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:∫0Tsin2πktTcos2πltTdt=0
k
l
k
l
t
0
T
2
k
t
T
2
l
t
T
0
(2)
These orthogonality relations follow from the following important trigonometric identities.
sinαsinβ=12(cosα−β−cosα+β)
α
β
1
2
α
β
α
β
(3)
These identities allow you to substitute a sum of sines and/or cosines for a product of them.
Each term in the sum can be integrating by noticing one of two important properties of sinusoids.
-
The integral of a sinusoid over an integer number of periods equals zero.
-
The integral of the square of a sinusoid over a period equals
12
1 2
.
To use these, let's multiply the Fourier series for a signal
(
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
.
∫0Tstcos2πltTdt=∫0T
a
0
cos2πltTdt+∑k=1∞
a
k
∫0Tcos2πktTcos2πltTdt+∑k=1∞
b
k
∫0Tsin2πktTcos2πltTdt
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
(4)
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∫0Tstcos2πltTdt
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
(5)
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.
Finding the Fourier series coefficients for the square wave is very simple.
sqTt
sqT
t
.
Mathmatically, this signal can be expressed as
sqTt={1 if 0≤t<T2−1 if T2≤t<T
sqT
t
1
0
t
T2
1
T2
t
T
The expressions for the Fourier coefficients have the common form
a
k
b
k
=2T∫0T2cos2πktTsin2πktTdt−2T∫T2Tcos2πktTsin2πktTdt
a
k
b
k
2
T
t
0
T
2
2 k t
T
2 k t
T
2
T
t
T
2
T
2 k t
T
2 k t
T
(6)
The cosine coefficients
a
k
a
k
are all zero, and the sine coefficients are
b
k
={4πk if k is odd0 if k is even
b
k
4
k
k is odd
0
k is even
(7)
Thus, the Fourier series for the square wave is
sqt=∑k∈13…4πksin2πktT
sq
t
k
k
1
3
…
4
k
2
k
t
T
(8)
Consequently, the square wave equals a sum of sinusoids, but only those having frequencies
equal to odd multiples of the fundamental frequency
1T1T.
The coefficients
decay slowly as the frequency index
kk increases.
This index corresponds to the
kk-th harmonic of the signal's period.
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)
Using our trigonometric 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 otherwise
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
(11)
Thus,
b
1
=12
b
1
1
2
b
2
=
b
3
=…=0
b
2
b
3
…
0
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
(12)
Thus, the Fourier series for the half-wave rectified sinusoid
has non-zero terms for the average, the fundamental, and the
even harmonics.
