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[?] Inverse Z-Transform of z/(z - a) with different region of convergence?

Gennaro Arguzzi
Posted 8 years ago

Hello everyone. I tried to get the inverse Z transform of z/(z - a) with different ROC.

InverseZTransform[z/(z - a), z, n, Assumptions -> Abs[z] > Abs[a]]

InverseZTransform[z/(z - a), z, n, Assumptions -> Abs[z] < Abs[a]]

Both cases give me the following output a^n. Actually the inverse Z transform is

(a^n) HeavisideTheta[n] when ROC is Abs[z] > Abs[a],

and is

(-a^n) HeavisideTheta[-n-1] when ROC is Abs[z] < Abs[a].

How can I get these outputs?

Thank you very much.
POSTED BY: Gennaro Arguzzi
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As Mathematica defines it InverseZTransform

The inverse Z transform of a function F(z) is given by the contour integral 1/(2\[Pi] i) \[ContourIntegral]F(z)z^(n-1) \[DifferentialD]z.

it seems to me meaningless - or even inadequate - to define a region of convergence like Abs[a] > Abs[z] because the contour has to walk a closed path around the singularity $a = z$ in the complex plane. It seems that Mathematica disregards assumptions of such type:

In[4]:= InverseZTransform[z/(z - a), z, n]
Out[4]= a^n
POSTED BY: Udo Krause

Hello @Udo Krause , see the table of my professor please:

http://www.iet.unipi.it/m.greco/esami_lab/bio/Trasformate%20zeta%20notevoli.pdf

On my book (it is a McGraw-Hill's book) there is the proof of what I said. I found an example here (examples 2 and 3): https://en.wikipedia.org/wiki/Z-transform#Example_2_.28causal_ROC.29
POSTED BY: Gennaro Arguzzi

Mathematica implements the unilateral Z-transform

In[20]:= ZTransform[a^n HeavisideTheta[n], n, z,  GenerateConditions -> True]
Out[20]= ConditionalExpression[-(z/(a - z)), Abs[z] > Abs[a]]

this is Professor's result, but that is not

In[26]:= ZTransform[-a^n HeavisideTheta[-n - 1], n, z, GenerateConditions -> True]
Out[26]= 0

nevertheless correct under unilateral defintion.
POSTED BY: Udo Krause

@Udo Krause can you explain me why the unilateral z transform of the anticausal sequences does not exist? Thank you very much.
POSTED BY: Gennaro Arguzzi

Okay, the assumptions give the area of definition of the ZTransform results.
POSTED BY: Udo Krause

Because the unilateral ZTransform $Z(f[n]) ) = \sum_{n=0}^{\infty}f[n]z^{-n}$ with $f[n]=-a^n u(-n-1)$ has $f[n]=0, n\geq0$ because the Heaviside function $u(-n-1) = 1$ only for $n<-1$.

But this still does not address your original question, why the InverseZTransform disregards the area of definition given in the assumptions. A dull and unbelievable explanation could be: it knows it has to handle only input wich was generated by the unilateral ZTransform[] - but that's so absurd that I wish I hadn't typed it ... just for discussion: You know you can compute contour integrals with Mathematica specifying the path on your own.
POSTED BY: Udo Krause
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