# fourier series

Document Sample

```					Learning Objectives
C H A P T E R
21
➣ Harmonic Analysis
FOURIER
➣ Periodic Functions
➣ Trigonometric Fourier
Series
SERIES
➣ Alternate      Forms   of
Trigonometric Fourier
Series
➣ Certain Useful Integral
Calculus Theorems
➣ Evalulation of Fourier
Constants
➣ Different     Types    of
Functional Symmetries
➣ Line    or    Frequency
Spectrum
➣ Procedure for Finding the
Fourier Series of a Given
Function
➣ Wave Analyzer
➣ Spectrum Analyzer
➣ Fourier Analyzer
➣ Harmonic Synthesis

With the help of Fourier Theorem, it is
possible to determine the magnitude, order
and phase of the several hormonics
present in a complex periodic wave
780         Electrical Technology

21.1. Harmonic Analysis
By harmonic analysis is meant the process of determining the magnitude, order and phase of
the several harmonics present in a complex periodic wave.
For carrying out this analysis, the following methods are available which are all based on
Fourier theorem:
(i) Analytical Method–the standard Fourier Analysis
(ii) Graphical Method– (a) by Superposition Method (Wedgemore’ Method) (b) Twenty four
Ordinate Method
(iii) Electronic Method–by using a special instrument called ‘harmonic analyser’
We will consider the first and third methods only.

21.2. Periodic Functions
A function f (t) is said to be periodic if f (t + T ) = f (t ) for all values of t where T is some
positive number. This T is the interval between two successive repetitions and is called the period
of f (t). A sine wave having a period of T = 2π / ω is a common example of periodic function.

21.3. Trigonometric Fourier Series
Suppose that a given function f (t) satisfies the following conditions (known as Dirichlet
conditions):
1. f (t) is periodic having a period of T.
2. f (t) is single-valued everywhere.
3. In case it is discontinuous, f (t) has a finite number of discontinuities in any one period.
4. f (t) has a finite number of maxima and minima in any one period.
The function f(t) may represent either a voltage or current waveform. According to Fourier
theorem, this function f (t) may be represented in the trigonometric form by the infinite series.
f (t ) = a0 + a1 cos ω 0 t + a2 cos 2ω 0 t + a3 cos 3ω 0 t +...+ an cos nω 0 t
+ b1 sin ω 0 t + b2 sin 2ω 0 t + b3 sin 3ω 0 t +...+bn sin nω 0 t
∞
= a0 +   ∑ (a
n =1
n       cos nω 0 t + bn sin nω 0 t )

... (i)
Putting ω 0 t = θ , we can write the above equation as under
f (θ) = a0 + a1 cos θ + a2 cos 2θ + a3 cos 3θ+...+an cos nθ + b1 sin θ + b2 sin 2θ + b3 sin 3θ+...+bn sin nθ
∞
= a0 +   ∑ (a
n=1
n   cos nθ + bn sin nθ)
... (ii)
Since ω 0 = 2π / T , Eq. (i) above can be written as

∞

∑Ga
F                 2πn            2πn     IJ
f (t ) = a0 +
H
n=1
n   cos
T
t + bn sin
T
t
K
... (iii)
Fourier Series               781
where ω 0 is the fundamental angular frequency, T is the period and a0, an and bn are constants which
depend on n and f (t). The process of determining the values of the constants a0, an and bn is called
Fourier Analysis. Also,         0       2 /T       2 f0 where f0 is the fundamental frequency.
It is seen from the above Fourier Series that the periodic function consists of sinusoidal
components of frequency 0, ω 0 , 2ω 0 ....n ω 0 . This representation of the function f(t) is in the
frequency domain. The first component a0 with zero frequency is called the dc component. The
sine and cosine terms represent the harmonics. The number n represents the order of the
harmonics.
When n = 1, the component (a1 cos ω 0 t + b1 sin ω 0 t ) is called the first harmonic or the
fundamental component of the waveform.
When n = 2, the component (a2 cos 2ω 0 t + b2 sin 2ω 0 t ) is called the second harmonic of the
waveform.
The nth harmonic of the waveform is represented by (an cos nω 0 t + bn sin nω 0 t ) . It has a
frequency of nω 0 i.e. n times the frequency of the fundamental component.

21.4. Alternate Forms of Trigonometric Fourier Series
Eq. (i) given above can be written as
f (t ) = a0 + (a1 cos ω 0 t + b1 sin ω 0 t ) + (a2 cos 2ω 0 t + b2 sin 2ω 0 t )+...+ (an cos nω 0 t + bn sin nω 0 t )

Let, an cos nω 0 t + bn sin nω 0 t = A n cos (nω 0 t − φ n )

= A n cos nω 0 t cos φ n + A n sin nω 0 t sin φ n

∴          an = An cos φ n and bn = An sin φ n

∴ An         an 2 bn 2 and          n     tan 1 bn / an

Similarly, let (an cos nω 0 t + bn sin nω 0 t = An sin (nω 0 t + Ψn ))

= An sin nω 0 t cos Ψn + An cos nω 0 t sin φ n                                            Fig. 21.1

As seen from Fig. 21.1, bn = An cos ψ n and an = An sin ψ n

An = an + bn and ψ = tan −1 an / bn
2         2
∴

The two angles φ n and ψ n are complementary angles.
Hence, the Fourier series given in Art. 21.2 may be put in the following two alternate forms
∞
f (t ) = A0 +   ∑A
n=1
n   cos (nω 0 t − φ n )

∞
or f (t ) = A0 +     ∑ A sin (nω t + ψ
n=1
n              0      n)
782               Electrical Technology

21.5. Certain Useful Integral Calculus Theorems
The Fourier coefficients or constants a0 , a1 , a2 ... an and b1 , b2 ..... bn can be evaluated by
integration process for which purpose the following theorems will be used.

(i)    z0
2π
sin nθdθ =
1
b
2π 1
cos nθ 0 = (1 − 1) = 0
n

(ii)    z  0
2π                  1                1
sin nθdθ = − | cos nθ|2 π = − (0 − 0) = 0
n
0
n
2                        1   2                                 1               1
(iii)               sin 2 n d                    (1 cos 2n )d                   |                        2
sin 2n |0
0                        2   0                                 2               2n

(iv)    z   0
2π
cos 2 nθdθ =
1
2   z
0
2π
(cos 2nθ + 1)dθ =
1 1
2 2n
sin 2nθ + θ = π
0
2π

2                                     1       2
(v)                sin m cos n d                                 {sin (m n)      sin (m n) }d
0                                     2       0

2π
1  1                 1
= −    cos(m + n)θ −     cos(m − n)θ = 0
2 m+n               m−n            0

(vi)    z   0
2π
cos mθ cos nθdθ =
1
2   z   2π

0
{cos(m + n) θ + cos(m − n)θ}dθ

2π
1 1                   1
=          sin(m + n) θ +     sin (m − n)θ = 0... for n ≠ m
2 m+n                m−n             0

2                                    1       2
(vii)                sin m sin n d                                {cos( m n)              cos(m n) } d
0                                     2       0

2π
1 1                     1
=          sin(m − n) θ −       sin(m + n)θ = 0... for n ≠ m
2 m−n                  m+n              0
where m and n are any positive integers.

21.6. Evaluation of Fourier Constants
Let us now evaluate the constants a0, an and bn by using the above integral calculus theorems
(i) Value of a0
For this purpose we will integrate both sides of the series given below over one period i.e.
for θ = 0 to θ = 2π.
f (θ) = a0 + a1 cos θ + a2 cos 2θ +...+ an cos nθ + b1 sin θ + b2 sin 2θ +...+bn sin nθ

∴
z0
2π
f (θ)dθ =   z0
2π
a0 dθ + a1          z   2π

0
cos θdθ + a2   z
0
2π
cos 2θdθ +...+ an       z   2π

0
cos nθdθ

+ b1      z0
2π
sin θdθ + b2   z
0
2π
sin 2θdθ +...+bn   z
0
2π
sin nθdθ
Fourier Series                  783

= a0 | θ|0π +0 + 0 +...0 + 0 + 0 +....+0 = 2πa0
2

∴                 a0 =
1
2π       z  0
2π
f (θ)dθ or =
1
2π      z
−π
π
f (θ)dθ

= mean value of f (θ) between the limits 0 to 2π i.e. over one cycle or period.

1           2
Also,             a0             [net area]0
2
If we take the periodic function as f(t) and integrate over period T (which corresponds to 2π ),

we get a0 =
1
T   z0
T
f (t )dt =
1
T    z   T /2

−T / 2
f (t )dt =
1
T   z   t1 +T

t1
f (t )dt

where t1 can have any value.
