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Given the $\mathcal Z$-transform of input $x[n]$ and output $y[n]$, how can I find the ROC of the system function $H(z) = Y(z)/X(z)$? I have $$X(z) = \frac{2z\left(z-\frac{10}{3}\right)}{\left(z-\frac 13\right)(z-3)},\quad\text{ROC}: \frac 13 < |z| < 3$$ and $$Y(z) = \frac{-1}{4}\frac{z}{\left(z-\frac 13\right)\left(z-\frac 14\right)},\quad\text{ROC}: |z| > \frac 13$$

If I substitute $\hat{X}(z) = 1/X(z)$, then $ H(z) = Y(z)\hat{X}(z)$. Then, I know that the ROC of $H(z)$ contains the intersection of the ROCs of $Y(z)$ and $\hat{X}(z)$. But how can I find the ROC of $\hat{X}(z)$? I don't know if the corresponding sequence $\hat{x}[n]$ is causal or not.

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The transfer function is

$$H(z)=\frac{Y(z)}{X(z)}=-\frac18\frac{z-3}{\left(z-\frac14\right)(z-\frac{10}{3})}\tag{1}$$

It has three possible ROCs:

  1. $|z|<\frac14$ (left-sided, unstable)
  2. $\frac14<|z|<\frac{10}{3}$ (two-sided, stable)
  3. $|z|>\frac{10}{3}$ (right-sided, unstable)

Only the second ROC, corresponding to a two-sided and stable impulse response, overlaps with the ROC of $X(z)$. The intersection, and, consequently, the ROC of $Y(z)$ would be $\frac13<|z|<3$ (equal to the ROC of $X(z)$). However, the pole at $z=3$ is removed by pole-zero cancellation, which results in the final ROC $|z|>\frac13$ for $Y(z)$.

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  • $\begingroup$ Thanks.But shouldn't the ROC of H(z) overlap with the ROC of 1/X(z)? Is the ROC of 1/X(z) same as that of X(z)? $\endgroup$ – ryan80 Oct 24 '16 at 19:33
  • $\begingroup$ @viz: In this case it's easier to look at the relation $Y(z)=H(z)X(z)$ to see that the ROC of $Y(z)$ is the intersection of the ROCs of $X(z)$ and $H(z)$ (apart from pole-zero cancellations). $\endgroup$ – Matt L. Oct 24 '16 at 20:27
  • $\begingroup$ @MattL. Sorry, I know this answer is old, but could you please elaborate on why the first ROC is unstable even though it is inside the unit circle? And similarly, why is 2 stable even though it extends beyond the unit circle? Further, why do you make the assumption that $H(z)$ is stable? $\endgroup$ – Val9265 Apr 1 '17 at 9:45
  • $\begingroup$ @Val9265: A system is stable if the ROC of its Z-transform includes the unit circle. The ROC $|z|<\frac14$ does not include the unit circle. The unit circle is outside the ROC in this case. $\endgroup$ – Matt L. Apr 1 '17 at 11:20

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