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Lecture #15:
THE BILATERAL Z-TRANSFORM
Motivation: Method for representing DT signals as superpositions of complex geometric (exponential) functions
Outline:
– Definition
– Properties
Review of last lecture
Solve linear difference equation for a causal exponential input
$\sum _{k=0}^{K}{a}_{k}y[n+k]=\sum _{l=0}^{L}{b}_{l}x[n+l]\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}x[n]={\text{Xz}}^{n}u[n]$
Solve homogeneous equation for n>0
$\sum _{k=0}^{K}{a}_{k}{y}_{h}[n+k]=0\begin{array}{cc}& \end{array}\text{by}\begin{array}{cc}& \end{array}\text{assu}\text{min}g\begin{array}{cc}& \end{array}{y}_{h}[n]={\mathrm{A\lambda}}^{n}$
Solve characteristic polynomial for λ.
$\sum _{k=0}^{K}{a}_{k}{\lambda}^{n+k}=0$
Solve for a particular solution for n>0
$\sum _{k=0}^{K}{a}_{k}{y}_{p}[n+k]=\sum _{l=0}^{L}{b}_{l}x[n+l]\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}x[n]={\text{Xz}}^{n}u[n]$
Assuming ${y}_{p}[n]={\text{Yz}}^{n}$ and solving for Y yields
$Y=\stackrel{\text{~}}{H}(z)X=\frac{\sum _{l=0}^{L}{b}_{l}{z}^{l}}{\sum _{k=0}^{K}{a}_{k}{z}^{k}}X$
Logic for an analysis method for DT LTI systems
I. THE BILATERAL Z-TRANSFORM
1/ Definition
The bilateral Z-transform is defined by the analysis formula
$\stackrel{\text{~}}{X}(z)=\sum _{n=-\infty}^{\infty}x[n]{z}^{-n}$
$\stackrel{\text{~}}{X}(z)$ 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
$x[n]=\frac{1}{\mathrm{2\pi j}}{\int}_{C}\stackrel{\text{~}}{X}(z){z}^{n-1}\text{dz}$
Since z is a complex quantity, $\stackrel{\text{~}}{X}(z)$ 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
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
$x[n]\stackrel{Z}{\iff}\stackrel{\text{~}}{X}(z)$
2/ Properties
a/ Linearity
${\text{ax}}_{1}[n]+{\text{bx}}_{2}[n]\stackrel{Z}{\iff}a{\stackrel{\text{~}}{X}}_{1}(z)+b{\stackrel{\text{~}}{X}}_{2}(z)$
The proof follows from the definition of the Z-transform as a sum.
$\begin{array}{}\stackrel{\text{~}}{X}(z)=\sum _{n=-\infty}^{\infty}({\text{ax}}_{1}[n]+{\text{bx}}_{2}[n]){z}^{-n}\\ \stackrel{\text{~}}{X}(z)=a\sum _{n=-\infty}^{\infty}{x}_{1}[n]{z}^{-n}+b\sum _{n=-\infty}^{\infty}{x}_{2}[n]{z}^{-n}\\ \stackrel{\text{~}}{X}(z)=a\stackrel{\text{~}}{{X}_{1}}(z)+b\stackrel{\text{~}}{{X}_{2}}(z)\end{array}$ $$
b/ Delay by k
$x[n-k]\stackrel{Z}{\iff}{z}^{-k}\stackrel{\text{~}}{X}(z)$
This result can be seen using the synthesis formula,
$\begin{array}{}x[n-k]=\frac{1}{\mathrm{2\pi j}}\int \stackrel{\text{~}}{X}(z){z}^{n-k-1}\text{dz}\\ x[n-k]=\frac{1}{\mathrm{2\pi j}}\int {z}^{-k}\stackrel{\text{~}}{X}(z){z}^{n-1}\text{dz}\end{array}$
c/ Multiply by n
$\text{nx}[n]\stackrel{Z}{\iff}-z\frac{d\stackrel{\text{~}}{X}(x)}{\text{dz}}$
This result can be seen using the analysis formula.
$\begin{array}{}\frac{d\stackrel{\text{~}}{X}(z)}{\text{dz}}=\sum _{n=-\infty}^{\infty}-\text{nx}[n]{z}^{-n-1}\\ -z\frac{d\stackrel{\text{~}}{X}(z)}{\text{dz}}=\sum _{n=-\infty}^{\infty}\text{nx}[n]{z}^{-n}\end{array}$
Most proofs of Z-transform properties are simple. Some of the important properties are summarized here.
R, R1, and R2 are the ROCs of $\stackrel{\text{~}}{X}(z)$ , $\stackrel{\text{~}}{{X}_{1}}(z)$ , and $\stackrel{\text{~}}{{X}_{2}}(z)$ , respectively. * Exceptions may occur at z = 0 and z = ∞.
II. Z-TRANSFORMS OF SIMPLE TIME FUNCTIONS
1/ Unit sample function
$\delta [n]=\{\begin{array}{c}1\begin{array}{cc}& \end{array}\text{if}\begin{array}{cc}& \end{array}n=0\\ 0\begin{array}{cc}& \end{array}\text{otherwise}\end{array}$
The Z-transform of the unit sample is
$Z\{\delta [n]\}=\sum _{n=-\infty}^{\infty}\delta [n]{z}^{-n}={z}^{0}=1$
for all values of z, i.e., the ROC is the entire z plane.
2/ Unit step function
$u[n]=\{\begin{array}{c}1\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}n\ge 0\\ 0\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}n<0\end{array}$
The unit step and unit sample functions are simply related.
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