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
k
t
T
k
1
b
k
2
k
t
T
(1)
The family of functions called
basis functions
cos2πktTsin2πktT
2
k
t
T
2
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
and
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.
∀
k
l
,k∈Zl∈Z:∫0Tsin2πktTcos2πltTd
t
=0
k
l
k
l
t
0
T
2
k
t
T
2
l
t
T
0
(2)
∫0Tsin2πktTsin2πltTdt={T2 if
(k=l)∧(k≠0)∧(l≠0)
0 if (k≠l)∨(k=0=l)
t
0
T
2
k
t
T
2
l
t
T
T
2
k
l
k
0
l
0
0
k
l
k
0
l
(3)
∫0Tcos2πktTcos2πltTdt={T2 if
(k=1)∧(k≠0)∧(l≠0)
T if k=0=l0 if k≠l
t
0
T
2
k
t
T
2
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
cos2πktT+∑k=1∞
b
k
sin2πktT
s
t
a
0
k
1
a
k
2
k
t
T
k
1
b
k
2
k
t
T
)
by
cos2πltT
2
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
.
∫0Tstcos2πltTd
t
=∫0T
a
0
cos2πltTd
t
+∑
k
=1∞
a
k
∫0Tcos2πktTcos2πltTd
t
+∑
k
=1∞
b
k
∫0Tsin2πktTcos2πltTd
t
t
0
T
s
t
2
l
t
T
t
0
T
a
0
2
l
t
T
k
1
a
k
t
0
T
2
k
t
T
2
l
t
T
k
1
b
k
t
0
T
2
k
t
T
2
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,
∀
l
,l≠0:
a
l
=2T∫0Tstcos2πltTd
t
l
l
0
a
l
2
T
t
0
T
s
t
2
l
t
T
All of the Fourier coefficients can be found similarly.
a
0
=1T∫0Tstdt
a
0
1
T
t
0
T
s
t
(6)
∀
k
,k≠0:
a
k
=2T∫0Tstcos2ktTdt
k
k
0
a
k
2
T
t
0
T
s
t
2
k
t
T
(7)
b
k
=2T∫0Tstsin2πktTdt
b
k
2
T
t
0
T
s
t
2
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={sin2πtT if 0≤t<T20 if T2≤t<T
s
t
2
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∫0T2sin2πtTsin2πktTdt
b
k
2
T
t
0
T
2
2
t
T
2
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.
∫0T2sin2πtTsin2πktTdt=12∫0T2cos2π(k−1)tT−cos2π(k+1)tTdt={12 if k=10 if kk
(k≥2)∧k∈N
t
0
T
2
2
t
T
2
k
t
T
1
2
t
0
T
2
2
k
1
t
T
2
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
.
The remainder of the cosine coefficients are easy to find, but
yield the complicated result
a
k
={(−2π)1k2−1 if k∈24…0 if k is odd
a
k
2
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.
