The Fourier Series
Subject: Signals and Systems
Prepared by: Dr. Ketki C Pathak
Electronics and Communication Engineering Dept
S.C.E.T., Surat
Introduction
A Fourier series is an expansion of a
periodic function f (t) in terms of an infinite sum
of cosines and sines

f (t )  a0   (an cos nt  bn sin nt )
n 1
In other words, any periodic function can be
resolved as a summation of
constant value and cosine and sine functions:

f (t )  a0   (an cos nt  bn sin nt )
n 1
 a0  (a1 cost  b1 sin t )
 (a2 cos 2t  b2 sin 2t )
 (a3 cos 3t  b3 sin 3t )  
dc
ac
The computation and study of Fourier series is known as harmonic
analysis and is extremely useful as a way to break up an arbitrary
periodic function into a set of simple terms that can be plugged in,
solved individually, and then recombined to obtain the solution to
the original problem or an approximation to it to whatever accuracy
is desired or practical.
f(t)
a0
Periodic Function
=
t
b1 sin t
a1 cos t
+
+
b2 sin 2t
a2 cos 2t
+
+
+ …

f (t )  a0   (an cos n0t  bn sin n0t )
n 1
2
 Fundementa l frequency
where 0 
T
1 t 0 T
a0  
f (t ) dt
T t0
2 t 0 T
an  
f (t ) cos n0t dt
T t0
2 t 0 T
bn  
f (t ) sin n0t dt
T t0
Notice that
1 t 0 T
a0  
f (t )dt
T t0
Area below graph over a period

T
 Average value
 dc component
[*The right sign of the area below graph must be
obeyed. +ve sign for area above x-axis & −ve sign
for area below x-axis]
Symmetry Considerations
 Symmetry functions:
(i)
even symmetry
(ii)
odd symmetry
Even symmetry
 Any function f (t) is even if its plot is symmetrical about the vertical
axis, i.e.
f (t )  f (t )
Even symmetry (cont.)
 The examples of even functions are:
f (t ) | t |
f (t )  t 2
t
t
f (t )  cos t
t
Even symmetry (cont.)
 The integral of an even function from −A to
+A is twice the integral from 0 to +A
f even (t )
A
t
−A
+A
f
A
A
even
(t )dt  2  f even (t )dt
0
Odd symmetry
 Any function f (t) is odd if its plot is antisymmetrical about the vertical
axis, i.e.
f (t )   f (t )
Odd symmetry (cont.)
 The examples of odd functions are:
f (t )  t 3
f (t )  t
t
t
f (t )  sin t
t
Odd symmetry (cont.)
 The integral of an odd function from −A to
+A is zero
f odd (t )
A
−A
+A
t
 f (t )dt  0
odd
A
Even and odd functions
The product properties of even and odd
functions are:
 (even) × (even) = (even)
 (odd) × (odd)
= (even)
 (even) × (odd) = (odd)
 (odd) × (even) = (odd)
Symmetry consideration
From the properties of even and odd functions, we can show that:
 for even periodic function;
T /2
4
an   f (t ) cos ntdt
T 0
bn  0
 for odd periodic function;
T /2
a0  an  0
4
bn   f (t ) sin ntdt
T 0
How?? [Even function]
f (t )
T

2
T /2
T /2
T
2
t
T /2
2
2
4
f (t ) sin ntdt  0
an 
f (t ) cos ntdt   f (t ) cos ntdt bn 


T T / 2
T T / 2
T 0
(even) × (even)
(even) × (odd)
||
||
(even)
(odd)
How?? [Odd function]
f (t )
T

2
t
T
2
T /2
2
a0 
f (t )dt  0

T T / 2
(odd)
T /2
2
an 
f (t ) cos ntdt  0

T T / 2
(odd) × (even)
||
(odd)
T /2
T /2
2
4
bn 
f
(
t
)
sin
n

tdt

f (t ) sin ntdt


T T / 2
T 0
(odd) × (odd)
||
(even)
The amplitude-phase form

f (t )  a0   (an cos n0t  bn sin n0t )
n 1 as the sine-cosine form
is known
 We can also express the Fourier series in the cosine
form only, that is

This form is called as the amplitude-phase form
0
n
0
n
n 1
f (t )  a   A cos(n t   )
 From this form, we can plot the amplitude spectrum,
phase spectrum,
vs. n.
vs. n and the
A
 It can be shown thatn the combination of cosine and sine function
can be expressed as a cosine function only:
n

 Comparing both sides of eqn:
a cos x  b sin x  A cos(x   )
 A(cos x cos  sin x sin  )
 ( A cos ) cos x  ( A sin  ) sin x
A cos  a .....(1)
A sin   b .....(2)
(1) 2  (2) 2 :
A cos   A sin   a  b
2
A (cos   sin  )  a  b
2
2
2
2
2
2
2
2
2
2
A2  a 2  b 2  A  a 2  b 2
(2)
:
(1)
A sin   b

A cos
a
b
1   b 
tan  
   tan 

a
 a 
 Hence
an cos n0t  bn sin n0t  An cos(n0t  n )
where
  bn 

phase, n  tan 
 an 
1
amplitude, An  a  b
2
n
2
n
A0  a0
Or in phasor/complex form:
  bn 

An n  an  jbn  a  b  tan 
 an 
2
n
2
n
1
Example 1
Determine the Fourier series of the following
waveform. Obtain the amplitude and phase
spectra.
Solution
First, determine the period & describe the one period
of the function:
T=2
1, 0  t  1
f (t )  
0, 1  t  2
2

We find that 0 
T
f (t  2)  f (t )
Then, obtain the coefficients a0, an and bn:
T
1
a0   f (t )dt
T 0
1
2

 1
1
1
1
  f (t )dt    1dt   0dt   (1  0) 
20
2 0
2
1
 2
2
Or
Area below graph 11 1
a0 


