In Signal
Decomposition , we have shown that we could express the
square wave as a superposition of pulses. This superposition
does not generalize well to other periodic signals: How can a
superposition of pulses equal a smooth signal like a sinusoid?
Because of the importance of sinusoids to linear systems, you
might wonder whether they could be added together to represent
a large number of periodic signals. You would be right and in
good company as well.
Euler and Gauss in particular worried about this problem,
and Fourier
got the credit even though tough mathematical issues were not
settled until later. They worked on what is now known as the Fourier
series.

Let
st
s
t
have period
T
T. We want to show that periodic signals, even those that have
constant-valued segments like a square wave, can be expressed as
sum of *harmonically *related sine waves.

st=
a
0
+∑
k
=1∞
a
k
cos2πktT+∑
k
=1∞
b
k
sin2πktT
s
t
a
0
k
1
∞
a
k
2
π
k
t
T
k
1
∞
b
k
2
π
k
t
T

(1)The family of functions called

*basis functions *
cos2πktTsin2πktT
2
π
k
t
T
2
π
k
t
T
form the foundation of the Fourier series. No matter what
the periodic signal might be, these functions are always
present and form the representation's building blocks. They
do depend on the signal's period

T
T, and are indexed by

k
k. The frequency of each term is

kT
k
T
. For

k=0
k
0
, the frequency is zero and the corresponding term

a
0
a
0
is a constant. The basic frequency

1T
1
T
is called the

fundamental frequency because
all other terms have frequencies that are integer multiples of
it. These higher frequency terms are called

harmonics
: The term at frequency

1T
1
T
is the fundamental and the first harmonic,

2T
2
T
the second harmonic, etc. Thus, larger values of the series
index correspond to higher harmonic-frequency sinusoids. The

Fourier coefficients
a
k
a
k
,

b
k
b
k
depend on the signal's waveform. Because frequency is linked to
index, the coefficients implicitly depend on frequency.

Assuming we know the period, knowing the Fourier coefficients is
equivalent to knowing the signal.

Assume for the moment that the Fourier series works. To find the Fourier
coefficients, we note the following orthogonality properties of sinusoids.

∫0Tsin2×πktTcos2×πltTdt=0
for all
k,l
t
0
T
2
π
k
t
T
2
π
l
t
T
0
for all k,l

(2)
∫0Tsin2×πktTsin2×πltTdt={T2 if
(k=l)∧(k≠0)∧(l≠0)
0 if (k≠l)∨(k=0=l)
t
0
T
2
π
k
t
T
2
π
l
t
T
T
2
k
l
k
0
l
0
0
k
l
k
0
l

(3)
∫0Tcos2×πktTcos2×πltTdt={T2 if
(k=1)∧(k≠0)∧(l≠0)
T if k=0=l0 if k≠l
t
0
T
2
π
k
t
T
2
π
l
t
T
T
2
k
1
k
0
l
0
T
k
0
l
0
k
l

(4)
To use these, let's multiply the Fourier series for a square wave (

st=
a
0
+∑k=1∞
a
k
cos2×πktT+∑k=1∞
b
k
sin2×πktT
s
t
a
0
k
1
∞
a
k
2
π
k
t
T
k
1
∞
b
k
2
π
k
t
T
) by

cos2×πltT
2
π
l
t
T
and integrate. The idea is that, because integration is linear, the integration
will sift out all but the term involving

a
l
a
l
.

∫0Tstcos2×πltTdt=∫0T
a
0
cos2×πltTdt+∑k=1∞
a
k
∫0Tcos2×πktTcos2×πltTdt+∑k=1∞
b
k
∫0Tsin2×πktTcos2×πltTdt
t
0
T
s
t
2
π
l
t
T
t
0
T
a
0
2
π
l
t
T
k
1
∞
a
k
t
0
T
2
π
k
t
T
2
π
l
t
T
k
1
∞
b
k
t
0
T
2
π
k
t
T
2
π
l
t
T

(5)
The first and third terms are zero; in the second, the only non-zero
term in the sum results when the indices

k
k
and

l
l
are equal (but not zero), in which case we obtain

a
l
T2
a
l
T
2
. If

k=0=l
k
0
l
, we obtain

a
0
T
a
0
T
. Consequently,

a
l
=2T∫0Tstcos2×πltTdt
, l
≠
0
a
l
2
T
t
0
T
s
t
2
π
l
t
T
, l
≠
0
All of the Fourier coefficients can be found similarly.

a
0
=1T∫0Tstdt
a
0
1
T
t
0
T
s
t

(6)
a
k
=2T∫0Tstcos2×πktTdt
, k
≠
0
a
k
2
T
t
0
T
s
t
2
π
k
t
T
, k
≠
0

(7)
b
k
=2T∫0Tstsin2×πktTdt
b
k
2
T
t
0
T
s
t
2
π
k
t
T

(8)
The expression for
a
0
a
0
is referred to as the average value of
st
s
t
. Why?

The average of a set of numbers is the sum divided by the number of terms.
Viewing signal integration as the limit of a Riemann sum, the integral
corresponds to the average.

Let's find the Fourier series representation for the half-wave rectified sinusoid.

st={sin2×πtT if 0≤t<T20 if T2≤t<T
s t
2
π
t
T
0
t
T
2
0
T
2
t
T

(9)
Begin with the sine terms in the series; to find

b
k
b
k
we must calculate the integral

b
k
=2T∫0T2sin2×πtTsin2×π×ktTdt
b
k
2
T
t
0
T
2
2
π
t
T
2
π
k
t
T

(10)
The key to evaluating such integrals is the classic trigonometric identities.

sinαsinβ=12(cosα−β−cosα+β)
α
β
1 2
α β
α β

(11)
cosαcosβ=12(cosα+β+cosα−β)
α
β
1 2
α β
α β

(12)
sinαcosβ=12(sinα+β+sinα−β)
α
β
1 2
α β
α β

(13)
Using these identities turns our integral of a product of sinusoids into a sum of integrals of
individual sinusoids, which are much easier to evaluate.

∫0T2sin2×πtTsin2×π×ktTdt=12∫0T2cos2×π(k−1)tTcos2×π(k+1)tTdt={12 if k=10 if kk
(k≥2)∧k∈N
t
0
T
2
2
π
t
T
2
π
k
t
T
1
2
t
0
T
2
2
π
k
1
t
T
2
π
k
1
t
T
1
2
k
1
0
k
k
2
k
k

(14)
Thus,

,

b
2
=
b
3
=...=0.
b
2
b
3
...
0
.

(16)
On to the cosine terms. The average value, which corresponds to
a
0
a
0
, equals
1π
1
π
The remainder of the cosine coefficients are easy to find, but yield the complicated result

a
k
={(−2π)1k2−1 if k=2,4,...0 if
k is odd
a
k
2
π
1
k
2
1
k=2,4,...
0
k is odd

(17)
Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average,
the fundamental, and the even harmonics. Plotting the Fourier coefficients reveals at what
component frequencies the half-wave rectified sinusoid has energy
(
Figure 1
).
Furthermore, this figure shows what the Fourier series sum looks like with these
coefficients as we add more and more terms. Presumably, you now believe more in the
Fourier series.