Re: ZTransform for a non-causal unstable signal. How to make Mathematica gives correct result?
- To: mathgroup at smc.vnet.net
- Subject: [mg108221] Re: ZTransform for a non-causal unstable signal. How to make Mathematica gives correct result?
- From: "Nasser M. Abbasi" <nma at 12000.org>
- Date: Thu, 11 Mar 2010 06:34:22 -0500 (EST)
- References: <hn7f5s$2u4$1@smc.vnet.net> <4B97560C.7040902@metrohm.com>
- Reply-to: "Nasser M. Abbasi" <nma at 12000.org>
>>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