doubt of intersection of ROC Z Transform

Question

dear friends of StackExchange. I have a doubt of the intersection of two ROC. I have H(Z), X(Z) and and i have to determine:

$$ \begin{align} Y(Z)= H(Z)X(Z)\end{align}$$ $$ \displaystyle $$

$ ROC $ H(Z) $ \cap $ $ ROC $ X(Z)
$$ \displaystyle $$ poles of $H(Z): p1= \left | -\frac{1}{4} \right | $ ; $p2= \left | \frac{3}{2} \right |$ $$ \displaystyle $$

The system is supposed stable. Ops, sorry i forgot to write it in the initial post, under this assumption the corresponding $ROC$ of $H(Z)$ is : $$\begin{align} \quad&\ \left | \frac{1}{4} \right | <z< \left | \frac{3}{2} \right | \end{align}$$ $$ \displaystyle $$ pole of $X(Z): p3= \left | 2 \right | $ $$ \displaystyle $$ For stability condition: $$\begin{align} \quad&\ z< \left |2 \right | \end{align}$$ The intersection area is: $$ \displaystyle$$

$Y(Z): ROC $ H(Z) $ \cap $ $ ROC $ Y(Z) = $$\begin{align} \quad&\ \left | \frac{1}{4} \right | < \left | z \right | < \left | \frac{3}{2} \right | \end{align}$$

i see it drawing the circumference and tracing the circumferences related to the poles and their corresponding ROC .

Answer 1

In order to answer this question we would need to know the ROCs of $H(z)$ and $Y(z)$. The pole radii only determine the limits of the possible ROCs. E.g., for $H(z)$ there are three different possible ROCs:

$$\begin{align}1.\quad&|z|>\frac{3}{2}\\ 2.\quad&\frac32>|z|>\frac14\\ 3.\quad&|z|<\frac14\end{align}\tag{1}$$

For $Y(z)$, there are two possible ROCs:

$$\begin{align}1.\quad&|z|<2\\ 2.\quad&|z|>2\end{align}\tag{2}$$

Note that each of these ROCs corresponds to a different time domain sequence.

Certain combinations of the ROCs in $(1)$ and $(2)$ overlap, others don't. Below are shown the combinations that result in an overlap of ROCs, and the resulting combined ROC. The first number is the number of the ROC of $H(z)$ given in $(1)$, the second number is the ROC of $Y(z)$ given in $(2)$:

$$\begin{align}1+2:\quad&|z|>2\\ 1+1:\quad&2>|z|> \frac32\\ 2+1:\quad&\frac32>|z|>\frac14\\ 3+1:\quad&|z|<\frac14\end{align}\tag{3}$$

All other combinations of ROCs result in no overlap.