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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- m T) = f (t) m = 1, 2, 3, ….. (16.1)
where T is the period. When m = 1, m T 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.
=
t T
t
t T
t
2
2
1
1
f (t )dt f (t) dt (16.2)
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) = a 0 + (^) ∑
∞
=
ω + ω n 1
(a n cos n t b n sin n t ) (16.3)
Equation (16.3) is called the trigonometric Fourier series and the constant C 0 , a n ,
and b n 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 ( t)dt sin( t)dt 0 T T
∫ n ω^^ =∫ n ω^ = ,^ cos(^ t)^ sin(^ t)dt^0 T
∫ n ω^ ⋅ m ω^ = ,^ cos(^ t)^ cos(^ t)^ dt^0 T
∫ n ω^ ⋅ m ω^ = for^ n^ ≠
m , sin( t) sin( t) dt 0
T
∫ n ω^ ⋅ m ω^ = for^ n^ ≠^ m.^ However,^ for^ n^ =^ m ,
( ) ( ) 2
T
cos tdt sin tdt T
2
T
2 ∫ n ω^^ =∫ n ω =.
Reference to equation (16.3), integration of the f (t) over a period T shall be
∫ T
f (t) dt =
∫ T
a 0 dt +^ ∑ (^) ∫ ∫
∞
=
ω + ω n (^1) T
n T
(a n cos n tdt b sin n tdt ) (16.4)
Equation (16.4) is equal to (^) ∫
T
f (t) dt = a 0 T. Thus, the constant a 0 is
a 0 =
T
∫ T
f (t) dt (16.5)
Note that a 0 is also the average value of function f (t).
f (t) = a 0 + (^) ∑ ∑
∞
=
∞
=
φ ω − φ ω n 1 n 1
(A n cos n cos n t) (A n sin n sin n t ) (16.11)
Equating the coefficient of equation (16.3) and (16.11), it gives rise to a n =
A n cosφ n and b n = − A n sinφ n. This shall also mean that
2 2
A n = a n +b n and
φ = −
−
n
n n a
b tan
1
. The relationship between amplitude and phase can also be
expressed in phasor form, which is A n ∠φ n =a n − j b n.
Based on the above discussion, a function
f (t) = A cos n t –Bsin n t =
−
A
B
A B cos t tan
2 2 1
n (16.12)
The plot of amplitude A n of harmonic versus n ω is called amplitude spectrum of
f (t) and the plot of phase φ n versus n ω is called phase spectrum of f (t). Both the
amplitude and phase spectra form the frequency spectrum of f (t).
Example 16.
A rectified half sine wave is defined over one period f (t) = Asinωt for 0 < t < T/
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 a 0 = (^) ∫ ω + ∫
T
T/ 2
T/ 2
0
0 dt T
Asin tdt T
π
ω
ω
− A
T
cos T
A
. The
cosine coefficient a n = α αα
π
∫ ω^ ω = ∫
π
sin cos d
A
Asin tcos tdt T
0
T/ 2
0
n n , after letting α = ωt.
The coefficient a n =
− + π − −
− − π
π
1 cos( 1 )
1
1 cos( 1 )
2
A
n
n
n
n
for n ≠ 1. Knowing that
cos( n ± 1)π = -1 when n is even and cos( n ± 1)π = 1 when n is odd. Thus, a n =
2 A
2 π −
n
for n = 2, 4, 6,….. and a n = 0 for n = 3, 5, 7,……. a 1 is found to be
α α α π
∫
π
sin sin s
A
0
n = 0.
The sine coefficient b n is (^) ∫
π
α α α π (^0)
sin sin d
A
n = A/2 for n = 1 and b n = 0 for n = 2, 3, 4,
5… Knowing the coefficient values, the rectified half-wave Fourier series shall be
f (t) = cos 6 t .................
2 A
cos 4 t 15
2 A
cos 2 t 3
2 A
sin t 2
A A
ω − π
ω − π
ω − π
sin t 2
A A
cos 2 t ( 4 1 )
2 A
1
2 ω π −
− (^) ∑
∞
=
n n n
. The plot of the rectified half-wave based on the
Fourier series is shown in Fig. 16.2.
