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. Oberdorfstr. 68 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>