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Re: ZTransform for a non-causal unstable signal. How to make Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108212] Re: ZTransform for a non-causal unstable signal. How to make Mathematica
  • From: dh <dh at metrohm.com>
  • Date: Wed, 10 Mar 2010 06:29:53 -0500 (EST)
  • References: <hn7f5s$2u4$1@smc.vnet.net>

Hi Nasser,
Mathematica calculates the unilateral Z transfrom. This is clearly zero in your 
case. But in your case you may get the bilateral by setting n -> -n in f:

f = (-(1/2)^(-n))*UnitStep[n - 1];
ZTransform[f, n, z]

Daniel

On 10.03.2010 07:45, Nasser M. Abbasi wrote:
> Hello;
>
> I am trying to find the Z transform using Mathematica for the signal
>
>      (-(1/2)^n)*UnitStep[-n - 1];
>
> This signal runs backward, from n=-1 to - infinity. It is not stable. But
> for |z|<1/2, the Z transform converges and can be found to be
>                 1/(1-  0.5 z^-1)
>
> Which is what I was hoping Mathematica to return, but it returns back zero !
>
> In[32]:= Clear[f, n, z]
>
> f = (-(1/2)^n)*UnitStep[-n - 1];
> ZTransform[f, n, z]
>
> Out[34]= 0
>
> I did the derivation by hand, and here is a link. I do not think I made a
> mistake myself in the hand derivation? but was hoping to use Mathematica to
> verify it. but not sure how to let Mathematica return back the same answer
> as I have.
>
> http://12000.org/tmp/030910/ztran.htm
>
> Am I missing something here? do I need to use some assumptions for
> Ztransform? did not have to when causal signals?
>
> --Nasser
>
>
>


-- 

Daniel Huber
Metrohm Ltd.
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CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh at metrohm.com>
Internet:<http://www.metrohm.com>



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