(ii) Value of an
For finding the value of an, multiply both sides of the Fourier Series by cos nθ and integrate
between the limits     0 to 2

∴    z0
2π
f (θ) cos nθdθ = a0                    z0
2π
cos nθdθ + a1                z  0
2π
cos θ cos nθd θ + a2         z0
2π
cos 2θ cos nθdθ

+an   z   2π

0
cos 2 nθdθ + b1                      z0
2π
cos θ cos nθdθ + b2                          z0
2π
sin 2θ cos nθdθ +...+ bn        z0
2π
sin nθ cos nθdθ

= 0 + 0 + 0 +...+ an
z 0
2π
cos 2 nθdθ + 0 + 0+...+0 = an                                 z0
2π
cos 2 nθdθ = πan

∴ an =
1
π    z  0
2π
f (θ) cos nθdθ = 2 ×
1
2π      z   2π

0
f (θ) cos nθ dθ

= 2 × average value of f (θ) cos nθ over one cycle of the fundamental.

1                                                         1
Also, an                               f ( ) cos n d                      2                        f ( )cos n d
2
If we take periodic function as f (t ), then different expressions for an are as under.

an =
2
T       z0
T
f (t ) cos
2πn
T
t dt =
2
T           z    T /2

−T / 2
f (t ) cos
2πn
T
t dt

Giving different numerical values to n, we get
a1 = 2 × average of f (θ) cos θ over one cycle                                                                                                                     ....n = 1
a2 = 2 × average value of f (θ) cos 2θ over one cycle etc.                                                                                                         .... n = 2
(iii) Value of bn
For finding its value, multiply both sides of the Fourier Series of Eq. (i) by sin nθ and
integrate between limits θ = 0 to θ = 2π.
784            Electrical Technology

∴   z0
2π
f (θ) sin nθdθ = a0                      z0
2π
sin nθdθ + a1                  z0
2π
cos θ sin nθd θ

+ a2      z   2π

0
cos 2θ sin nθdθ +...+an                        z0
2π
cos nθ sin nθdθ

+ b1      z   2π

0
sin θ sin nθdθ + b2                   z0
2π
sin 2θ sin nθdθ +....+ bn                 z0
2π
sin 2 nθdθ

= 0 + 0 + 0 +...+0 + 0 +...+ bn                           z0
2π
sin 2 nθdθ = bn                    z   2π

0
sin 2 nθ dθ = bn π

∴
z0
2π
f (θ)sin θdθ = bn × π

∴       bn =
1
π      z   2π

0
f (θ) sin nθdθ = 2 ×
1
2π       z
0
2π
f (θ) sin nθdθ

= 2 × average value of f (θ) sin nθ over one cycle of the fundamental.
∴ b1 = 2 × average value of f (θ) sin θ over one cycle                                                                                                                          ... n = 1
b2 = 2 × average value of f (θ) sin 2θ over one cycle                                                                                                                           ... n = 2

Also, bn =
2
t      z   T

0
f (t )sin
2πn
T
t dt +
2
T                z   T /2

T /2
f (t )sin
2πn
T
t dt

2 T
=
T 0          z
f (t )sin nω 0 t dt =
2 T /2
T T /2
f (t )sin nω 0 tdt              z
Hence, for Fourier analysis of a periodic function, the following procedure should be
(i ) Find the term a0 by integrating both sides of the equation representing the periodic
function between limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).

∴ a0 =
1
2π      z   2π

0
f (θ)dθ =
1
T    z  0
T
f (t )dt =
1
T    z    T /2

−T / 2
f (t )dt =
1
T    z   t1 +T

t1
f (t )dt

= average value of the function over one cycle.
(ii) Find the value of an by multiplying both sides of the expression for Fourier series by
cos nθ and then integrating it between limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).

∴            an =
1
π       z   2π

0
f (θ) cos nθdθ = 2 ×
1
2π     z 0
2π
f (θ) cos nθdθ

Since π = T/2, we have

an =
2
T   z   T

0
f (t ) cos
2πn
T
t dt =
2
T                z    T /2

−T / 2
f (t ) cos
2πn
T
t dt =
2
T              z t1
t1 +T
f (t ) cos
2πn
T
t dt

=
2
T   z   T

0
f (t ) cos nω 0 tdt =
2
T   z   T /2

−T / 2
f (t ) cos nω 0 tdt =
2
T       zt1
t1 +T
f (t ) cos nω 0 tdt

= 2 × average value of f (θ) cos nθ over one cycle of the fundamental.
Fourier Series       785
Values of a1, a2, a3 etc. can be found from above by putting n = 1, 2, 3 etc.
(iii) Similarly, find the value of bn by multiplying both sides of Fourier series by sin nθ and
integrating it between the limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).

∴ bn =
1
π       z0
2π
f (θ) sin nθ = 2 ×
1
2π     z
0
2π
f (θ) sin nθdθ

=
2
T    z
0
T
f (t ) sin
2πn
T
t dt =
2
T      z   T /2

−T / 2
f (t )sin
2πn
T
t dt =
2
T      z t1
t1 +T
f (t )sin
2πn
t
t dt

=
2
T   z
0
T
f (t )sin n ω 0 tdt =
2
T   z   T /2

−T / 2
f (t )sin nω 0 t dt =
2
T   z
t1
t1 + T
f (t )sin nω 0 t dt

2πn
= 2 × average value of f (θ) sin nθ of f (t) sin                                    t or f (t ) sin nω 0 t over one cycle of
T
the fundamental.
Values of b1, b2, b3 etc. can be found from above by putting n = 1, 2, 3 etc.

21.7. Different Types of Functional Symmetries
A non-sinusoidal wave can have the following types of symmetry:
1. Even Symmetry
The function f (t) is said to possess even symmetry if f (t) = f (– t).
It means that as we travel equal amounts in time to the left and right of the origin (i.e. along
the + X-axis and –X-axis), we find the function to have the same value. For example in
Fig. 21.2 (a), points A and B are equidistant from point O. Here the two function values are equal
and positive. At points C and D, the two values of the function are again equal, though negative.
Such a function is symmetric with respect to the vertical axis. Examples of even function are: t2,

Fig. 21.2
786         Electrical Technology

cos 3t, sin2 5t (2 + t2 + t4) and a constant A because the replacement of t by (– t) does not change
the value of any of these functions. For example, cos ωt = cos (−ωt ) .
This type of symmetry can be easily recognised graphically because mirror symmetry exists
about the vertical or f (t) axis. The function shown in Fig. 21.2 has even symmetry because when
folded along vertical axis, the portions of the graph of the function for positive and negative time
fit exactly, one on top of the other.
The effect of the even symmetry on Fourier series is that the constant bn = 0 i.e. the wave has
no sine terms. In general, b1, b2, b3 ... bn = 0. The Fourier series of an even function contains only
a constant term and cosine terms i.e.
∞                          ∞
2πn
f (t ) = a0 +    ∑
n=1
an cos nω 0 t = a0 +   ∑a
n=1
n   cos
T
t

The value of an may be found by integrating over any half-period.

∴
2 π
an =
π 0
2. Odd Symmetry
z
f (θ) cos nθdθ =
4
T              z 0
T /2
f (t ) cos nωt dt

A function f (t) is said to possess odd symmetry if f (−t ) = − f (t ) .
It means that as we travel an equal amount in time to the left or right from the origin, we find
the function to be the same except for a reversal in sign. For example, in Fig. 21.3 the two points
A and B are equidistant from point O. The two function values at A and B are equal in magnitude
but opposite in sign. In other words, if we replace t by (– t), we obtained the negative of the given
function. The X-axis divides an odd function into two halves with equal areas above and below the
X-axis. Hence, a0 = 0.

Examples of odd functions are: t, sin t, t cos 50 t (t + t 3 + t 5 ) and t (1 + t 2 )

A sine function is an odd function because sin (− ωt ) = − sin ωt .

Fig. 21.3
An odd function has symmetry about the origin rather than about the f(t) axis which was the
case for an even function. The effect of odd symmetry on a Fourier series is that it contains no
constant term or consine term. It means that a0 = 0 and an = 0 i.e. a1, a2, a3....an = 0. The Fourier
series expansion contains only sine terms.
∞
∴ f (t ) =   ∑ b sin nω t
n =1
n           0

The value of bn may be found by integrating over any half-period.
Fourier Series         787

∴ bn =
2 π
π 0   zf (θ) sin nθdθ =
4 T /2
T 0    z
f (t ) sin nωt dt

3. Half-wave Symmetry or Mirror-Symmetry or Rotational Symmetry
A function f (t) is said to possess half-wave symmetry if f (t) = − f (t ± T / 2) or − f (t) = f (t ± T / 2) .
It means that the function remains the same if it is shifted to the left or right by half a period
and then flipped over (i.e. multiplied by – 1) in respect to the t-axis or horizontal axis. It is called
mirror symmetry because the negative portion of the wave is the mirror image of the positive
portion of the wave displaced horizontally a distance T/2.
In other words, a waveform possesses half symmetry only when we invert its negative half-
cycle and get an exact duplicate of its positive half-cycle. For example, in Fig. 21.4 (a) if we invert
the negative half-cycle, we get the dashed ABC half-cycle which is exact duplicate of the positive
half-cycle. Same is the case with the waveforms of Fig. 21.4 (b) and Fig. 21.4 (c). In case

Fig. 21.4
of doubt, it is helpful to shift the inverted half-cycle by a half-period to the left and see if it super-
imposes the positive half-cycle. If it does so, there exists half-wave symmetry otherwise not. It is
seen that the waveform of Fig. 21.4 (d) does not possess half-wave symmetry. It is so because
when its negative half-cycle is inverted and shifted by half a period to the left it does not
superimpose the positive half-cycle.
