Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
Course: Instrumentation Instrumentation Subject: Engineering Mathematics Lesson: Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis Lesson Developer: Ms.Shweta Dua College/Dept: College/Dept: Bhaskaracharya College of Applied Sciences University of Delhi
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
Table of Contents Chapter 3: Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis 3.1 Fourier Series for even functions 3.1.1 Properties of even function 3.1.2 Expansion of even function 3.2Fourier Series for odd functions 3.2.1 Properties of odd function 3.2.2 Expansion of odd function 3.3 Half range Expansion 3.3.1 Sine Half range Expansion 3.3.2 Cosine Half range Expansion 3.4 Harmonic Analysis 3.5 Summary 3.6 Exercise 3.7 Glossary 3.8References
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
Learning ObjectivesAfter going through this chapter, you will be able to :
Explain the Fourier Series for even and odd functions Know the half range expansion Discuss the harmonic analysis
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis In the previous two chapters, we have studied the Fourier Series, its three classes, properties, convergence and Fourier Series for functions having any period. Now we study the Fourier Series for even and odd functions, half range expansions and harmonic analysis.
3.1 Fourier Series for even functionsA function f(x) is an even function if :f(-x)=f(x) for all x. e.g- x2n is an even function if :f(-x) = (-x) 2n = [(-x)2]n = x2n = f(x)
3.1.1 Properties of even functions
The graph of even function is symmetrical about y axis.
The area under the graph of an even function f(x) from - to from 0 to .
i.e
is double the area
f ( x)dx = 2 f ( x)dx
0
Even function contains only even powers of x. The product of two even functions is always even i.e (even function)(even function) = even function The sum of two even functions is even i.e even function + even function = even function
3.1.2 Expansion of an even function
–
If f(x) is an even function on [-L,L], then : L
L
L
0
f ( x)dx = 2 f ( x)dx
Fourier Coefficients are : L
a0 =
f ( x)dx - - - - - - - - - - - - - - - - - - - - - - - - - - - (1) 0
L
n x )dx an = f ( x )Cos ( L
; n = 1,2….
--------------------------- (2)
0
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis L
n x )dx bn = f ( x ) Sin( L
; n = 1,2…..
---------------------------
(3)
0
= 0 Fourier Series is given by :
anCos
f(x) = a0
n1
n x L
-----------------------------
(4)
Fourier Series of an even function contains only cosine terms & hence called Fourier Cosine Series.
3.2 Fourier Series for odd functions Afunction f(x) is odd function if :f(-x) = -f(x) for all x e.g x
2n+1
is an odd function if :-
f(-x) = (-x) = (-1) =-x
2n+1
2n+1
.x
2n+1
2n+1
= -f(x)
3.2.1 Properties of odd function
The graph of odd function is symmetrical about the origin.
