ECEN 314: Signals and Systems
Solutions to HW 8
1. Note that
H
(
s
) =
10
s
+10
. The steady state response is then:
y
(
t
)
=
2

H
(
j
10)

cos(10
t
+
∠
H
(
j
10))
=
1
.
414 cos(10
t

45
◦
)
2.
Note that the period of
x
(
t
) is
T
= 1.
The simplest way to do this problem is as
follows:
2 cos 10
πt
+ sin 6
πt
=
e
j
10
πt
+
e

j
10
πt
+
1
2
j
e
j
6
πt

e

j
6
πt
¶
=
e
j
2
π
(5)
t
+
e
j
2
π
(

5)
t
+
1
2
j
e
j
2
π
(3)
t

e
j
2
π
(

3)
t
¶
One can easily see that
x
5
=
x

5
= 1 and
x
3
=

x

3
=

j/
2. By direct inspection of
x
(
t
)
one easily gets
a
v
= 0,
a
5
= 2 and
b
3
= 1.
We can also verify the above result by using the standard integrals as follows:
x
k
=
1
/
2
Z

1
/
2
x
(
t
)
e

j
2
πkt
dt
=
1
/
2
Z

1
/
2
e
j
10
πt
+
e

j
10
πt
+
1
2
j
(
e
j
6
πt

e

j
6
πt
)
¶
e

j
2
πkt
dt
=
sin
π
(5

k
)
π
(5

k
)
+
sin
π
(5 +
k
)
π
(5 +
k
)
+
j
2
sin
π
(3 +
k
)
π
(3 +
k
)

sin
π
(3

k
)
π
(3

k
)
¶
=
1
k
=
±
5
j/
2
k
=

3

j/
2
k
= 3
0
else
We will now look at the alternative form of the Fourier series.
a
v
=
1
/
2
Z

1
/
2
(
2 cos 10
πt
+ sin 6
πt
)
dt
= 0
.
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 Spring '08
 HALVERSON
 Fourier Series, Steady State, Sin, dt, 0 k, 1 j

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