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:

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

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

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

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

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

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

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

$$= 1-\sum_{n=0}^\infty (a^{-1}z)^n=1-\frac{1}{1-a^{-1}z}=\frac{z}{z-a}=\frac{1}{1-az^{-1}}$$

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

$$x[n]=4\delta[n+2]+2\delta[n]+3\delta[n-1]$$

In general, we can use the time shifting property

$${\cal Z}[\delta[n+n_0]]=z^{n_0}$$

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.

$$X(z)=\frac{1}{1-az^{-1}}$$

By a long division, we get

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

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

$$x[n]=a^n u[n]$$

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

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

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

$$x[n]=-a^n u[-1-n]$$