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Z transform - choice of z?

Started by Unknown September 23, 2005
>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?