Solution of ECE 316 Test #9 S03 3/26/03
( ) =
1
2
1.
z
−
2 z
1
3
then, by finding the partial fraction expansion of this improper fraction in z two different ways, its inverse z transform,
[ ]
can be written in two different forms, h
[ ] =
A
1
2
n
+
B
−
1
3
n
u
[ ]
and h
[ ] = δ [ ] +
C
1
2
n
−
1
+
D
−
1
3
n
−
1
u
[ n
−
1
]
.
Find A, B, C, and D (2 pts each).
H
( )
= z
z
− z
1
2
3
1
3
= z
5
−
1
2
2
+ z
5
+
1
3
⇒
H
( ) = z
3 z
5
−
1
2
+ z
2 z
5
+
1
3
⇒ h
[ ] =
3
5
1
2
n
+
2
5
−
1
3
n
u
[ ]
A
= =
5
0 6
2
5
( ) =
z
− z
2
1
2
1
3
= z
2 z
2 −
1 z
−
6
1
6 h
[ ] = δ [ ] +
3
10
1
2
n
−
1
−
2
15
−
1
3
n
−
1
u
[ n
−
1
]
C
=
3
=
10
0 3
= −
2
= −
15
0 1333
K
1
6
z
− z
+
1
1
2
1
3
1
6
9
5 z
−
1
2
− z
4
5
+
1
3
Circle two correct answers.
2.
The
ω
(2 pts) Which of these lines in the s plane maps into the unit circle in the z plane?
axis and The line segment 0
2
π σ
=
0
T s
1