The classic Fourier series as derived originally expressed a periodic signal (period
TT) in terms of harmonically related sines and cosines.
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), each representing a
signal's spectrum.
The
Fourier coefficients,
ak
ak
and
bk
bk,
express the real and imaginary parts respectively of the spectrum while the coefficients
ck
ck
of the complex Fourier series express the spectrum as a magnitude and phase.
Equating the
classic Fourier series to the
complex Fourier series, an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each.
ck=12(ak−jbk)
ck
1
2
ak
bk
Derive this relationship between the coefficients of the two Fourier series.
Write the coefficients of the complex Fourier series in Cartesian form as
ck=Ak+jBk
ck
Ak
Bk
and substitute into the expression for the complex Fourier series.
∑k=−∞∞ckej2πktT=∑k=−∞∞(Ak+jBk)ej2πktT
k
ck
2kt
T
k
Ak
Bk
2kt
T
Simplifying each term in the sum using Euler's formula,
(Ak+jBk)ej2πktT
=
(Ak+jBk)(cos2πktT+jsin2πktT)
=
Akcos2πktT−Bksin2πktT+j(Aksin2πktT+Bkcos2πktT)
Ak
Bk
2kt
T
=
Ak
Bk
2kt
T
2kt
T
=
Ak
2kt
T
Bk
2kt
T
Ak
2kt
T
Bk
2kt
T
We now combine terms that have the same frequency index in magnitude.
Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry:
c−k=ck*
ck
ck
or
A−k=Ak
Ak
Ak
and
B−k=−Bk
Bk
Bk
.
After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes
2Akcos2πktT−2Bksin2πktT
2
Ak
2kt
T
2
Bk
2kt
T
.
To obtain the classic Fourier series, we must have
2Ak=ak
2
Ak
ak
and
2Bk=−bk
2
Bk
bk
.
Just as with the complex Fourier series, we can find the
Fourier coefficients using the
orthogonality properties of sinusoids.
Note that the cosine and sine of harmonically related frequencies, even the same frequency, are orthogonal.
∫0Tsin2πktTcos2πltTdt=0 ,
k∈Z
l∈Z
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)
cosαcosβ=12(cosα+β+cosα−β)
sinαcosβ=12(sinα+β+sinα−β)
α
β
1
2
α
β
α
β
α
β
1
2
α
β
α
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
integrated 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, for example, multiply the Fourier series for a signal by the cosine of the
lth
lth
harmonic
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,
a
l
=2T∫0Tstcos2πltTdt ,
l≠0
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
k
=2T∫0Tstcos2πktTdt ,
k≠0
b
k
=2T∫0Tstsin2πktTdt
a
0
1
T
t
0
T
s
t
k
k
0
a
k
2
T
t
0
T
s
t
2
k
t
T
b
k
2
T
t
0
T
s
t
2
k
t
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.
What is the Fourier series for a unit-amplitude square wave?
We found that the complex Fourier series coefficients are given by
ck=2jπk
ck
2
k
.
The coefficients are pure imaginary, which means
ak=0
ak
0
.
The coefficients of the sine terms are given by
bk=−(2Imck)
bk
2
ck
so that
bk={4πk if
k odd
0 if
k even
bk
4
k
k odd
0
k even
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
(6)
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
(7)
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
(8)
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
(9)
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
={−(2π1k2−1) if k∈24…0 if k odd
a
k
2
1
k
2
1
k
2
4
…
0
k odd
(10)
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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