Pole/Zero Plots on the Z-Plane, ROC

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

On Mar 7, 10:55=A0am, "Nasser M. Abbasi" <n...@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 =3D z/( =A0(z-1/2)*(z+3/4) )
> Then they show the ROC. But I do not understand how they got the ROC to b=
e
> as shown
>
> http://cnx.org/content/m10556/latest/
>
> If I do partial fractions, I get
>
> =A0 =A0 =A0 =A0 =A0 =A0 H(z)=3D -(4/5) =A01/(1-2 z) =A0+ 4/5 =A01/(1+4/3 =
z).
>
> So, I get ROC: =A0|z|<1/2 =A0from 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 shou=
ld
> 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>=3D0
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

"HardySpicer" <gyansorova@gmail.com> wrote in message 
news:99fe0d62-b7bf-435c-b475-95edd257feca@c34g2000pri.googlegroups.com...
On Mar 7, 10:55 am, "Nasser M. Abbasi" <n...@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 

On Mar 7, 2:02&#4294967295;pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> "HardySpicer" <gyansor...@gmail.com> wrote in message
>
> news:99fe0d62-b7bf-435c-b475-95edd257feca@c34g2000pri.googlegroups.com...
> On Mar 7, 10:55 am, "Nasser M. Abbasi" <n...@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 &#4294967295;|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
>
> &#4294967295; &#4294967295; 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

On 6 Mar, 22:55, "Nasser M. Abbasi" <n...@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

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

|z|> - 3/4 is trivial.  Did you really mean to write that?

"Michael Plante" <michael.plante@gmail.com> wrote in message 
news:5rKdneT6-9aPtwnWnZ2dnUVZ_tCdnZ2d@giganews.com...
> Nasser M. Abbasi wrote:
>>So, I get ROC:  |z|<1/2  from the first term, and |z|> - 3/4 from the
>
> |z|> - 3/4 is trivial.  Did you really mean to write that?
>

no, that does not make sense. I meant |z|>3/4. Why did I write that?

On this problem, which is example 4 below,

"function = z/(  (z-1/2)*(z+3/4) )
http://cnx.org/content/m10556/latest/"

I think the problem is that this H(z) can have 2 ROC's, and the diagram 
shown above is for an ROC  |z|>3/4, but it is also possible to choose ROC 
|z|<1/2, and so, the diagram shown is correct, becuase they selected |z|>3/4 
and I was only thinking the ROC must be |z|<1/2.

H(z) does not have a unique ROC. So, my mistake, I forgot that. Better not 
forget this for the exam.

--Nasser 

On Mar 8, 2:37&#4294967295;pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> "Michael Plante" <michael.pla...@gmail.com> wrote in message
>
> news:5rKdneT6-9aPtwnWnZ2dnUVZ_tCdnZ2d@giganews.com...
>
> > Nasser M. Abbasi wrote:
> >>So, I get ROC: &#4294967295;|z|<1/2 &#4294967295;from the first term, and |z|> - 3/4 from the
>
> > |z|> - 3/4 is trivial. &#4294967295;Did you really mean to write that?
>
> no, that does not make sense. I meant |z|>3/4. Why did I write that?
>
> On this problem, which is example 4 below,
>
> "function = z/( &#4294967295;(z-1/2)*(z+3/4) )http://cnx.org/content/m10556/latest/"
>
> I think the problem is that this H(z) can have 2 ROC's, and the diagram
> shown above is for an ROC &#4294967295;|z|>3/4, but it is also possible to choose ROC
> |z|<1/2, and so, the diagram shown is correct, becuase they selected |z|>3/4
> and I was only thinking the ROC must be |z|<1/2.
>
> H(z) does not have a unique ROC. So, my mistake, I forgot that. Better not
> forget this for the exam.
>
> --Nasser

There is nothing wrong with the web page. The system is stable that
you describe. The poles are within the unit circle!!


Hardy

"HardySpicer" <gyansorova@gmail.com> wrote in message 
news:4a70141d-680f-4031-93d3-


"There is nothing wrong with the web page. The system is stable that
you describe. The poles are within the unit circle!!

Hardy"

You seem to be fixated by poles inside a unit circle meaning stability.

Look at   x(n)=(1/2)^n  u(n), where u(n) is the unit step function. This has 
X(z)=1/(1- 1/2 z^-1), with pole at z=1/2
Look at   x(n)=(1/2)^n  u(-n-1), this has X(z)=1/(1- 2 z), with pole at 
z=1/2

Both have a pole inside the unit circle, so are you saying they are both 
stable systems?

--Nasser 

On Mar 9, 12:02&#4294967295;pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> "HardySpicer" <gyansor...@gmail.com> wrote in message
>
> news:4a70141d-680f-4031-93d3-
>
> "There is nothing wrong with the web page. The system is stable that
> you describe. The poles are within the unit circle!!
>
> Hardy"
>
> You seem to be fixated by poles inside a unit circle meaning stability.
>
> Look at &#4294967295; x(n)=(1/2)^n &#4294967295;u(n), where u(n) is the unit step function. This has
> X(z)=1/(1- 1/2 z^-1), with pole at z=1/2
> Look at &#4294967295; x(n)=(1/2)^n &#4294967295;u(-n-1), this has X(z)=1/(1- 2 z), with pole at
> z=1/2
>
> Both have a pole inside the unit circle, so are you saying they are both
> stable systems?
>
> --Nasser

You are seriously messed up here.


>Look at   x(n)=(1/2)^n  u(-n-1),

Is stable if time is going forward. since 0.5^n dies out.

0.5^-n grows bigger of course in negative time - unstable. I assume u(-
n-1) means uncausal in your nomeclature?


Therefore adopt a convention ie let all z-transforms be expressed in
positive powers of z. So for a casual TF F(z)  with poles within
mod(z)=1 are stable and unstable if outside.
For a TF F(z^-1) it is  defined as uncausal and does not exist for
time n>=0! For n<0 it will be stable iff its poles of z  lie OUTSIDE
the unit circle.
Now you won't come across the uncausal variety much except in Wiener
filtering and the like where we split a TF that exists for BOTH n<0
and n>=0 into causal and uncausal TFs..

Hardy