Chapter 16
Fourier Series Analysis
16.0 Introduction
Many electrical waveforms are period but not sinusoidal. For analysis purposes,
such waveform can be represented in series form based on the original work of
Jean Baptise Joseph Fourier.
The application of Fourier-series method includes signal generators, power
supplies, and communication circuits. Fourier series decomposes non-sinusoidal
waveform into series of sinusoidal components of various frequencies. With this
property, frequency-domain representation or spectrum for periodic waveform is
developed. The spectral concept ties the relationship between time-domain and
frequency-domain properties of waveform. In this chapter, we shall the various
methods to generate Fourier series and the application of Fourier series in ac
steady-state circuit analysis.
16.1 Fourier Series
The period waveform of function f(t) is repetition over time such that
f(t-mT) = f(t) m = 1, 2, 3, …..
(16.1)
where T is the period. When m = 1, mT becomes T, which is the smallest T and it
is termed as fundamental period.
Theoretically equation (16.1) is true for value of t ranges from - ∞ to ∞. But
in practice, the waveform lasts only for a finite amount time. The assumption can
be true if the period T is small as compared with duration of repeating waveform.
The net area under a periodic waveform f(t) over any period is independent of
where the period begins. Thus, the integration of the f(t) over at any begin point is
equal.
t1 + T
t 2 +T
t1
t2
∫ f (t )dt = ∫ f ( t)dt
(16.2)
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16.2 Trigonometric Fourier Series
Fourier series state that almost any periodic waveform f(t) with fundamental
frequency ω can be expanded as an infinite series in the form
∞
f(t) = a0 + ∑ (a n cos nωt + b n sin nωt )
(16.3)
n =1
Equation (16.3) is called the trigonometric Fourier series and the constant C0, an,
and bn are dependent on f(t). All the oscillatory components are integer multiple of
fundamental angular frequency ω or harmonics. Fourier series can also be
expressed in exponential form, in which we will deal with later.
By including an infinite number of harmonics, Fourier series can represent
any “well-behaved” period function. This well-behaved periodic function is
defined by Dirichlet’s condition, which states the function must be single-valued,
must have a finite number of maxima, minima, and discontinuities per period and
the integral ∫T | f ( t ) | dt must be finite. Put in another word. When Dirichlet’s
condition hold, the infinite series summation converges to the value of f(t)
wherever the waveform is continuous.
