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From: Nasser M. Abbasi on 6 Mar 2010 16:55

On this page below, titled "Understanding Pole/Zero Plots on the Z-Plane",
if you look at example 4 (near the end of the page), they show this H(z)
function = z/( (z-1/2)*(z+3/4) )
Then they show the ROC. But I do not understand how they got the ROC to be
as shown

http://cnx.org/content/m10556/latest/

If I do partial fractions, I get

H(z)= -(4/5) 1/(1-2 z) + 4/5 1/(1+4/3 z).

So, I get ROC: |z|<1/2 from the first term, and |z|> - 3/4 from the second
term.

So, it is a left sided signal (notice it is z not z^-1), and the ROC should
be inside the circle, not to the outside as shown.

Is this page wrong, or Am I missing something?

(I am learning poles/zeros and z-transforms, yet again).

--Nasser


From: HardySpicer on 6 Mar 2010 17:25
On Mar 7, 10:55 am, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
> On this page below, titled "Understanding Pole/Zero Plots on the Z-Plane",
> if you look at example 4 (near the end of the page), they show this H(z)
> function = z/(  (z-1/2)*(z+3/4) )
> Then they show the ROC. But I do not understand how they got the ROC to be
> as shown
>
> http://cnx.org/content/m10556/latest/
>
> If I do partial fractions, I get
>
>             H(z)= -(4/5)  1/(1-2 z)  + 4/5  1/(1+4/3 z).
>
> So, I get ROC:  |z|<1/2  from the first term, and |z|> - 3/4 from the second
> term.
>
> So, it is a left sided signal (notice it is z not z^-1), and the ROC should
> be inside the circle, not to the outside as shown.
>
> Is this page wrong, or Am I missing something?
>
> (I am learning poles/zeros and z-transforms, yet again).
>
> --Nasser

All I can tell you is that the TF you quote is a stable TF with both
poles within the unit circle.

What do you mean left handed signal? Are you confusing non-causality
with stability?
Two sided z-transforms are a little different in that it depends how
you define z. I normally say that F(z) is stable for f(k),k>=0
provided all poles lie within mod(z)<1.

You can also define a non-causal signal f(k),k<0. Understand causal
signals first. For example the TF z/(z-2) is unstable for k>0 but
stable for k<0!


Hardy
From: Nasser M. Abbasi on 6 Mar 2010 20:02

"HardySpicer" <gyansorova(a)gmail.com> wrote in message
news:99fe0d62-b7bf-435c-b475-95edd257feca(a)c34g2000pri.googlegroups.com...
On Mar 7, 10:55 am, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
> On this page below, titled "Understanding Pole/Zero Plots on the Z-Plane",
> if you look at example 4 (near the end of the page), they show this H(z)
> function = z/( (z-1/2)*(z+3/4) )
> Then they show the ROC. But I do not understand how they got the ROC to be
> as shown
>
> http://cnx.org/content/m10556/latest/
>
> If I do partial fractions, I get
>
> H(z)= -(4/5) 1/(1-2 z) + 4/5 1/(1+4/3 z).
>
> So, I get ROC: |z|<1/2 from the first term, and |z|> - 3/4 from the second
> term.
>
> So, it is a left sided signal (notice it is z not z^-1), and the ROC
> should
> be inside the circle, not to the outside as shown.
>
> Is this page wrong, or Am I missing something?
>
> (I am learning poles/zeros and z-transforms, yet again).
>
> --Nasser

"All I can tell you is that the TF you quote is a stable TF with both poles
within the unit circle."

But that is the point. It is NOT stable. I get ROC |Z|<1/2 and |Z|>-3/4,
therefore the ROC does NOT include the unit circle, hence not BIBO stable.
Just becuase the poles inside the unit circle does not mean it is stable.
the ROC has to include the unit circle for it to be stable.

How did you determine it is stable? May be I am doing something wrong. WHat
is the ROC that you get for

z/( (z-1/2)*(z+3/4) )

?

