It's sort of a dumb question but I must ask it. If we have access to the z-transform of a sequence, the DFT is the sampled version of z-transform evaluated at the unit circle. Consider the opposite case: if we have the DFT, is there *any* information we can deduce about the z-transform of the sequence (or the location of poles and zeroes)? _____________________________ Posted through www.DSPRelated.com

# Relationship between z and Fourier transforms

Started by ●July 30, 2013

Reply by ●July 30, 20132013-07-30

On Tue, 30 Jul 2013 19:51:15 -0500, "commsignal" <58672@dsprelated> wrote:>It's sort of a dumb question but I must ask it. If we have access to the >z-transform of a sequence, the DFT is the sampled version of z-transform >evaluated at the unit circle. Consider the opposite case: if we have the >DFT, is there *any* information we can deduce about the z-transform of the >sequence (or the location of poles and zeroes)?You know the values on the unit circle. ;)>_____________________________ >Posted through www.DSPRelated.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com

Reply by ●July 30, 20132013-07-30

On 7/30/13 6:14 PM, Eric Jacobsen wrote:> On Tue, 30 Jul 2013 19:51:15 -0500, "commsignal"<58672@dsprelated> > wrote: > >> It's sort of a dumb question but I must ask it. If we have access to the >> z-transform of a sequence, the DFT is the sampled version of z-transform >> evaluated at the unit circle. Consider the opposite case: if we have the >> DFT, is there *any* information we can deduce about the z-transform of the >> sequence (or the location of poles and zeroes)? > > You know the values on the unit circle. ;)only the values of N points equally spaced on the unit circle. remember the totality of information in the DFT is only N numbers. what goes into the Z transform or the DTFT (Discrete-Time Fourier Transform) is an infinite number of numbers (albeit countably infinite). so, if you are willing to make some assumptions (essentially one assumption) about what the samples are outside of the original N samples passed to the DFT, then you know what the Z transform is, from the DFT data. i think that usually the assumption is that outside those N samples, all of the other samples are zero. but you can make other assumptions (like the samples are periodic) and that doesn't change the DFT, it only changes the result of the ZT or the DTFT. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Reply by ●July 30, 20132013-07-30

On Tue, 30 Jul 2013 19:51:15 -0500, commsignal wrote:> It's sort of a dumb question but I must ask it. If we have access to the > z-transform of a sequence, the DFT is the sampled version of z-transform > evaluated at the unit circle. Consider the opposite case: if we have the > DFT, is there *any* information we can deduce about the z-transform of > the sequence (or the location of poles and zeroes)? > > _____________________________ > Posted through www.DSPRelated.comIf it's a truly repetitive signal (or one that is only defined over a finite span) then in theory the DFT tells you everything. If it is not repetitive, then the DFT tells you everything about the signal after it's been truncated and possibly windowed -- but information was lost in the windowing and truncating part. I'm not sure that the concept of poles and zeros has a great deal of meaning in terms of a signal. While I think you could make a case for saying it's still valid to talk about them, I feel that the concept of poles and zeros more appropriately belong to signals, not systems. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com

Reply by ●July 30, 20132013-07-30

On Tue, 30 Jul 2013 18:26:35 -0700, robert bristow-johnson <rbj@audioimagination.com> wrote:>On 7/30/13 6:14 PM, Eric Jacobsen wrote: >> On Tue, 30 Jul 2013 19:51:15 -0500, "commsignal"<58672@dsprelated> >> wrote: >> >>> It's sort of a dumb question but I must ask it. If we have access to the >>> z-transform of a sequence, the DFT is the sampled version of z-transform >>> evaluated at the unit circle. Consider the opposite case: if we have the >>> DFT, is there *any* information we can deduce about the z-transform of the >>> sequence (or the location of poles and zeroes)? >> >> You know the values on the unit circle. ;) > >only the values of N points equally spaced on the unit circle. > >remember the totality of information in the DFT is only N numbers. what >goes into the Z transform or the DTFT (Discrete-Time Fourier Transform) >is an infinite number of numbers (albeit countably infinite). > >so, if you are willing to make some assumptions (essentially one >assumption) about what the samples are outside of the original N samples >passed to the DFT, then you know what the Z transform is, from the DFT >data. i think that usually the assumption is that outside those N >samples, all of the other samples are zero. but you can make other >assumptions (like the samples are periodic) and that doesn't change the >DFT, it only changes the result of the ZT or the DTFT. > > >-- > >r b-j rbj@audioimagination.com > >"Imagination is more important than knowledge." > >It's kind of a trick question. As O&S points out, if one takes the position that the time series repeats with period N, then there is no value of z for which the z-transform converges. The DFS (or DFT, by similarity) represents samples of the z-transform on the unit circle with equal angular spacing between them, of only one instance of the input sequence, i.e., of only the input vector by itself with no periodic extension. If you know the DFT, you know equally-spaced discrete samples on the unit circle of the z-transform of the input vector. I don't know if much can be said beyond that. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com

