Inverse Z-Transform – SmartDATA Lab

Motivation

One of the challenges with Z-transform is that that inverse is not unique. A single z-transform representation can correspond have multiple time-domain signals. However, if you know the z-transform’s region of convergence, you can unique identify the inverse.

Region of Convergence

Since the Z-transform is a sum, then $X(z)=\infty$ if any value in the sum is $\infty$ . This is not invertible (i.e., we cannot get back to $x[n]$ ). Hence, we can only invert for values of $z$ in which $X(z) \neq \infty$ . This set of $z$ values is known as the region of convergence. The region of convergence exists in the complex plane.

In the complex plane, the region of convergence begins at a pole. If the pole is causal, the region of convergence exists for all values with a radius greater than $\sigma$ (the magnitude of the pole) from the origin.

If the pole is anti-causal, the region of convergence exists for all values with a radius of less than $\sigma$ (the magnitude of the pole) from the origin. If the system has multiple poles, the total region of convergence is the intersection of the region of convergence for each pole.

An Example

Consider

$x[n] = (1/2)^n u[n] \; .$

The Z-transform is defined by

$X(z) = \frac{1}{1-(1/2)z^{-1}} \; .$

The time-domain signal $x[n]$ is causal, so the ROC is $|z| > 1/2$ .

Similarly, consider

$x[n] = -(1/2)^n u[-n-1] \; .$

The Z-transform is defined by

$X(z) = \frac{1}{1-(1/2)z^{-1}} \; .$

The time-domain signal $x[n]$ is anti-causal, so the ROC is $|z| < 1/2$ .

Show Proof

Consider

$x[n] = -(1/2)^n u[-n-1] \; .$

The time-reversal of $x[n]$ is

$x[-n] = -(1/2)^{-n} u[n-1] = -(2)^{n} u[n-1] \; .$

The Z-transform of $x[-n]$ is $X(z^{-1})$ due to the time-reversal property. To compute the Z-transform of a time-domain signal, we can make one additional modification:

$x[-(n+1)] = -(2)^{n+1} u[n] = -2 (2)^{n} u[n] \; .$

By applying the time-shifting property, the Z-transform of $x[-(n+1)]$ is $X(z^{-1}) z^{+1}$ .

Hence, by using the Z-transform table, we can establish that

$X(z^{-1}) z^{+1} = -2 \frac{1}{1 - 2 z^{-1}} \; .$

This can be simplified to

$<span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="https://smartdata.ece.ufl.edu/wp-content/ql-cache/quicklatex.com-3edf431baa77f6e2a5c3ebbcf9333b04_l3.png" height="45" width="567" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}X(z^{-1}) &= -2 \frac{z^{-1}}{1- 2 z^{-1}} \X(z) &= -2 \frac{z}{1 - 2 z} \&= -2 \frac{1}{z^{-1} - 2} \&= \frac{1}{1 - (1/2) z^{-1}}\end{align*}" title="Rendered by QuickLaTeX.com"/>$

Partial Fraction Decomposition

One of the most common ways to perform an inverse Z-transform representation is to decompose it into easy-to-invert components. Partial fraction decomposition for $N > M$ (i.e., more poles than zeros) can be expressed as

$\begin{align*} H(z) &= \frac{b_0}{a_0} \frac{\prod_{m=1}^{M} \left( 1 - z_{m} z^{-1} \right) }{\prod_{m=1}^{N} \left( 1 - p_{m} z^{-1} \right) } \\ &= \frac{b_0}{a_0} \sum_{m=1}^N \frac{A_m}{1 - p_{m} z^{-1} } \; , \end{align*}$

where $p_{m}$ are scalar constants representing our poles and $A_m$ are scalar constants for each term. After performing the partial fraction decomposition, we can complete the inverse Z-transform for each component in $H(z)$ .