# Region of Convergence

In attached 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.

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 $$y[n]$$.
• $x[n] = a^n u[n]$ 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. – Fat32 Apr 2 '19 at 9:47