# Z transform notation Question

Started by 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?

```
```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?
>

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.

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.
>
>
>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.
>

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.

```