UNIT – I Fourier Series: Introduction to Fourier series, Fourier series for Discontinuous functions, Fourier series for even and odd function, Half range series Fourier Transform: Definition and properties of Fourier transform, Sine and Cosine transform
CHAPTER 1
Fourier Series
Introduction: In many physical and engineering problems, especially in the study of periodic function, it is necessary to express a real valued function in series of sines and cosines, which can be expressed in the form.
a0 a1 cos x a2 cos 2 x ...... b1 sin x b2 sin 2 x ...... 2 where
a0 , a1 , a2 ,.... and b1 , b2 ,.... are some constants.
Periodic functions: A function f x is said to be periodic if f x l f x , x being a real number. If l is the least positive number called the period of the function f x . For example: sinx, cos x,sec x, cos ecx is a periodic function with period 2 . Also, sin nx, cos nx are periodic function with period
2 . n
Trigonometric Series: A series of the form
a0 an cos nx bn sin nx n 1
is called a trigonometric series. Some useful Definite Integrals: Let m and n be any two integers and the interval c, c 2 . Then the following definite integrals are as follows: 3
4
Engineering Mathematics - II 2
(i)
(ii)
(iii)
(iv)
2
2
2
2
(v) (vi) (vii)
(viii) (ix) (x) (xi)
2
cos nx sin nx dx n
0, n 0
cos nx dx 0, n 0 2
1 sin m n x sin m n x sin mx sin nx dx 2 mn mn
0, m n
2
1 sin m n x sin m n x cos mx cos nx dx 2 mn mn
0, m n
1 2 1 cos 2nx dx , n 0 2 2 1 2 2 cos nx dx 1 cos 2nx dx , n 0 2 2 2 1 cos m n x cos m n x sin mx cos nx dx 2 m n m n 0, m n sin 2 nx dx
2
sin 2 nx sin nx cos nx dx 2n 0, n 0 ax e ax e sin bx dx a 2 b2 a sin bx b cos bx e ax ax e bx dx cos a cos bx b sin bx a 2 b2 n sin n 0 and cos n 1 , n N . 2
Fourier Series: The Fourier Series for the function f x in the interval , 2 is given by
f x
a0 an cos nx bn sin nx 2 n 1
where
a0
1
2
f x dx
Fourier Series
and
an
1
2
f x cos nx dx
bn
1
2
f x sin nx dx
5
The values of a0 , an , bn are known as Euler’s Coefficients or Fourier Coefficients. Proof: Let f x be represented in the interval , 2 by the Fourier Series:
f x a0 an cos nx bn sin nx
…..(i)
n 1
To find the coefficients a0 , an , bn . We assume that equation (i) can be integrated term
by term in the given interval , 2 .
To find a0 : integrated both sides of (i), we get 2
f x dx
a0 2
2
dx
2
2 a cos nx dx bn sin nx dx n n 1 n 1
a0 2 0 0 2 a0 .
Hence a0
1
2
f x dx
To find an : multiply both sides of (i) by cos nx and integrating. 2
2
f x cos nx dx
a0 2
2
cos nx dx
2 a cos nx cos nx dx bn sin nx cos nx dx n n 1 n 1
0 an 0 Hence an
1
2
f x cos nx dx .
To find bn : multiply both sides of (i) by sin nx and integrating
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Engineering Mathematics - II 2
f x sin nx dx
a0 2
2
sin nx dx
2 a cos nx sin nx dx n bn sin nx sin nx dx n1 n 1 0 0 bn 1 2 f x sin nx dx . Hence bn 2
Deductions: (i) Taking 0 , then interval becomes 0 x 2 and (i) becomes
a0
1
2
f x dx, an
0
1
2
0
f x cos nx dx, bn
1
2
f x sin nx dx
0
(ii) taking , then interval becomes x and (i) becomes
a0
1
f x dx, an
1
f x cos nx dx, bn
Dirichlet’s conditions:
1
f x sin nx dx
(RGPV Feb 2007)
Any function f x can be expressed as a Fourier series
f x
a0 an cos nx bn sin nx 2 n 1 n 1
in the interval 0, 2 or , , where a0 , an , bn are constants provided that f x satisfies the following conditions: (i)
f x is periodic.
(ii)
f x and its integrals are finite and single valued.
(iii)
f x has a finite number of discontinuities.
