ECEN 314: Signals and Systems
Solutions to HW 9
1. (3.34) The frequency response of the system is
H
(
jω
)
=
∞
Z
∞
(
e

4
t
u
(
t
) +
e
4
t
u
(

t
)
)
e

jωt
dt
=
0
Z
∞
e
t
(4

jω
)
dt
+
∞
Z
0
e

t
(4+
jω
)
dt
=
1
4

jω
+
1
4 +
jω
(a) Here
T
= 1 and
ω
o
= 2
π
and
a
k
= 1 for all
k
. The FS coefficients of the output are:
b
k
=
a
k
H
(
jkω
o
) =
1
4

j
2
kπ
+
1
4 +
j
2
kπ
(b) Here
T
= 2 and
ω
o
=
π
and
a
k
=
‰
0
,
k even
1
,
k odd
The FS coefficients of the output are:
b
k
=
a
k
H
(
jkω
o
) =
‰
0
,
k even
1
4

jkπ
+
1
4+
jkπ
,
k odd
(c) Here
T
= 1,
ω
o
= 2
π
and
a
k
=
1
/
2
,
k
= 0
0
,
k even, k
6
= 0
sin
πk
2
πk
,
k odd
The FS coefficients of the output are:
b
k
=
a
k
H
(
jkω
o
) =
1
/
4
,
k
= 0
0
,
k even, k
6
= 0
sin
πk
2
πk
•
1
4

jk
2
π
+
1
4+
jk
2
π
‚
k odd
1
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2. (3.35)
The fourier series coefficients of
y
(
t
) are
b
k
=
H
(
jkω
o
)
a
k
, where
ω
o
is the fundamental fre
quency of
x
(
t
) and
a
k
are the FS coefficients of
x
(
t
).
Since
y
(
t
) =
x
(
t
), then
b
k
=
a
k
for all
k
. Noting that
H
(
jω
) = 0 for

ω

<
250, and since
ω
o
= 2
π/
(
π/
7) = 14, we can be sure that for

k
 ≤
¥
250
14
ƒ
= 17,
b
k
=
a
k
= 0.
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 Spring '08
 HALVERSON
 Fourier Series, Frequency, Cos, Fundamental frequency, FS coefficients

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