The infinite series has orthogonal property meaning that that the integral over
one period of the product of any two different terms vanishes. Thus,
∫ cos(nωt )dt = ∫ sin (nωt )dt = 0 , ∫ cos(nωt ) ⋅ sin(mωt )dt = 0 , ∫ cos(nωt ) ⋅ cos(mωt )dt = 0 for n ≠
T
T
∫ sin(nωt ) ⋅ sin (mωt )dt = 0
m,
T
T
for
n
≠
m.
However,
for
n
=
T
T
∫ cos (nωt )dt = ∫ sin (nωt )dt = 2 .
2
2
T
T
Reference to equation (16.3), integration of the f(t) over a period T shall be
∫
f ( t )dt =
T
∫ a 0 dt
T
Equation (16.4) is equal to
∞
+ ∑ (a n ∫ cos nωtdt + b n ∫ sin nωtdt )
n =1
∫ f ( t)dt
T
(16.4)
T
= a0T. Thus, the constant a0 is
T
a0 =
1
T
∫ f ( t)dt
(16.5)
T
Note that a0 is also the average value of function f(t).
- 258 -
m,
The coefficient an is determined by multiplying equation (16.3) with cosmωt
and integrating the equation for a period T. This equation used is
∫ cos(mωt ) f ( t )dt
T
∞
= ∫ C 0 dt + ∑ (a n ∫ cos mωt ⋅ cos nωtdt + b n ∫ cos mωt ⋅ sin nωtdt ) (16.6)
n =1
T
T
T
∞
Knowing that
∑ ∫ cos mωt ⋅ sin nωtdt = 0 and ∫ cos mωt ⋅ cos nωtdt = 0
n =1 T
∞
T
2
∑ ∫ cos mωt ⋅ cos nωtdt =
n =1 T
∫ cos(mωt) f ( t)dt = a
T
an =
for all n ≠ m, the
T
n
for n = m. Thus, equation (16.6) shall be
T
. The coefficient am shall follow equation (16.7).
2
2
2
cos(mωt) f ( t )dt = ∫ cos(nωt) f ( t )dt for n = m
∫
TT
TT
(16.7)
The coefficient bn is determined by multiplying equation (16.3) with sinmωt and
integrating the equation for a period T. The equation used is
∫ sin(mωt) f (t )dt =
∫a
T
T
∞
0
dt
+ ∑ (a n ∫ sinmωt ⋅ cos nωtdt + b n ∫ sin nωt ⋅ sin mωtdt )
n =1
T
T
(16.8)
∞
Knowing that
∑ ∫ sin mωt ⋅ sin nωtdt = 0 and ∫ sin mωt ⋅ cos nωtdt = 0
n =1 T
∞
∑ ∫ sin mωt ⋅ sin nωtdt =
n =1 T
bn
for all n ≠ m, the
T
T
for n = m. Thus, equation (16.6) shall be
2
∫ sin(mωt) f ( t )dt =
T
T
. The coefficient bn shall follow equation (16.7).
2
bn =
2
2
sin(mωt) f ( t )dt = ∫ sin( nωt) f ( t )dt for n = m
∫
TT
TT
(16.9)
Alternative form of equation (16.3) is the amplitude-phase form, which is
∞
f(t) = a0 + ∑ (A n cos nωt + φ n )
(16.10)
n =1
Knowing that cos(α+β) = cosαcosβ-sinαsinβ, equation (16.10) shall become
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∞
∞
n =1
n =1
f(t) = a0 + ∑ (A n cos φ n cos nωt ) − ∑ (A n sin φ n sin nωt )
(16.11)
Equating the coefficient of equation (16.3) and (16.11), it gives rise to an =
A n cos φ n and bn = − A n sin φ n . This shall also mean that A n = a 2n + b 2n and
b 
φ n = − tan −1  n  . The relationship between amplitude and phase can also be
 an 
expressed in phasor form, which is A n ∠φ n = a n − jb n .
Based on the above discussion, a function
f(t) = A cosnt –Bsinnt = A 2 + B 2 cos nt + tan −1