T
2
2
2
2
an   f (t ) cos n0tdt
T 0
sin n
 sin nt 
  1 cos ntdt   0dt  


n
 n  0
0
1
1
2
1
Notice that n is integer which leads sin n  0 ,
since sin   sin 2  sin 3    0
Therefore, an  0.
2
2
bn   f (t ) sin n0tdt
T 0
 cos nt  1  cos n
  1sin ntdt   0dt  


n  0
n

0
1
1
2
1
Notice that cos  cos 3  cos 5    1
cos 2  cos 4  cos 6    1
or
cos n  (1)n
1  (1) n 2 / n

Therefore, bn 
n
 0
, n odd
, n even
Finally,

f (t )  a0   (an cos n0t  bn sin n0t )
n 1
1  1  (1) n 
 
 sin nt
2 n 1  n 
1   2 
  
 sin nt
2 n 1  n 
n odd
1 2
2
2
  sin t 
sin 3t 
sin 5t  
2 
3
5
In amplitude-phase form,
1   2 
f (t )    
 sin nt
2 n 1  n 
n odd
1   2 
  
 cos(nt  90)
2 n 1  n 
n odd
1 2
2
  cos(t  90) 
cos(3t  90)
2 
3
2

cos(5t  90)  
5
Amplitude spectrum:
Phase spectrum:
Some helpful identities
sin x  cos(x  90)
sin( x)   sin x
 sin x  cos(x  90)
cos( x)  cos x
 cos x  cos(x  180)
For n integers,
sin n  0
cos n  (1)n
sin 2n  0
cos 2n  1
Notes:
 The sum of the Fourier series terms can evolve (progress) into the
original waveform
 From Example 1, we obtain
1 2
2
2
f (t )   sin t 
sin 3t 
sin 5t  
2 
3
5
 It can be demonstrated that the sum will
lead to the square wave:
(a)
(b)
2

(c)
2

2
sin t

sin t 
2
sin 3t
3
(d)
sin t 
2
2
sin 3t 
sin 5t
3
5
2

sin t 
2
2
2
sin 3t 
sin 5t 
sin 7t
3
5
7
(e)
2

sin t 
2
2
2
2
sin 3t 
sin 5t 
sin 7t 
sin 9t
3
5
7
9
(f)
1 2
2
2
 sin t 
sin 3t   
sin 23t
2 
3
23
Example 2
Given f (t )  t ,
1  t  1
f (t  2)  f (t )
Sketch the graph of f (t) such that  3  t  3.
Then compute the Fourier series expansion of f (t).
Plot the amplitude and phase spectra until the forth
harmonic.
Solution
The function is described by the following graph:
T=2
2

We find that 0 
T
Then we compute the coefficients:
a0  an  0 since f(t) is an odd function.
1
1
4
4
bn   f (t ) sin n0tdt   t sin ntdt
T 0
20
cos nt
 t cos nt 
 2 
 2
dt

n  0
n

0
1
1
 2 cos n
 sin nt 

 2 2 2 
n
 n  0
1
2 cos n 2 sin n
2(1) n
2(1) n 1

 2 2 
0 
n
n
n
n
Hence,

f (t )  a0   (an cos n0t  bn sin n0t )
n 1
2(1) n 1

sin nt
n
n 1
2
1
2
1
 sin t  sin 2t 
sin 3t 
sin 4t  


3
2
2
1
 cos(t  90)  cos(2t  90)



2
1

cos(3t  90) 
cos(4t  90)  
3
2
Amplitude spectrum:
0.64
0.32
0.21
0.16
Phase spectrum:
Example 3
2  t , 0  t  2
Given v(t )  
, 2t 4
 0
v(t  4)  v(t )
(i) Sketch the graph of v (t) such that
0  t  12.
(ii) Compute the trigonometric Fourier series of v (t).
(iii) Express the Fourier series of v (t) in the
amplitude-phase form. Then plot the amplitude
and phase spectra until the forth harmonic.
Solution
(i) The function is described by the following graph:
v (t)
2
0
2
4
6
T=4
2 

We find that 0 
T
2
8
10
12
t
(ii) Then we compute the coefficients:
2
4


1
1
a0   v(t )dt   (2  t )dt   0dt 
T 0
4 0
2

4
2
1
1
t 
1
  (2  t )dt  2t   
40
4
2 0 2
2
Or
 2 2 1
a0 

4
2
1
2
2
4
2
4
2
1
an   v(t ) cos n0tdt   (2  t ) cos n0tdt   0
T 0
20
2
2
1  (2  t ) sin n0t  1 sin n0t
 
dt
  
2
n 0
 0 2 0 n 0
2
2
1  cos n0t  1  cos 2n0
 0   2 2  
2  n 0  0
2n 202
2(1  cos n ) 2[1  (1) n ]  0



2 2
2 2
2 2
4
/
n

n
n

, n even
, n odd
4
2
4
2
1
bn   v(t ) sin n0tdt   (2  t ) sin n0tdt   0
T 0
20
2
2
2


