A major goal of this book is to develop tools which will enable us to study the frequency content of signals. An important first step is the Fourier Series.
The Fourier Series enables us to completely characterize the frequency content of a periodic signal. A periodic signal x(t)x(t) can be expressed in terms of the Fourier Series, which is given by:
x
(
t
)
=
a
0
+
∑
n
=
1
∞
a
n
cos
(
n
Ω
0
t
)
+
∑
n
=
1
∞
b
n
sin
(
n
Ω
0
t
)
x
(
t
)
=
a
0
+
∑
n
=
1
∞
a
n
cos
(
n
Ω
0
t
)
+
∑
n
=
1
∞
b
n
sin
(
n
Ω
0
t
)
(1)
where
Ω
0
=
2
π
T
Ω
0
=
2
π
T
(2)
is the fundamental frequency of the periodic signal.
Examination of Equation 1 suggests that periodic signals can be represented as a sum of suitably scaled cosine and sine waveforms at frequencies of Ω0,2Ω0,3Ω0,...Ω0,2Ω0,3Ω0,.... The cosine and sine terms at frequency nΩ0nΩ0 are called the nthnth harmonics. Evidently, periodic signals contain only the fundamental frequency and its harmonics. A periodic signal cannot contain a frequency that is not an integer multiple of its fundamental frequency.
In order to find the Fourier Series, we must compute the Fourier Series coefficients. These are given by
a
0
=
1
T
∫
t
0
t
0
+
T
x
(
t
)
d
t
a
0
=
1
T
∫
t
0
t
0
+
T
x
(
t
)
d
t
(3)
a
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
a
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
(4)
b
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
b
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
(5)
From our discussion of even and odd symmetric signals, it is clear that if x(t)x(t) is even, then x(t)sin(nΩ0t)x(t)sin(nΩ0t) must be odd and so bn=0bn=0. Also if, x(t)x(t) has odd symmetry, then x(t)cos(nΩ0t)x(t)cos(nΩ0t) also has odd symmetry and hence an=0an=0 (see exercise (Reference)). Moreover, if a signal is even, since x(t)cos(nΩ0t)x(t)cos(nΩ0t) is also even, if we use the fact that for any even symmetric periodic signal v(t)v(t),
∫

T
/
2
T
/
2
v
(
t
)
d
t
=
2
∫
0
T
/
2
v
(
t
)
d
t
∫

T
/
2
T
/
2
v
(
t
)
d
t
=
2
∫
0
T
/
2
v
(
t
)
d
t
(6)
then setting t0=T/2t0=T/2 in Equation 4 gives,
a
n
=
4
T
∫
0
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
a
n
=
4
T
∫
0
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
(7)
This can sometimes lead to a savings in the number of integrals that must be computed. Similarly, if x(t)x(t) has odd symmetry, we have
b
n
=
4
T
∫
0
T
/
2
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
b
n
=
4
T
∫
0
T
/
2
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
=
1
,
2
,
...
(8)
Example 2.1 Consider the signal in Figure 1. This signal has even symmetry, hence all of the bn=0bn=0. We compute a0a0 using,
a
0
=
1
T
∫
t
0
t
0
+
T
x
(
t
)
d
t
a
0
=
1
T
∫
t
0
t
0
+
T
x
(
t
)
d
t
(9)
which we recognize as the area of one period, divided by the period. Hence, a0=τ/Ta0=τ/T. Next, using Equation 4 we get
a
n
=
2
T
∫

τ
/
2
τ
/
2
cos
(
n
Ω
0
t
)
d
t
a
n
=
2
T
∫

τ
/
2
τ
/
2
cos
(
n
Ω
0
t
)
d
t
(10)
Note how the limits of integration only go from τ/2τ/2 to τ/2τ/2 since x(t)x(t) is zero everywhere else. Evaluating this integral leads to
a
n
=
2
τ
T
s
i
n
n
Ω
0
τ
/
2
n
Ω
0
τ
/
2
,
n
=
1
,
2
,
...
a
n
=
2
τ
T
s
i
n
n
Ω
0
τ
/
2
n
Ω
0
τ
/
2
,
n
=
1
,
2
,
...
(11)
Figure 2 shows the first few Fourier Series coefficients for τ=1/2τ=1/2 and T=1T=1. If we attempt to reconstruct x(t)x(t) based on only a limited number (say, NN) of Fourier Series coefficients, we have
x
^
(
t
)
=
a
0
+
∑
n
=
0
N
a
n
cos
(
n
Ω
0
t
)
x
^
(
t
)
=
a
0
+
∑
n
=
0
N
a
n
cos
(
n
Ω
0
t
)
(12)
Figures Figure 3 and Figure 4 show x^(t)x^(t) for N=10N=10, and N=50N=50, respectively.
The ringing characteristic is known as Gibb's phenomenon and disappears only as NN approaches ∞∞.
The following example looks at the Fourier series of an oddsymmetric signal, a sawtooth signal.
Example 2.2 Now let's compute the Fourier series for the signal in Figure 5. The signal is oddsymmetric, so all of the anan are zero. The period is T=3/2T=3/2, hence Ω0=4π/3Ω0=4π/3.
Using Equation 5, the bnbn coefficients are found by computing the following integral,
b
n
=
8
3
∫

1
/
2
1
/
2
t
sin
(
4
π
n
t
/
3
)
d
t
b
n
=
8
3
∫

1
/
2
1
/
2
t
sin
(
4
π
n
t
/
3
)
d
t
(13)
After integrating by parts, we get
b
n
=
3
sin
(
2
π
n
/
3
)
(
π
n
)
2

2
cos
(
2
π
n
/
3
)
π
n
,
n
=
1
,
2
,
...
b
n
=
3
sin
(
2
π
n
/
3
)
(
π
n
)
2

2
cos
(
2
π
n
/
3
)
π
n
,
n
=
1
,
2
,
...
(14)
These are plotted in Figure 6 and approximations of x(t)x(t) using N=10N=10 and N=50N=50 coefficients are shown in Figures Figure 7 and Figure 8, respectively.
References