UploadFile_6417

130 Chapter 4 THE z-TRANSFORM
or
1
Y + 1 − 1 4
(z) =
z−1
1 − 3 2 z−1 + + 1 − 2z −1
(4.27)
1
2 z−2 1 − 3 2 z−1 + 1 2 z−2
Finally,
Y + 2 − 9 4
(z) =
z−1 + 1 2 z−2
(1 − 1 2 z−1 )(1 − z −1 )(1 − 1 4 z−1 )
Using the partial fraction expansion, we obtain
2
1
Y + 1
(z) =
1 − + 3
1
2 z−1 1 − z + 3
(4.28)
−1 1 − 1 4 z−1
After inverse transformation the solution is
[( ) 1 n
y(n) = + 2 2 3 + 1 3
( 1
4) n ]
u(n) (4.29)
Forms of the solutions The preceding solution is the complete response
of the difference equation. It can be expressed in several forms.
• Homogeneous and particular parts:
[( ) n 1
y(n) = + 2 u(n) +
2 3]
1 ( ) n 1
u(n)
3 4
} {{ } } {{ }
Homogeneous part Particular part
The homogeneous part is due to the system poles, and the particular
part is due to the input poles.
• Transient and steady-state responses:
[ ( ) n ( n ]
1 1 1 2
y(n) = + u(n) +
3 4 2)
3 u(n)
} {{ } } {{ }
Transient response
Steady-state response
The transient response is due to poles that are inside the unit circle,
whereas the steady-state response is due to poles that are on the unit
circle. Note that when the poles are outside the unit circle, the response
is termed an unbounded response.
• Zero-input (or initial condition) and zero-state responses:
In equation (4.27) Y + (z) has two parts. The first part can be interpreted
as
Y ZS (z) =H(z)X(z)
□
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.