1  (2  t ) cos n0t
1 cos n0t
 
dt
  
2
n0
 0 2 0 n0
2
1  sin n0t 

  2 2 
n0 2  n 0  0
1
sin 2n0
1
2




2 2
n0
2 n 0
n0 n
1
since sin 2n0  sin n  0
Hence,

v(t )  a0   (an cos n0t  bn sin n0t )
n 1
1   2[1  (1) n ]  nt  2
 nt 
  
cos
sin


2 2
2 n 1  n 
 2  n
 2 

 a0   An cos(n0t  n )
n 1
(iii) Where
An n  an  jbn
2

, n even
  j n

2  2

 
 j  , n odd
 n  n

2

90
, n even

n
 

2
4
1  n 



 1   tan 
 , n odd
2 2

 n n 
 2 

We obtained that
A0  a0  0.5
A11  0.75  57.5
A2 2  0.32  90
A33  0.22  78.0
A4 4  0.16  90

v(t )  a0   An cos(n0t  n )
n 1
 t

 0.5  0.75 cos  57.5   0.32 cost  90
2

 3t

 0.22 cos
 78.0   0.16 cos2t  90  
 2

Amplitude spectrum:
Phase spectrum:
0.75
0.5
π/2
0.32
π
3π/2
2π
0.22
0.16
−57.5°
−90°
0
π/2
π
3π/2 2π
−78.0°
−90°
Example 4
Given
 1 ,  2  t  1

f (t )   t ,  1  t  1
1 ,
1 t  2

f (t  4)  f (t )
Sketch the graph of f (t) such that  6  t  6.
Then compute the Fourier series expansion of f (t).
Solution
The function is described by the following graph:
f (t)
1
−6
−4
−2
−1
0
T=4
2 

We find that  
T
2
2
4
6
t
Then we compute the coefficients. Since f (t) is
an odd function, then
2
2
a0   f (t )dt  0
T 2
and
2
2
an   f (t ) cos ntdt  0
T 2
2
2
2
4
bn   f (t ) sin ntdt   f (t ) sin ntdt
T 2
T 0
1
2


4
   t sin ntdt   1sin ntdt 
4 0
1

cos nt
 t cos nt 
 cos nt 
 

dt  


n

n

n


0 0

1
1
2
1
cos n  sin nt  cos 2n  cos n

 2 2  
n
n
 n  0
1
cos 2n sin n
2 cos n 4 sin( n / 2)

 2 2 

2 2
n
n
n
n
since sin 2n  sin n  0
Finally,
a0 
f (t )    (an cos nt  bn sin nt )
2 n 1
nt
 2 cos n 
 
 sin
n 
2
n 1 

(1) n 1
nt
 2
sin
n
2
n 1

♣
Example 5
Compute the Fourier series expansion of f (t).
Solution
The function is described by
1 , 0  t  1

f (t )  2 , 1  t  2
1 , 2  t  3

T=3
f (t  3)  f (t )
T=3
and
2 2
0 

T
3
Then we compute the coefficients.
3
1
2
3
 2
2
2
8
a0   f (t )dt    1dt   2dt   1dt   (1  0)  2(2  1)  (3  2) 
T 0
3 0
3
1
2
 3
Or, since f (t) is an even function, then
3
3/ 2
1
3/ 2
 4
2
4
4
 3  8
a0   f (t )dt   f (t )dt    1dt   2dt   (1  0)  2  1 
T 0
T 0
3 0
 2  3
1
 3
Or, simply
3
2
2  Total area below graph  2
8
   4 
a0   f (t )dt   
in a period
T 0
T 
3
 3
3
3/ 2
2
4
an   f (t ) cos ntdt   f (t ) cos ntdt
T 0
T 0
1
3/ 2

4
   1cos ntdt   2 cos ntdt 
3 0
1

4  sin nt  4  2 sin nt 
 
 

3  n  0 3  n 1
1
3/ 2
4 
 3n


sin n  2 sin
 sin n 

3n 
2


4 
3n


2
sin

sin
n



3n 
2

;
2

3
2 
2n 
2
2n

sin
 2 sin n  sin

n 
3 
n
3
and bn  0
since f (t) is an even function.
Finally,
a0 
f (t )    (an cos nt  bn sin nt )
2 n 1
4   2
2n 
2nt
  
sin
 cos
3 n 1  n
3 
3
4 2  1
2n 
2nt
    sin
 cos
3  n 1  n
3 
3
♣
Parseval’s Theorem
 Parserval’s theorem states that the average power in
a periodic signal is equal to the sum of the average
power in its DC component and the average powers
in its harmonics
f(t)
Pdc
Pavg
a0
=
Pa1
t
Pb1
b1 sin t
a1 cos t
+
+
Pa2
+
Pb2
+ …
+
a2 cos 2t
b2 sin 2t
 For sinusoidal (cosine or sine) signal,
2
Vpeak



2
2
V
Vrms 
1 peak
2
P


R
R
2 R
 For simplicity, we often assume R = 1Ω,
which yields
P V
2
rms
1 2
 Vpeak
2
 Therefore, the total power of all harmonics is
Pavg  Pdc  Pa1  Pb1  Pa2  Pb2  
1 2 1 2 1 2 1 2
 a  a1  b1  a2  b2  
2
2
2
2


1
1
 Pavg  a02   (an2  bn2 )  a02   An2
2 n 1
2 n 1
2
0
2
 And since P  Vrms
,


1
1
Vrms  a02   (an2  bn2 )  a02   An2
2 n 1
2 n 1
Circuit applications
Steps for applying Fourier series:
1.
Express the excitation as a Fourier series
2.
Transform the circuit from the time domain to the frequency
domain
3.
Find the response of the dc and ac components in the Fourier
series
4.
Add the individual dc and ac responses using the superposition
principle
Example 6
The signal in Figure 6.1 is applied to the circuit in
Figure 6.2.
(i)
(ii)
(iii)
(iv)
Find the Fourier series of is(t).
Plot the amplitude spectrum for the dc
component and the first three non-zero
harmonics of is(t).
Find the load current iL(t).
Find the average power dissipated in the
resistor RL.
is (t )
Figure 6.1
iL (t )
1H
is (t )
1
Figure 6.2
RL  1 
Solution
(i)
The current source is described by
is (t )  | t | ,    t  
 t , 0t 

 t ,    t  0
and
T  2
, 0  1
Compute the Fourier series. Since is(t) is an even
function,


