Sunday, July 19, 2009

Region of Convergence

Whether the z-transform X(z) of a signal x[n] exists or not depends on the complex variable z=es as well as the signal itself. X(z) exists if and only if the argument z is inside the region of convergence (ROC) in the z-plane. The ROC is determined by $\vert z\vert=\vert e^s\vert=e^{\sigma}$, the magnitude of variable z, as shown in the following examples. (Recall the ROC for Laplace transform is determined by $\sigma=Re[s]$, the real part of s.) This formula is always needed in the examples:

x[n]=an u[n] is

\begin{displaymath}X(e^{j\omega})=\sum_{n=-\infty}^\infty a^n u[n] e^{-j\omega n} =\sum_{n=0}^\infty (ae^{-j\omega})^n \end{displaymath}

This summation does not converge unless $X(e^{j\omega})$ exist:
\begin{displaymath}X(z)=\sum_{n=-\infty}^\infty a^n u[n] z^{-n} =\sum_{n=0}^\infty (az^{-1})^n \end{displaymath}

Similar to the discrete Fourier transform, for this integral to converge, or for z-transform X(z) to exist, it is necessary to have \vert a\vert$" align="middle" border="0" height="37" width="147"> so that the z-transform is
\vert a\vert \end{displaymath}" height="55" width="418">

As a special case where a=1, x[n]=u[n] and we have
1 \end{displaymath}" height="47" width="256">

When $Re[s]=\sigma=0$, we have $z=e^{j\omega}$ and the z-transform X(z) becomes discrete Fourier transform $X(e^{j\omega})$.
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