Signals and Systems - Electrical Engineering

550 C H A P T E R 9: The Z-Transform

where the second term is not found in the Z-transforms table. To write it so that each of the terms in

the expansion are in the Z-transforms table, we need to obtain values forAandBin the expansion

X(z)=

A

1 −az−^1

+

Bz−^1

( 1 −az−^1 )^2

(9.34)

so that Equations (9.33) and (9.34) are equal. We find thatA=r 1 +r 2 , whileB−Aa=−r 1 aor

B=ar 2. With these values we find the inverse to be

x[n]=[(r 1 +r 2 )an+nr 2 an]u[n]=nanu[n]

as expected.

To illustrate the computation of the inverse Z-transform from the residues in the case of multiple

poles, consider the transfer function

X(z)=

0.5z−^1

1 −0.5z−^1 −0.25z−^2 +0.125z−^3

The following script is used.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Inverse Z-transform --- multiple poles

%%%%%%%%%%%%%%%%%%%%%%%%%%%%

b = [0 0.5 0 0 ];

a = [1 -0.5 -0.25 0.125]

[r,p,k] = residuez(b,a) % partial fraction expansion, poles and gain

zplane(b,a) % plot of poles and zeros

n = 0:99; xx = p(1). ˆ n; yy = xx.∗n;

x1 = (r(1) + r(2)).∗xx + r(2).∗yy + r(3)∗p(3). ˆ n; % inverse computation

The poles and the zeros and the inverse Z-transform are shown in Figure 9.10—there is a double pole

at 0.5. The residues and the corresponding poles are

r = -0.2500

0.5000

-0.2500

p = 0.5000

0.5000

-0.5000

Computationally, our method and MATLAB’s are comparable but the inverse transform in our

method is found directly from the table, while in the case of MATLAB’s you need to change the

expansion to get it into the forms found in the tables.

9.5.4 Solution of Difference Equations

In this section we will use the shifting in time property of the Z-transform in the solution of difference

equations with initial conditions. You will see that the partial fraction expansion used to find the

inverse Z-transform is like the one used in the inverse Laplace transform.