(iv)
f x has a finite number of maxima and minima.
when these conditions are satisfied, the Fourier series converges to f x at every point of continuity. At a point of discontinuity, the sum of the series is equal to the mean of the limits on the right and left. i.e.
1 f x 0 f x 0 2
where f x 0 and f x 0 be the right and left limit.
Fourier Series
7
Parseval’s theorem:- Let the Fourier series
f x
a0 an cos nx bn sin nx, 2 n 1 n 1
If f x converges uniformly to f x at every point of the interval 0, 2 then
1
2
0
2 a2 f x dx 0 an 2 bn 2 . 2 n 1
Illustrative Examples Example 1. Find a Fourier Series to represent x x 2 from x to x . Hence show
1 1 1 1 2 .... 12 22 32 42 12 Sol: Let
(RGPV Feb 2005, June 2006, Feb 2010)
f x x x2 , x .
Fourier Series over the interval , be
f x
a0 an cos nx bn sin nx 2 n 1 n 1
..…(i)
Now,
a0
x x dx
1
2
2 2 1 x 2 x3 2 3 3
an
x x cos nx dx
1
2
1 x cos nx dx x 2 cos nx dx
1
x
2
cos nx dx,
since x cos nx is an odd function.
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Engineering Mathematics - II
1 2 sin nx sin nx x dx 2 x n n
1 sin nx cos nx sin nx x2 2x 2 2 3 n n n 1 sin n 0, cos n 1n 2 4 cos n n 4 n 2 1 n bn
x x sin nx dx 1
2
1 sin x nx x 2 sin nx dx 1 since x 2 sin nx is an odd function. x sin nx dx
cos nx 1 cos nx x dx n n2
1 cos nx sin nx x 3 n n 1 cos n cos n n n 2 n 1 n Hence (i) becomes
xx 2
2 3
2
4 n 1
1 n2
n
cos nx 2 n 1
1 n
n
sin nx
cos x cos 2 x cos 3x sin x sin 2 x sin 3x 4 2 2 ... 2 ... 2 3 2 3 2 3 1 1 2 cos x cos 2 x cos 3x sin x sin 2 x sin 3 x 4 2 ... 2 ... 2 2 3 2 3 2 3 1 1
Fourier Series
9
Putting x 0 , we get
2
1 1 1 4 2 2 2 ... 3 1 2 3 2 1 1 1 2 2 2 ... 1 2 3 12 0
2 Example 2. Find a Fourier Series to represent the function f x x x in the interval
x . Hence show that Sol: Let
2 6
1
1 1 1 ... 22 32 42
(RGPV Feb 2006)
f x x x2 , x .
Fourier Series over the interval , be
f x
a0 an cos nx bn sin nx 2 n 1 n 1
…..(i)
Now,
a0
x x dx
1
2
1 x 2 x3 2 2 3 2 3 an
x x cos nx dx
x x cos nx dx
2
1
2
1
2
0
Since x cos nx is an odd function.
x 2 cos nx dx
2 sin nx cos nx sin nx x2 2 x 2 2 3 n n n 0 2 cos n 4 n 2 2 1 2 n n
10
bn
Engineering Mathematics - II
x x sin nx dx
1
2
1
x sin nx dx
2
since x 2 sin nx is an odd function.