B

A
(16.12)
The plot of amplitude An of harmonic versus nω is called amplitude spectrum of
f(t) and the plot of phase φn versus nω is called phase spectrum of f(t). Both the
amplitude and phase spectra form the frequency spectrum of f(t).
Example 16.1
A rectified half sine wave is defined over one period f(t) = Asinωt for 0 < t < T/2
and f(t) = 0 for T/2 < t < T as shown in Fig. 16.1. Find the Fourier series of this
wave form.
Figure 16.1: A half-wave rectifier
Solution
The dc voltage shall be a0
cosine coefficient an =
1
T
1
=
T
1
−A
ωT  A
∫0 A sin ωtdt + T T∫/ 20dt = ωT  cos 2 − 1 = π . The
T/2
T/2
∫ A sin ωt cos nωtdt =
0
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T
π
A
sinα cos nαdα , after letting α = ωt.
π ∫0
The coefficient an = −
A 1 − cos(n − 1)π 1 − cos(n + 1)π 
−
 for n ≠ 1. Knowing that
2π 
n −1
n +1
cos(n ± 1)π = -1 when n is even and cos(n ± 1)π = 1 when n is odd. Thus, an =
−
2A
for n = 2, 4, 6,….. and an = 0 for n = 3, 5, 7,……. a1 is found to be
π(n 2 − 1)
π
A
sin α sin nαsα = 0.
π ∫0
π
The sine coefficient bn is
A
sin α sin nαdα = A/2 for n = 1 and bn = 0 for n = 2, 3, 4,
π ∫0
5… Knowing the coefficient values, the rectified half-wave Fourier series shall be
f(t)
=
A A
2A
2A
2A
+ sin ωt −
cos 2ωt −
cos 4ωt −
cos 6ωt − ................. =
π 2
3π
15π
35π
∞
A A
2A
+ sin ωt − ∑
cos 2nωt . The plot of the rectified half-wave based on the
2
π 2
n =1 π( 4n − 1)
Fourier series is shown in Fig. 16.2.
Figure 16.2: The plot of f(t) =
A A
2A
2A
2A
+ sin ωt −
cos 2ωt −
cos 4ωt −
cos 6ωt
3π
15π
35π
π 2
16.3 Exponential Fourier Series
Another way of expressing Fourier series is in exponential form. It is done by
applying Euler’s rule to equation (16.3). The equation shall be
f(t) = a0 +
1 ∞
1 ∞
(a n − jb n )e jnωt + ∑ (a n + jb n )e − jnωt
∑
2 n =1
2 n =1
(16.13)
Letting a0 = c0ej0t and summing over both positive and negative values of n, the
compact expression shall be
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n =∞
f(t) =
∑c e
n = −∞
jnωt
(16.14)
n
Equation (16.14) has an advantage that the trigonometric expansion because cn
represents all series coefficients. Thus, exponential series is preferred over the
trigonometric series for analytical investigation.
Like the trigonometric series, the exponential series are orthogonal in the
T
T
0
0
sense that ∫ e jnωt e − jmωt dt = 0 for n ≠ m and ∫ e jnωt e − jmωt dt = T for n = m.
Multiply equation (16.14) by e − jmωt yields
T
∫ f ( t )e
− jmωt
n =∞
dt =
∑ c ∫e
n = −∞
0
T
jnωt
n
e − jmωt dt
(16.15)
0
Knowing all terms with n ≠ m varnish and remaining term with n = m reduces to
cmT, thus,
T
∫ f ( t )e
− jmωt
dt = c m T
(16.16)
0
Hence the coefficient cn shall be
T
cn =
1
f ( t )e − jnωt dt
∫
T0
(16.17)
Equation (16.17) holds for all values of n including n = 0. When n = 0, the
equation reduces to equation (16.5).
For n ≠ 0, insert e − jnωt = cosnωt - jsinnωt to equation (16.17), it becomes
T
cn =
T
1
1
f ( t ) cos nωtdt − ∫ f ( t ) j sin nωtdt
∫
T0
T0
(16.18)
Exchanging the sign of n, it gives rise to c-n = c *n .
Comparing equation (16.18) with equation (16.7) and (16.9), its concludes
that an = 2Re[cn] and bn = -2 Im[cn] for n ≥ 1.
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Based on the above analysis, a new set of coefficients shall be defined, which
are c0 = a0, c n =
a n − jb n
a + jb n
, and c −n = c *n = n
.
2
2
Example 16.2
Determine the complex Fourier series for the waveform shown in Fig. 16.3.
Figure 16.3: The square wave
Solution
T
−T / 4
T/4
T/2

1
1
The coefficient c0 = ∫ f ( t ) =  ∫ − Adt + ∫ Adt − ∫ Adt  = 0
T0
T  −T / 2
−T / 4
T/4

−T / 4
T/4
T/2

1
The coefficient cn =  ∫ − Ae jnωt dt + ∫ Ae jnωt dt − ∫ Ae jnωt dt 
T  −T / 2
−T / 4
T/4

−
T
/
4
T
/
4
T
/
2

A  e jnωt
e jnωt
e jnωt

+
−
= −
T  jnω −T / 2 jnω −T / 4 jnω T / 4 


A
=
− e − jnπ / 2 + e - jnπ + e jnπ / 2 − e - jnπ / 2 − e jnπ + e jnπ / 2
jnωT
A
=
− 2e − jnπ / 2 + e - jnπ − e jnπ + 2e jnπ / 2
jnωT
A
=
[4 j sin(nπ / 2) − 2 j sin(nπ)]
j 2 nπ
A
= [2 sin(nπ / 2) − sin(nπ)]
nπ
for n even
0

