# 0.14 Lecture 15:the bilateral z-transform

Method for representing DT signals as superpositions of complex geometric (exponential) functions.

Lecture #15:

THE BILATERAL Z-TRANSFORM

Motivation: Method for representing DT signals as superpositions of complex geometric (exponential) functions

Outline:

• Review of last lecture
• The bilateral Z-transform

– Definition

– Properties

• Inventory of transform pairs
• Conclusion

Review of last lecture

Solve linear difference equation for a causal exponential input

$\sum _{k=0}^{K}{a}_{k}y\left[n+k\right]=\sum _{l=0}^{L}{b}_{l}x\left[n+l\right]\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}x\left[n\right]={\text{Xz}}^{n}u\left[n\right]$

Solve homogeneous equation for n>0

$\sum _{k=0}^{K}{a}_{k}{y}_{h}\left[n+k\right]=0\begin{array}{cc}& \end{array}\text{by}\begin{array}{cc}& \end{array}\text{assu}\text{min}g\begin{array}{cc}& \end{array}{y}_{h}\left[n\right]={\mathrm{A\lambda }}^{n}$

Solve characteristic polynomial for λ.

$\sum _{k=0}^{K}{a}_{k}{\lambda }^{n+k}=0$

Solve for a particular solution for n>0

$\sum _{k=0}^{K}{a}_{k}{y}_{p}\left[n+k\right]=\sum _{l=0}^{L}{b}_{l}x\left[n+l\right]\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}x\left[n\right]={\text{Xz}}^{n}u\left[n\right]$

Assuming ${y}_{p}\left[n\right]={\text{Yz}}^{n}$ and solving for Y yields

$Y=\stackrel{\text{~}}{H}\left(z\right)X=\frac{\sum _{l=0}^{L}{b}_{l}{z}^{l}}{\sum _{k=0}^{K}{a}_{k}{z}^{k}}X$

Logic for an analysis method for DT LTI systems

• $\stackrel{\text{~}}{H}\left(z\right)$ characterizes system  compute $\stackrel{\text{~}}{H}\left(z\right)$ efficiently.
• In steady state, response to ${\text{Xz}}^{n}$ is $\stackrel{\text{~}}{H}\left(z\right){z}^{n}$ .
• Represent arbitrary x[n] as superpositions of ${\text{Xz}}^{n}$ on z.
• Compute response y[n] as superpositions of $\stackrel{\text{~}}{H}\left(z\right){\text{Xz}}^{n}$ on z.

I. THE BILATERAL Z-TRANSFORM

1/ Definition

The bilateral Z-transform is defined by the analysis formula

$\stackrel{\text{~}}{X}\left(z\right)=\sum _{n=-\infty }^{\infty }x\left[n\right]{z}^{-n}$

$\stackrel{\text{~}}{X}\left(z\right)$ is defined for a region in z — called the region of convergence — for which the sum exists.

The inverse transform is defined by the synthesis formula

$x\left[n\right]=\frac{1}{2\pi j}{\int }_{C}\stackrel{\text{~}}{X}\left(z\right){z}^{n-1}\text{dz}$

Since z is a complex quantity, $\stackrel{\text{~}}{X}\left(z\right)$ is a complex function of a complex variable. Hence, the synthesis formula involves integration in the complex z domain. We shall not perform this integration in this subject. The synthesis formula will be used only to prove theorems and not to compute time functions directly.

a/ Approach

An inventory of time functions and their Z-transforms will be developed by

• Using the Z-transform properties,
• Determining the Z-transforms of elementary DT time functions,
• Combining the results of the above two items.

b/ Notation

We shall use two useful notations — Z{x[n]} signifies the Z-transform of x[n]and a Z-transform pair is indicated by

$x\left[n\right]\stackrel{Z}{⇔}\stackrel{\text{~}}{X}\left(z\right)$

2/ Properties

a/ Linearity

${\text{ax}}_{1}\left[n\right]+{\text{bx}}_{2}\left[n\right]\stackrel{Z}{⇔}a{\stackrel{\text{~}}{X}}_{1}\left(z\right)+b{\stackrel{\text{~}}{X}}_{2}\left(z\right)$

The proof follows from the definition of the Z-transform as a sum.

$\begin{array}{}\stackrel{\text{~}}{X}\left(z\right)=\sum _{n=-\infty }^{\infty }\left({\text{ax}}_{1}\left[n\right]+{\text{bx}}_{2}\left[n\right]\right){z}^{-n}\\ \stackrel{\text{~}}{X}\left(z\right)=a\sum _{n=-\infty }^{\infty }{x}_{1}\left[n\right]{z}^{-n}+b\sum _{n=-\infty }^{\infty }{x}_{2}\left[n\right]{z}^{-n}\\ \stackrel{\text{~}}{X}\left(z\right)=a\stackrel{\text{~}}{{X}_{1}}\left(z\right)+b\stackrel{\text{~}}{{X}_{2}}\left(z\right)\end{array}$ 

b/ Delay by k

$x\left[n-k\right]\stackrel{Z}{⇔}{z}^{-k}\stackrel{\text{~}}{X}\left(z\right)$

This result can be seen using the synthesis formula,

$\begin{array}{}x\left[n-k\right]=\frac{1}{2\pi j}\int \stackrel{\text{~}}{X}\left(z\right){z}^{n-k-1}\text{dz}\\ x\left[n-k\right]=\frac{1}{2\pi j}\int {z}^{-k}\stackrel{\text{~}}{X}\left(z\right){z}^{n-1}\text{dz}\end{array}$

c/ Multiply by n

$\text{nx}\left[n\right]\stackrel{Z}{⇔}-z\frac{d\stackrel{\text{~}}{X}\left(x\right)}{\text{dz}}$

This result can be seen using the analysis formula.

$\begin{array}{}\frac{d\stackrel{\text{~}}{X}\left(z\right)}{\text{dz}}=\sum _{n=-\infty }^{\infty }-\text{nx}\left[n\right]{z}^{-n-1}\\ -z\frac{d\stackrel{\text{~}}{X}\left(z\right)}{\text{dz}}=\sum _{n=-\infty }^{\infty }\text{nx}\left[n\right]{z}^{-n}\end{array}$

Most proofs of Z-transform properties are simple. Some of the important properties are summarized here.

R, R1, and R2 are the ROCs of $\stackrel{\text{~}}{X}\left(z\right)$ , $\stackrel{\text{~}}{{X}_{1}}\left(z\right)$ , and $\stackrel{\text{~}}{{X}_{2}}\left(z\right)$ , respectively. * Exceptions may occur at z = 0 and z = ∞.

II. Z-TRANSFORMS OF SIMPLE TIME FUNCTIONS

1/ Unit sample function

$\delta \left[n\right]=\left\{\begin{array}{c}1\begin{array}{cc}& \end{array}\text{if}\begin{array}{cc}& \end{array}n=0\\ 0\begin{array}{cc}& \end{array}\text{otherwise}\end{array}$

The Z-transform of the unit sample is

$Z\left\{\delta \left[n\right]\right\}=\sum _{n=-\infty }^{\infty }\delta \left[n\right]{z}^{-n}={z}^{0}=1$

for all values of z, i.e., the ROC is the entire z plane.

2/ Unit step function

$u\left[n\right]=\left\{\begin{array}{c}1\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}n\ge 0\\ 0\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}n<0\end{array}$

The unit step and unit sample functions are simply related.

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