2
2
1 t 

a0   is (t )dt 
tdt    

T 0
2 0
  2 0 2
T /2
2
or
1  Area below graph  1  1
 
         
a0   
  over 0  t      2
 2
T /2
4
an   is (t ) cos n0tdt
T 0



4
2  t sin nt 
2 sin nt

t cos ntdt  
 
dt


2 0
  n 0  0 n

2  sin n 2  cos nt 

  2 

n
  n 0
2(cos n  1)

n 2
2


4
/
n

2[( 1)  1]


2
n
 0
n
, n odd
, n even
bn  0
Therefore, the Fourier series of is(t) is


 4 
is (t )      2  cos nt
2 n 1  n  
n odd


 4 
    2  cos(nt  180)
2 n 1  n  
n odd

4
4
  cos(t  180) 
cos(3t  180)
2 
9
4

cos(5t  180)  
25
(ii)
Hence,
I s0 

I sn 
2
4
n
2
I sn
I sn
1.571
180 , n  1,3,5,
1.273
0.141
0.051
n
n
(iii)
To compute iL(t), separate to dc and ac analysis:
dc analysis (n = 0)
S/c the inductor, hence
I L0 
1
I s0
1
I s0
2

 /2
2


4
ac analysis (n ≥ 1)
R  j L
IL 
Is
RL  ( R  j  L )
or
R  j n L
I Ln 
I sn
RL  ( R  jn L)
, n  n0  n
 1  jn  4

 2 180 
 

 2  jn  n 

4
n 2
1  n 2  1

1 n
 tan n  tan
 180 
2
4n 
2

I L0  0.785,
I L1  0.805198.43, I L3  0.124195.26,
I L5  0.048190.49
Hence,
iL (t )  0.785  0.805 cos(t  198.43)
 0.124 cos(3t  195.26)  0.048 cos(5t  190.49)

(iv)
Average power absorbed by Rl is
PI
2
L ( rms )
RL
 2 1  2 
  I L0   I Ln  RL
2 n 1 

1
2
 0.785  (0.8052  0.1242  0.0482 )  0.949 W
2
Example 7
The signal in Figure 7.1 is applied to the circuit in
Figure 7.2.
(i)
Find the Fourier series of vs(t).
(ii) Find the output voltage vo(t).
(iii) Plot the amplitude spectrum for the dc
component and the first three non-zero harmonics
of the output voltage vo(t).
(iv) Find the r.m.s value of vo(t) .
vs (t )
Figure 7.1
1
vs (t )

1
vo (t )
1F

Figure 7.2
Solution
(i)
The voltage source is described by
 1 , 0  t 1
vs (t )  
 1 ,  1  t  0
and
T  2 , 0  
Compute the Fourier series. Since vs(t) is an odd
function,
a0  an  0
T /2
4
bn   vs (t ) sin n0tdt
T 0
4
 cos nt 
  sin ntdt  2 

20
n


0
1
2(1  cos n )

n
2[1  (1) n ] 4 / n


n
 0
1
, n odd
, n even
Therefore, the Fourier series of vs(t) is

4
 4 


vs (t )   
 cos(nt  90)
 sin nt   
n 1  n 
n 1  n 

n odd
n odd
4
cos(3t  90)
 cos(t  90) 
3

4
cos(5t  90)  

5
4
Hence,
Vs0  0
4
Vsn 
  90 , n  1,3,5,
n
(ii)
To compute vo(t), separate to dc and ac analysis:
dc analysis (n = 0)
O/c the capacitor, hence
I 0
1
Vs0  0

1
Vo0  0

ac analysis (n ≥ 1)
R  (1 / jC )
1  jRC
Vo 
Vs 
Vs
R  [ R  (1 / jC )]
1  j 2RC
or
1  jn RC
Von 
Vsn
1  j 2n RC
, n  n0  n
 1  jn  4


 
  90 

 1  j 2n  n
4

n
1  n 2 2
1
1

tan
n


tan
2n  90
2 2
1  4n 


Vo0  0,
Vo1  0.660  98.61, Vo3  0.213  93.02,
Vo5  0.128  91.82
Hence,
vo (t )  0.660 cos(t  98.61)  0.213cos(3t  93.02)
 0.128 cos(5t  91.82)  
(iii) Vo
0.660
Von
n
0.213
n
0.128
n
(iv)
−98.61°
−93.02°
−91.82°

1
Vo ( rms)  Vo20  Vo2n
2 n1
1

(0.6602  0.2132  0.1282 )  0.705 V
2
Exponential Fourier series
 Recall that, from the Euler’s identity,
e jx  cos x  j sin x
yields
e e
cos x 
2
jx
 jx
and
e jx  e  jx
sin x 
j2
Then the Fourier series representation becomes
a0 
f (t )    (an cos nt  bn sin nt )
2 n 1
 e jnt  e  jnt 
a0    e jnt  e  jnt 
  bn 

   an 
2 n 1  
2
j2



 e jnt  e  jnt 
a0    e jnt  e  jnt 
  jbn 

   an 
2 n 1  
2
2



a0   an  jbn  jnt  an  jbn   jnt 
   
e  
e

2 n 1 
2 
2



a0   an  jbn  jnt   an  jbn   jnt
  
e   
e
2 n 1 
2 
2 
n 1 
an  jbn
cn 
2
Here, let we name
a0
and c0  . Hence,
an  jbn
, c n  2
2
a0   an  jbn  jnt   an  jbn   jnt
f (t )    
e   
e
2 n 1 
2 
2

n 1 
c0
c−n
cn


n 1
n 1


n 1
n  1
 c0   cn e jnt   c n e  jnt
 c0   cn e jnt   cn e jnt
1


n  
n 1
n  
  cn e jnt  c0   cn e jnt   cn e jnt
Then, the coefficient cn can be derived from
an  jbn
cn 
2
T
T
12
j 2

f (t ) cos ntdt 
f (t ) sin ntdt


2T 0
2T 0
T
T

1
   f (t ) cos ntdt  j  f (t ) sin ntdt 
T 0
0

T
1
  f (t )[cos nt  j sin nt ]dt
T 0
T
1
  f (t )e  jnt dt
T 0
 In fact, in many cases, the complex Fourier series is easier to obtain
rather than the trigonometrical Fourier series
 In summary, the relationship between the complex and
trigonometrical Fourier series are:
T
T
1
c0  a0   f (t )dt
T 0
1
cn   f (t )e  jnt dt
T 0
an  jbn 1
cn 
 An n
2
2
an  jbn
c n 
2
an2  bn2 An
cn 