0
x sin nx dx
2 cos nx sin nx x 2 n n 0 2 cos n 2 n 1 n n Hence (i) becomes x x2
2 3
4
1
n 1
n2
n
cos nx 2
1
n 1
n
n
sin nx
2
cos 2 x cos 3 x sin 2 x sin 3 x 4 cos x 2 ... 2 sin x ... 2 3 2 3 2 3 2 sin 2 x sin 3 x cos x cos 2 x cos 3x 4 2 ... 2 sin x ... 2 2 3 2 3 2 3 1
2
Which is the required Fourier Series. Now putting x and x in equation (2), we have
2
2
1 1 4 1 2 2 ... 3 2 3
2 and
2
1 1 4 1 2 2 ... 3 2 3
Adding (3) and (4), we have
2 2 1 1 8 1 2 2 ... 3 2 3 2 1 1 1 2 2 ... 6 2 3 2 2
..…(3)
..…(4)
Fourier Series
11
Example 3 Find Fourier Series represented of f x x sin x, 0 x 2 . (RGPV June 2004, June 2007, Dec 2008) Sol: Let f x x sin x, 0 x 2 The Fourier series of f x in 0, 2 be a0 an cos nx bn sin nx 2 n 1 n 1
f x
…..(1)
Now
a0 an
2
2
1
0
1
0
2
2
1
1
0
0
1 2 1 2
f x dx x sin x dx
2
x cos x sin x 0 2
f x cos nx dx x sin x cos nx dx
2
2
0
0
1
x 2sin x cos nx dx x sin n 1 x sin n 1 x dx 2 cos A sinB sin A B sin A B 2
1 cos n 1 x cos n 1 x sin n 1 x sin n 1 x x 2 2 n 1 n 1 2 n n 1 1 0 cos 2 n 1 cos 2 n 1 2 n 1 n 1 1 1 2 if n 1 2 , n 1 n 1 n 1
1 2
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Engineering Mathematics - II
But if n 1 , then
a1
2
1 2
1
2
f x sin nx dx
0
1
1 2
x sin 2 x dx
1 cos 2 x sin 2 x x 2 4 2 0
2
1
2
0
1 2 bn
x sin x cos x dx
0
2
0
x sin x sin nx dx
2
0
x cos n 1 x cos n 1 x dx 2
1 sin n 1 x sin n 1 x cos n 1 x cos n 1 x x 2 2 2 n 1 n 1 1 1 n n 0
cos 2 n 1 cos 2 n 1 1 1 2 2 2 2 n 1 n 1 n 1 n 1
1 2
1 1 1 1 0, 2 2 2 2 n 1 n 1 n 1 n 1 But if n 1 , then
b1
1 2
1
2
0
1 2 1 2
x sin x sin x dx
2
2
0
0
x 2sin 2 x dx x 1 cos 2 x dx 2
1 sin 2 x x 2 cos 2 x x x 2 2 2 4 0
1 2
4 2 1 1 2 .2 2 4 4
if n 1
Fourier Series
13
Hence Fourier Series (1) becomes
f x
a0 a1 cos x b1 sin x an cos nx 0 2 n2
bn 0
1 2 x sin x 1 cos x sin x 2 cos nx 2 n 2 n 1
1 2 2 1 cos x sin x 2 cos 2 x 2 cos 3 x ... 2 2 1 3 1 Example 4. Find the Fourier Series to represent the function, if
, x 0 f x 0 x x, Hence deduce that
2 8
Sol: Let f x a0
1 1 1 ... 12 32 52
a n 1
n
(RGPV Dec 2004, Feb 2007, Dec 2008)
cos nx bn sin nx
where
1 f x dx 2 1 0 f x dx f x dx 0 2 1 0 1 dx x dx 2 2 0
a0
1 0 1 x2 x 2 2 2 0
2
2 4 4
2 f x cos nx dx 2 1 0 cos nx dx x cos nx dx 0
an
…..(1)
14
Engineering Mathematics - II 0 1 sin nx sin nx cos nx x 1 2 n n n 0
1 cos n 1 0 2 n2 n cos n 1 n2
If n is even cos n 1 then an 0 i.e. a2 a4 .... 0 If n is odd cos n 1 then an
2 2 2 i.e. a1 2 , a3 etc. 2 n .1 .32
and
2 f x sin nx dx 2 1 0 sin nx dx x sin nx dx 0
bn
0 1 cos nx cos nx sin nx x 1 2 n n 0 n
1 cos n cos n 1 2 cos n 0 n n n n 1 1 1 if n is even cos n 1 then bn , i.e. b2 , b4 etc. 2 4 n 3 3 3 3 If n is odd cos n 1 then bn , i.e. b1 , b3 , b5 etc. 1 3 5 n
Now putting values in (1) then required Fourier Series is
21 1 1 cos x 2 cos 3 x 2 cos 5 x ... 2 4 1 3 5 1 3 3 1 sin x 2 sin 2 x 3 sin 3 x ...
f x
To find the sum: At x 0 the series converges to
f 0 0 f 0 0 2
0 2 2
…..(2)
Fourier Series
15
and at x the series converges to
f 0 f 0 0 2 2 The L.H.S. of (2) is
at x 0 and is 0 at x 2
Therefore putting x 0 in R.H.S. of (2) we get
2
4
2 1 1 1 2 2 ... 2 1 3 5
or
2 8
1 1 1 ... 12 32 52
x Example 5 Obtain the Fourier Series for f x e in the interval x .
