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Given the z-transform of a sequence:

$$ X(z) = \frac{1-\frac{1}{2}z^{-1}}{1+\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}} , |z| > \frac{1}{2} $$

after partial fraction decomposition I get,

$$ X(z) = \frac{4}{1+\frac{1}{2}z^{-1}} + \frac{-3}{1+\frac{1}{4}z^{-1}} $$

Now I was thinking about the ROC (region of convergence). I know from the given that $|z| > \frac{1}{2}$, but now I have a pole at $\frac{1}{4}$. What will my ROC be / look like?

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Well, as you mentioned, from the given $|z|>\frac{1}{2},$ so there is no need to worry about $z_0=\frac{1}{4}$ which does not belong to that region. – leshik Jan 14 '13 at 2:37
up vote 1 down vote accepted

Your $X(z)$ has two poles, at $z=1/2$ and $z=1/4$, hence there are, a priori, three possible ROC: inside the inner circle ($|z|<1/4$), the annulus between the two circles ($1/4<|z|<1/2$) and the region outside the exterior circle ($|z|>1/2$). But you are already told that you must consider the later ROC. So all it's right, you just go ahead.

(You had reasons to worry -to suspect that something was wrong- if you had obtained a pole at, say $z=3/4$)

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Thanks, it's comforting to hear that confirmation. – Kyle Weller Jan 14 '13 at 3:22

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