2
2
or
cn  cn
 The average power and the rms value in the term of Fourier complex
coefficient are


P1    cn  c  2 cn
2
2
0
n  
Frms 

c
n  
2
n
2
n 1

 c  2  cn
2
0
n 1
2
Example 8
For the given function,
(i)
Obtain the complex Fourier series
(ii) Plot the amplitude and the phase spectra
of the complex Fourier series for −5 ≤ n ≤ 5
(iii) Calculate the average power and the rms value
of the signal
v(t )
e 2
1
 4
 2
0
2
4
t
Solution
(i)
Since T  2 , 0  1 . Hence
T
1
c0   v(t )dt
T 0
2
1
t

e dt

2 0
1 t 2 e 2  1

e 0 
2
2
 
T
1
 jn0t
cn   v(t )e
dt
T 0
2
2
1
1
t  jnt
(1 jn ) t

e e dt 
e
dt


2 0
2 0
2

1 e



2  1  jn  0
(1 jn ) t
e 2 (1 jn)  1 e 2 e  j 2 n  1
e 2  1



2 (1  jn)
2 (1  jn)
2 (1  jn)
since e j 2n  cos 2n  j sin 2n  1  0  1
2
2
e 1
e 1
cn n  0 

 c0
2 (1  jn) n 0
2
Therefore, the complex Fourier series of v(t) is

v(t )   cn e
n  
jn0t
2
e  1 jnt
 
e
n   2 (1  jn)

*Notes: Even though c0 can be found by substituting
cn with n = 0, sometimes it doesn’t works (as shown
in the next example). Therefore, it is always better to
calculate c0 alone.
(ii)
e 2  1

1   n  
cn  n 
0  tan 

2
 1 
2 1  n

85

 tan 1 n
2
1 n
The complex frequency spectra are
(ii)
The average power is (assume R = 1 Ω)

P1    cn
2
n  
 852  2(60.12  382  26.9 2  20.6 2  16.7 2 )
 20.19 kW
2
and since P1   Vrms
Vrms  20.19k  142.09 V
Example 9
Obtain the complex Fourier series of the function in
Example 1. Then plot the complex frequency spectra
for −4 ≤ n ≤4 and calculate the rms value of the signal.
Solution
0  
T
1
1
1
1
c0   f (t )dt   1dt 
T 0
20
2
T
1
2
1
1
 jnt
cn   f (t )e
dt   1e  jnt dt   0
T 0
20
1
 jnt
1

1e
j
 jn
 
(e
 1)
 
2   jn  0 2n
But
e jn  cos n  j sin n  cos n  (1)n
j
 jn
Thus, cn 
(e
 1)
2n
 j / n
j
n

[( 1)  1]  
2n
 0
, n odd
, n even
*Here notice that cn n0  c0 .


1
 j  jnt
jn0t
   
e
Therefore, f (t )   cn e
2 n    n 
n  
n0
n odd
For n even,
For n odd,
0.5
cn  0,n  0
 90, n  0
1
cn 
, n  
n
 90, n  0
1
c0 
2
The average rms value is
Frms 

c
n  
2
n
 0.52  2(0.322  0.112 )  0.692
Summary

f (t )  a0   (an cos n0t  bn sin n0t )
 Sine-cosine
form n 1
T
T
2
2
1
2
0 
, a0   f (t )dt , an   f (t ) cos n0tdt , bn   f (t ) sin n0tdt
T 0
T
T 0
T 0
T

 Amplitude-phase
form
f (t )  a  A cos(n t   )
0
Ann  an  jbn

n 1
n
0
 Exponential (or complex) form

c0  a0 ,
an  jbn
cn 
,
2
f (t )   cn e jn0t
cn  c
*
n
n  
n
Prob.17.37, pg.803
If the sawtooth waveform in Fig.17.9 is the voltage
source vs(t) in the circuit of Fig.17.22, find the response
vo(t).
Prob.17.37, pg.803
If the periodic current waveform in Fig.17.73(a) is
applied to the circuit in Fig.17.73(b), find vo.
is (t )

3
3
0
−1
1
2
(a)
3
t

(b)
Problem
vs (t )
Rs  1 
C 1 F
3
2
vs (t )
1
Ro  1 

vo (t )

0
(i)
(ii)
(ii)
1
2
3
4
5
6
t
Compute the output voltage vo (t).
Plot the amplitude and phase spectra of vo (t)
for the first three nonzero harmonics.
Calculate the average power dissipated in
the resistor Ro.
QUIZ 1
Given the Fourier series expansion of the
voltage signal vs(t) is

 10 
vs (t )  5   
 sin nt
n 1  n 
1H
vs (t )
1
1H
1

vo (t )

Find vo(t) in the amplitude-phase form.
FOURIER SERIES
Definition of a Fourier
series
A Fourier series may be defined as an expansion of a function in a series
of sine's and cosine’s such as
a0 
f ( x)    (an cos nx  bn sin nx).
2 n 1
(1)
The coefficients are related to the periodic function f(x)
by definite integrals in equation 1.
Henceforth we assume f satisfies the following conditions:
(1) f(x) is a periodic function;
(2) f(x) has only a finite number of finite discontinuities;
(3) f(x) has only a finite number of extreme values, maxima and minima in the
interval [0,2].
Fourier series are named in honour of Joseph Fourier (1768-1830), who made
important contributions to the study of trigonometric series, in connection with the
solution of the heat equation
Periodic Function

