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Z transform notation Question

Started by westocl April 29, 2010
Hello all.

Could somebody help me interpret the meaning of this Z tranform notation.

If H(z) represented a complex frequency spectrum lets say a lowpass filter
response centered at center frequency k/M . (I hope im saying this right,
but the point is that the spectrum need not be symetric).

That is to say H(-z) represents that same spectrum except it now resides at
center frequency -k/M.

How is the notation H(-z^-1) interpreted?  Would that be the spectrum
located at the center frequency -k/M but frequency inverted such that the
shape of the magnitude (what ever it may have been flips from left to right
on the frequency axis)?  

How is H(-z^-1) interpreted in relation to H(z) of the origional spectrum
in terms of magnitude and phase?

thanks in advance
westocl wrote:
> Hello all. > > Could somebody help me interpret the meaning of this Z tranform notation. > > If H(z) represented a complex frequency spectrum lets say a lowpass filter > response centered at center frequency k/M . (I hope im saying this right, > but the point is that the spectrum need not be symetric). > > That is to say H(-z) represents that same spectrum except it now resides at > center frequency -k/M. > > How is the notation H(-z^-1) interpreted? Would that be the spectrum > located at the center frequency -k/M but frequency inverted such that the > shape of the magnitude (what ever it may have been flips from left to right > on the frequency axis)? > > How is H(-z^-1) interpreted in relation to H(z) of the origional spectrum > in terms of magnitude and phase? > > thanks in advance
Uhh, it doesn't quite work that way... A sinusoidal signal in the sampled time domain corresponds to a point z = e^(j*th) in the z domain; i.e. if you have a signal with a frequency of 0.1 radian/sample, that turns into a z-domain signal with poles at e^(0.1 * j) and e^(-0.1 * j). What are you trying to _do_? H(-z^-1) would be a system where the stability boundary is at |z| = 1, like the regular z domain, but where systems inside the boundary are unstable, and where DC is at z = -1. I don't see any advantage to it in normal usage. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
On 29 Apr, 17:47, Tim Wescott <t...@seemywebsite.now> wrote:

> What are you trying to _do_? H(-z^-1) would be a system where the > stability boundary is at |z| = 1, like the regular z domain, but where > systems inside the boundary are unstable, and where DC is at z = -1. &#2013266080;I > don't see any advantage to it in normal usage.
One way to think of H(z^-1) is as H(z), but time-reversed. One way to think of H(-z) is as H(z) modulated by a unit sinusoid of frequency fs/2 (so its values are just + / - 1). Combine the two and you have a modulated, time-reversed version of H(z). Sometimes "time"-reversal is useful in image processing if you want to run the filter H(z) forwards in "time" (e.g.left-to-right across the image) and then backwards in "time" (e.g. right-to-left across the image). Modulation can be useful if you're interested in complementary filters where | H1(z) |^2 +| H2(z)|^2 = K, a constant. With the constraint H1(z) = H2(-z) and H1 being low-pass, you can get some nice properties for sub-band coding / decomposition. Ciao, Peter K.
Peter K. wrote:
> On 29 Apr, 17:47, Tim Wescott <t...@seemywebsite.now> wrote: > >> What are you trying to _do_? H(-z^-1) would be a system where the >> stability boundary is at |z| = 1, like the regular z domain, but where >> systems inside the boundary are unstable, and where DC is at z = -1. I >> don't see any advantage to it in normal usage. > > One way to think of H(z^-1) is as H(z), but time-reversed.
Depending on your viewpoint (or your chosen region of convergence). You can choose to view 1/(z-d), |d| > 1 as a stable non-causal system whose impulse response occurs strictly before the excitation, or you can choose to view it as an unstable system.
> One way to think of H(-z) is as H(z) modulated by a unit sinusoid of > frequency fs/2 (so its values are just + / - 1). > Combine the two and you have a modulated, time-reversed version of > H(z). > > Sometimes "time"-reversal is useful in image processing if you want to > run the filter H(z) forwards in "time" (e.g.left-to-right across the > image) and then backwards in "time" (e.g. right-to-left across the > image). > > Modulation can be useful if you're interested in complementary filters > where | H1(z) |^2 +| H2(z)|^2 = K, a constant. > > With the constraint H1(z) = H2(-z) and H1 being low-pass, you can get > some nice properties for sub-band coding / decomposition.
Check to see how the author of the text you're reading wants you to view system transfer functions with poles |d| > 1. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
>Peter K. wrote: >> On 29 Apr, 17:47, Tim Wescott <t...@seemywebsite.now> wrote: >> >>> What are you trying to _do_? H(-z^-1) would be a system where the >>> stability boundary is at |z| = 1, like the regular z domain, but where >>> systems inside the boundary are unstable, and where DC is at z = -1.
I
>>> don't see any advantage to it in normal usage. >> >> One way to think of H(z^-1) is as H(z), but time-reversed. > >Depending on your viewpoint (or your chosen region of convergence). > >You can choose to view 1/(z-d), |d| > 1 as a stable non-causal system >whose impulse response occurs strictly before the excitation, or you can >choose to view it as an unstable system. > >> One way to think of H(-z) is as H(z) modulated by a unit sinusoid of >> frequency fs/2 (so its values are just + / - 1). >> Combine the two and you have a modulated, time-reversed version of >> H(z). >> >> Sometimes "time"-reversal is useful in image processing if you want to >> run the filter H(z) forwards in "time" (e.g.left-to-right across the >> image) and then backwards in "time" (e.g. right-to-left across the >> image). >> >> Modulation can be useful if you're interested in complementary filters >> where | H1(z) |^2 +| H2(z)|^2 = K, a constant. >> >> With the constraint H1(z) = H2(-z) and H1 being low-pass, you can get >> some nice properties for sub-band coding / decomposition. > >Check to see how the author of the text you're reading wants you to view >system transfer functions with poles |d| > 1. > >-- >Tim Wescott >Control system and signal processing consulting >www.wescottdesign.com >
H(e^*jw) --> H(e^(j(pi-w))) : So spectrum flip and shift by pi. So an LPF will become a HPF. But as others have said ROC/stability needs thought. Best, -k
On 29 Apr, 21:21, Tim Wescott <t...@seemywebsite.now> wrote:
> Peter K. wrote: > > > One way to think of H(z^-1) is as H(z), but time-reversed. > > Depending on your viewpoint (or your chosen region of convergence).
Definitely!
> You can choose to view 1/(z-d), |d| > 1 as a stable non-causal system > whose impulse response occurs strictly before the excitation, or you can > choose to view it as an unstable system.
Exactly. You need to specify the ROC as well.
> Check to see how the author of the text you're reading wants you to view > system transfer functions with poles |d| > 1.
Text? I'm making it up as I go along. Ciao, Peter K.