Thus, the coefficient for cn is c n =  2A
 nπ sin(nπ / 2) for n odd
[
[
Let n = 1, c1 =
]
2A
2A
. This implies that c-1 is also equal to .
π
π
Let n = 3, c3 = −
2A
2A
. This implies that c-3 is also equal to −
.
3π
3π
- 263 -
]
Let n = 5, c5 =
2A
2A
. This implies that c-5 is also equal to
.
5π
5π
Let n = 7, c7 = −
2A
2A
. This implies that c-7 is also equal to −
.
7π
7π
n =∞
Expansion of function f(t) according to equation (16.13)
∑c e
n = −∞
n
jnωt
for n = -7 to n =
7 yields
2 A − j 7 ωt 2 A − j 5 ωt
e
e
+
7π
5π
2 A − j 3 ωt 2 A − j ωt 2 A j 3 ωt 2 A j ωt 2 A j 5 ωt 2 A j 7 ωt
−
e
+ e − e + e +
e
−
e
+..
3π
π
3π
π
5π
7π
4A
4A
4A
4A
cos ωt −
cos 3ωt +
=
cos 5ωt −
cos 7ωt +…….
π
3π
5π
7π
4A ∞ (−1) ( n−1) / 2
=
∑ n cos nωt .
π n=1
f(t) = ….. −
n = odd
The plot of the square wave function based on Fourier series is shown in Fig. 16.4.
Figure 16.4: The plot of square wave f(t) =
4A ∞ (−1) ( n−1) / 2
∑ n cos nωt for n =1, 3, 5, 7
π n=1
n = odd
16.4 Symmetry Considerations
The analysis done so far pointed out that the Fourier series mostly consists of
either sine terms or cosine terms. One may ask if there a method that can be used
in advance to avoid tedious mathematical process. Such method does exist based
on recognizing the existence of symmetry, which are even symmetry, odd
symmetry, and half-wave symmetry.
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16.4.1 Even Symmetry
A function f(t) is even if its plot is symmetrical about the vertical axis; that is
f(t) = f(-t)
(16.19)
Examples of the even function are t2, t4, and cos t. Figure 16.5 shows an example
of periodic even function.
Figure 16.5: An even periodic function
T/2
A main property of an even function f(t) is that
∫
−T / 2
T/2
f ( t )dt = 2
∫ f (t )dt .
The
0
coefficient of Fourier series for even function shall be
a0 =
an =
2
T
T/2
4
T
T/2
∫ f (t )dt
0
∫ f ( t) cos(nωt )dt
0
bn = 0
(16.20)
16.4.2 Odd Symmetry
A function f(t) is said to be odd if the plot is anti-symmetrical about the vertical
axis; that is
f(-t) = -f(t)
(16.21)
T/2
The main characteristic of an odd function is
∫ f (t )dt = 0 . The coefficients of the
−T / 2
Fourier series are
- 265 -
T
1
a0 = ∫ f ( t )dt
T0
an = 0
bn =
4
T
T/2
∫ f (t ) sin(nωt )dt
(16.22)
0
16.4.3 Half-Wave Symmetry
A function is half-wave (odd) symmetric if
 T
f  t -  = − f (t)
 2
(16.23)
This shall mean that each half-cycle is the mirror image of the next half-cycle. An
example of half-wave symmetrical function is shown in Fig. 16.6.
Figure 16.6: Half-wave symmetrical function
The coefficient of the Fourier series shall be
a0 = 0
 4 T/2
f ( t ) cos(nωt )dt for n odd

