"Bhaskar Bhattacharya" <bbhattac@mathworks.com> writes:> You can derive at least one version of the Z-transform from the DFT. > Compute the inverse DFT and get the samples (of the same length as the > DFT). Once you know the samples, you know the Z-transform, right?Not for my definition of z-transform. My definition requires a (countably) infinite set of input samples. So how are you going to extend the (finite) N samples you get from the DFT into a countably infinite set? Set everything else to zero? That _assumes_ they were zero to begin with - perhaps not a good assumption. I mean, come on, do we really have to make this into an academic horse-beating? You can't know what you don't know. --Randy> > - Bhaskar Bhattacharya > > "commsignal" <58672@dsprelated> wrote in message > news:PqadnVGJ5roewmXMnZ2dnUVZ_uydnZ2d@giganews.com... >> 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 >-- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com

# Relationship between z and Fourier transforms

Started by ●July 30, 2013

Reply by ●August 1, 20132013-08-01

Reply by ●August 1, 20132013-08-01

Randy Yates <yates@digitalsignallabs.com> writes:> [...] > from the DFT ...Should be "from the inverse DFT..." -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com

Reply by ●August 1, 20132013-08-01

Randy Yates <yates@digitalsignallabs.com> wrote:> "Bhaskar Bhattacharya" <bbhattac@mathworks.com> writes:>> You can derive at least one version of the Z-transform from the DFT. >> Compute the inverse DFT and get the samples (of the same length as the >> DFT). Once you know the samples, you know the Z-transform, right?> Not for my definition of z-transform. My definition requires a > (countably) infinite set of input samples. So how are you going to > extend the (finite) N samples you get from the DFT into a countably > infinite set? Set everything else to zero? That _assumes_ they were zero > to begin with - perhaps not a good assumption.> I mean, come on, do we really have to make this into an academic > horse-beating? You can't know what you don't know.But isn't this exactly the same as the Fourier transform on a finite number of points? If you do a continuous transform with delta functions at each sample point, you again have the question of how to extend it past the finite N samples. Any transform that, when inverted, goes through those N points is a valid solution. Adding periodic boundary conditions conveniently gives a finite number of frequencies instead of an infinite number. In logic terms, you make the other points "don't care" and choose the simplest solution. Seems like that would work for the Z-transform, too. -- glen

Reply by ●August 1, 20132013-08-01

On Wednesday, July 31, 2013 9:51:57 PM UTC-7, Tim Wescott wrote:> ... > I detect either a minor semantic difference or some really deep > philosophical difference. > > To me, "ignorant" implies cognition, which I do not impute to any > algorithms. >...> Tim WescottMore seriously this time, I think our semantic difference is minor. I use the definition of "ignorant" as meaning "lacking information", which is the reason for the ambiguity, as opposed to "stupid" which I would use to describe a state of limited cognition. Dale B. Dalrymple

Reply by ●August 1, 20132013-08-01

On Thu, 01 Aug 2013 08:55:44 -0700, dbd wrote:> On Wednesday, July 31, 2013 9:51:57 PM UTC-7, Tim Wescott wrote: >> ... >> I detect either a minor semantic difference or some really deep >> philosophical difference. >> >> To me, "ignorant" implies cognition, which I do not impute to any >> algorithms. >> > ... >> Tim Wescott > > More seriously this time, I think our semantic difference is minor. I > use the definition of "ignorant" as meaning "lacking information", which > is the reason for the ambiguity, as opposed to "stupid" which I would > use to describe a state of limited cognition.Hmmm. I can neither agree or disagree. I think my brain is going to have to grow a bit over this... -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●August 1, 20132013-08-01