Odd function contains only odd powers of x. The area under the graph of an odd function f(x) from - π to +π is zero
i.e
f ( x)dx = 0
Sum of two odd functions is always odd
i.e odd function + odd function = odd function
The product of two odd functions is even
i.e (odd function)(odd function) = even function
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis The product of an odd function and even function is odd
i.e (odd function)(even function) = odd function
3.2.2 Expansion of an odd function
–
If f(x) is an odd function on [-L,L], then : L
f ( x)dx =
0
L
Fourier Coefficients are :a0 = 0 an = 0 L
n x bn = f ( x) Sin dx L
; n = 1,2…..----------------------------- (5)
0
Fourier Series is given by :
f(x) =
bn Sin
n x
n 1
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (6)
L
Fourier Series of an odd function contains only Sine terms & hence called Fourier Sine Series Question Obtain the Fourier Series expansion of the following function f(x) = x2 ; -2 ≤ x ≤ 2 Solution
–
Here 2L = 4
Therefore, L = 2 Since the given function is an even function Using equation (1), a0 =
=
L
1 L 1
f ( x)dx 0 2
x dx 2 2
0
1 x3 2 4 = [ ]0 = 2 3 3 Using equation (2), L
n x an = f ( x ) Cos ( )dx L
0
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis 2
x Cos( 2
=
n x 2
0
)dx
Using integration by parts, = [ x
( n / 2)
4
=
n
Sin(n x / 2) (n / 2)
0
(n / 2)
16( 1) n n
] - 2 x
Sin(n x / 2) (n / 2)
2
dx
]02
Cosn
n 2 2
2
2
2 0
Cos(n x / 2)
[ x
16
=
=
Sin( n x / 2)
2
[since Cosn = (-1)n]
2
Therefore, Fourier Series expansion of the given function is –
f(x) = a0
anCos
n x L
n1
f(x) =
4 3
16
2
n1
( 1) n n2
Cos (
n x 2
)
3.3 Half range Expansions When Fourier Series is defined on an interval [0,L] instead of [-L,L], then it is said to be half range Fourier Series. It can be divided into two types :-
3.3.1 Sine Half range Expansion – A Fourier Series will be generated that involves only sine terms. It is represented as :f(x) =
bn Sin( n x ) ------------------------------------
L
dx
where bn =
(7)
; n is a positive integer -------------------------------(8)
3.3.2 Cosine Half range Expansion – A Fourier Series will be generated that involves only cosine terms. It is represented as :f(x) = a0
anCos (
n x L
) ------------------------------------ (9)
-----------------------------------------
where a 0 =
(10)
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
n x
an =
; n is a positive integer ---------------------------
L
Question Find Half range Fourier Sine Series if :f(t) = t 2 0
bn = dx
2 t dt
=
Integration by parts,
] dt
= [t2 .-Cos 2
3t 3 = {[ Cos ] 0 + ) }dt n 2
3t 3 = {[ Cos ] 0 + [t )- )] }dt n 2
3t 3 3 3 = {[ Cos ] 0 + [t )] 0 - ) [-Cos )] 0 } n By taking the limits,
54
= { Cosn + 3 3 (Cosn 1)} n When n is odd :-
bn =
18 n
-
72
[Using Cosn = -1]
n3 3
When n is even :-
bn = -
18
[Using Cosn = 1]
n
So, Fourier Sine Series is :f(t) = 18(
1
4 3
)Sin(