a n =  T ∫0
0
for n even

 4 T/2
f ( t ) sin (nωt )dt for n odd

b n =  T ∫0
0
for n even

(16.24)
16.5 Circuit Applications of Fourier Series
In practice, many circuits are driven by non-sinusoidal periodic function. To
obtain the steady state response of the circuit to a non-sinusoidal periodic function
- 266 -
requires the application of Fourier series, ac phasor analysis, and superposition
principle. The procedure usually involves three steps, which are
•
•
•
•
Express the excitation as Fourier series.
Transform the circuit from time domain to the frequency domain.
Find the response of dc and ac components in the Fourier series.
Add the individual dc and ac response using the superposition principle.
Let’s use two examples to illustrate the procedures.
Example 16.3
The circuit shown in Fig. 16.7 has a non-sinusoidal vs(t) source that has Fourier
series vs(t) =
1 2 ∞ 1
+ ∑ sin (nπt ) for n = 2k -1. Find the voltage vo(t) at inductor and
2 π k =1 n
the corresponding amplitude spectrum.
Figure 16.7: An ac circuit
Solution
The output voltage vo(t) is v o ( t ) =
vo(t) shall be vo ( t ) =
jω n L
v s ( t ) . From the input vs(t), that ωn = nπ,
R + jω n L
j 2 nπ
vs ( t ) . The dc component shall be zero after substituting
5 + j 2 nπ
ωn = 0 into vo(t).
The phasor of sine component of the ac portion is
Thus, the output vo(t) shall be vo(t) =
4
25 + 4n π
2
2
∠ − tan −1 (2nπ / 5) ⋅
- 267 -
2
∠ − 90 0 .
nπ
2nπ∠90 0
25 + 4n 2 π 2 ∠ tan −1 (2nπ / 5)
⋅
2
∠ − 90 0 =
nπ
In
time-domain
4
25 + 4n π
2
2
[
the
output
voltage
]
shall
be
vo(t)
=
cos nπt − ∠ tan −1 (2nπ / 5) .
∞
The overall output vo(t) =
∑
n =1
n = odd
4
25 + 4n π
2
2
[
]
cos nπt − ∠ tan −1 (2nπ / 5) V.
The first four harmonics of the output voltage are vo(t) = 0.50 cos(π-51.490) + 0.21
cos(3π - 75.140) + 0.13 cos(5π - 80.960) + 0.10 cos(7π - 83.10) +……… V.
The amplitude spectrum of the output voltage is shown in Fig. 16.8.
Figure 16.8: Amplitude spectrum of the output voltage
Example 16.4
Find the response io(t) in the circuit shown in Fig. 16.9 if the input voltage v(t) has
2(−1) n
the Fourier series expansion v(t) = 1+ ∑
(cos nt − n sin nt ) .
2
n =1 1 + n
∞
Figure 16.9: ac series-parallel circuit
Solution
- 268 -
∞
2(−1) n
2(−1) n
cos
n
t
−
n sin nt and
∑
2
2
n =1 1 + n
n =1 1 + n
∞
Rewrite the input voltage function v(t) = 1+ ∑
convert the input voltage function to amplitude-phase form using equation (16.12).
Thus, A =
A 2 + B2 =
2(−1) n
2(−1) n n
B
and
B
=
−
. The phase φn = - tan −1   = tan −1 n . An =
2
2
1+ n
1+ n
A
n
2(−1)
1 + n2
.
The input voltage v(t) shall then be equal to v(t) = 1+
∞
The phasor form is 1+
∑
2(−1)
n
1+ n
n =1
2
∞
2(−1) n
n =1
1+ n
∑
2
(cos nt + tan −1 n) .
∠ tan −1 n .
From the equation ω = 1 rad/s and ωn = n rad/s.
The total impedance of the circuit Z = 4 + jωn2||4 = 4+

jω n 8
8 + jω n 8
=
.
4 + jω n 8 2 + jω n
2(−1) n ∠ tan −1 n  2 + jω n
.
⋅
8
+
j
ω
8
n =1
1+ n2
n