It may be noted that half-wave symmetry may be present in a waveform which also shows
either odd symmetry or even symmetry:
For example, the square waveform shown in Fig. 21.4 (a) possesses even symmetry whereas
the triangular waveform of Fig. 24.4 (b) has odd symmetry. All cosine and sine waves possess half-
wave symmetry because

2πFG  T    IJ 2π  FG        IJ 2π        2π    T     FG2π   IJ      FG  2π     IJ
cos
T  H
t±
2     K
= cos
T    H         K
t ± π = − cos
T
t; sin
T
t±
2
= sin
HT     K       H
t ± π = − sin
T
t
K
It is worth noting that the Fourier series of any function which possesses half-wave symmetry
has zero average value and contains only odd harmonics and is given by
788          Electrical Technology

∞
F 2πn              2πn I
∞                        ∞

∑G a cos T t + b sin T tJ = ∑(a cos nω t + b sin nω t) = ∑(a cos nθ + b sin nθ)
f (t ) =
H
n=1
n
K             n
n=1
n      0   n   0
n=1
n   n

odd                                                odd                      odd

where, an =
0
4
T   z
2πn
T
T /2
dt =
2 π
f (θ) cos nθdθ
f (t ) cos
π 0                  z                  ... n odd

bn =
4 T /2
T 0
4. Quarter-wave Symmetry
z
f (t )sin
2 πn
T
dt =
2 π
π 0
f (θ)sin nθdθ z                         ... n odd

An odd or even function with rotational symmetry is said to possess quarter-wave symmetry.
Fig. 21.5 (a) possesses half-wave symmetry as well as odd symmetry. The wave shown in Fig.
21.5 (b) has both half-wave symmetry and even symmetry.
The mathematical test for quarter-wave symmetry is as under:
Odd quarter-wave f (t ) = − f (t + T / 2) and f (− t ) = − f (t )
Even quarter-wave f (t ) = − f (t + T / 2) and f (t ) = f (−t )

Fig. 21.5
Since each quarter cycle is the same in a way having quarter-wave symmetry, it is sufficient
to integrate over one quarter period i.e. from 0 to T/4 and then multiply the result by 4.
(i) If f(t) or f (θ) is odd and has quarter-wave symmetry, then a0 is 0 and an is 0. Hence, the
Fourier series will contain only odd sine terms.
∞
2πnt             ∞
1 2π
∴ f (t ) = ∑ bn sin                        or f (θ) = ∑ bn sin nθ, where bn = ∫ f (θ)sin nθd θ
n =1                T              n =1                   π 0
odd                                  odd

It may be noted that in the case of odd quarter-wave symmetry, the integration may be carried
over a quarter cycle.

∴ an =
4
π    z0
π/2
f (θ) cos nθdθ                                                   ... n odd

=        z
8 T /4
T 0
f (t )sin nωtdt                                                        ... n odd

(ii) If f(t) or f (θ) is even and, additionally, has quarter-wave symmetry, then a0 is 0 and bn
is 0. Hence, the Fourier series will contain only odd cosine terms.
Fourier Series      789

∴    f (θ) =   ∑
∞

n =1
odd
an cos nθdθ =
∞

∑a
n =1
odd
n   cos nω 0 tdt; where an =
1
π   z
0
2π
f (θ) cos nθdθ

In this case an may be found by integrating over any quarter period.

an =   z
4 π/2
π 0
f (θ) sin nθdθ                                                                            ... n odd

=   z
8 T /4
T 0
f (t ) cos nωt dt                                                                      ... n odd

21.8. Line or Frequency Spectrum
A plot which shows the amplitude of each frequency component in a complex waveform is
called the line spectrum or frequency spectrum (Fig. 21.6). The amplitude of each frequency
component is indicated by the length of
the vertical line located at the
corresponding frequency. Since the
spectrum represents frequencies of the
harmonics as discrete lines of appropriate
height, it is also called a discrete
spectrum. The lines decrease rapidly for
waves having convergent series. Waves
with discontinuities such as the sawtooth
and square waves have spectra whose
amplitudes decrease slowly because their
series have strong high harmonics. On
the other hand, the line spectra of
waveforms without discontinuities and
with a smooth appearance have lines                                Fig. 21.6
which decrease in height very rapidly.
The harmonic content and the line spectrum of a wave represent the basic nature of that wave
and never change irrespective of the method of analysis. Shifting the zero axis changes the
symmetry of a given wave and gives its trigonometric series a completely different appearance but
the same harmonics always appear in the series and their amplitude remains constant.

Fig. 21.7

Fig. 21.7 shows a smooth wave alongwith its line spectrum. Since there are only sine terms
in its trigonometric series (apart from a0 = π ), the harmonic amplitudes are given by bn.
790        Electrical Technology

21.9. Procedure for Finding the Fourier Series of a Given Function
1. Step No. 1
If the function is defined by a set of equations, sketch it approximately and examine for
symmetry.
2. Step No. 2
Whatever be the period of the given function, take it as 2π (Ex. 20.6) and find the Fourier
series in the form
f (θ) = a0 + a1 cos θ + a2 cos 2θ +...+ an cos nθ + b1 sin θ + b2 sin 2θ +...+bn sin nθ
3. Step No. 3
The value of the constant a0 can be found in most cases by inspection. Otherwise it can be
found as under:

a0 =

4. Step No. 4
1
2π       z0
2π
f (θ)dθ =
1
2π   z
−π
π
f (θ)dθ

If there is no symmetry, then a0 is found as above whereas the other two fourier constants can
be found by the relation.

an =
1
π   z 0
2π
f (θ) cos nθdθ =
1
π    z   π

−π
f (θ) cos nθdθ

bn =

5. Step No. 5
z
0
2π
f (θ)sin nθdθ =
1
π   z   π

−π
f (θ)sin nθdθ

If the function has even symmetry i.e. f (θ) = f (−θ) , then bn = 0 so that the Fourier series
will have no sine terms. The series would be given by

f (θ) = a0 +        ∑a
∞

n =1
n    cos nθdθ where an =
1
π    z0
2π
f (θ) cos nθ =
2
π   z   π

0
f (θ) cos nθdθ

6. Step No. 6
If the given function has odd symmetry i.e. f (−θ) = − f (θ) then a0 = 0 and an = 0. Hence,
there would be no cosine terms in the Fourier series which accordingly would be given by

z                                  z
∞

∑ b sin ω t ; where b                                        1       2π                        2        π
f (θ) =             n               0                             =                 f (θ) sin nθdθ =                      f (θ) sin nθdθ
n
n =1                                                        π   0                             π    0

7. Step No. 7
If the function possesses half-wave symmetry i.e. f (θ) = − f (θ ± π) or f (t ) = − f (t ± T / 2) ,
then a0 is 0 and the Fourier series contains only odd harmonics. The Fourier series is given by
∞
f (θ) =         ∑ a (cos nθ + b sin nθ)
n =1
n                         n

odd
Fourier Series                      791

where       an =

8. Step No. 8
1
π   z 0
2π
f (θ) cos nθdθ... n odd, bn = 2
π       z   0
π
f (θ) sin nθdθ                                        ... n odd

If the function has even quarter-wave symmetry then a0 = 0 and bn = 0. It means the Fourier
series will contain no sine terms but only odd cosine terms. It would be given by
1   2                               2                              4            /2
f( )          an cos n ;where an                         f ( )cos n d                 f ( )cos n d                            f ( )cos n d
0                                    0                                 0
n 1
odd
... n odd
9. Step No. 9
If the function has odd quarter-wave symmetry, then a0 = 0 and an = 0. The Fourier series will
contain only odd sine terms (but no cosine terms).
1   2                        2                            4        /2
∴ f( )              bn sin n ;where bn
0
f ( )sin n d
0
f ( )sin n d
0
f ( )sin n d
n 1
odd

... n odd
10. Step No. 10
Having found the coefficients, the Fourier series as given in step No. 2 can be written down.
11. Step No. 11
The different harmonic amplitudes can be found by combining similar sine and cosine terms
i.e. An = an + bn
2    2

where An is the amplitude of the nth harmonic.
Table No. 21.1
Wave form                        Appearance                                     Equation

A.      Sine wave                                                                       f (t) = A = A sin ω t

FG 1 + 1 sin ωt − 2 cos 2ωtIJ
B.      Half-wave rectified
sine wave
f (t ) = A
Hπ 2            3π         K
2             2
−       cos 4ωt −     cos 6 ωt
15π           35π

2
−       cos 8ωt.....
63π

2A     2
C.      Full-wave rectified                                                             f (t ) =      (1 − cos 2ωt
sine wave                                                                                   π     3

2           2
−      cos 4ωt −    cos 6ωt.....)
