Re: ZTransform for a non-causal unstable signal. How to make Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg108233] Re: ZTransform for a non-causal unstable signal. How to make Mathematica
- From: dh <dh at metrohm.com>
- Date: Thu, 11 Mar 2010 06:36:39 -0500 (EST)
- References: <hn7f5s$2u4$1@smc.vnet.net> <4B97560C.7040902@metrohm.com> <014b01cac084$e1fa1700$2c8cfea9@computerh20djr>
Hi Nasser, the result from Mathematica is: -(2/(-2 + z)) The inverse z transform of this: -2^n UnitStep[-1 + n] now you can write this as: - (1/2)^(-n) UnitStep[n-1] what is consistent with where we started. However, if we back transform your "hand" result, we get: (-(1/2))^n Binomial[-1, n] what does not look like your initial expression. Daniel On 10.03.2010 20:07, Nasser M. Abbasi wrote: >>> On 10.03.2010 07:45, Nasser M. Abbasi wrote: >>> >>> I am trying to find the Z transform using Mathematica for the signal >>> >>> (-(1/2)^n)*UnitStep[-n - 1]; >>> > > >> 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 >> > > > I had a second look at the above solution, but the result obtained still > does not match my hand calculation actually, Here is the output from the > above code: > > -(2/(-2 + z)) > > Which can be rewritten as > > 1 > ------------- > 1- 1/2 z > > In addition, when using GenerateConditions -> True option, I get that |z|>2 > > *But* the result I should get is > > 1 > ------------- > 1- 1/2 z^-1 > > With |z|<1/2. Notice the z^-1 vs z in the denominator. > > so, more tweaking is needed I think. getting close, but Just changing > the "n" from +n to "-n" did not seem to do the trick? > > I need to look more into this. But at least now I know what the correct > definition of Ztransform used by Mathematica is. All of this could have > been avoided if the Ztransform was defined to be 2-sided. Now each time > I have a left sided signal, I have to play tricks to feed it to the > Ztransform. Also, If I have 2 sided signal, I have to break it up and > play more tricks, which can lead to errors. > > --Nasser > > > > -- Daniel Huber Metrohm AG International Headquarters Oberdorfstr. 68, CH-9101 Herisau / Switzerland Phone +41 71 353 8606, Fax +41 71 353 89 01 Mail <mailto:dh at metrohm.com> Web <http://www.metrohm.com