Reply by ●July 30, 20132013-07-30

"commsignal" <58672@dsprelated> writes:> [...] > the DFT is the sampled version of z-transform...The z-transform is already in the discrete domain. It, and the z-transform, operate on sequences. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com

Reply by ●July 30, 20132013-07-30

Randy Yates <yates@digitalsignallabs.com> writes:> "commsignal" <58672@dsprelated> writes: >> [...] >> the DFT is the sampled version of z-transform... > > The z-transform is already in the discrete domain. It, and the > z-transform, operate on sequences.Clarification: The z-transform and the DFT both operate on sequences. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com

Reply by ●July 31, 20132013-07-31

>Randy Yates <yates@digitalsignallabs.com> writes: > >> "commsignal" <58672@dsprelated> writes: >>> [...] >>> the DFT is the sampled version of z-transform... >> >> The z-transform is already in the discrete domain. It, and the >> z-transform, operate on sequences. > >Clarification: The z-transform and the DFT both operate on sequences. >-- >Randy Yates >Digital Signal Labs >http://www.digitalsignallabs.com >No, z-transform is computed from a discrete-time sequence, but it is not discrete in itself. It is a function of z = r.exp(j \omega) where r and \omega can take any values. It has to be sampled at the unit circle to get the DFT. _____________________________ Posted through www.DSPRelated.com

Reply by ●July 31, 20132013-07-31

On Tue, 30 Jul 2013 20:59:37 -0500, Tim Wescott <tim@seemywebsite.please> wrote:>On Tue, 30 Jul 2013 19:51:15 -0500, commsignal wrote: > >> It's sort of a dumb question but I must ask it. If we have access to the >> z-transform of a sequence, the DFT is the sampled version of z-transform >> evaluated at the unit circle. Consider the opposite case: if we have the >> DFT, is there *any* information we can deduce about the z-transform of >> the sequence (or the location of poles and zeroes)? >> >> _____________________________ >> Posted through www.DSPRelated.com > >If it's a truly repetitive signal (or one that is only defined over a >finite span) then in theory the DFT tells you everything. > >If it is not repetitive, then the DFT tells you everything about the >signal after it's been truncated and possibly windowed -- but information >was lost in the windowing and truncating part.I disagree on the bit about losing information. It is not necessary to assume periodicity to derive the behavior of the DFT. You don't have any information about the signal outside of the window, but that's always true, whether you assume periodicity or not, with practical signals.>I'm not sure that the concept of poles and zeros has a great deal of >meaning in terms of a signal. While I think you could make a case for >saying it's still valid to talk about them, I feel that the concept of >poles and zeros more appropriately belong to signals, not systems.I agree with this. Unless the "signal" somehow is characteristic of a system, which would be pretty rare, I'd think, the utlity of determining the poles and zeros of a signal is pretty limited. Even when doing channel analysis for a signal, one could arguably use poles and zeros to analyze things like multipath reflections and channel frequency selectivity, but that's seldom done because it's not that useful.>-- >Tim Wescott >Control system and signal processing consulting >www.wescottdesign.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com

Reply by ●July 31, 20132013-07-31

"commsignal" <58672@dsprelated> writes:>>Randy Yates <yates@digitalsignallabs.com> writes: >> >>> "commsignal" <58672@dsprelated> writes: >>>> [...] >>>> the DFT is the sampled version of z-transform... >>> >>> The z-transform is already in the discrete domain. It, and the >>> z-transform, operate on sequences. >> >>Clarification: The z-transform and the DFT both operate on sequences. >>-- >>Randy Yates >>Digital Signal Labs >>http://www.digitalsignallabs.com >> > > No, z-transform is computed from a discrete-time sequence, but it is not > discrete in itself.That is true.> It is a function of z = r.exp(j \omega) where r and \omega can take > any values.Yes.> It has to be sampled at the unit circle to get the DFT.OK, I made a mistake. I thought you meant "discrete-time Fourier Transform" when you said DFT. That is because the DFT and the z-transform are, in general unrelated. You cannot relate these two functions except in special cases (I think - I'd have to work it out to be more sure). The domain of one is a subsequence of the domain of the other and they will in general give different results. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com