15           35
792       Electrical Technology

44          1
D.    Rectangular or                         f (t ) =      (sin ωt + sin 3ωt .....)
π           3
square wave
1         1
+ sin 5ωt + sin 7ωt +.... )
5         7

4A          1
f (t ) =      (cos ωt − cos 3ωt +
π           3

1
cos 5ωt........)
5

A 2A          1
E.    Rectangular or square                  f (t ) =    +   (sin ωt + sin 3ωt +
2 π           3
wave pulse
1          1
sin 5ωt + sin 7ωt +....)
5          7

8A         1          1
F. Triangular wave                           f (t ) =     (sin ωt − sin 3ωt +
π2         9          25

1
sin 5ωt −          sin 7ωt +....)
49

8A            1
f (t ) =        (cos ωt + cos 3ωt +
π2            9

1            1
cos 5ωt +    cos 7ωt +.....)
25           49

A 4A          1
G. Triangular pulse                          f (t ) =    +   (sin ωt − sin 3ωt
2 π2          9

1           1
+      sin 5ωt −    sin 7ωt +.....)
25           49

2A              1
H.    Sawtooth wave                          f (t ) =      (sin olegat − sin 2ωt +
π              2

1          1
sin 3ωt − sin 4ωt +....)
3          4

A π FG           1         1
I.    Sawtooth pulse              f (t ) =
π 2  H
− sin ω 0 t − sin 2ωt − sin 3ωt
2         3
1               IJ
− sin 4ωt
4                K
Fourier Series      793

FG sin ωt − 1 sin 5ωt
3 3A
π H
J.    Trapezoidal wave                                        f (t ) =     2         25

+ sin 7ωt......J
1              I
49             K
21.10. Wave Analyzer
A wave analyzer is an instrument designed to measure the individual amplitude of each harmonic

Fig. 21.8
component in a complex waveform. It is the simplest form of analysis in the frequency domain and
can be performed with a set of tuned filters and a voltmeter. That is why such analyzers are also
called frequency-selective voltmeters. Since
such analyzers sample the frequency spectrum
successively, i.e. one after the other, they are
called non-real-time analyzers.
The block diagram of a simple wave
analyzer is shown in Fig. 21.8. It consists of a
tunable fundamental frequency selector that
detects the fundamental frequency f1 which is
the lowest frequency contained in the input
waveform.
Once tuned to this fundamental frequency,
a selective harmonic filter enables switching to
multiples of f1. After amplification, the output
is fed to an a.c. voltmeter, a recorder and a                       Wave analyzer
frequency counter. The voltmeter reads the
r.m.s amplitude of the harmonic wave, the recorder traces its waveform and the frequency counter
gives its frequency. The line spectrum of the harmonic component can be plotted from the above
data.
For higher frequencies (MHz) heterodyne wave analyzers are generally used. Here, the input
complex wave signal is heterodyned to a higher intermediate frequency (IF) by an internal local
oscillator. The output of the IF amplifier is rectified and is applied to a dc voltmeter called
heterodyned tuned voltmeter.
794        Electrical Technology

The block diagram of a wave analyzer using the heterodyning principle is shown in Fig. 21.9.

Fig. 21.9
The signal from the internal, variable-frequency oscillator heterodynes with the input signal
in a mixer to produce output signal having frequencies equal to the sum and difference of the
oscillator frequency f0 and the input frequency f1. Generally, the bandpass filter is tuned to the ’sum
frequency’ which is allowed to pass through. The signal coming out of the filter is amplified,
rectified and then applied to a dc voltmeter having a decibel-calibrated scale. In this way, the peak
amplitudes of the fundamental component and other harmonic components can be calculated.

21.11. Spectrum Analyzer
It is a real-time instrument i.e. it simultaneously displays on a CRT, the harmonic peak

Fig. 21.10
values versus frequency of all wave
components in the frequency range of the
analyzer. The block diagram of such an
analyzer is shown in Fig. 21.10.
As seen, the spectrum analyzer uses a
CRT in combination with a narrow-band
tuned by varying the frequency of the
voltage-tuned variable-frequency oscillator
which also controls the sawtooth generator
that sweeps the horizontal time base of the                       Spectrum analyzer
CRT deflection plates. As the oscillator is
Fourier Series         795
swept through its frequency band by the sawtooth generator, the resultant signal mixes and beats
with the input signal to produce an intermediate frequency (IF) signal in the mixer. The mixer
output occurs only when there is a corresponding harmonic component in the input signal which
matches with the sawtooth generator signal. The signals from the IF amplifier are detected and
further amplified before applying them to the vertical deflection plates of the CRT. The resultant
output displayed on the CRT represents the line spectrum of the input complex or nonsinusoidal
waveform.

21.12. Fourier Analyzer
A Fourier analyzer uses digital signal processing technique and provides information
regarding the contents of a complex wave which goes beyond the capabilities of a spectrum
analyzer. These analyzers are based on the calculation of the discrete Fourier transform using an
algorithm called the fast Fourier transformer. This algorithm calculates the amplitude and phase of
each harmonic component from a set of time domain samples of the input complex wave signal.

Fig. 21.11
A basic block diagram of a Fourier analyzer is
shown in Fig. 21.11. The complex wave signal
applied to the instrument is first filtered to remove
out-of-band frequency components. Next, the signal
is applied to an analog/digital (A/D) convertor
which samples and digitizes it at regular time
intervals until a full set of samples (called a time
record) has been collected. The microprocessor then
performs a series of computations on the time data
to obtain the frequency-domain results i.e.
amplitude versus frequency relationships. These
results which are stored in memory can be
displayed on a CRT or recorded permanently with a
recorder or plotter.                                                 Fourier Analyzer
Since Fourier analyzers are digital
instruments, they can be easily interfaced with a computer or other digital systems. Moreover, as
compared to spectrum analyzers, they provide a higher degree of accuracy, stability and repeatability.

21.13. Harmonic Synthesis
It is the process of building up the shape of a complex waveform by adding the instantaneous
values of the fundamental and harmonics. It is a graphical procedure based on the knowledge of the
different components of a complex waveform.
796        Electrical Technology

Example 21.1. A complex voltage waveform contains a fundamental voltage of r.m.s. value
220 V and frequency 50 Hz alongwith a 20% third harmonic which has a phase angle lagging by
3π / 4 radian at t = 0. Find an expression representing the instantaneous complex voltage v.
Using harmonic synthesis, also sketch the complex waveform over one cycle of the fundamental.
Solution. The maximum value of the fundamental voltage is = 200 × 2 = 310 V. Its angular
velocity is ω = 2π × 50 = 100π rad/s. Hence, the fundamental voltage is represented by 310 sin
100π t.
The amplitude of the third harmonic = 20% of 310 = 62V. Its frequency is 3 × 50 = 150 Hz.
Hence, its angular frequency is = 2π × 150 = 300 π rad/s. Accordingly, the third harmonic voltage
can be represented by the equation 62 sin (300 πt − 3π / 4) . The equation of the complex voltage
is given by υ = 310 sin 100 πt + 62 sin(300 πt − 3π / 4) .

Fig 21.12
In Fig. 21.12 are shown one cycle of the fundamental and three cycles of the third harmonic
component initially lagging by 3 π / 4 radian or 135°. By adding ordinates at different intervals,
the complex voltage waveform is built up as shown.
Incidentally, it would be seen that if the negative half-cycle is reversed, it is identical to the
positive half-cycle. This is a feature of waveforms possessing half-wave symmetry which contains
the fundamental and odd harmonics.
Example 21.2. For the nonsinusoidal wave shown in Fig. 21.13, determine (a) Fourier
coefficients a0, a3 and b4 and (b) frequency of the fourth harmonic if the wave has a period of 0.02
second.
Solution. The function f (θ) has a constant value from
θ = 0 to θ = 4 π / 3 radian and 0 value from θ = 4 π / 3

1 ( net area per cycle) 2π = 1 6 × 4 π = 4      FG         IJ
(a) a0 =
2π
0
2π     3            H          K
a3 =
1
π   z
0
2π
f (θ) cos 3θdθ =
1
π   z
0
4π/3
6 cos 3θdθ

Fig. 21.13
Fourier Series    797
4π/3
6 sin 3θ                        2
=                               =     (sin 4 π) = 0
π 3              0              π

b4 =
1
π   z0
2π
f (θ)sin 4θdθ =
1
π    z0
4π/3
6 sin 4θdθ

6 − cos 4θ
4π /3
FG
3 cos 16 π           −3              IJ
9
=
π    4               0
=−
2π    3 H   − cos 0 =
2π
(−0.5 − 1) =
K
4π
(b) Frequency of the fourth harmonic = 4 f0 = 4 / T = 4 / 0.02 = 200 Hz.
Example 21.3. Find the Fourier series of the “half sinusoidal” voltage waveform which
represents the output of a half-wave rectifier. Sketch its line spectrum.
Solution. As seen from Fig. 21.14 (a), T = 0.2 second, f0 = 1/T = 1/0.2 = 5 Hz and
ω 0 = 2 f 0 = 10π rad/s. Moreover, the function has even symmetry. Hence, the Fourier series
will contain no sine terms because bn = 0.