On Thu, 01 Aug 2013 14:40:52 +0000, glen herrmannsfeldt wrote:> Randy Yates <yates@digitalsignallabs.com> wrote: >> "Bhaskar Bhattacharya" <bbhattac@mathworks.com> writes: > >>> You can derive at least one version of the Z-transform from the DFT. >>> Compute the inverse DFT and get the samples (of the same length as the >>> DFT). Once you know the samples, you know the Z-transform, right? > >> Not for my definition of z-transform. My definition requires a >> (countably) infinite set of input samples. So how are you going to >> extend the (finite) N samples you get from the DFT into a countably >> infinite set? Set everything else to zero? That _assumes_ they were >> zero to begin with - perhaps not a good assumption. > >> I mean, come on, do we really have to make this into an academic >> horse-beating? You can't know what you don't know. > > But isn't this exactly the same as the Fourier transform on a finite > number of points?The problem is ill defined. If you take a signal of infinite extent and truncate it, then you can't take the Fourier or the z transform of the original signal, because information has been lost.> If you do a continuous transform with delta functions > at each sample point, you again have the question of how to extend it > past the finite N samples. Any transform that, when inverted, goes > through those N points is a valid solution.Equally valid or equally invalid.> Adding periodic boundary conditions conveniently gives a finite number > of frequencies instead of an infinite number. In logic terms, you make > the other points "don't care" and choose the simplest solution. > > Seems like that would work for the Z-transform, too.Well, yes, and if you put periodic boundary conditions on a signal and take the z transform, then you end up with something a lot like the DFT. But you can't truncate a signal and then take the z transform of it, or any lossless transformations of the truncated signal, without having some additional knowledge of what the signal was before truncation. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●August 1, 20132013-08-01

On 8/1/13 9:46 AM, Tim Wescott wrote:> On Thu, 01 Aug 2013 14:40:52 +0000, glen herrmannsfeldt wrote: > >> Randy Yates<yates@digitalsignallabs.com> wrote: >>> "Bhaskar Bhattacharya"<bbhattac@mathworks.com> writes: >> >>>> You can derive at least one version of the Z-transform from the DFT. >>>> Compute the inverse DFT and get the samples (of the same length as the >>>> DFT). Once you know the samples, you know the Z-transform, right? >> >>> Not for my definition of z-transform. My definition requires a >>> (countably) infinite set of input samples. So how are you going to >>> extend the (finite) N samples you get from the DFT into a countably >>> infinite set? Set everything else to zero? That _assumes_ they were >>> zero to begin with - perhaps not a good assumption. >> >>> I mean, come on, do we really have to make this into an academic >>> horse-beating? You can't know what you don't know. >> >> But isn't this exactly the same as the Fourier transform on a finite >> number of points? > > The problem is ill defined. If you take a signal of infinite extent and > truncate it, then you can't take the Fourier or the z transform of the > original signal, because information has been lost. > >> If you do a continuous transform with delta functions >> at each sample point, you again have the question of how to extend it >> past the finite N samples. Any transform that, when inverted, goes >> through those N points is a valid solution. > > Equally valid or equally invalid. > >> Adding periodic boundary conditions conveniently gives a finite number >> of frequencies instead of an infinite number. In logic terms, you make >> the other points "don't care" and choose the simplest solution. >> >> Seems like that would work for the Z-transform, too. > > Well, yes, and if you put periodic boundary conditions on a signal and > take the z transform, then you end up with something a lot like the DFT. > But you can't truncate a signal and then take the z transform of it, or > any lossless transformations of the truncated signal, without having some > additional knowledge of what the signal was before truncation.perhaps another way to define the "missing samples" might be singular value decomposition. i mean, suppose you have a finite number of samples that appear to be h[n] = e^(-alpha*n) * cos(omega*n + theta) for 0 <= n < N = 0 otherwise can't (using the OP's language) you infer a pole at z = e(-alpha + j*omega) from that finite number of samples? i think there is another answer for the OP regarding the relationships of all these transforms (need mono-spaced font): Continuous Time Discrete Time .-----------------------.----------------------. | | | Discrete Frequency: | Fourier Series | DFS (a.k.a. DFT) | | | | | -- | -- | | | | Continuous Frequency: | Fourier Transform | DTFT | | | | | -- | -- | | | | Cont. Freq with | Laplace Transform | Z Transform | real "sigma" | | | '-----------------------'----------------------' ain't that the relationship between Z and Laplace and Fourier whatever? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Reply by ●August 1, 20132013-08-01