Sol: Suppose that
n 1
n 1
f x a0 an cos nx bn sin nx Now
1 1 x f x dx e dx 2 2 1 x 1 sinh e e e 2 2 1 1 an f x cos nx dx e x cos nx dx a0
1 ex 1.cos nx n sin nx 2 1 n 1 e cos n e cos n 2 1 n 2 1 sinh cos n e e 2 1 n 1 n 2 n
…..(1)
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Engineering Mathematics - II
bn
1
f x sin nx dx
1
e
x
sin nx dx
1 ex 1.sin nx n cos nx 2 1 n 1 n e cos n n e cos n 2 1 n
n cos n 1 n 2
e
e
2 1 n sinh n
1 n 2
Now putting values of a0 , an & bn in (1) we get
f x e x
ex
sinh
2 1 sinh
n 1
n
1 n 2
cos nx n sin nx
2sinh x 1 1 1 cos x sin x cos 2 x 2sin 2 x ... 2 2 5
Which is the required Fourier Expansion. Example 6. Find a series of sin es and cosines of multiples of x which will represent f x in the interval , then
x 0 0 f x 1 4 x 0 x Sol: Let f x a0
a n 1
n
cos nx bn sin nx
…..(1)
Fourier Series
1 2
a0
0
0
17
1 1 f x dx 2 f x dx 2 f x dx 1 2
0
0
1 x dx 4
2 2 1 . . 8 2 16 1 an f x cos nx dx
…..(2)
1
0
0
1 f x cos nx dx f x cos nx dx
0
1
0
1 x cos nx dx 4
1 sin nx cos nx x. 1. n n 2 0 4 1 1 0 2 cos n 1 4 n 1 n 2 1 1 4n
Now if n is odd
1
n
1
1 1 1 2 2 4n 2n 1 1 a1 , a2 2 ,.... 2 2.3 an
And if n is even
1 an
n
1
1 1 0, a2 a4 ... 0 4n 2
Similarly we have
bn
4n
1
n
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Engineering Mathematics - II
b1
4
, b2
4.2
, b3
4.3
, b4
4.4
......
Now putting values in (1) we get
f x a0 a1 cos x b1 sin x a2 cos 2 x b2 sin 2 x ......
2
1 cos x sin x 0 sin 2 x . 16 2 4 4.2 1 sin 3x ..... 2 cos 3x 4.3 2.3
Example 7. Find the Fourier Series expression of the function f x given by
f x x for x
(RGPV Dec 2005, Dec 2007)
Sol: Let the Fourier Series be
f x
a0 an cos nx bn sin nx 2 n 1 n 1
Now,
a0
1
f x dx
0
0
x dx x dx 1
1
x dx
0 x2 1 2 2 1 x 2 2 2 0 2 2
an
1
1
f x cos nx dx
x cos nx dx
1 0 x nx dx cos 0 x cos nx dx
..…(1)
Fourier Series
2
0
19
x cos nx dx
2 sin nx cos nx 2 2 n x 1 1 cos n 1 2 2 2 n n n 0 n
bn
1
1
f x sin nx dx
x sin nx dx
1 0 x nx dx sin 0 x sin nx dx 0
Putting the value of a0 , an , bn in (1), we get
f x
2
2
2 n 1 1 cos nx 2 n 1 n
4 1 1 cos x 2 cos 3 x 2 cos 5 x ...... 3 5
Even and Odd functions: A function
f x is said to be an even function if
f x f x , e.g. cos x, x 2 , x3 sin x, etc. A function f x is said to be an odd function if
f x f x , e.g. sin x, x3 , x3 cos x, x 2 sin x etc. The Fourier Series of f x in the interval , is given by
n 1
n 1
f x a0 an cos nx bn sin nx For even function:
2
a0
f x dx
an
2
0
f x cos nx dx 0
…..(1)
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Engineering Mathematics - II
and
bn 0
from (1), The Fourier Series becomes
f x a0 an cos nx n 1
Hence if f x is an even function then Fourier Series contains only cosine term. For Odd function:
a0 0, an 0 and bn
2
f x sin nx dx 0
from (1), the Fourier series becomes
f x bn sin nx n 1
Hence if f x is an odd function then Fourier series contains only Sine term. Half Range Fourier Series: If a half range series for a function f x is desired, then the function is defined in the interval 0, l , (i.e. half of the interval l , l and is said to be half range.) Thus we can obtain the Fourier series either cosine series or sin e series only. Cosine Series: If
f x is an even function defined on interval l x l. Then
cosine Fourier series in half range interval 0,l becomes f x where
a0
a0 n x an cos 2 n 1 l
2 l 2 l n x f x dx and an f x cos dx, n 1, 2,3,... l 0 l 0 l
Sine Series: If f x is an odd function defined on interval l x l , then Sine Fourier series in half range interval 0,l becomes n x f x bn sin l n 1
Fourier Series
where
bn
2 l n x f x sin dx, l 0 l
n 1, 2,3,......