Any function that satisfies
f (t )  f (t  T )
where T is a constant and is called the
period of the function.
A function f(x) which satisfies the relation
f(x) = f(x + T) for all real x and some fixed T
is called Periodic function. The smallest
positive number T, for which this relation
holds, is called the period of f(x).
Euler’s Formula
The Fourier series for the function f(x) in the interval  x    2
Is given by 
a0
  an cos nx  bn sin nx
2 n 1
f ( x) 
a0 
  2
1
 
f ( x)dx
an 
1
  2
 
bn 
1
  2
 
f ( x) cos nxdx
f ( x) sin nxdx
These values of a0,an&bn are known as Euler’s Formula.
Problems

Q1.Obtain the Fourier series for f(x)=e-x in the interval 0<x<2
.

Sol. We know that, f ( x)  a0  a cos nx  b sin nx
 n
n
2 n 1

a0
e 
  an cos nx  bn sin nx
2
n 1
1 2  x
a0 
 e
x

a0 

1

0
 ex
1  e 2 x

2
0
2
1
an 
 0
an 
1
2
e  x  cos nx  n sin nx 
2
0
 ( n  1)
e  x cos nxdx
 1  e  2  1
. 2
an  


 n 1
 1  e  2  1
 1  e  2  1
 & a2  
 ,....
 a1  



2

5
1
2
bn 
 
bn 
1
2
x


e

sin
nx

n
cos
nx
0
 n2  1
0

e  x sin nxdx

 1  e 2 
n
. 2
bn  


 n 1
 1  e  2  1
 1  e  2  2
. & b2  
. .......
 b1  

2




 5
Putting the values of a0,an&bn in the Fourier Series
1  e 2  1  1
1
1
2
3
 1

e 
   cos x  cos 2 x  cos 3x  .....    sin x  sin 2 x  sin 3x  ..... 
 2  2
5
10
5
10
 2

x
Answer
Q2. Find a Fourier series to represent x-x2 from x=- to x= 
Sol. We know that,
a0 
f ( x)    an cos nx  bn sin nx
2 n 1
a0 
x  x    an cos nx  bn sin nx
2 n 1
2
a0 
1

 
( x  x 2 ) dx
1 x2
x3 
a0 

 2
3 
 2 2
a0 
3
1 
2
an 
(
x

x
) cos nxdx



1
sin nx
  cos nx 
  sin nx  


an   x  x 2
 1  2 x  


2

 3 
2

n
 n 
 n  


 4 1
an 
n2
4
4
 a1  2 ; a2  2 .......
1
2
1 
bn   ( x  x 2 ) sin nxdx
 
1
  sin nx 
 cos nx  
2   cos nx 
bn   x  x 
  1  2 x   2   (2) 3 

 n 
 n 
 n  
n

 2(1) n
bn 
n

 b1 
2
2
; b2
;.......
1
2
Putting this value in Fourier Series,
we get
 2
 cos x cos 2 x
  sin x sin 2 x

xx 
 4 2 

.....

2


.....
  1

3
22
2
 1
2
Answer
Functions having point of
discontinuity
In the interval (α, α+2π)
f(x)= Φ(x), α<x<c
= Ψ(x),c<x< α+2π
  2
1 c
a0    ( x)dx    ( x)dx

c
  
  2
1 c
an    ( x) cos nxdx    ( x) cos nxdx

c
  
  2
1 c
bn    ( x) sin nxdx    ( x) sin nxdx

c
  
Problems
Q1. Find the Fourier Series expansion for f(x), if
f(x)=-π<x<0
x, 0<x< π.
1
1
1

Deduce that 2  2  2       
1
3
5
8
2


a0
f ( x )

an cos nxdx 
bn sin nxdx
We know
2
n1
n1
Sol.
Then
a0 
1



0


0
[  (  ) dx   xdx]
2
x
0


[|  |  |
|0 ]

2
1
---A

1


(  2 

2
2
)
2
therefore

1 0
an  [  (  ) cos nxdx   x cos nxdx]



1

1



[ 
sin nx 0
x sin nx cos nx 


]
2
n 
n
n
0
[0 
1
1
cos
n


]
2
2
n
n
1
(cos n  1)
2
n
0
2
2
2
 a1 
, a2  0, a3  
, a4  0, a5  
2
2
 1
 3
  52
therefore,

1 0
bn  [  ( ) sin nx   x sin nxdx]
0
 
1  cos nx 0
cos nx sin nx 
 [
 x

]
2

n
n
n 0

1 

 [ (1  cos n )  cos n ]
 n
n
1
 (1  2 cos n )
n
1
1
 b1  3, b2 
, b3  1, b4 
,.......
2
4
Hence putting the values of a’s and b’s in equation –A
 2
cos 3x cos 5 x
sin 2 x 3 sin 3x sin 4 x

f ( x)     cos x  2  2  ......   3 sin x 


 .......
4 
3
5
2
3
4

-----B
Hence this is the required result.
Putting x=0 in equation B
We get,  2 
1
1

f (0)    1  2  2  ...... 
4  3
5

f (x )
 f (0 is
 discontinuous
0)  andf (at
0 x=0,
0)  0
-------C
1

 f (0)  [ f (0  0)  f (0  0)] 
2
2
Hence C equation takes the form


2


4

2 1 1
1




.....

