Recall that in Chapter 1, we defined the power of a periodic signal as
p
x
=
1
T
∫
t
0
t
0
+
T
x
2
(
t
)
d
t
p
x
=
1
T
∫
t
0
t
0
+
T
x
2
(
t
)
d
t
(1)
where TT is the period. Using the complex form of the Fourier series, we can write
x
(
t
)
2
=
∑
n
=

∞
∞
c
n
e
j
n
Ω
0
t
∑
m
=

∞
∞
c
m
e
j
m
Ω
0
t
*
x
(
t
)
2
=
∑
n
=

∞
∞
c
n
e
j
n
Ω
0
t
∑
m
=

∞
∞
c
m
e
j
m
Ω
0
t
*
(2)
where we have used the fact that x(t)2=x(t)x(t)*x(t)2=x(t)x(t)*, i.e. since x(t)x(t) is real x(t)=x(t)*x(t)=x(t)*. Applying (Reference) and (Reference) gives
x
(
t
)
2
=
∑
n
=

∞
∞
c
n
e
j
n
Ω
0
t
∑
m
=

∞
∞
c
m
*
e

j
m
Ω
0
t
=
∑
n
=

∞
∞
∑
m
=

∞
∞
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
=
∑
n
=

∞
∞
c
n
2
+
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
x
(
t
)
2
=
∑
n
=

∞
∞
c
n
e
j
n
Ω
0
t
∑
m
=

∞
∞
c
m
*
e

j
m
Ω
0
t
=
∑
n
=

∞
∞
∑
m
=

∞
∞
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
=
∑
n
=

∞
∞
c
n
2
+
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
(3)
Substituting this quantity into Equation 1 gives
p
x
=
1
T
∫
t
0
t
0
+
T
∑
n
=

∞
∞
c
n
2
+
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
=
∑
n
=

∞
∞
c
n
2
+
1
T
∫
t
0
t
0
+
T
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
p
x
=
1
T
∫
t
0
t
0
+
T
∑
n
=

∞
∞
c
n
2
+
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
=
∑
n
=

∞
∞
c
n
2
+
1
T
∫
t
0
t
0
+
T
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
(4)
It is straightforward to show that
1
T
∫
t
0
t
0
+
T
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
=
0
1
T
∫
t
0
t
0
+
T
∑
n
≠
m
c
n
c
m
*
e
j
(
n

m
)
Ω
0
t
d
t
=
0
(5)
This leads to Parseval's Theorem for the Fourier series:
p
x
=
∑
n
=

∞
∞
c
n
2
p
x
=
∑
n
=

∞
∞
c
n
2
(6)
which states that the power of a periodic signal is the sum of the magnitude of the complex Fourier series coefficients.