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

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

  • $\begingroup$ Please include a question. $\endgroup$
    – 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.


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