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I have a z transform $Y(z)=(z^2-z) /(z^2+1.3z+0.3)$

In this case i have two poles ,one at -0.3 and one at -1.0

I want to find inverse z transform of Y(z)

Following is my matlab code:

clc
clear all
close all
syms z n
num=[1 -1 0]
den=[1 1.3 0.3]
[r,p,k]=residuez(num,den)

We know that ztransform of Ξ΄(n) is 1 and ztransform of u[n] is 1/(1-z^-1)

Keeping value of r,p and k in mind,our partial fraction becomes: $$Y(z)= \frac{2.86}{z-(-1)} -\frac{1.86}{z-(-0.3)}$$

keeping in view that ztransform of u[n] is 1/(1-z^-1)

$y[n]=2.86(-1)^nu(n) - 1.86(-0.3)^nu(n)$
This above line is inverse ztransform for ROC |z|>1

$y[n]=-2.86(-1)^nu(-n-1)+1.86(-0.3)^nu(-n-1)$

This above line is inverse ztransform for ROC |z|<0.3

My question is that,how can i find inverse z transform for ROC 0.3<|z|<1

I have also attached a snapshot of a table of common z transform and i have higlighted two cases that i have used

enter image description here

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Since you have two poles, you have to decompose $0.3 <|z|<1$ into two ROCs, $R_A$ and $R_B$ such that: $$R_A \cap R_B \subseteq \{0.3 <|z| < 1\}$$

Your poles are $z=-0.3$, $z=-1$, so you have two choices for each ROC:

  • $|z| > 1$
  • $|z| < 1$

and

  • $|z| > 0.3$
  • $|z| < 0.3$

The requested ROC is given by $R_A \cap R_B = \{|z| < 1\}\cap\{|z| > 0.3\} = \{0.3 < |z| < 1\}$

In this case (and using your table),

$$y[𝑛]=-2.86(βˆ’1)^{𝑛}𝑒[-𝑛-1]βˆ’1.86(βˆ’0.3)^{𝑛}𝑒[𝑛]$$

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  • $\begingroup$ What do you mean by RA n RB? Do you intersection and overlap between two ROCs? $\endgroup$ – engr Mar 2 at 6:26
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    $\begingroup$ I understand you have a good reference (book) since you posted a nice table of Z transform pairs. Check the linearity property of the Z transform. That will answer your question. $\endgroup$ – GKH Mar 2 at 6:36
  • $\begingroup$ "In this case (and using your table)," which formulas/cases have you used from my attached table to write your final expression of y[n]?? $\endgroup$ – engr Mar 2 at 6:52
  • $\begingroup$ Both the highlighted ones (in yellow). $\endgroup$ – GKH Mar 2 at 8:23
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Let me give you a simple tip:

1) ROC of a right sided signal is outer to the outermost pole in its Z transform

e.g. $\mathcal{Z}[(2)^nu(n)] = \frac{z}{z-2}$ and ROC: $|z|>2 $

2) ROC of a left sided signal is inner to the innermost pole in its Z transform

e.g. $\mathcal{Z}[-(2)^nu(-n-1)] = \frac{z}{z-2}$ and ROC: $|z|<2 $

3) ROC of a two sided signal is the common area or intersection of the ROCs of its right sided term and that of the left sided term. If there is no such an intersecting ROC, then Z transform of such a two sided signal does not exist.

e.g.1) $x(n) = (\frac{1}{3})^nu(n)-(\frac{1}{2})^nu(-n-1)$

$X(z) = \dfrac{z}{z-\frac{1}{3}}+\dfrac{z}{z-\frac{1}{2}}$

First term in the above equation is right sided (ROC is $|z|>\frac{1}{3}$) and second term is left sided (ROC is $|z|<\frac{1}{2}$). So the intersection is ROC: $\frac{1}{3}<|z|<\frac{1}{2}$

enter image description here

e.g.2) $x(n) = -(\frac{1}{3})^nu(-n-1)+(\frac{1}{2})^nu(n)$

$X(z) = \dfrac{z}{z-\frac{1}{3}}+\dfrac{z}{z-\frac{1}{2}}$

First term is left sided (So ROC is $|z|<\frac{1}{3}$) and second term is right sided (ROC is $|z|>\frac{1}{2}$). But there is no intersection of these two individual ROCs and hence Z transform for e.g.2 signal doesnot exist.

enter image description here

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