"What do you mean left handed signal? Are you confusing non-causality with
stability?"

Left handed signal is one which has x(n)=0 for all values of n above some
value, say N.


--Nasser


From: HardySpicer on 6 Mar 2010 23:02
On Mar 7, 2:02 pm, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
> "HardySpicer" <gyansor...(a)gmail.com> wrote in message
>
> news:99fe0d62-b7bf-435c-b475-95edd257feca(a)c34g2000pri.googlegroups.com...
> On Mar 7, 10:55 am, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
>
>
>
> > On this page below, titled "Understanding Pole/Zero Plots on the Z-Plane",
> > if you look at example 4 (near the end of the page), they show this H(z)
> > function = z/( (z-1/2)*(z+3/4) )
> > Then they show the ROC. But I do not understand how they got the ROC to be
> > as shown
>
> >http://cnx.org/content/m10556/latest/
>
> > If I do partial fractions, I get
>
> > H(z)= -(4/5) 1/(1-2 z) + 4/5 1/(1+4/3 z).
>
> > So, I get ROC: |z|<1/2 from the first term, and |z|> - 3/4 from the second
> > term.
>
> > So, it is a left sided signal (notice it is z not z^-1), and the ROC
> > should
> > be inside the circle, not to the outside as shown.
>
> > Is this page wrong, or Am I missing something?
>
> > (I am learning poles/zeros and z-transforms, yet again).
>
> > --Nasser
>
> "All I can tell you is that the TF you quote is a stable TF with both poles
> within the unit circle."
>
> But that is the point. It is NOT stable. I get ROC |Z|<1/2 and  |Z|>-3/4,
> therefore the ROC does NOT include the unit circle, hence not BIBO stable..
> Just becuase the poles inside the unit circle does not mean it is stable.
> the ROC has to include the unit circle for it to be stable.
>
> How did you determine it is stable? May be I am doing something wrong. WHat
> is the ROC that you get for
>
>     z/( (z-1/2)*(z+3/4) )
>
> ?
>
> "What do you mean left handed signal? Are you confusing non-causality with
> stability?"
>
> Left handed signal is one which has x(n)=0 for all values of n above some
> value, say N.
>
> --Nasser

The poles of the TF are at z= +0.5 andz= -0.75 which are both within
the unit circle in the z-plane.

Of course if you define the 1/z plane then it is unstable because the
poles lie outside of the 1/z plane.
assumeing that the impulse response responds in positive time f(k)=0
for k<0 then the convention used is that the z-plane is used and not
the 1/z plane.

If you don't believe me then find the impulse resonse of your system
for k>=0 and you will find that it dies out with time and does not
grow bigger.
In other words there is convergence. Any causal TF with poles within
the unit circle of the z-plane is stable. Poles need not be ON the
unit circle as you are suggesting.

Hardy
From: Rune Allnor on 7 Mar 2010 04:44
On 6 Mar, 22:55, "Nasser M. Abbasi" <n...(a)12000.org> wrote:

> Is this page wrong, or Am I missing something?

I haven't looked through the maths, so I can't tell if there
are errors or flaws on that page.

However, it seems you have missed some essential point about
Z transforms (ZTs). I don't know which, since I am ridiculously
poor at reading other people's minds, but I can guess:

1) The exact details of the z transform depend on context.
Unfortunatey, there are two conventions in use, one from
seismology, where the ZT is expressed in terms of z,
and another, everywhere else, where the ZT is expressed
in terms of z^-1. If you are not aware of thus, make sure
you review the definition of the ZT to determine which
form you are working with.

2) With the ZT it is the pole location wrt the unit circle
that determines the stability of the system. With the
Laplace transform (LT) stability is discussed wrt the
real s axis. Make sure you do not bring in left/right
half-plane arguments, that belong in the LT world, into
a discussion about ZTs.

Rune
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