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From: Michael Plante on 7 Mar 2010 17:34
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?

From: Nasser M. Abbasi on 7 Mar 2010 20:37

"Michael Plante" <michael.plante(a)gmail.com> wrote in message
news:5rKdneT6-9aPtwnWnZ2dnUVZ_tCdnZ2d(a)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


From: HardySpicer on 8 Mar 2010 17:26
On Mar 8, 2:37 pm, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
> "Michael Plante" <michael.pla...(a)gmail.com> wrote in message
>
> news:5rKdneT6-9aPtwnWnZ2dnUVZ_tCdnZ2d(a)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

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


Hardy
From: Nasser M. Abbasi on 8 Mar 2010 18:02

"HardySpicer" <gyansorova(a)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


From: HardySpicer on 9 Mar 2010 18:06
On Mar 9, 12:02 pm, "Nasser M. Abbasi" <n...(a)12000.org> wrote:
> "HardySpicer" <gyansor...(a)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

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
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