The limits of integration would not be taken from t = 0 to t = 0.2 second, but from t = – 0.5
to t = 0.15 second in order to get fewer equations and hence fewer integrals. The function can be
written as
f (t ) = Vm cos 10πt                                                   −0.05 < t < 0.05
=0                                                               0.05 < t < 0.15

a0 =
1
T   z   0.15

−0.05
f (t )dt =
1
0.2
LM
N   z   0.05

−0.05
Vm cos 10 πtdt +           z   0.15

0.05
0. dt
OP
Q
=
Vm sin 10 πt
0.2 10π
0.05

−0 .05
=
Vm
π
2V
an = m
0.2                  z   0.05

−0.05
cos 10 πt.cos 10 πnt dt

The expression we obtain after integration cannot be solved when n = 1 although it can be
solved when n is other than unity. For n = 1, we have

a1 = 10Vm        z   0.05

−0.05
cos 2 10 πt dt =
Vm
2

When n ≠ 1 , then an = 10Vm                                z    0 .05

−0.05
cos 10 πt.cos 10 π ntdt

=
10Vm
2        z   0 .05

−0.05
[cos 10 π(1 + n)t cos 10 π(1 − n)t ]dt =
2Vm cos(πn / 2)
π
.
(1 − n2 )
... n ≠ 1

2Vm cos π 2Vm −1 2Vm       2V cos 3π / 2          2V cos 2π   2V
a2 =      .     =   .  =    ; a3 = m .          = 0; a4 = m .      =− m
π 1− 4    π −3   2π        π   1− 3 2             π 1− 4 2   15π

2Vm cos 3π 2Vm
a5 = 0; a6 =               .      =     and so on
π 1 − 62   35π
Substituting the values of a0, a1, a2, a4 etc. in the standard Fourier series expression given in
Art. 20.3. we have
f (t )       a0     a1 cos 2         0t            a2 cos 2          0t     a4 cos 4            0t    a6 cos 6    0t   ....
798        Electrical Technology

Fig. 21.14

Vm Vm            2V            2V           2V
=      +   cos 10πt + m cos 20 πt − m cos 40πt + m cos 60 πt....
π   2            3π           15π           35

FG 1 + 1 cos ω t + 2 cos 2ω tπ − 2 cos 4ω t + 2 cos 6ω t...IJ
= Vm
Hπ 2               0
3π           15π
0
35π            K          0                0

The line spectrum which is a plot of the harmonic amplitudes versus frequency is given in
Fig. 21.14(b).
Example 21.4. Determine the Fourier series for the square voltage pulse shown in
Fig. 21.15 (a) and plot its line spectrum.            (Network Theory, Nagpur Univ. 1992)

FG 2πt IJ
Solution. The wave represents a periodic function of θ or ωt or
HTK        having a period

extending over 2π radians or T seconds. The general expression for this wave can be written as
f (θ) = a0 + a1 cos θ + a2 cos 2θ + a3 cos 3θ +...

...+ b1 sin θ + b2 sin 2θ + b3 sin 3θ +...

Fig. 21.15

(i) Now,   f (θ) = V; θ = 0 to θ = π; f (θ) = 0, from θ = π to θ = 2π

∴           a0 =
1
2π   z
0
2π
(θ)d θ =
1
2π
R
S
T   z
0
π
f (θ)dθ +   z
π
2π
f (θ)dθ
U
V
W
Fourier Series               799

or       a0 =
1
2π
R
S
T   z z
π

0
Vdθ +
2π

π
0dθ =
U
V
W
V π
2π
| θ|0 +0 = V2 π × π =
V
2

(ii) an =
1
π   z   2π

0
f (θ) cos nθdθ =
1
π
R
S
T   z0
π
V cos nθdθ +            z   2π

π
0 × cos nθdθ
U
V
W
=
V
π   z   π

0
cos nθdθ + 0 =
V
nπ
π
|sin nθ|0 = 0 ... whether n is odd or even

(iii) bn =
1
π       z
0
2π
f (θ) sin nθdθ =
1
π
R
S
T   z    π

0
V sin nθdθ +            z π
2π
0 × sin nθdθ
U
V
W
=
V
π       z   π

0
sin nθdθ + 0 =
V − cos nθ
π    n
π

0
=
V
nπ
(− cos nπ + 1)

Now, when n is odd, (1 − cos nπ ) = 2 but when n is even, (1 − cos nπ ) = 0.

2V                 V                          V       2V
∴ b1 =         ... n = 1; b2 =    × 0 = 0 ... n = 2; b3 =    ×2 =    ... n = 3 and so on.
π                 2π                         3π      3π
Hence, substituting the values of a0 , a1 , a2 etc. and b1 , b2 etc. in the above given Fourier
series, we get
V 2V         2V          2V              E 2V            1            1                                  FG                                         IJ
f (θ) =    +
2 π
sin θ +
3π
sin 3θ +
5π
sin 5θ +... = +
2  π
sin ω 0 t + sin 3ω 0 t + sin 5ω 0 t +...
3            5                                   H                                          K
It is seen that the Fourier series contains a constant term V/2 and odd harmonic components
whose amplitudes are as under:
2V
Amplitude of fundamental or first harmonic =
π
2V
Amplitude of second harmonic =
3π
2V
Amplitude of third harmonic =      and so on.
5
The plot of harmonic amplitude versus the harmonic frequencies (called line spectrum) is
shown in Fig. 21.15 (b).
Example 21.5. Obtain the Fourier series for the square wave pulse train indicated in
Fig. 21.16.                            (Network Theory and Design, AMIETE June, 1990)
Solution. Here T = 2 second, ω 0 = 2π / T = π rad/s. The given function is defined by
f (t ) = 1        0 < t < 1 = 0 and
= 0;            1<t<2

a0 =
1
T   z   T

0
f (t )dt =
1
2   z   2

0
1. dt =
1
2
LM
N   z z 1

0
1. dt +
1
2        OP
0. dt =
Q
1
2
Even otherwise by inspection a0 = (1 + 0)/2 = 1/2

an =
2
T    z0
T
f (t ) cos nω 0 tdt =             z  2
2
2

0
1. cos nπtdt =
LM
N   z
0
1
z
1.cos nπtdt + (0). cos nπtdt
1
2             OP
Q
800           Electrical Technology

1
1                                   sin nπt
=   ∫0
cos nπtdt =
nπ 0
=0

bn =
2
T      z
0
T
z
f (t )sin nω 0 tdt = [ 1.sin nπt dt ] +
0
1
z
1
2
(0) sin nπtdt ] =   z
0
1
sin nπt dt

1
− cos nπt                               1 − cos nπ
=                                       =
nπ                           0           nπ

∴                   bn = 2 / nπ ..... when n is odd; = 0
... when n is even

Fig. 21.16

2       1                       1       2            1                1
∴               f (t )              a0                    sin 3n t                           sin t     sin 3 t          sin 5 t , etc.
n 1
n                       2                    3                5
odd

Example 21.6. Find the trigonometric Fourier series for the square voltage waveform shown
in Fig. 21.17(a) and sketch the line spectrum.
Solution. The function shown in Fig. 21.17 (a) is an odd function because at any time f(– t)
= – f (t). Hence, its Fourier series will contain only sine terms i.e. an = 0. The function also
possesses half-wave symmetry, hence, it will contain only odd harmonics.
As seen from Art. 21.7 (2) the Fourier series for the above wave is given by

f (θ) =
odd
∞

∑ b sin nθt
n =1
n                     where bn =
1
π   z0
2π
f (θ) sin nθdθ

=
1
π
R
S
T   z0
π
V sin nθdθ +                  z
π
2π
− V sin nθdθ =
U
V
W
V
πn
π
− cos nθ 0 +
V
πn
2π
cos nθ π

V
=      {(− cos nπ + cos 0) − cos πn}
πn

V                                   2V
=      {(1 − cos nπ ) + (1 − cos nπ)} =    (1 − cos nπ)
πn                                  πn
Now, 1 – cos nπ = 2 when n is odd
Fourier Series              801
and                                         =0               when n is even
2V     4V
∴ b1 =             ×2=    ... putting n = 1; b2 = 0 ... putting n = 2
π      π

2V     4V
b3 =        ×2=    ... putting n = 3; b4 = 0... putting n = 4
π3     3π

2V       4V
b5 =         ×2 =    .. putting n = 5 and so on.
π.5      5π

Fig. 21.17
Hence, the Fourier series for the given waveform is
4V                 4V                    4V
f( )                 sin         sin 3                 sin 5       ...
3                     5

4V    FG      1        1                  4V     2π    1  IJ6π    1FG  10 π                                       IJ
=
π     H
sin ω 0 t + 3ω 0 t + sin 5ω 0 t +... =
3        5                   π
sin
T
t + sin
3   KT     5 H
t + sin
T
t +...
K
The line spectrum of the function is shown in Fig. 21.17 (b). It would be seen that the
harmonic amplitudes decrease as 1/n, that is, the third harmonic amplitude is 1/3 as large as the
fundamental, the fifth harmonic is 1/5 as large and so on.
