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.
Key point:
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)
∫0Tsin2πktTsin2πltTdt=T2ifk=l∧k≠0∧l≠00ifk≠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
∫0Tcos2πktTcos2πltTdt=T2ifk=l∧k≠0∧l≠0Tifk=0=l0ifk≠l
t
0
T
2
k
t
T
2
l
t
T
T
2
k
l
k
0
l
0
T
k
0
l
0
k
l
These orthogonality relations follow from the following
important trigonometric identities.
sinαsinβ=12cosα-β-cosα+β
α
β
1
2
α
β
α
β
(3)
cosαcosβ=12cosα+β+cosα-β
α
β
1
2
α
β
α
β
sinαcosβ=12sinα+β+sinα-β
α
β
1
2
α
β
α
β
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
unit-amplitude sinusoid over a period
TT equals
T2
T 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)
∀k,k≠0:
a
k
=2T∫0Tstcos2πktTdt
k
k
0
a
k
2
T
t
0
T
s
t
2
k
t
T
b
k
=2T∫0Tstsin2πktTdt
b
k
2
T
t
0
T
s
t
2
k
t
T
Problem 1
The expression for
a
0
a
0
is referred to as the average value of
st
s
t
.
Why?
[
Click for Solution 1 ]
Solution 1
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.
[
Hide Solution 1 ]
Example 1
Finding the Fourier series coefficients for the square wave is
very simple.
sqTt
sqT
t
.
Mathematically, this signal can be expressed as
sqTt=1if0≤t<T2-1ifT2≤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
=4πkifk is odd0ifk 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
1T
1
T
. The coefficients decay slowly as the frequency
index
kk increases. This index
corresponds to the
kk-th harmonic
of the signal's period.
Example 2
Let's find the Fourier series representation for the half-wave
rectified sinusoid.
st=sin2πtTif0≤t<T20ifT2≤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∫0T2cos2πk-1tT-cos2πk+1tTdt=12ifk=10otherwise
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
=-2π1k2-1ifk∈24…0ifk 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.
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