 12 32 52

Q2. Find the Fourier Series to represent the function f (x) given by
f ( x)  x for 0  x   , and=2  x for   x  2
1
1
1
2
Deduce that 2  2  2  .... 
1
3
5
8
Sol. We know that,


a0
f ( x )

an cos nxdx 
bn sin nxdx
2
n1
n1


-----A
2
1 
 a0 
xdx   ( 2  x ) dx



  0
1  x2 
x 2 2 
 
 2x 

  2 0
2  
 an 

2
1 
x
cos
nxdx

( 2  x ) cos nxdx




  0
1  x sin nx cos nx 
sin nx nx 2 


(
2


x
)
 2

 
n
n2 0
n
n  
n
n

1   1
1 
1
 1 
 2
 2 


2


  n
n  n
n2 

2   1  1
 an 


 
n2

n
2
1  
bn 
x sin nxdx   ( 2  x ) sin nxdx




  0
 bn  0
Required Fourier Series
Put x=a in equation B
We get,

2  (1) n  1
f ( x)    
 cos nx
2
2 n 1   n


-------B

 ( 1) n  1
f ( 0) 



2
 
n2
n 1 

  
2
2
2


 2  2  ...... 
4
2
3
5
2
1
1
1

 2  2  2  ...... 
8
1
3
5
2

Half Range Series
The Fourier series which contains terms sine or cosine only is
known as half range Fourier sine series or half range Fourier
cosine series.
The function will be defined in range of 0 to  but in order to
obtain half range Fourier cosine series or half range Fourier
sine series we extend the range of the function f(x)( ,  ) or
in general (-l,l). So, that the function is either converted in
form of even function of even or odd function.
Case-1 Half range Fourier cosine series:
For the half range Fourier cosine series of the function f(x) in
the range (0,l), we extend the function f(x) over the range (-l,l).
So that the function become even function.

a0
an cos nx
f ( x) 

2 n 1
l
an 
Where, a0 
2 l
f ( x)dx

0
l
2 l
 nx 
f
(
x
)
cos

dx

0
l
 l 
Case-2 Half range Fourier sine series:
For half range fourier sine series of function f(x),in the
range(0,l), we extend the function f(x) over the range (-l,l); so,
that the function becomes odd function.
 nx 
f ( x )   bn sin

l


n 1
2 l
 nx 
bn 
f
(
x
)
sin

 dx

0
l
 l 

Problems
Q1. Find the Fourier cosine series for the function
0 x 
f(x)=x2 in the range
.
Sol.
The given function f(x)=x2 is a even function.
So, we apply case 1 a

an cos nx
0
f
(
x
)



i.e.
2
l
n 1
an 
2 l
 nx 
f ( x ) cos
 dx

l 0
 l 

1
a0 
x dx
 
a0 
2
 

0

x 2 dx
2
x3
3
2 2

3
a0 
a0
2


0

1
a0 
x dx
 
a0 
2
2

 

0
x 2 dx
 
2

x3 0
3
2 2
a0 
3
a0 
1

an 
x cos nxdx
 
an 
2
2


x cos nxdx


2
0
2  x 2 sin nx 2 x cos nx  2 sin nx   
an  




n
n2
n3
0
an 
2  2 cos n 

 
n2

4( 1) n
an 
n2
2
( 1) n cos nx
x 
 4
3
n2
n 1
2

Hence the required result is ,
2 
1
1

x   4cos x  2 cos 2 x  2 cos 3x  ....... 
3 
2
3

2
Q2. Obtain the Fourier expansion of xsinx as a cosine series in (0, ).
Hence deduce 1  1  1  ......     2
Sol.
Let
1 3
35
2

57
a0 
x sin xdx


a0 
2
4
0

x cos x   1 sin x  0

a0  2
2

an 
x sin x cos nxdx


an 
xsin(n  1) x  sin( n  1) x dx


0
1

0
an 
1    cos(n  1) x cos(n  1) x    sin(n  1) x sin(n  1) x  


  1

 x
2
   n 1
n  1   (n  1)
(n  1) 2  0
 cos(n  1) cos(n  1) 
an  

(n  1)
n 1 
 n 1
When,
n  1, a1 
1

2

x sin x cos xdx


0
a1 
 0
a1 
  cos 2 x 
  sin 2 x  
x

1



 
2
2


0
a1 
1    cos 2   1


 
2
2

x sin 2 xdx
1
1
 cos 2 x cos 3x cos 4 x

x sin x  1  cos x  2


 .....
2
35
57
 1 3

Putting x=

2
,then
Hence,

1
1
1

 1  2 

 ......
2
1 3 3  5 5  7

1
1
1
 2


 ......  
1 3 3  5 5  7
4
Even and Odd Functions:Even function:-
1. The function
is said to be even function if
2. The function f(x) is said to be Odd function if
How To Determine Whether The Function Is
Even Or Odd
 The graph of odd function is symmetric about origin.
 The graph of even function is symmetric about Y-axis.
 First we have to check whether the domain of the function is
symmetric about the y-axis.
y
sin x : 2π  x 
5π
2
y
f(x)=sin x
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
f(x)=sin x
1
1
x
x
-3π
-5π/2
-2π
-3π/2
-π
-π/2
π/2
π
3π/2
2π
5π/2
3π
-3π
-5π/2
-2π
-3π/2
-π
-π/2
π/2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
-8
-8
-9
-9
π
3π/2
2π
5π/2
3π
Key Point
Products of functions
(even)×(even) = (even)
(even)×(odd) = (odd)
(odd)×(odd) = (even)
Sums of functions
(even) + (even) = (even)
(even) + (odd) = (neither)
(odd) + (odd) = (odd)
If Function is in this form
If
Or
and
then the function is
even function
then the function is
odd function
Fourier series for even & odd function
Case1:The Fourier series for the even function F(x) in the interval (-L,+L) is given by
Case2:-
If the function f(x) is an odd function then the fourier series (-L,+L) is given by