Example 21.7. Determine the Fourier series for the square voltage waveform shown in Fig.
21.17 (a). Plot its line spectrum.
Solution. This is the same question as given in Ex. 21.6 but has been repeated to
illustrated a singhtly different technique. As seen from Fig. 21.17(a) T = 2π , hence,
ω 0 = 2πf0 = 2π / T = 2π / π = 1. Over one period the function can be defined as

f (t ) = V 0 < t < 5
= −V , π < t < 2π

1    2                                   1   2                     1                           1
bn                       f (t)sin n 0tdt                    f (t )sin ntdt            f (t)sin ntdt           f (t )sin ntdt
0                                       0                         0

=
1
π    z0
π
V sin ntdt +
1
π   z
π
2π
(−V ) sin ntdt =
V − cos nt
π    n
π

0
+
V cos nt
π n
2π

π
802          Electrical Technology

V                     V
= −        (cos nπ − cos 0) +    (cos 2nπ − cos nπ)
nπ                    nπ
2V
Since cos 0 is 1 and cos 2nπ = 1 ∴ bn =                       (1 − cos nπ)
nπ
When n is even, cos nπ = 1                 ∴ bn = 0
2V          4V
When n is odd, cos nπ = – 1 ∴ bn            (1 1)      , n 1, 3, 5...
n           n
Substituting the value of bn, the Fourier series become

4V F                                I
∞                       ∞

∑                       ∑ n sin nt = π G sin t + 3 sin 3t + 5 sin 5t +...J
4V          4V     1                      1          1
f (t ) =
n =1
nπ
sin nt =
π   n =1
H                                K
odd                     odd

Since ω 0 = 1 , the above expression in general terms becomes
4V     FG     1            1                       IJ
f (t ) =
π      H
sin ω 0 t + sin 3ω 0 t + sin 5ω 0 t +...
3            5                        K
The line spectrum is as shown in Fig.
21.17 (b).
Example 21.8. Determine the
Fourier series of the square voltage
waveform shown in Fig. 21.18.
Solution. As compared to Fig. 21.17
(a) given above, the vertical axis of figure
has been shifted by π / 2 radians.
Replacing t by (t + π / 2) in the above
equation, the Fourier series of the
waveform shown in Fig. 21.18 becomes                                               Fig. 21.18

4V LM FG IJπ  1          π  FG IJ
1         π           FG    IJ OP
f (t ) =
π       H K
sin t +
N      2
+ sin 3 t +
3          2   H K
+ sin 5 t + +...
5         2            H     K Q
4V F
=
π
GH cos t − 1 cos 3t + 1 cos 5t...IJK
3          5
Example 21.9. Determine the trigonometric Fourier series for the half-wave rectified sine
wave form shown in Fig. 21.19 (a) and sketches line spectrum.

Fig. 21.19
Fourier Series             803
Solution. The given waveform shows no symmetry, hence its series would contain both sine
and cosine terms. Moreover, its average value is not obtainable by inpsection, hence it will have
to be found by integration.
Here, T                 2             0    2 /T                 1 . Hence, equation of the given waveform is V = Vm sin
ωt = Vm sin t .
The given waveform is defined by f (t ) = Vm sin t, 0 < t < π = 0, π < t < 2π

a0 =
1
2π         z   2π

0
f (t )dt =
1
2π
LM
N   z
0
π
Vm sin tdt +               zπ
2π
dt =
1
2π   z   0
π
Vm sin tdt
OP
Q
Vm          π    V                   V
=      |− cos t|0 = − m (cos π − cos 0) = m
2π               2π                   π

an =
1
π  z       2π

0
f (t ) cos ntdt =
1
π   z   0
π
Vm sin t cos ntdt

=
Vm
2π     z    π

0
[sin(n + 1)t − sin(n − 1)t ]dt =
Vm − cos(n + 1)t cos(n − 1)t
2π     n+1
+
n −1
π

0

Vm
−
LM
cos(n + 1)π cos(n − 1)π
+            +
1
−
1                                                                  OP
=
2π      N
n+1         n−1        n+1 n−1                                                                  Q
when n is even, cos (n + 1)π = −1 and cos(n − 1)π = −1

an =
Vm   1
−
1LM+
1
−
1
=−
2Vm                                   OP
∴
N
2π n + 1 n − 1 n + 1 n − 1    π(n 2 − 1)                               Q                                                              ... n = 2, 4, 6 etc.

when n is odd and ≠ 1, cos (n + 1)π = 1 and cos (n − 1)π = 1
Vm     1    FG
1    1    1                                                    IJ
∴       an =
2π
−     +
H+    −
n +1 n −1 n +1 n −1
=0
K                                                         ... n = 3, 5, 7 etc.

1                                                Vm                                        Vm
when n 1, a1                                   Vm sin t.cos tdt                                     sin t cos tdt                          sin 2tdt   0
0                                                   0                                  2    0

Hence, an = 0                                                                    ...n = 1, 3, 5...

bn =
1
π   z      0
2π
f (t ) sin ntdt =
1
π
LM
N   z
0
π
Vm sin t sin ntdt +            z
π
2π
(0) sin ntdt
OP
Q
Vm π
=
π 0           z
sin t sin ntdt = 0 for n = 2, 3, 4, 5 etc.
However, the expression indeterminate for n = 1 so that b1 has to be evaluated separately.

b1 =
1 π
π  0        z          V π           V
Vm sin t.sin tdt = m sin 2 tdt = m
π 0           2
The required Fourier series for the half-wave rectified voltage waveform is
z
Vm Vm         2V
∑G n
∞
F cos nt IJ
f (t ) =
π
+
2
sin t − m
π                                H
n =1          − 1K   2

even

Vm    π    FG 2         2         2                                                                             IJ
=
π    2     H
1 + sin t − cos 2t − cos 4t −
3        15        35
cos 6t...
K
804           Electrical Technology

FG
Vm    π           2             2            2                                   IJ
=
Hπ
1 + sin ω 0 t − cos 2ω 0 t − cos ω 0 t −
2           3            15           35
cos 6ω 0 t −...
K
V F π
=
T H
G1 + 2 sin θ − 2 cos 2θ − 15 cos 4θ − 35 cos 6θ−...IJK
m
3
2           2

The line spectrum is shown in Fig. 21.19 (b) which has a strong fundamental term with
rapidly decreasing amplitudes of the higher harmonics.
Example 21.10. Find the trigonometrical Fourier series for the full wave rectified voltage
sine wave shown in Fig. 21.20.
Solution. Since the given function has even symmetry, bn = 0 i.e. it will contain no sine terms
in its series.
The equation of the sinusoidal sine wave given by V = Vm sin θ . In other words,
f (θ) = Vm = sin θ .

a0 =
1 2π
2π 0     z
f (θ)dθ =
1 2π
2π 0               z
2V π
Vm sin θdθ = m
2π 0
sin θdθ

It is so because the two parts 0 − π and π − 2 are identical.
z
Vm          π  2V
∴       a0 =      |− cos θ|0 = m
π              π

an =
1
π   z0
2π
f (θ) cos nθdθ =
2Vm
π    z
0
π
sin θ cos nθdθ

1
Now, sin A cos B =                    [sin( A + B) + sin( A − B)]
2

∴ an =
2Vm
π          z0
π
[sin (1 + n)θ + sin (1 − n)θ]dθ

Fig. 21.20

Vm cos(1 + n)θ cos(1 − n)θ
= −                  +
π   (1 + n)     (1 − n)

LM
−Vm cos(1 + n)π cos(1 − n)π
+            −
1
−
1                            OP ... when n is odd
=0=
π     1+ n     N  1− n       1+ n 1− n                           Q
Fourier Series   805
However, when n is even, then
Vm                1             1                1           1          2Vm        1                1            4Vm
an
1 n 1 n                      1 n 1 n                              1 n 1 n                  ( n 2 1)
∞
cos 2θ
∑ (n
4Vm
∴     f (θ) = a0 −
π       n=2
2
− 1)
even

2Vm 4Vm 1           1  FG     1                                                             IJ
∴    f (θ) =
π
−
π 3           15   H
cos 2 θ − cos 4θ +
35
cos 6θ +...
K
Example 21.11. Determine the Fourier series for the waveform shown in Fig. 21.21 (a) and
sketch its line spectrum.
Solution. Its is seen from Fig. 21.21 (a) that the waveform equation is f (θ) = (Vm / π)θ . The
given function f (θ) is defined by

Vm
f( )                       , 0

= 0,                      2
Since the function possesses neither even nor odd symmetry, it will contain both sine and
cosine terms.
