Signals and Systems
Fall 2013
Solution to Homework Assignment #6
1.
Find the
z
transform of each of the following sequences.
(a)
f
(
k
)
=
2
δ
(
k
)
−
δ
(
k
−
2)
=
=
−
=
otherwise
0
2
,
1
0
,
2
k
k
We then obtain the
z
-transform as follows
2
2
2
1
2
1
( )
2
for
0
z
F z
z
z
z
−
= −
=
≠
(b)
f
(
k
)
=
≥
≤
≤
<
−
5
,
2
5
0
,
1
0
,
0
5
k
k
k
k
.
5
( )
( )
(
5)
2
(
5)
k
s
s
s
f k
u
k
u
k
u
k
−
⇒
=
−
−
+
−
The
z
-transform can then be obtained as,
6
5
5
5
4
1
1
2
1
( )
for
2
1
1
2
(
1)(
2)
z
z
z
z
z
F z
z
z
z
z
z
z
z
z
z
−
+
=
−
×
+
×
=
>
−
−
−
−
−
.
(c)
f
(
k
)
=
{2,2,0,
−
1,
−
1,
−
1,2,2,0,
−
1,
−
1,
−
1,2,2,0,
−
1,
−
1,
−
1,
⋅⋅⋅
}.
[Hint:
Express
f
(
k
) as the sum of shifted, identical, finite sequences.]
Let
g
(
k
)
=
{2,2,0,
−
1,
−
1,
−
1, 0,0,0,
0
,
0
,0,0,0,0,
0
,
0
,0
⋅⋅⋅
}.
Then,
6
12
18
1
1
1
( )
( )
{ ( )}
{ ( )}
{ ( )}
f k
g k
g k
g k
g k
E
E
E
=+
+
+
+
It then follows that,
6
12
18
6
6
0
0
6
1
1
1
( )
( )
( )
( )
( )
1
1
1
( )
( )
( )
( )
1
(
)
1
m
m
m
m
F z
G z
G z
G z
G z
z
z
z
G z
G z
G z
G z
z
z
z
∞
∞
=
=
=
+
+
+
+
=
=
=
=
−
∑
∑
6
6
( )
( )
1
z
F z
G z
z
⇒
=
−
where
5
4
2
3
4
5
5
2
1
1
1
2
2
1
( )
2
for
0
z
z
z
z
G z
z
z
z
z
z
z
+