We have discussed right- and left-sided signals, but what about two-sided signals? Consider the figure of an annular region of convergence of a Z-transform:

%\img{images/roc/twosided/twosided.ps}

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The causal part of the signal is represented by the region outside of $R_1$, and the anti-causal part of the signal corresponds to the region inside $R_2$. The intersection is the region of convergence for the transform. If, for a given signal, the anti-causal and causal regions do not overlap, the Z-transform is not defined.

\begin{example}

Find the Z-transform for $f(n) = \lr{\frac{1}{2}}^n u(n) - \lr{\frac{3}{4}}^n u(-n-1)$.

\begin{align*}
\hat{F}(z) &= \frac{1}{1 - \frac{1}{2}z^{-1}} + \frac{1}{1-\frac{3}{4}z^{-1}} \quad \abs{z} \lt \frac{3}{4} \mbox{ and } \abs{z} \gt \frac{1}{2} \\
\hat{F}(z) &= \frac{z}{z - \frac{1}{2}} + \frac{z}{z-\frac{3}{4}} \\
\hat{F}(z) &= \frac{2z\lr{z - \frac{5}{8}}}{\lr{z-\frac{1}{2}}\lr{z-\frac{3}{4}}} \\ 
\end{align*}

%\img{images/roc/examples/twosided/twosided.ps}

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The poles are where the denominator is zero, $z = \frac{1}{2}$, and $z = \frac{3}{4}$. The zeros come from the numerator, $z = 0$, and $z=\frac{5}{8}$. Below is the pole-zero diagram for $\hat{F}(z)$.

%\img{images/roc/examples/twosided/polezeros.ps}

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The region of convergence is bounded by the poles. The zeros play no role. However, there can be pole/zero cancellations. Recall that the Z-transform can only exist when there is a non-trivial overlap for the region of convergence. For example, the signal $q(n) = \lr{\frac{3}{4}}^n u(n) - \lr{\frac{1}{2}}^n u(-n-1)$ has no overlap for the region of convergence, hence its Z-transform doesn't exist.

\end{example}

Two-sided signals