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In attached ROC image why does the ROC have these values for $$ X(z) = \frac{1}{1-\frac{1}{3}z^{-1}} - \frac{1}{1-2z^{-1}} ~~~~~,~~~~~ 1/3 < |z| < 2 $$

and for $$ Y(z) = \frac{5}{1-\frac{1}{3}z^{-1}} - \frac{5}{1- \frac{2}{3}z^{-1}} ~~~~~,~~~~~ 2/3< |z| $$

Also why is ROC for $X(z)$ between 2 values.

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According to the properties of Z-transform, the ROC of sum of two signals is the intersection of their individual ROCs.

If you look at the signal $$ x[n] = (\frac{1}{3})^n u[n] + (2)^n u[-n-1] $$

you see that it consists of two signals; one right sided $(\frac{1}{3})^n u[n]$ with a pole at $z = 1/3$ and whose Z-transform is $X(z) = 1/(1 - (1/3)z^{-1})$ and with region of convergence ROC1: $ |z| > 1/3$ and the other is a left sided one $(2)^n u[-n-1]$ with a pole at $z=2$ whose Z-transform is $-1/(1-2z^{-1})$ and with ROC2 as $ |z| < 2$. Therefore, the intersection of their ROCs is $$ROC_x = ROC1 \cap ROC2 = \{ 1/3 < |z| \} \cap \{|z| < 2 \} = \{ 1/3 < |z| < 2 \} $$

similarly for the signal $y[n]$.

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  • $\begingroup$ Hi thanks for your answer. Can you please comment on the following: 1) how to determine the > or < sign for ROC. 2)Also, what determines whether a system is right- or left-sided. 3)and when is a^n u[n] term & a^n [-n-1] are used ? $\endgroup$ – Leo Apr 2 at 0:25
  • $\begingroup$ $x[n] = a^n u[n]$ is a causal, right sided sequence whereas $x[n] = a^n u[-n-1]$ is a anti-causal left sided sequence. Right sided sequences have ROC |z|>pole_max and left sided sequences have ROC |z|<pole_min. $\endgroup$ – Fat32 Apr 2 at 9:47

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