Introduction to Signal Processing
Hayder Radha
 1 
ECE 366
Homework Set #5 Solutions
ECE 366 – Introduction to Signal Processing
Spring 2009
Michigan State University
Department of Electrical and Computer Engineering
[1]
Problem 6.1310
(a)
From Exercise E6.1a in the book, we have:
( )
(
)
2
2
1
1
( 1)
4
cos
3
n
n
x t
n t
n
π
π
∞
=
−
=
+
∑
.
We also have,
( )
2
x t
t
=
for
1
1
t
−
<
< +
with a period of
0
2
T
=
and hence a fundamental frequency
0
ω
π
=
.
We know that this is a power signal with a power
x
P
that can be evaluated using
the standard formula for average power. For a periodic signal, we have:
( )
0
2
0
1
x
T
P
x
t
dt
T
=
∫
.
Parseval’s theorem states that the signal power
x
P
can be evaluated using the
Fourier series amplitudes:
2
2
0
1
1
2
x
n
n
P
C
C
∞
=
=
+
∑
.
Here, we need to show that both of the above two expressions for
x
P
lead to the
same value.
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Introduction to Signal Processing
Hayder Radha
 2 
ECE 366
Now, using the standard power expression:
( )
(
)
0
1
2
2
2
0
1
1
1
1
2
5
x
T
P
x
t
dt
t
dt
T
+
−
=
=
=
∫
∫
.
Now, using the second expression for
x
P
:
2
2
2
2
2
0
2
2
1
1
1
( 1)
1
1
4
2
3
2
n
x
n
n
n
P
C
C
n
π
∞
∞
=
=
⎛
⎞
⎛
⎞
−
⎛
⎞
=
+
=
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
∑
∑
.
Hence, we have:
2
2
4
2
4
4
1
1
1
( 1)
1
1
16
1
8
3
2
9
n
x
n
n
P
n
n
π
π
∞
∞
=
=
⎛
⎞
−
⎛
⎞
=
+
=
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∑
∑
.
Now using the expression given by the problem statement regarding the
summation of the infinite series:
4
4
1
1
90
n
n
π
∞
=
=
∑
, leads to:
4
4
4
4
1
1
1
8
1
8
10
8
9
9
90
90
x
n
P
n
π
π
π
∞
=
+
=
+
=
+
=
∑
Hence, the power evaluation based on the Fourier series terms leads to:
18
1
90
5
x
P
=
=
.
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 Spring '08
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 Fourier Series, Signal Processing, Complex number, Hayder Radha

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