Average value of the wave over one cycle is Vm/4 or a0 = Vm/4. It is so because the average
value over the first half cycle is Vm/2 and over the second half cycle is 0 hence, the average value
(Vm / 2) + 0 Vm
for full cycle is =                     =
2        4

an =
1
π   z
0
2π
f (θ)nθdθ =
1
π
LM
N   z
0
π
(Vm / π )θ cos nθdθ +        zπ
2π
(0) cos nθdθ
OP
Q
=
Vm
π2     z   0
π
θ cos nθdθ =
Vm cos nθ
π2             n2
+
θ
n
π
V
sin nθ = 2m 2 (cos nπ − 1)
0 π n
∴         an = 0 when n is odd because cos nπ − 1 = 0

= −2Vm / π 2n2 when n is even

1 2π                  1 π                         2π
bn =
π ∫0 f (θ) sin θd θ = π ∫0 (Vm / π)θ sin nθd θ + ∫π (0) sin nθd θ

=
1
π   z   π

0
FG V IJ θ sin nθdθ = V
H πK
m
π
m
2
sin nθ
n   2
θ
− cos nθ
n
π

0
=
−Vm
πn
cos nπ

∴      bn = −Vm | πnwhenn is even bn = +Vm | π n when n is odd
Substituting the values of various constants in the general expression for Fourier series, we
get
806          Electrical Technology

Vm 2Vm          2Vm             2Vm
f (θ) =       − 2 cos θ −        cos 3θ −        cos 5θ...
4  π          (3π) 2
(5π) 2

Vm        V          V
+      sin θ − m sin 2θ + m sin 3θ....
π        2π         3π

Fig. 21.21
The amplitudes of even harmonics are given directly by bn but amplitudes of odd harmonics
are given by An = an + bn
2    2                                                                                     (Art. 21.4)
For example,

A1 = (2Vm / π 2 ) 2 + (Vm / π) 2 = 0.377 Vm

F 2V I + FG V IJ = 0.109 V
2        2
A3 =   GH (3π) JK H 2π K
m
2
m
m

F 2V I + FG V IJ = 0.064 V and so on.
=G
2        2

H (5π) JK H 5π K
m         m
A5           2                      m

The line spectrum is as shown in Fig. 21.21 (b).
Example 21.12. Find the Fourier series for the sawtooth waveform shown in Fig. 21.22 (a).
Sketch its line spectrum.
Solution. Using by the relation y = mx, the equation of the function becomes f (t) = 1, t or
f (t) = t.
T = 2, ω 0 = 2π / t = 2π / 2 = π

By inspection it is clear that a0 = 2/2 = 1, an =
2
T   z
0
T
z   2
f (t ) cos nω 0 t = t.cos nπt. dt
0

Since we have to find the integral of two functions, we use the technique of integration by
parts i.e.

z             z z FGH z
uvdx = u vdx −
du
dx
vdx dx
IJ
K
Fourier Series               807

∴               an = t.       z0
2
cos nπtdt −              z FGH z
0
2
1.
2

0
cos nπtdt dt
IJ
K
2                            2
t           cos nπt                                                               1
=       sin nπt +                                                = 0+                            (cos 2nπ − cos 0)
nπ        0   (nπ ) 2                               0                        (nπ ) 2

Since cos 2nπ = cos 0 for all values of n, hence an = 0

Fig. 21.22

bn =
2
T    z   0
T
f (t ) sin nω 0 tdt =                 z
0
T
t sin nπtdt

Employing integration by parts, we get,

z   2
bn = t sin nπt dt −
0                           z z 2

0
FG1.
H     0
2
sin nπtdt dt
IJ
K
z
2                                                                       2           2
− cos nπt                                     2 − cos nπt                     −t            sin nπt                    sin 2nπ          2
= t.             dt −                                                            =    .cos nπt +                        =               −      cos 2nπ
nπ     0
0            nπ                     nπ         0   (nπ ) 2           0       (nπ )   2       nπ

The sine term is 0 for all values of n because sign of any multiple of 2π is 0. Since value of
cosine term is 1 for any multiple of 2π , we have, bn = −2 / nπ .

∞                                                               ∞

∑ b sin nω t = a                                                   ∑
2              1
∴         f (t ) = a0 +                    n                   0                0   −                      sin nπt
n=1
π       n=1
n

2   FG    1         1                                                                 IJ
= 1−
π    H
sin πt + sin 2πt + sin 3πt +...
2         3                                                                  K
The line spectrum showing the amplitudes of various harmonics is shown in Fig. 21.22 (b).
Example 21.13. Determine the trigonometric Fourier series of the triangular waveform
shown in Fig. 21.23.
Solution. Since the waveform possesses odd symmetry, hence a0 = 0 and an = 0 i.e. there
would be no cosine terms in the series. Moreover, the waveform has half-wave symmetry. Hence,
series will have only odd harmonics. In the present case, there would be only odd sine terms. Since
the waveform possesses quarter-wave symmetry, it is necessary to integrate over only one quarter
period of finding the Fourier coefficients.
808           Electrical Technology

Fig. 21.23

∞
f (t ) =    ∑ b sin nω t
n =1
n                   0

odd

where         bn =
8
T   z   t/4

0
f (t )sin nω 0 t

The quarter-wave of the given waveform can be represented by equation of a straight line.
Slope of the straight line is Vm /(T / 40)                                                 4Vm / T .
Hence, using Y = mx, we have

f (t ) =
FG 4V IJ t 0 < t < T / 4 ∴ b
HTK
m
n   =
8
T     z    T /4

0
FG 4V IJ t.sin nω t = 32V
HTK
m
T
0     2
m
z
0
T /4
t.sin nω 0 tdt

Using the theorem of integration by parts, we have

bn =
32Vm
T   2
LMt
N    z
0
t/4
sin nω 0 t dt −    z z T /4

0
FG1.
H
T /4

0
sin nω 0 t dt dt
IJ OP
K Q
T /4
32Vm           t            cos n           0T      /4       sin n         0t                   32Vm         T cos n 0T / 4             sin n     0T / 4
2                                                                      2                          2                                              2
T              4               n           0                  (n   0)                             T          4     n 0                       (n   0)

Now, ω 0 = 2π / T or ω 0T = 2π ∴ nω 0 T / 4 = nπ / 2

∴ cos nω 0 T / 4 = cos nπ / 2 = 0 when n is odd

32Vm                                T   32V       nπ 8Vm       nπ
∴    bn =                               sin nω 0          = 2 m 2 sin    = 2 2 sin
n   2
ω 2T 2
0
4 n (2 π )     2  n π       2

8Vm                               −8V
∴    bn =            ....n = 1, 5,9,13 .... bn = 2 m .... n = 3, 7,11,15,...
n 2 π2                            n π2
Substituting this value of bn, the Fourier series for the given waveform becomes

8Vm                                  1              1                                  1
f (t )             2
sin            0t        2
3   0t        2
sin 5           0t             sin 7   0t   ...
3              5                               72
Example 21.14. Determine the Fourier series of the triangular waveform shown in Fig. 21.24.
Fourier Series                      809
Solution. Since the function has even
symmetry, bn = 0. Moreover, it also has half-
wave symmerty, hence, a0 = 0. The Fourier series
∞

can be written as f (t ) =                         ∑a
n =1
n   cos nω 0 t where

2      T
an                  f (t ) cos n        0 t dt.
T      0
Fig. 21.24
4Vm   T
The function is given by the relation f (t )                                                              t
T     4

It is so because for the interval 0 ≤ t ≤ T / 2, the slope of the line is                                                                          4 Vm / T .

2       T                                         4        T /4
∴ an                             f (t )cos n 0t dt                                         f (t )cos n 0t dt
T       0                                         T    0

16Vm             T /2                T                                    16Vm           T /2                         4Vm        T /2
t       cos n 0t dt                                              t cos n 0t dt                               cos n 0t dt
T2           0                   4                                      T2          0                              T         0

T /2                            T /2
−16Vm                  1                 t                                                       4V sin nω 0 t
=                            ⋅ cos nω 0 t +      sin nω 0 t                                        +
T2             4n ω 0
2 2                 nω 0                                                     T    nω 0              0
0

Substituting ω = 2π / T , we get

16Vm                 T2                                        T2                           2V
an                 2                   2 2
(cos n         1)               (sin n )                    sin n
T                4 n                                           4 n                           n

Now, sin nπ = 0 for all values of n, cos nπ = 1 when n is even add – 1 when n is odd.
8Vm
∴          an =
π 2 n2
∞                                                          ∞
cos nω 0 t
∑                                                           ∑
8Vm                                  8Vm
∴ f (t ) =                                    cos nω 0 t =
n= 0        π n
2 2
π   2
n =1            n2
odd                                                          odd

8Vm                                 1                          1                            1
=            2
cos          0t          cos 3           0t          .cos 5           0t          cos 7         0t   ...
9                          25                           49
Alternative Solution
We can deduce the Fourier series from Fig. of Ex. 21.11 by shifting the vertical axis by π / 2
radians to the right. Replacing t by (t + π / 2) in the Fourier series of Ex. 21.11, we get
8Vm                                               1                                        1                                     1                        n
f(t)             2
sin      0       t                   2
sin 3         0      t                2
sin 5    0    t                  2
sin 7     0    t       ...
2             3                              2         5                          2        7                          2
8Vm                             1                           1                            1
2
cos          0t         cos 3            0t          cos 5              0t        cos         0t       ...
