Suppose I have $x[n]$ and $y[n]$, and I calculate their respective Z-transforms $X(z)$ and $Y(z)$ as well as their respective ROCs.
Calculating $H(z)$ is as simple as calculating the quotient of $\frac{Y(z)}{X(z)}$, but how do I get $H(z)$'s ROC?
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Sign up to join this communitySuppose I have $x[n]$ and $y[n]$, and I calculate their respective Z-transforms $X(z)$ and $Y(z)$ as well as their respective ROCs.
Calculating $H(z)$ is as simple as calculating the quotient of $\frac{Y(z)}{X(z)}$, but how do I get $H(z)$'s ROC?
The ROC of $Y(z)=X(z)H(z)$ is the intersection of the ROCs of $X(z)$ and $H(z)$, unless there are pole-zero cancellations, which change the ROC of the product. In the given example we have to take into account the pole-zero cancellation at $z_0=2$. We can define a new function $X'(z)=X(z)(1-2z^{-1})$ with the pole at $z_0=2$ removed. We do not need to define a new $H'(z)$ with the corresponding zero removed, because that zero doesn't influence the ROC anymore (since the corresponding pole has already been removed). The ROC of $Y(z)$ is now the intersection of the ROCs of $X'(z)$ ($|z|>\frac13$) and $H(z)$. Consequently, the only possible ROC for $H(z)$ is $|z|>\frac23$. This means that the corresponding system is causal and stable, because the ROC is outside the pole with the largest magnitude and the ROC contains the unit circle.