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132 Chapter 4 THE z-TRANSFORM
which agrees with the response given in (4.29). In Example 4.14 we computed
x IC (n) analytically. However, in practice, and especially for largeorder
difference equations, it is tedious to determine x IC (n) analytically.
MATLAB provides a function called filtic, which is available only in
the Signal Processing toolbox. It is invoked by
xic = filtic(b,a,Y,X)
in which b and a are the filter coefficient arrays and Y and X are the initialcondition
arrays from the initial conditions on y(n) and x(n), respectively,
in the form
Y =[y(−1), y(−2),..., y(−N)]
X =[x(−1), x(−2),..., x(−M)]
If x(n) =0, n ≤−1 then X need not be specified in the filtic function.
In Example 4.14 we could have used
>> Y = [4, 10]; xic = filtic(b,a,Y)
xic =
1 -2
to determine x IC (n).
□ EXAMPLE 4.15 Solve the difference equation
y(n) = 1 [x(n)+x(n − 1) + x(n − 2)]+0.95y(n − 1) − 0.9025y(n − 2),
3
n ≥ 0
where x(n) =cos(πn/3)u(n) and
y(−1) = −2, y(−2) = −3; x(−1) = 1, x(−2) = 1
First determine the solution analytically and then by using MATLAB.
Solution
Taking a one-sided z-transform of the difference equation
Y + (z) = 1 3 [X+ (z)+x(−1) + z −1 X + (z)+x(−2) + z −1 x(−1) + z −2 X + (z)]
+0.95[y(−1) + z −1 Y + (z)] − 0.9025[y(−2) + z −1 y(−1) + z −2 Y + (z)]
and substituting the initial conditions, we obtain
1
Y + 3
(z) =
+ 1 3 z−1 + 1 3 z−2
1.4742 +
1 − 0.95z −1 +0.9025z −2 X+ 2.1383z−1
(z)+
1 − 0.95z −1 +0.9025z −2
Clearly, x IC(n) =[1.4742, 2.1383]. Now substituting X + (z) =
1 − 0.5z−1
1 − z −1 + z −2
and simplifying, we will obtain Y + (z) asarational function. This simplification
and further partial fraction expansion can be done using MATLAB.
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