Z transform - choice of z?

>From scratching around on the internet and reading Hamming's Digital
Filters book it appears that the z transform is used pretty much only
with z = exp(iw), which means it works like a discrete Fourier
transform.

I was wondering whether the z transform is ever used with z =
exp(iw+y). This seems to be remarkably useful - I wonder if it is
impractical to compute for a given signal, therefore not used in
practice.

spasmous@yahoo.com wrote:

>>From scratching around on the internet and reading Hamming's Digital
> Filters book it appears that the z transform is used pretty much only
> with z = exp(iw), which means it works like a discrete Fourier
> transform.
> 
> I was wondering whether the z transform is ever used with z =
> exp(iw+y). This seems to be remarkably useful - I wonder if it is
> impractical to compute for a given signal, therefore not used in
> practice.
> 
In control systems the z transform is used as z = exp(sT), with T = 
sampling rate & s being the frequency variable in the Laplace transform. 
  It has some conversion issues that the mathematicians quibble about, 
but it works just fine so there must be some way of dancing around them...

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

<spasmous@yahoo.com> wrote in message
news:1127513981.711359.232890@g44g2000cwa.googlegroups.com...
> >From scratching around on the internet and reading Hamming's Digital
> Filters book it appears that the z transform is used pretty much only
> with z = exp(iw), which means it works like a discrete Fourier
> transform.
>
> I was wondering whether the z transform is ever used with z =
> exp(iw+y). This seems to be remarkably useful - I wonder if it is
> impractical to compute for a given signal, therefore not used in
> practice.
>
The exact mapping from s to z is z=exp(sT) where s=sigma+j omega. So this
gives a complex z.

Any other transforms are approximates ie Bilinear TF,forward or backward
difference etc. The reason that z=exp(sT) is not normally used is that to go
from s to z would involve a log multi valued function - so we use a transfer
function approximation. Everything in z is approximate - only analogue is
exact (assuming we start from analogue that is).Maybe if we sampled a
different way we could get a different expression from z to s which was
rational - until that time we are stuck with the above.

Shytot

Shytot wrote:
> <spasmous@yahoo.com> wrote in message
> news:1127513981.711359.232890@g44g2000cwa.googlegroups.com...
> > >From scratching around on the internet and reading Hamming's Digital
> > Filters book it appears that the z transform is used pretty much only
> > with z = exp(iw), which means it works like a discrete Fourier
> > transform.
> >
> > I was wondering whether the z transform is ever used with z =
> > exp(iw+y). This seems to be remarkably useful - I wonder if it is
> > impractical to compute for a given signal, therefore not used in
> > practice.
> >
> The exact mapping from s to z is z=exp(sT) where s=sigma+j omega. So this
> gives a complex z.
>



Right. Here's the thing: I have a finite length time signal sampled at,
say, 256 times-points that is composed of different frequencies which
decay (exponentially) at different rates. I can FFT the signal to see
the frequency components however due to the decay the peaks are
broadened. Both the frequency spectrum and decay rates are interesting
parameters.

Seems the Z transform offers a way of prising the freq and decay out of
the signal.

In fact I'm a little confused about the uniqueness of the Z transform.
Eg. for the case of a pure sinusoidal variation undergoing exp decay,
the FFT will give you a broad spectrum indicating several freqs are
present. The Z transform *should* give you a single peak in the complex
plane telling you the freq and decay rate. However, it is also valid to
represent the signal as a collection of non-decaying frequencies as per
the FFT. How could the Z transform distinguish the two cases?