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124 Chapter 4 THE z-TRANSFORM
□ EXAMPLE 4.12 Given that
is a causal system, find
H(z) =
z +1
z 2 − 0.9z +0.81
a. its transfer function representation,
b. its difference equation representation, and
c. its impulse response representation.
Solution The poles of the system function are at z =0.9̸ ± π/3. Hence the ROC of
this causal system is |z| > 0.9. Therefore the unit circle is in the ROC, and the
discrete-time Fourier transform H(e jω ) exists.
a. Substituting z = e jω in H(z),
H(e jω )=
b. Using H(z) =Y (z)/X(z),
e jω +1
e j2ω − 0.9e jω +0.81 = e jω +1
(e jω − 0.9e jπ/3 )(e jω − 0.9e −jπ/3 )
Y (z)
X(z) = z +1
z 2 − 0.9z +0.81
Cross multiplying,
(
z
−2
z −2 )
=
z −1 + z −2
1 − 0.9z −1 +0.81z −2
Y (z) − 0.9z −1 Y (z)+0.81z −2 Y (z) =z −1 X(z)+z −2 X(z)
Now taking the inverse z-transform,
or
y(n) − 0.9y(n − 1) + 0.81y(n − 2) = x(n − 1) + x(n − 2)
y(n) =0.9y(n − 1) − 0.81y(n − 2) + x(n − 1) + x(n − 2)
c. Using the MATLAB script,
>> b = [0,1,1]; a = [1,-0.9,0.81]; [R,p,C] = residuez(b,a)
R =
-0.6173 - 0.9979i
-0.6173 + 0.9979i
p =
0.4500 + 0.7794i
0.4500 - 0.7794i
C =
1.2346
>> Mp = (abs(p))’
Mp =
0.9000 0.9000
>> Ap = (angle(p))’/pi
Ap =
0.3333 -0.3333
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