UploadFile_6417

Solutions of the Difference Equations 129
Then the sample shifting property is given by
Z + [x(n − k)] = Z [x(n − k)u(n)]
or
=
=
∞∑
x(n − k)z −n =
n=0
∑−1
m=−k
∞∑
m=−k
x(m)z −(m+k)
[ ∞
]
∑
x(m)z −(m+k) + x(m)z −m
m=0
z −k
Z + [x(n − k)] = x(−1)z 1−k +x(−2)z 2−k +···+x(−k)+z −k X + (z) (4.26)
This result can now be used to solve difference equations with nonzero
initial conditions or with changing inputs. We want to solve the difference
equation
N∑
M∑
1+ a k y(n − k) = b m x(n − m), n≥ 0
k=1
subject to these initial conditions:
m=0
{y(i),i= −1,...,−N} and {x(i),i= −1,...,−M}.
We now demonstrate its solution using an example.
□ EXAMPLE 4.14 Solve
where
y(n) − 3 2 y(n − 1) + 1 y(n − 2) = x(n), n ≥ 0
2
subject to y(−1) = 4 and y(−2) = 10.
( ) 1 n
x(n) = u(n)
4
Solution
Taking the one-sided z-transform of both sides of the difference equation, we
obtain
Y + (z) − 3 2 [y(−1) + z−1 Y + (z)] + 1 2 [y(−2) + z−1 y(−1) + z −2 Y + 1
(z)] =
1 − 1 4 z−1
Substituting the initial conditions and rearranging,
Y + (z)
[1 − 3 2 z−1 + 1 ]
1
2 z−2 =
1 − +(1− 1 2z−1 )
4 z−1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.