Figure 16.2: The plot of f (t) = cos 6 t 35
2 A
cos 4 t 15
2 A
cos 2 t 3
2 A
sin t 2
A A
ω π
ω − π
ω − π
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) = a 0 + (^) ∑ ∑
∞
=
− ω
∞
=
ω − + + n 1
nt n n n 1
nt n n (a b )e 2
(a b )e 2
(^1) j j
j j (16.13)
Letting a 0 = c 0 e
j 0t
and summing over both positive and negative values of n , the
compact expression shall be
Based on the above analysis, a new set of coefficients shall be defined, which
are c 0 = a 0 ,
a b c
n n n
− j
= , and
a b c c
Example 16.
Determine the complex Fourier series for the waveform shown in Fig. 16.3.
Figure 16.3: The square wave
Solution
The coefficient c 0 =
T
0
(t ) T
f (^)
−
− −
T/ 4
T/ 2
T/ 4
T/ 4
T/ 2
T/ 4
Adt Adt Adt T
The coefficient c n =
−
−
ω
−
ω ω
T/ 4
T/ 2
T/ 2
T/ 4
nt
T/ 4
T/ 4
n t nt Ae dt Ae dt Ae dt T
(^1) j j j
ω
ω
ω
ω
−
− ω
−
ω T/^2
T/ 4
T/ (^4) nt
T/ 4
T/ (^4) nt
T/ 2
n t e e e
T
A
jn jn jn
j j j
= [ ]
n / 2 n n/ 2 n / 2 n n/ 2 e e e e e e T
A (^) − π π π π π π − + + − − + ω
j -j j -j j j
jn
= [ ]
n / 2 n n n / 2 2 e e e 2 e T
A (^) − π π π π − + − + ω
j -j j j
jn
= [ 4 sin( / 2 ) 2 sin( )]
A
π − π π
j n j n j n
= [ 2 sin( / 2 ) sin( )]
A
π − π π
n n n
Thus, the coefficient for c n is
π π
sin( / 2 )for odd
2A
0 for even
c n n n
n
n
Let n = 1, c 1 =
π
2 A
. This implies that c-1 is also equal to
π
2 A
Let n = 3, c 3 =
π
2 A
. This implies that c-3 is also equal to
π
2 A
Let n = 5, c 5 =
5 π
2 A
. This implies that c-5 is also equal to
5 π
2 A
Let n = 7, c 7 =
π
2 A
. This implies that c-7 is also equal to
π
2 A
Expansion of function f (t) according to equation (16.13) (^) ∑
= ∞
=−∞
ω
n
n
nt c e
j
n for^ n^ = -7 to^ n^ =
7 yields
f (t) = …..
7 t e 7
2 A (^) − ω
π
j 5 t e 5
2 A (^) − ω
π
j
3 t e 3
2 A (^) − ω
π
j
t e
2 A −ω
π
j 3 t e 3
2 A ω
π
j
t e
2 A ω
π
j 5 t e 5
2 A (^) ω
π
j 7 t e 7
2 A ω
π
j
= cos 5 t
4 A
cos 3 t 3
4 A
cos t
4 A
ω π
ω + π
ω − π
cos 7 t 7
4 A
ω π
= cos t
4 A ( 1 )
odd
1
( 1 )/ 2 ω
π
∑
∞
=
=
− n n n
n
n
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) = cos t
4 A ( 1 )
odd
1
( 1 )/ 2 ω
π
∑
∞
=
=
− n n n
n
n for n =1, 3, 5, 7
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.
a 0 = (^) ∫
T
0
(t) dt T
f
a n = 0
b n = (^) ∫ ( ω)
T/ 2
0
(t)sin t dt T
f n (16.22)
16.4.3 Half-Wave Symmetry
A function is half-wave (odd) symmetric if
(t ) 2
T
f t - =− f
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
a 0 = 0
( )
ω
∫
0 for even
(t)cos tdtfor odd T
a
T/
0
n
f n n n
( )
ω
∫
0 for even
(t)sin tdtfor odd T
b
T/
0
n
f n n
n (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
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.
The circuit shown in Fig. 16.7 has a non-sinusoidal v s(t) source that has Fourier
series v s(t) = (^) ∑ ( )
∞
=
π π
k 1
sin t
n n
for n = 2k -1. Find the voltage v o(t) at inductor and
the corresponding amplitude spectrum.