t 18 2 t 1 4 ) Sin( ) 18( )Sin( t ) ............... 3 2 3 3 27 3
QuestionFind Fourier Sine Series for f(x) = 1 ; 0 ≤ x ≤ 2 SolutionHere, L= 2 Using equation (8)
bn = dx Institute of Lifelong Learning, University of Delhi
(11)
Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis 2
=
Sin( 0
n x 2
)dx
Cos (n x / 2)
=[
( n / 2)
]02
By taking the limits, =
=
2 n 2 n
=
4 n
[1 Cosn ] [1 (1) n ]
0 ; n= even
; n = odd
So, Fourier Sine Series is :-
f(x) =
4
=
n
bn Sin( [Sin (
n x ) L
x 1 3 x 1 5 x ) Sin( ) Sin( ) ......] 2 3 2 5 2
Question Find Fourier Cosine Series for f(x) = 1- in 0 < x < L Using equation (10),
Solution L
a0 = f ( x )dx
0
L
=
x
(1 L )dx 0
=
[ x -
x 2 2 L
] L0
By taking the limits, a0 =
an =
=
( n x )
L
x L
) (
n x L
)
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis Integration by parts,
x
=
[
=
[
L x
L
= [
) (
)
x L
)
n x L
L
L
)
n
n x
L
L
n
1
Sin( 0
+
L
1 L L n
n x L ) dx] L n
{(Cos
n x L
)(
L n
)}] L0
n x L 1 L2 + Cos ) ]0 ( 2 2 L n L n
n x
L
L
By taking the limits, =
L
[
n 2 2
2
an =
2
n
2
(1) n (
L
1 )( )] n 2 2 L
[Cos n =
[1 (1) n ]
Using equation (9), f(x) = a0
anCos
n x L
So, Fourier Cosine Series for f(x)is :-
1
f(x)=
2
2
[ 2
1 ( 1) n
n 1
n
2
] Cos
n x L
Question Find Fourier Cosine Series of the following functionx 2; 0 ≤ x ≤ 2
f(x) =
4 ; 2≤ x≤ 4
Solution
–
Here L= 4
Using equation (10) L
a0 = f ( x)dx
0
=
1 4
2
4
[ x dx 4dx] 0
2
2
1 x3 2 = [( )0 (4 x)42 ] 4 3 By taking the limits,
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis =
8 3
( n x )
an =
=
L
2 4
2
2
[ x Cos(
n x 4
0
4
)dx 4Cos(
n x 4
2
)dx]
Integration by parts, =
=
=
1 2
[{ x 2
Sin (n x / 4)
(n / 4)
2 xCos (n x / 4) (n / 4)
2
2Sin(n x / 4) (n / 4)
3
}02 4{
Sin (n x / 4) (n / 4)
}42
1 4Sin(n / 2) 4Cos (n / 2) 2Sin(n / 2) 4Sin(n / 2) [ ] 2 (n / 4) (n / 4) 2 (n / 4)3 (n / 4)
32
[Cos ( 2
n
n 2
2
)(
2 n
) Sin(
n 2
)]
So, Fourier Cosine Series for f(x)is :f(x) = a0
=
8 3
anCos (
32 2
n x
1
L
n [Cos( n 1
2
)
n 2
)(
2 n
) Sin(
n 2
)]Cos (
n x 4
)
3.4 Harmonic AnalysisIt is applicable when the function is not given by a formula but by a table of corresponding values. The process of expanding the Fourier Series for a function given by such values of the function is called Harmonic Analysis. It is a technique of finding the harmonic elements of a given function. Fourier Coefficients are given by :a0= Mean value of f(x) an =
2[Mean value of f(x)Cosnx]
bn =
2[Mean value of f(x)Sinnx]
Fourier Series of f(x) is given by :-
f(x) = a0 + Cos x + Sin nx +………….
Cos 2x + ……+
Cos nx +……+
Sin x +
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Sin 2x +…..+
Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis Question In a machine, displacement f(x) of a point is given for a certain angle x°. Find the Fourier Seriesupto second harmonic. x
0
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
f(x)
1.8
1.1
0.3
0.16
0.5
1.3
2.16
1.25
1.3
1.52
1.76
2
Solution x
f(x)
Sinx
Sin2x
