Re: ZTransform for a non-causal unstable signal. How to make Mathematica gives correct result?
- To: mathgroup at smc.vnet.net
- Subject: [mg108216] Re: ZTransform for a non-causal unstable signal. How to make Mathematica gives correct result?
- From: "Nasser M. Abbasi" <nma at 12000.org>
- Date: Wed, 10 Mar 2010 06:30:39 -0500 (EST)
- References: <hn7f5s$2u4$1@smc.vnet.net> <4B97560C.7040902@metrohm.com>
- Reply-to: "Nasser M. Abbasi" <nma at 12000.org>
Thanks Daniel; I really really did look at the definition, quickly, and saw it as saying -infinity to infinity, but now I looked again, with larger magnification, and it is 0 to infinity. Clearly I need better glasses or make the fonts larger on my screen. I think this is a bit unfortunate how the Z transform is defined, because the "more" standard or common way is from -infinity to infinity. But my error, ok. I'll live with this definition, as long as I can get the correct result. thanks, --Nasser ----- Original Message ----- From: "dh" <dh at metrohm.com> Sent: Wednesday, March 10, 2010 12:19 AM Subject: [mg108216] Re: ZTransform for a non-causal unstable signal. How to make Mathematica gives correct result? > 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> >