Figure 16.7: An ac circuit
Solution
The output voltage v o(t) is v (t)
R L
L
v (t) s n
n o
ω
j
j
. From the input v s(t), that ω n = n π,
v o(t) shall be (t)
o (t^ ) v s j n
j n v
π
=. The dc component shall be zero after substituting
ωn = 0 into v o(t).
The phasor of sine component of the ac portion is
0 90
n π
Thus, the output v o(t) shall be v o(t) =
( )
π∠
− 25 4 tan 2 / 5
2 2 1
0
n n
n (^) 0 90
n π
∠− ( π )⋅
− tan 2 / 5 25 4
2 2
n n
Rewrite the input voltage function v (t) = 1+ (^) ∑ ∑
∞
=
∞
n 1
2 1
2 sin t 1
cos t 1
n n n
n n
n
n
n
and
convert the input voltage function to amplitude-phase form using equation (16.12).
Thus, A = 2
n
n
and B = 2
n
n
n
−. The phase φ n = - n
1 1 tan A
B
tan
− − =
. A n =
2 2
A + B =
2 1
n
n
The input voltage v (t) shall then be equal to v (t) = 1+ (cos t tan )
1
2
n n n n
n −
∞
=
∑.
The phasor form is 1+ n
n n
n 1
1
2
tan 1
∞
=
∑.
From the equation ω = 1 rad/s and ω n = n rad/s.
The total impedance of the circuit Z = 4 + j ω n 2||4 = 4+
n
n
ω
j
j
n
n
2
j
j
The current flows in the circuit shall be I =
2 ( 1 ) tan 1 n 1
2
1
n
n
n
j
j
n
n
∞
=
−
The output current shall be i o(t)
4 + ω 2
j (^) n 8 8
2 ( 1 ) tan 1 n 1
2
1
n
n
n
j
j
n
n
∞
=
−
2 ( 1 ) tan 1 n 1
2
1
n
n
n^ j
n
∞
=
−
Setting ω n = 0, the dc current shall be
4 0 x 4
The ac component shall be
2 ( 1 ) tan
n 1
2
1
n^ jn
n
n
∑
∞
=
−
∑
∞
=
−
−
n 1
2 1
1
1 4 tan
2 ( 1 ) tan
n n
n
n = (^) ∑
∞
n 1
2 2 1
n
n
In time-domain format, ac component is (^) ∑
∞
n 1
2
cos t 2 1
n n
n
Thus, the current i o(t) shall be i o(t) = +
∑
∞
n 1
2
cos t 2 1
n n
n
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 cos( t v n )
n 1
∑ n n^ ω −θ
∞
=
i (t) = Idc + I cos( t in )
n 1
∑ n n^ ω −θ
∞
=
In Chapter 10 AC Power Analysis, we learnt that the average power is
Pavg = (^) ∫
T
0
P(t) dt T
Thus, the average power expressed in amplitude-phase Fourier series shall be
Pavg = V I dt
T
1 T
∫ 0 dc dc
+ VI cos( t )cos( t )dt
T
1 T
0 v n 1
∑ ∫ n n n^ ω −θ n n ω −θ in
∞
=
Equation (16.28) which is the average power, shall finally become
Pavg = VdcIdc + (^) ∑
∞
=
θ − θ n 1
I cos( v ) 2
Vn n i (16.29)
Example 16.
Determine the average power supplied to the circuit shown in Fig. 16.10 if i (t) = 2
+ cos(t + 10
0
) + 6 cos(3t + 35
0
) A.
Figure 16.10: An ac parallel circuit
Solution
The impedance of the circuit Z is Z =
2 ω
j
1 + 20 ω
j
Hence the voltage v (t) is v (t) = ZxI =
1 + 20 ω
10 I
j
− 1 400 tan 20
10 I
2 1
For dc component, I = 2 A and ω = 0, v (t) = 20 V.
For I = 10 cos(t+
0
) and ω = 1, v (t) =
1 400 tan ( 20 )
10 x 10 10 1
0
−
0
= 5 cos(t –
0
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)
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 i s(t) = 1 + cos 3 t
n 1
2 n n
∑
∞
=
A.
16.6. In the circuit, the Fourier series expansion of vs(t) is vs(t) = 3 +
∑
∞
=
π π (^) n 1
sin( t )
n n
16.7. If the periodic voltage shown in the figure is applied to the circuit, find i o(t).