∞
The current flows in the circuit shall be I = 1 + ∑

The output current shall be io(t)
=
4
⋅
4 + jω n 2
∞
∞

2(−1) n ∠ tan −1 n  2 + jω n 
2(−1) n ∠ tan −1 n 
1
=
.
1
+
⋅
1
+
⋅
 ∑



∑
2
2
8
+
j
ω
8
4
+
j
ω
4
=
=
n
1
n
1
1
+
n
1
+
n
n
n




Setting ωn = 0, the dc current shall be
1
1
1
= .
=
4 + jω n 4 4 + j 0 x 4 4
∞
The
ac
component
shall
∑
be
n =1
∞
=∑
n =1
2(−1) n ∠ tan −1 n
1 + n 2 ⋅ 4∠ tan −1 n
∞
(−1) n
n =1
2 1 + n2
=∑
2(−1) n ∠ tan −1 n
1 + n2
⋅
1
4 + jn 4
.
(−1) n
∞
In time-domain format, ac component is
∑
n =1
Thus, the current io(t) shall be io(t) =
1
+
4
∞
∑
n =1
2 1 + n2
(−1) n
2 1 + n2
cos nt .
cos nt A.
16.6 Average Power of Period Functions
As shown in equation (16.1) for the amplitude-phase Fourier series of a periodic
function, the voltage and current functions at the terminal of the network are
∞
v(t) = Vdc +
∑V
n =1
n
cos(nωt − θ vn )
- 269 -
(16.25)
∞
i(t) = Idc +
∑I
n =1
n
cos(nωt − θ in )
(16.26)
In Chapter 10 AC Power Analysis, we learnt that the average power is
T
Pavg =
1
P( t )dt
T ∫0
(16.27)
Thus, the average power expressed in amplitude-phase Fourier series shall be
Pavg =
∞
1 T
1 T
V
I
dt
+
∑
dc dc
∫
∫0 Vn I n cos(nωt − θ vn ) cos(nωt − θ in )dt
T 0
n =1 T
(16.28)
Equation (16.28) which is the average power, shall finally become
Pavg = VdcIdc +
1 ∞
∑Vn I n cos(θ v − θ i )
2 n =1
(16.29)
Example 16.5
Determine the average power supplied to the circuit shown in Fig. 16.10 if i(t) = 2
+ cos(t + 100) + 6 cos(3t + 350) A.
Figure 16.10: An ac parallel circuit
Solution
1
10
=
.
j 2ω 1 + j 20ω
10I
10I
Hence the voltage v(t) is v(t) = ZxI =
=
1 + j 20ω
1 + 400ω 2 ∠ tan −1 (20ω)
The impedance of the circuit Z is Z = 10 ||
For dc component, I = 2 A and ω = 0, v(t) = 20 V.
For I = 10 cos(t+100) and ω = 1, v(t) =
10 x10∠10 0
1 + 400∠ tan
0
77.14 ).
- 270 -
−1
(20 )
= 5∠-77.140 = 5 cos(t –
For I = 6 cos(3t + 350) A and ω = 3, v(t) =
10 x 6∠35 0
1 + 400 x 9∠ tan
−1
(20 x3)
= 1∠-54.040 =
0
cos(3t – 54.04 ).
The overall voltage v(t) = 20 + 5 cos(t – 77.140) + cos(3t – 54.040).
The average power supplied to the circuit follows equation
Pavg = VdcIdc +
[
[
1
1 ∞
(5)(10) cos 77.14 0 − (−10 0 )
Vn I n cos(θ vn − θ in ) = 20(2) +
∑
2
2 n =1
]
+
]
1
(1)(6) cos 54.04 0 − (−35 0 ) = 40 + 1.247 + 0.05 = 41.5 W.
2
The supply voltage is v(t) = 20 + 5 cos(t – 77.140) + cos(3t – 54.040).
The supply current is i(t) = 2 + cos(t + 100) + 6 cos(3t + 350) A.
The supply power is 20(2) +
[
]
[
1
1
(5)(10) cos 77.14 0 − (−10 0 ) + (1)(6) cos 54.04 0 − (−35 0 )
2
2
]
= 40 + 1.247 + 0.05 = 41.5 W.
This average power Pavg is same as the supply power since capacitor absorbed no
power.
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Exercises
16.1. Find the Fourier series of the waveform shown in the figure. Plot the
amplitude and phase.
16.2. Find the Fourier series of the waveform shown in the figure. Plot the
amplitude and phase. (17.4)
16.3. Determine the fundamental frequency and specify the type of symmetry
present in the functions in the figures. (17.18)
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16.4. Determine the Fourier series representation of the function shown in the
figure. (17.25)
∞
16.5. Find i(t) in the circuit given that is(t) = 1 +
1
∑n
n =1
2
cos 3nt A.
16.6. In the circuit, the Fourier series expansion of vs(t) is vs(t) = 3 +
4 ∞ 1
∑ sin(nπt ) . (17.33)
π n =1 n
16.7. If the periodic voltage shown in the figure is applied to the circuit, find io(t).
(17.39)
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16.8. The signal shown in the figure is applied to the circuit. Find output voltage
vo(t).
Bibliography
1. A Bruce Carlson, “Circuits”, Brooks/Cole 2000.
2. James W. Nilsson and Susan A. Riedel, “Electric Circuits”, 6th Edition,
Prentice Hall 2001.
3. Charles K. Alexander and Matthew N.O Sadiku, “Electric Circuits”, 2nd
edition, McGraw-Hill 2004.
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