f(x)Sinx
f(x)Sin2x
Cosx
Cos2x
f(x)Cosx
f(x)Cos2x
0
1.8
0
0
0
0
1
1
1.8
1.8
30°
1.1
0.5
0.87
0.55
0.95
0.87
0.5
0.95
0.55
60°
0.3
0.87
0.87
0.26
0.26
0.5
-0.5
0.15
-0.15
90°
0.16
1
0
0.16
0
0
-1
0
-0.16
120°
0.5
0.87
-0.87
0.43
-0.43
-0.5
-0.5
-0.25
-0.25
150°
1.3
0.5
-0.87
0.65
-1.13
-0.87
0.5
-1.13
0.65
180°
2.16
0
0
0
0
-1
1
-2.16
2.16
210°
1.25
-0.5
0.87
-0.62
1.08
-0.87
0.5
-1.08
0.62
240°
1.3
-0.87
0.87
-1.13
1.13
-0.5
-0.5
-0.65
-0.65
270°
1.52
-1
0
-1.52
0
0
-1
0
-1.52
300°
1.76
-0.87
-0.87
-1.53
-1.53
0.5
-0.5
0.88
-0.88
330°
2
-0.5
-0.87
-1
-1.74
0.87
0.5
1.74
1
-3.75
-1.41
0.25
3.17
15.15
a0 =
15 .15
= 1.26
12
a1 = 2 ×
a2 = 2 ×
0.25 12
= 0.04
3.17
b1 = 2 × (
b2 = 2 × (
12
= 0.52
3.75 12
) = -0.62
1.41 12
) = -0.23
So, Fourier Series of f(x) upto second harmonic is given by :-
f(x) = a0 + Cos x + Cos 2x + Sin x + Sin 2x = 1.26 + 0.04Cosx + 0.52Cos2x - 0.62Sinx - 0.23Sin2x
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis QuestionFind the harmonics a0, a1, a2, b1, b2 of the Fourier Series expansion of the following tabulated function x◦ 0◦ f(x) 2.34 Solution x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330°
a0=
30◦ 3.01
f(x) 2.34 3.01 3.69 4.15 3.69 2.20 0.83 0.51 0.88 1.09 1.19 1.64 25.22
25.22 12
a1 = 2 x
a2 = 2 x
b1 = 2 x
b2 = 2 x
60◦ 3.69
Sinx 0 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5
90◦ 4.15
Sin2x 0 0.87 0.87 0 -0.87 -0.87 0 0.87 0.87 0 -0.87 -0.87
120◦ 3.69
150◦ 2.20
f(x)Sinx 0 1.50 3.21 4.15 3.21 1.10 0 -0.25 -0.76 -1.09 -1.03 -0.82 9.209
180◦ 0.83
f(x)Sin2x 0 2.61 3.21 0 -3.21 -1.91 0 0.44 0.76 0 -1.03 -1.42 -0.547
210◦ 0.51
240◦ 0.88
Cosx 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87
270◦ 1.09
Cos2x 1 0.5 -0.5 -1 -0.5 0.5 1 0.5 -0.5 -1 -0.5 0.5
300◦ 1.19
330◦ 1.64
f(x)Cosx 2.34 2.61 1.84 0 -1.84 -1.91 -0.83 -0.44 -0.44 0 0.59 1.42 3.353
= 2.1015
3.353 12
= 0.559
3.115 12 9.209 12
= -0.519
= 1.535
0.547 12
= -0.091
So, Fourier Series of f(x) upto second harmonic is given by :-
f(x) = a0 + Cos x + Cos 2x + Sin x + Sin 2x = 2.1015 + 0.559Cosx – 0.519Cos2x + 1.535Sinx-0.091Sin2x
3.5 SummaryAn even function is defined as a function which is symmetrical about y axis i.e it is a mirror image. Fourier Series expansion of even function on [-L,L] is : L
L
L
0
f ( x)dx = 2 f ( x)dx
Fourier Coefficients are given by:-
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f(x)Cosx 2.34 1.5 -1.84 -4.15 -1.84 1.1 0.83 0.25 -0.44 -1.09 -0.59 0.82 -3.115
Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis L
a0 = f ( x )dx
0
L
n x an = f ( x )Cos dx L
; n = 1,2……………….
0
L
n x bn = f ( x) Sin dx L
;
n = 1,2……………….
0
= 0 Fourier Series is given by :
f(x) = a0
an Cos n 1
n x L
An odd function is defined as a function which is symmetrical about the origin. Fourier Series expansion of odd function on [-L,L] is : L
f ( x)dx =
0
L
Fourier Coefficients are :a0 = 0 an = 0 L
n x bn = f ( x) Sin dx L
; n = 1,2……………….