On Thu, 01 Aug 2013 11:20:46 -0700, robert bristow-johnson wrote:> On 8/1/13 9:46 AM, Tim Wescott wrote: >> On Thu, 01 Aug 2013 14:40:52 +0000, glen herrmannsfeldt wrote: >> >>> Randy Yates<yates@digitalsignallabs.com> wrote: >>>> "Bhaskar Bhattacharya"<bbhattac@mathworks.com> writes: >>> >>>>> You can derive at least one version of the Z-transform from the DFT. >>>>> Compute the inverse DFT and get the samples (of the same length as >>>>> the DFT). Once you know the samples, you know the Z-transform, >>>>> right? >>> >>>> Not for my definition of z-transform. My definition requires a >>>> (countably) infinite set of input samples. So how are you going to >>>> extend the (finite) N samples you get from the DFT into a countably >>>> infinite set? Set everything else to zero? That _assumes_ they were >>>> zero to begin with - perhaps not a good assumption. >>> >>>> I mean, come on, do we really have to make this into an academic >>>> horse-beating? You can't know what you don't know. >>> >>> But isn't this exactly the same as the Fourier transform on a finite >>> number of points? >> >> The problem is ill defined. If you take a signal of infinite extent >> and truncate it, then you can't take the Fourier or the z transform of >> the original signal, because information has been lost. >> >>> If you do a continuous transform with delta functions at each sample >>> point, you again have the question of how to extend it past the finite >>> N samples. Any transform that, when inverted, goes through those N >>> points is a valid solution. >> >> Equally valid or equally invalid. >> >>> Adding periodic boundary conditions conveniently gives a finite number >>> of frequencies instead of an infinite number. In logic terms, you make >>> the other points "don't care" and choose the simplest solution. >>> >>> Seems like that would work for the Z-transform, too. >> >> Well, yes, and if you put periodic boundary conditions on a signal and >> take the z transform, then you end up with something a lot like the >> DFT. But you can't truncate a signal and then take the z transform of >> it, or any lossless transformations of the truncated signal, without >> having some additional knowledge of what the signal was before >> truncation. > > > perhaps another way to define the "missing samples" might be singular > value decomposition. > > i mean, suppose you have a finite number of samples that appear to be > > > h[n] = e^(-alpha*n) * cos(omega*n + theta) for 0 <= n < N > > = 0 otherwise > > > can't (using the OP's language) you infer a pole at > > z = e(-alpha + j*omega) > > from that finite number of samples? > > i think there is another answer for the OP regarding the relationships > of all these transforms (need mono-spaced font): > > > > Continuous Time Discrete Time > .-----------------------.----------------------. > | | > | > Discrete Frequency: | Fourier Series | DFS (a.k.a. DFT) > | > | | > | > | -- | -- > | > | | > | > Continuous Frequency: | Fourier Transform | DTFT > | > | | > | > | -- | -- > | > | | > | > Cont. Freq with | Laplace Transform | Z Transform > | > real "sigma" | | > | >'-----------------------'----------------------'> > > ain't that the relationship between Z and Laplace and Fourier whatever?Your classification system works for me (if not for my newsreader, which seems to have mangled your nice table). -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●August 1, 20132013-08-01

On 8/1/13 12:17 PM, Tim Wescott wrote:> > Your classification system works for me (if not for my newsreader, which > seems to have mangled your nice table). >careful, Tim. you might inadvertently have taken a side in "_That_ argument". note the upper right corner of the table. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Reply by ●August 1, 20132013-08-01

On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:> On 8/1/13 12:17 PM, Tim Wescott wrote: >> >> Your classification system works for me (if not for my newsreader, >> which seems to have mangled your nice table). >> >> > careful, Tim. you might inadvertently have taken a side in "_That_ > argument". note the upper right corner of the table.The only side I take in _that_ argument is that both sides are being remarkably inflexible. The math just is, and if you're getting the right results you are, ipso facto, using it correctly. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com