebsite.com> writes:

> On Wed, 28 Jan 2009 07:36:39 -0500, Randy Yates wrote:
>
>> "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.
>
> He's right.  You _can_ use the z transform for analyzing systems that
> contain delays that are all integer multiples of some characteristic
> time.

I wasn't too clear on the point of my disagreement, which wasn't the
"delayed signals" statement, but rather the "irrespective of whether or
not they are sampled" statement.

The independent variable in the "signal" (or system, which is merely a
signal representing the system's impulse response) transformed by the
z-transform must be taken from a countably infinite set. That, by
definition, means it's not a continuous-time transformation.

> [...]
> The 't' in the z-transform is the kT in z = e^(skT).

There is no such substitution of z that I'm aware of.

Perhaps what you were thinking of is that if we sample a continuous-time
system x_a(t),

x(t) = \sum_{n=-\infty}^{+\infty} x_a(nT) \delta(t - nT),

then the Laplace transform of x(t) is equivalent to the z-transform of
x_a(nT) when z is evaluated at e^{sT}.

But this isn't using the z-transform to analyze a continuous-time
system, it's using the z-transform to analyze a discrete-time system
that was sampled from a continuous-time system.

>>> z = e^(sT) is an analogue transformation.
>>
>> This is a mapping from C to C (C = the complex plane) - it is not a
>> transform.
>
> OK, it's an analog mapping.  It still shows how you can use the z
> transform to analyze continuous-time systems.

I don't see any such utility for the mapping. What I do see is that it
allows us to relate the Laplace transform of a sampled system to the
z-transform of a sampled system. In neither case is a continous-time
system involved in the normal sense of the term.
--
%  Randy Yates                  % "So now it's getting late,
%% Fuquay-Varina, NC            %    and those who hesitate
%%% 919-577-9882                %    got no one..."
%%%% <yates@ieee.org>           % 'Waterfall', *Face The Music*, ELO
http://www.digitalsignallabs.com

On Wed, 28 Jan 2009 10:32:33 -0800 (PST), maury <maury001@core.com>
wrote:

[Snipped by Lyons]

I wish you guys would stop clowning around here.
Woody Allen's situation is nothin' to laugh about.

While married to Mia Farrow, Woody and Mia adopted
a teenage Asian girl.  Next, Mia finds nude
pictures of the girl taken by Woody.

So then, Woody divorces Mia and marries the
Asian gal.  And you guys thought *YOU* had
Mother-in-Law problems!   :-)  :-)

See Ya',
[-Rick-]