0
Fourier Series is given by :
f(x) =
b Sin n
n1
n x L
Half rangeFourier SeriesExpansionis defined on an interval [0,L]. Sine Half range Expansionis represented as : f(x) =
bn Sin
n x L
where bn =
dx
; n is a positive integer
Cosine Half range Expansionis represented as :f(x) = a0
anCos
n x L
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis where a 0 =
n x
an =
; n is a positive integer
L
Harmonic Analysisis applicable when the function is not given by a formula but by a table of corresponding values. The process of expanding the Fourier Series for a function given by such values of the function is called Harmonic Analysis. It is a technique of finding the harmonic elements of a given function. Fourier Coefficients are given by :a0 =
Mean value of f(x)
an =
2[Mean value of f(x)Cosnx]
bn =
2[Mean value of f(x)Sinnx]
Fourier Series of f(x) is given by :-
f(x) = a0 + Cos x + Sin nx +………….
Cos 2x + ……+
Cos nx +……+
Sin x +
Sin 2x +…..+
3.6 Exercise Question 1 Find the Fourier Cosine Series of the following periodic function :x2 ;
f(x) =
4
8
Answer
3
32 2
0≤x≤2 ;
2≤x≤4
1
n n 1
2
[Cos (
n 2
)(
2 n
) Sin(
n 2
)]Cos (
n x 4
)
Question 2 Find the Fourier Sine Series of the following periodic function :f(x) =
2
;
2
t
Answer (
2
;0
2
1
2
2
9
1)S int Sin2t (
1
)Sin3t .............. 3
Question 3 Express f(x) as a Fourier S eriesupto second harmonic. x f(x)
0° 2.34
30° 3.01
60° 3.69
90° 4.15
120° 3.69
150° 2.20
180° 0.83
210° 0.51
240° 0.88
Answer 2.105 + 0.559Cosx – 0.519Cos2x + 1.535Sinx - 0.091Sin2x
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270° 1.09
300° 1.19
330° 1.64
Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
Short Questions
–
Question 1How do you differentiate between even function & odd function? Question 2What is a half range expansion ? What are its types ? Question 3What do you understand by harmonic analysis ?
Multiple Choice Questions 1) An even function satisfies (a) x(t) = -x(t) (b) x(t) = x(t-T/2) (c) x(t) = x(-t) (d) x(t) = x(t+T/2) Answer (c) 2) An odd function satisfies (a) x(-t) = -x(t) (b) x(t) = x(t-T/2) (c) x(t) = x(t+T/2) (d) x(t) = -x(-t) Answer (a) 3) Even function x Even function (a) Odd function (b) Even function (c) Can be even or odd (d) Neither even nor odd Answer (b) 4) Odd function + Odd function (a) Odd function (b) Even function (c) Neither even nor odd (d) Can be even or odd Answer (a)
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis
f ( x)dx =
5) If f(x) is an even function, then
(a) 0
(b)
f ( x)dx 0
(c) 2 f ( x)dx 0
(d) None Answer (c)
6) If f(x) is an odd function, then
f ( x)dx =
(a) 2 f ( x )dx 0
(b) 0
(c)
f ( x)dx 0
(d) None Answer (b)
3.7 Glossary
–
1) Even function It is defined as a function which is symmetric with respect to y axis i.e it is a mirror image. A function is even if f(-x) = f(x) –
2) Harmonic Analysis- It is the process of expanding the Fourier Series for a function given by a table of corresponding values of the function. 3) Half Range Fourier Series - When Fourier Series is defined on an interval [0,L] instead of [-L,L], then it is said to be half range Fourier Series. 4) Odd function - It is defined as a function which has rotational symmetry with respect to the origin, i.e its graph remains unchanged after rotation of 180 degrees about the origin.
3.8 References 1) Advanced Engineering Mathematics,H.KDass,S.Chand and Company Ltd. 2) Higher Engineering Mathematics,B.VRamana,TataMcGraw Hill Education Pvt. Ltd.
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Fourier Series: Even and Odd Functions, Half Range Expansions and Harmonic Analysis 3) Advanced Engineering Mathematics, Erwin Kreyszig, Wiley India Edition. 4)Advanced Engineering Mathematics, RK Jain and SRK Iyengar, Narosa Publishing House
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