4.2.2. Finite and Infinite Signals

\(\newcommand{\op}[1]{\mathsf #1}\) \(\newcommand{\ztarrow}{\stackrel{\op Z}{\longrightarrow}}\)

Consider a finite signal (different from zero in a finite number of samples):

\[x[n] = \begin{matrix}2& \underline{3}& 5& 3& 1 \end{matrix}\]

where we use the convention that the origin (\(n=0\)) is denoted with the underlining and signal values that are not given are equal to zero. The Z-transform equals:

\[X(z) = 2 z^1 + 3 + 5 z^{-1} + 3 z^{-2} + z^{-3}\]

with ROC the entire complex plane without the origin. Now consider an infinite signal

\[x[n] = \begin{matrix}\underline{1}& 0.8& 0.8^2& 0.8^3& 0.8^4& \cdots\end{matrix}\]

with Z-transform:

\[X(z) = \sum_{n=0}^{\infty} 0.8^n z^{-n} = \sum_{n=0}^{\infty} \left(0.8 z^{-1}\right)^n\]

Remember your geometric sequences (meetkundige reeks) from math class? Without proof we state:

\[X(z) = \frac{1}{1-0.8 z^{-1}},\quad |z|>0.8\]

the ROC is \(|z|>0.8\). The ROC follows from the observation that the geometric sequence only converges in case \(|0.8 z^{-1}|<1\). Note that not all infinite signals result in a finite Z-transform (in fact, most infinite signals do not have bounded Z-transform).

The second example can be summarized as:

\[0.8^n u[n] \ztarrow \frac{1}{1-0.8 z^{-1}},\quad |z|>0.8.\]