# ROC of transfer function

given:

$$H(z) = \frac{4z(z-1)}{z-0.5}$$

I would say, when all poles are in the unit circles, the impulse response is right sided and causal.

• Please include a question. – Juancho Sep 23 '17 at 17:30

For that transfer function, you have 2 regions of convergence (ROCs): $R_1 = \{|z|>0.5\}$ and $R_2=\{|z|<0.5\}$.
$R_1$ includes the unit circle, so the corresponding impulse response corresponds to a BIBO-stable filter. Since it is an outer disc, it is also right-sided.
Conversely, $R_2$ corresponds to a left-sided impulse response, and an unstable filter.
The extra $z$ in the numerator means that the impulse responses will be shifted 1 sample to the left, so the impulse responses are not causal.