Answer
Half Range Series
The Fourier series which contains terms of sine or cosine only
is known as half range Fourier sine series or half range Fourier
cosine series.
The function will be defined in range of 0 to  but in order to
obtain half range Fourier cosine series or half range Fourier
sine series we extend the range of the function f(x)( ,  ) or
in general (-l,l). So, that the function is either converted in
form of even function of even or odd function.
Case-1 Half range Fourier cosine series:
For the half range Fourier cosine series of the function f(x) in
the range (0,l), we extend the function f(x) over the range (-l,l).
So that the function become even function.

a0
an cos nx
f ( x) 

2 n 1
l
an 
2 l
a

Where,0 l 0 f ( x)dx
2 l
 nx 
f
(
x
)
cos

dx

0
l
 l 
Case-2 Half range Fourier sine series:
For half range fourier sine series of function f(x),in the
range(0,l), we extend the function f(x) over the range (-l,l); so,
that the function becomes odd function.
 nx 
f ( x )   bn sin

l


n 1
2 l
 nx 
bn 
f
(
x
)
sin

 dx

0
l
 l 

Problems
Q1. Find the Fourier cosine series for the function
f(x)=x2 in the range0  x   .
Sol.
The given function f(x)=x2 is a even function.
So, we apply case 1 a

an cos nx
0
f
(
x
)



i.e.
2
l
n 1
1 π
a

x 2 dx


π
0
π
2 π 2
a

 x dx
0
π 0
2  3 π
a

x

0

0
3π 
2π 2
a

0
3

1
a0 
x dx
 
a0 
2


 
0
x 2 dx
 
2

x3 0
3
2 2
a0 
3
2 l
 nx 
an   f ( x ) cos
 dx
l 0
 l 
a0 
and,
2
1

an 
x cos nxdx
 
an 
2
2


x cos nxdx

2
0
2  x 2 sin nx 2 x cos nx  2 sin nx   
an  




n
n2
n3
0
an 
2  2 cos n 

 
n2

4( 1) n
an 
n2
2
( 1) n cos nx
x 
 4
3
n2
n 1
2

Hence the required result is ,
2 
1
1

x   4cos x  2 cos 2 x  2 cos 3x  ....... 
3 
2
3

2
Q2. Obtain the Fourier expansion of xsinx as a cosine series in (0, ).
Hence deduce 1  1  1  ......     2
Sol.
Let
1 3
35
2

57
a0 
x sin xdx


a0 
2
4
0

x cos x   1 sin x  0

a0  2
2

an 
x sin x cos nxdx


an 
xsin(n  1) x  sin( n  1) x dx


0
1

0
an 
1    cos(n  1) x cos(n  1) x    sin(n  1) x sin(n  1) x  


  1

 x
2
   n 1
n  1   (n  1)
(n  1) 2  0
 cos(n  1) cos(n  1) 
an  

(n  1)
n 1 
 n 1
When,
n  1, a1 
1

2

x sin x cos xdx


0
a1 
 0
a1 
  cos 2 x 
  sin 2 x  
x

1



 
2
2


0
a1 
1    cos 2   1


 
2
2

x sin 2 xdx
1
1
 cos 2 x cos 3x cos 4 x

x sin x  1  cos x  2


 .....
2
35
57
 1 3


Putting x=
2
,then

1
1
1

 1  2 

 ......
2
1 3 3  5 5  7

1
1
1
 2


 ......  
Hence,
1 3 3  5 5  7
4
Harmonic Analysis
The process of finding the Fourier series corresponding to the
function when the function
values is known as
  xby
  numerical
 2c
The Fourier series for the function f(x) in
harmonic analysis.

a0
 nx 
 nx 
f
(
x
)


a
cos

b
sin





n
n
the interval 2 
isgiven
c by c 
n 1 
1   2c
f ( x ) dx
c 
1   2c
 nx 
an  
f ( x ) cos
dx
c 
 c 
Wher a0 
e,
bn 
1   2c
 nx 
f
(
x
)
sin

dx
c 
c


If the given function is not the explicit function of the
independent variable x and the function is defined by the
numerical values then the formula of a0,an and bn are given by
  2c
a0 = 2x[Mean value of f(x)
in the ,interval
(
)]
 nx 

,


2
c
f ( x)  cos
an = 2x[Mean value
of  c 
in the interval(
 nx 
f ( x )  sin

 c 
bn = 2x[Mean value of
)]
 ,  2c
in the interval (
a1 cosx bn sin x

c
c
)]
In formula 1 the
first
is
a cos
2x term
b sin 2of
x expansion

c
first
harmonic.
aknown
cos 3x as
b sin
3x orcfundamental

c second
c term
The
is known as second
harmonic and the term
2
3
2
3
is known as third harmonic and so on…..
Problems

Q1. In a machine the displacement y of a given point is given for a
certain
angle as follows.

y
0
7.9
30
8
60
7.2
90
5.6
120
3.6
150
1.7
180
0.5
210
0.2
240
0.9
270
2.5
300
4.7
350
6.8
sin 2
0,2
Findathe coefficient
of
in the Fourier series representing
the

0
y
  an cos nπx  bn sin nπx .
above
2 variations.
n 1
Sol.
The Fourier series for the function y in the given interval 0-360o or
(
) is given by
-: THANKYOU :-
Presented By :Shiv Prasad Gupta
Naveen Kumar
Avinash
Quiz1
Determine the Fourier series of the following
waveform.
v (t )
4
QUIZ 2
Find i(t) if v(t) is the signal as given in QUIZ 1.
1   2 
1   2 
v(t )     
 sin nt    
 cos(nt  90)
2 n 1  n 
2 n 1  n 
n odd
n odd
QUIZ 3
Find the output voltage vo(t) if the current
source is(t) is given by

 2 
is (t )  1   
 cos(nt  90)
n 1  n 
3
is (t )
5
0.5 F

vo (t )

QUIZ 4
Using time differentiation technique, find the
Fourier transform of f(t).