9                           25                           49
810            Electrical Technology
Example 21.15. Obtain the Fourier series representation of the sawtooth waveform shown
in Fig. 21.25 (a) and plot its spectrum.

Fig. 21.25
Solution. By inspection, we know that the average value of the wave is zero over a cycle
because the height of the curve below and above the X-axis is the same hence, a0 = 0. Moreover,
it has odd symmetry so that an = 0 i.e. there would be no cosine terms. The series will contain
only sine terms.

1   2
∴ f( )             bn sin n where bn                       f ( )sin n d
0
n 1

The slope of curve is m = V / π
∴ we get, f (θ) = (V / π) θ.
If we are the limit of integration form − π to + π then

1       V                      V        1                                 2V
bn                  sin n d              2   2
sin n           cos n            cos n
n                            n                 n
The above result has been obtained by making use of integration by parts as explained earlier.
cos nπ is positive when n is even and is negative when n is odd and thus the signs of the
coefficients alternate. The required Fourier series is
2V                1             1               1
f( )           sin          sin 2         sin 3           sin 4     ...
2             3               4

2V           1           1            1
or f (t )           sin   0tsin 2 0t    sin 3 0t     sin 4 0t ...
2           3            4
As seen, the coefficients decrease as 1/n so that the series converges slowly as shown by the
line spectrum of Fig. 21.25 (b).
The amplitudes of the fundamental of first harmonic, second harmonic, third harmonic and
fourth harmonic are (2 / π),(2V / 2π),(2V / 3π) and (2V / 4 π ) respectively.

Tutorial Problem. 21.1
1. Determine the Fourier series for the triangular waveform shown in Fig. 21.26 (a)
(Network Theory and Design, AMIETE June 1990)
Fourier Series         811

Fig. 21.26

⎡1 4               1             1                 ⎤
⎢ 2 − π2 cosω0 t + 32 cos3ω0 t + 52 cos5ω0 t + ....⎥
⎣                                                  ⎦
2. Find the values of the Fourier coefficients a0, an and bn for the function given in fig. 21.26 (b).

⎡     2        7    2πn       7 ⎛        2πn ⎞ ⎤
⎢a0 = 3 ;an = nπ sin 3 ;bn = nπ ⎜ 1 − cos 3 ⎟ ⎥
⎣                               ⎝            ⎠⎦
3. Determine the trigonometric series of the triangular waveform shown in Fig. 21.27. Sketch its line
spectrum.

Fig. 21.27

Vm      4Vm                 1            1
2
cos   0   t      cos 3 0 t 2 cos5 0 t ...
2                          32           5
4. Determine the Fourier series for the sawtooth waveform shown in Fig. 21.28.

Fig. 21.28                                        Fig. 21.29
812        Electrical Technology

⎡      Vm Vm             1           1               ⎤
⎢f(t) = 2 + π (sinω0 t + 2 sin2ω0t + 3 sin3ω0t + ....⎥
⎣                                                    ⎦
5. Represent the full-wave rectified voltage sine waveform shown in Fig. 21.29 by a Fourier series.

⎡      2Vm ⎛ 2                2          2             ⎞⎤
⎢f(t) = π + ⎜ 1 + 3 cos2ωt − 15 cos4ωt + 35 cos6ωt ... ⎟⎥
⎣           ⎝                                          ⎠⎦
6. Obtain trigonometric Fourier series for the wave form shown in Fig. 21.30.

Fig. 21.30                                        Fig. 21.31

⎡        2Vm ⎛          1            1            1              ⎞⎤
⎢f(t) = − π ⎜ sinω0 t + 2 sin2ω0 t + 3 sin3ω0 t + 4 sin4ω0 t ... ⎟⎥
⎣            ⎝                                                   ⎠⎦
7. Find the Fourier series for the sawtooth waveform shown in Fig. 21.31.

⎡       2⎛            1          1              ⎞⎤
⎢f(t) = π ⎜ −sin2πt + 2 sin4πt − 3 sin6πt + ... ⎟⎥
⎣         ⎝                                     ⎠⎦
8. For the waveform of Fig. 21.32, find the Fourier
series terms up to the 5th harmonic.
(Network Theory Nagpur Univ. 1993)

⎡      16 ⎛        2       ⎞ 1              ⎤
⎢V(t) = π ⎜ sint + 3 sin3t ⎟ + 5 sin5t + ...⎥
⎣         ⎝                ⎠                ⎦
Fig. 21.32
9. Determine Fourier series of a repetitive triangular wave as shown in Fig. 21.33.

Fig. 21.33
(a) What is the magnitude of d.c. component?
(b) What is the fundamental frequency?
Fourier Series         813
(c) What is the magnitude of the fundamental?
(d) Obtain its frequency spectrurm.                            (Network Theory Nagpur Univ.1993)
[(a) 5 V (b) 1 Hz (c) 10 / π volt ]
10. Determine the Fourier series of voltage responses obtained at the o/p of a half wave rectifier
shown in Fig. 21.34. Plot the discrete spectrum of the waveform.
(Elect. Network Analysis Nagpur Univ. 1993)

Fig. 21.34

⎡       Vm Vm           2Vm          2Vm           2Vm            ⎤
⎢ V(t) = π + 2 cosω0 t + 3π cos10πt − 15 cos20πt + 35π cos30πt ...⎥
⎣                                                                 ⎦
11. Determine the Fourier coefficients and plot amplitude and phase spectral.
(Network Analysis Nagpur Univ. 1993)
⎡                  4Vm
⎢a3 = 0,bn = 0,a1 = π ,a2 = 0
⎣
−4Vm             4V ⎤
a3 =        ,a4 = 0,a5 = m ⎥
3π               π ⎦

Fig. 21.35

OBJECTIVE TESTS – 21
1. A given function f (t) can be represented by           (d) all of the above.
a Fourier series if it                              2. In a Fourier series expansion of a periodic
(a) is periodic                                        function, the coefficient a0 represents its
(b) is single valued                                   (a) net area per cycle
(c) has a finite number of maxima and                   (b) d.c value
minima in any one period
814        Electrical Technology

(c) average value over half cycle                  10. If the Fourier coefficient bn of a periodic
(d) average a.c value per cycle                        function is zero, then it must possess ______
3.   If in the Fourier series of a periodic function,       symmetry.
the coefficient a0 is zero, it means that the          (a) even
function has                                           (b) even quarter-wave
(a) odd symmetry                                       (c) odd
(b) even quarter-wave symmetry                         (d) either (a) and (b).
(c) odd quarter-wave symmetry                      11. A complex voltage waveform is given by
(d) any of the above.                                  V = 120 sin ωt + 36(3ωt + π / 2) + 12 sin
4.   A periodic function f (t) is said to possess            (5ωt + π ). It has a time period of T seconds.
odd quarter-wave symmetry if                           The percentage fifth harmonic contents in the
(a) f (t) = f (–t)                                     waveform is
(b) f (–t) = –f (t)                                    (a) 12                   (b) 10
(c) f (t) = –f (t + T/2)                               (c) 36                   (d) 5
(a) both (b) and (c).                              12. In the waveform of Q. 11 above, the phase
5.   If the average value of a periodic function            displacement of the third harmonic represents
over one period is zero and it consists of             a time interval of ___ seconds.
only odd harmonics then it must be                     (a) T/12                 (b) T/3
possessing _______ symmetry.                           (c) 3T                   (d) T/36
(a) half-wave                                      13. When the negative half-cycle of a complex
(b) even quarter-wave                                  waveform is reversed, it becomes identical to
(c) odd quarter-wave                                   its positive half-cycle. This feature indicates
(d) odd.                                               that the complex waveform is composed of
6.   If in the Fourier series of a periodic function,       (a) fundamental
the coefficient a0 = 0 and an = 0, then it must        (b) odd harmonics
be having _______ symmetry.                            (c) even harmonics
(a) odd                                                (d) both (a) and (b)
(b) odd quarter-wave
(e) both (a) and (c)
(c) even
14. A periodic waveform possessing half-wave
(d) either (a) or (b).                                 symmetry has no
7.   In the case of a periodic function possessing
(a) even harmonics
half-wave symmetry, which Fourier
(b) odd harmonics
coefficient is zero ?
(a) an                   (b) bn                        (c) sine terms
(c) a0                   (d) none of above.            (d) cosine terms
8.   A periodic function has zero average value         15. The Fourier series of a wave form possessing
over a cycle and its Fourier series consists of        even quarter-wave symmetry has only
only odd cosine terms. What is the symmetry            (a) even harmonics
possessed by this function.                            (b) odd cosine terms
(a) even                                               (c) odd sine terms
(b) odd                                                (d) both (b) and (c).
(c) even quarter-wave                              16. The Fourier series of a waveform possessing
(d) odd quarter-wave                                   odd quarter-wave symmetry contains only
9.   Which of the following periodic function               (a) even harmonics
possesses even symmetry ?                              (b) odd cosine terms
(a) cos 3t               (b) sin t                     (c) odd cosine terms
(c) t . cos 50 t         (d) (t + t2 + t5).            (d) none of above