next up previous
Next: Solving LCCDEs by Unilateral Up: Z_Transform Previous: System Algebra and Block

Unilateral Z-Transform

The unilateral z-transform of an arbitrary signal $x[n]$ is defined as

\begin{displaymath}{\cal UZ}[x[n]]=X(z)\stackrel{\triangle}{=}
\sum_{n=-\infty}^\infty x[n]u[n] z^{-n}=\sum_{n=0}^\infty x[n] z^{-n}
\end{displaymath}

When the unilateral z-transform is applied to find the transfer function $H(z)={\cal UZ}[h[n]]$ of an LTI system, it is always assumed to be causal, and the ROC is always the exterior of a circle. The unilateral z-transform of any signal $x[n]=x[n]u[n]$ is identical to its bilateral Laplace transform. However, if $x[n] \ne x[n]u[n]$, the two z-transforms are different. Some of the properties of the unilateral z-transform different from the bilateral z-transform are listed below.

Example:

\begin{displaymath}x[n]=a^{n+1}u[n+1] \end{displaymath}

This signal is right sided starting at $n=-1$ (i.e., $x[n] \ne x[n]u[n]$). By definition, the bilateral z-transform of $x[n]$ is

\begin{displaymath}{\cal Z}[x[n]]=\sum_{n=-1}^\infty a^{n+1}z^{-n}
=z+a\sum_{n=0}^\infty a^n z^{-n}=z+\frac{a}{1-az^{-1}}
=\frac{z}{1-az^{-1}}
\end{displaymath}

It was assumed that $\vert z\vert > \vert a\vert$. The unilateral z-transform of this signal is

\begin{displaymath}{\cal UZ}[x[n]]=\sum_{n=0}^\infty a^{n+1}z^{-n}
=a\sum_{n=0}^\infty a^n z^{-n}=\frac{a}{1-az^{-1}}
\end{displaymath}

If we assume zero initial condition $y[-1]=0$,


next up previous
Next: Solving LCCDEs by Unilateral Up: Z_Transform Previous: System Algebra and Block
Ruye Wang 2014-10-28