# Using ROC to find stability of system in specific example

I've started learning about finding the ROC from the transfer function, but I'm confused about an example.

$$H(Z) = \frac{2Z + 1}{Z^2 + Z - \frac{5}{16}}$$

I understand the poles lie at $$z = \frac{-5}{4}$$and $$z = \frac{1}{4}$$, and I assumed the system was non-causal because even though it has more poles than zeros, the unit circle does not lie to the exterior of the outermost pole(if I'm right), but because the ROC includes the unit circle, the system was stable. However, the solution to the problem states the system is both unstable and non-casual so I was hoping someone could explain this to me? I've only seen examples giving the ROC when both poles have been of the form $$(z-a)$$ however here one is $$(z + b)$$, so I thought the ROC was $$\frac{5}{4} > |Z| > \frac{1}{4}$$? I imagine this is where I have went wrong.

WRONG. The Z-transform transfer function will always have equal number of poles and zeros. Your poles are at $$z = -5/4$$ and $$z = 1/4$$. Zeros are at $$z = -1/2$$ and $$z = \infty$$. Since ROC will always be concentric circle region without including poles, there are 3 possible ROC for the given transfer function. $$ROC_1 = 5/4 \lt |z| \lt \infty$$ $$ROC_2 = 1/4 \lt |z| \lt 5/4$$ $$ROC_3 = 0 \le |z| \lt 1/4$$ $$ROC_1$$ represents an unstable but causal system (since it can include $$z \rightarrow \infty$$ so terms like $$z^{-k} \,\,with\,\, k \ge 0$$ will only be present (only negative or zero powers of $$z$$). And because only negative powers of $$z$$ are present, system is non-zero for $$n \ge 0$$. So it is causal. Example - $$H(z) = a_0 + a_1z^{-1}+a_2z^{-2}+a_3z^{-3}+a_4z^{-4}+...$$
$$ROC_2$$ represents stable but non-causal system because it includes unit circle $$|z| = 1$$ but does not include $$z \rightarrow \infty$$. Terms with $$z$$ raised to positive power can be there along with negative power of $$z$$. So you can rule out $$z=0$$ and $$z \rightarrow \infty$$.Example - $$H(z) = a_0 + a_1z^{-1}+a_2z^{-2}+a_3z^{-3}+a_4z^{-4}+a_{-1}z^{1}+a_{-2}z^{2}+a_{-3}z^{3}$$
$$ROC_3$$ represents an unstable and non-causal system because it does not include unit circle and terms with only positive power of $$z$$ are present (because ROC includes $$z=0$$ and does not include $$z \rightarrow \infty$$. Example - $$H(z) = a_0 + a_{-1}z^{1}+a_{-2}z^{2}+a_{-3}z^{3}+...$$
In your case, the ROC might be $$ROC_3$$.