Signals and Systems - Electrical Engineering

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

in terms of one-sided Z-transforms. Another valid reason to study the one-sided Z-transform in more

detail is its use in solving difference equations with initial conditions.

Recall that the one-sided Z-transform is defined as

X 1 (z)=Z(x[n]u[n])=

∑∞

n= 0

x[n]u[n]z−n (9.9)

in a region of convergenceR 1. Also recall that the computation of the two-sided Z-transform using

the one-sided Z-transform is given in Equation (9.6).

9.4.1 Computing the Z-Transform with Symbolic MATLAB

Similar to the computation of the Laplace transform, the computation of the Z-transform can be

done using the symbolic toolbox of MATLAB. The following is the necessary code for computing the

Z-transform of

h 1 [n]=0.9u[n]

h 2 [n]=u[n]−u[n−10]

h 3 [n]=cos(ω 0 n)u[n]

h 4 [n]=hsigna 1 [n]h 2 [n]

The results are shown at the bottom. (As in the continuous case, in MATLAB theheavisidefunction is

the same as the unit-step function.)

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

% Z-transform computation

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

syms n w0

h1 = 0.9. ˆ n; H1 = ztrans(h1)

h2 = heaviside(n) - heaviside(n-10); H2 = ztrans(h2)

h3 = cos(w0∗n)∗heaviside(n); H3 = ztrans(h3)

H4 = ztrans(h1∗h3)

H1 = 10/9/(10/9∗z - 1)∗z

H2 = 1 + 1/z + 1/z ˆ 2 + 1/z ˆ 3 + 1/z ˆ 4 + 1/z ˆ 5 + 1/z ˆ 6 + 1/z ˆ 7 + 1/z ˆ 8 + 1/z ˆ 9

H3 = (z - cos(w0))∗z/(z ˆ 2 - 2∗z∗cos(w0) + 1)

H4 = 10/9∗(10/9∗z - cos(w0))*z/(100/81∗z ˆ 2 - 20/9∗z∗cos(w0) + 1)

The functioniztranscomputes the inverse Z-transform. We will illustrate its use later on.

9.4.2 Signal Behavior and Poles

In this section we will use the linearity property of the Z-transform to connect the behavior of the

signal with the poles of its Z-transform.