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Region of Convergence and Examples

Whether the z-transform $X(z)$ of a signal $x[n]$ exists depends on the complex variable $z=e^s$ 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, which is composed of all $z$ values for the summation of the Z-transform to converge. The ROC of the Z-transform is determined by $\vert z\vert=\vert e^s\vert=e^{\sigma}$ (a circle), the magnitude of variable $z$, while the ROC for the Laplace transform is determined by $\sigma=Re[s]$, (a vertical line), the real part of $s$. This formula is always needed in the examples:

\begin{displaymath}\sum_{n=0}^\infty x^n=\frac{1}{1-x},\;\;\;\;\mbox{for $\vert x\vert<1$} \end{displaymath}

Example 1: The Z transform of a right sided signal $x[n]=a^n u[n]$ is

\begin{displaymath}X(z)=\sum_{n=-\infty}^\infty a^n u[n] z^{-n}
=\sum_{n=0}^\infty (az^{-1})^n=\frac{1}{1-az^{-1}}=\frac{z}{z-a} \end{displaymath}

For this summation to converge, i.e., for $X(z)$ to exist, it is necessary to have $\vert az^{-1} \vert<1$, i.e., the ROC is $\vert z\vert > \vert a\vert$. As a special case when $a=1$, $x[n]=u[n]$ and we have

\begin{displaymath}{\cal Z}[u[n]]=\frac{1}{1-z^{-1}}, \;\;\;\;\vert z\vert>1 \end{displaymath}

Example 2: The Z-transform of a left sided signal $x[n]=-a^nu[-n-1]$ is:

$\displaystyle X(z)$ $\textstyle =$ $\displaystyle -\sum_{n=-\infty}^\infty a^nu[-n-1]z^{-n}=-\sum_{n=-\infty}^{-1} (az^{-1})^n$  
  $\textstyle =$ $\displaystyle 1-\sum_{n=0}^\infty (a^{-1}z)^n=1-\frac{1}{1-a^{-1}z}=\frac{z}{z-a}=\frac{1}{1-az^{-1}}$  

For the summation above to converge, it is required that $\vert a^{-1}z\vert<1$, i.e., the ROC is $\vert z\vert<\vert a\vert$. Comparing the two examples above we see that two different signals can have identical z-transform, but with different ROCs.

Example 3: Find the inverse of the given z-transform $X(z)=4z^2+2+3z^{-1}$. Comparing this with the definition of z-transform:

X(z)=\sum_{n=-\infty}^\infty x[n]z^{-n}=x[-2]z^2+x[-1]z^1+x[0]+x[1]z^{-1}+x[2]z^{-2}

we get

\begin{displaymath}x[n]=4\delta[n+2]+2\delta[n]+3\delta[n-1] \end{displaymath}

In general, we can use the time shifting property

\begin{displaymath}{\cal Z}[\delta[n+n_0]]=z^{n_0} \end{displaymath}

to inverse transform the $X(z)$ given above to $x[n]$ directly.

Example 4: Sometimes the inverse transform of a given $X(z)$ can be obtained by long division.

\begin{displaymath}X(z)=\frac{1}{1-az^{-1}} \end{displaymath}

By a long division, we get

\begin{displaymath}1\div (1-az^{-1})=1+az^{-1}+a^2z^{-2}+\cdots \end{displaymath}

which converges if the ROC is $\vert z\vert > \vert a\vert$, i.e., $\vert az^{-1} \vert<1$ and we get

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

. Alternatively, the long division can also be carried out as:

\begin{displaymath}1\div (-az^{-1}+1)=-a^{-1}z-a^{-2}z^2-\cdots \end{displaymath}

which converges if the ROC is $\vert z\vert<\vert a\vert$, i.e., $\vert a^{-1}z\vert<1$ and we get

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

next up previous
Next: Zeros and Poles of Up: Z_Transform Previous: Conformal Mapping between S-Plane
Ruye Wang 2012-01-28