Example 8. Find Fourier Series representation of f x x in the interval l x l. Or Find cosine Fourier series of f x x in 0, l .
(RGPV 2001)
Sol: Let f x x , which is an even function in the interval l , l
cos ine Fourier series of f x in half range interval 0,l becomes f x
a0 n x an cos 2 n 1 l
…..(1)
Now,
2 l 2 l f x dx x dx l 0 l 0 2 l x dx l 0 2 l2 . l l 2
a0
x x,
and
an
2 l n x f x cos dx l 0 l 2 l n x x cos dx 0 l l
l n x n x sin sin 2 l l l dx x n 0 n l l l 0 l
n x cos 2 l 0 2 l n l 0
0 x l
21
22
Engineering Mathematics - II
2 l2 . 2 2 cos n cos 0 l n 2l n 2 2 1 1 n when n is odd, an
4l n 2 2
when n is even, an 0 Hence Fourier series (1) becomes
n x cos l n 1 x 3 x 5 x cos cos cos l 4l l l ...... 2 2l 2 2 2 1 3 5
f x
l 4l 2 2
1
n
2
Example 9. A periodic function of period 4 is defined as f x x , 2 x 2. Find its Fourier Series expansion. Sol: Taking l 2 in the above example 8 and proceed.
(RGPV Dec 2002)
2 Example 10. Find a Fourier series to represent f x x in the interval l x l.
(RGPV June 2005, Feb 2010) Sol: Let f x x is an even function in the interval l , l . 2
f x
a0 Now,
a0 n x an cos 2 n 1 l
2 l 2 l f x dx x 2 dx l 0 l 0 l
2 x3 2l 2 3 l 3 0
2 l n x f x cos dx l 0 l 2 l n x dx x 2 cos l 0 l
an
..…(1)
Fourier Series
23
l
n x n x n x sin cos sin 2 l 2x l 2 l x2 n l n 2 2 n 3 3 l l2 l3 0
4l 2 n 1 2 2 n
Hence the Fourier series (1) becomes
l 3 4l 2 x 2 3 2
n 1
1 n
2
n
cos
n x l
2 x 1 3 x x 1 1 cos cos cos ...... 2 2 2 1 2 3 l l l 3 2 2 x 1 3 x l 4l 1 x 1 x 2 2 2 cos 2 cos 2 cos ...... 3 1 2 3 l l l
3
2
4l l 2 3
2 Example 11. Obtain the half range sine series for f x x x in the interval
0 x .
(RGPV June 2005)
n
Sol :- Let the Fourier Sine series be
f x bn sin nx n 1
where
bn
2
f x sin nx dx 0
x x sin nx dx 2
2
0
2 cos nx cos nx 2 2 x x x dx n 0 0 n 2 sin nx sin nx 0 0 2 x 2 2 2 dx 0 n 0 n
24
Engineering Mathematics - II
2 cos nx 0 2 3 n
0
4 cos n 1 n3 4 n 3 1 1 n
Hence the required Fourier half range Sine series is
n 4 1 1 sin nx 3 n 1 n 4 2 2 2 3 sin x 3 sin 3 x 3 sin 5 x ...... 3 5 1
x x2
8 1 1 1 sin x 3 sin 3 x 3 sin 5 x ...... 3 3 5 1
Example 12. Express f x x as a: (i) Half range Sine series in 0 x 2.
(RGPV June 2006)
(ii) Half range cosine series in 0 x 2.
(RGPV Jan 2007, June 2009)
Sol: (i)
The half range Fourier Sine series be
f x bn sin n 1
where
bn
n x 2
..…(1)
2 2 n x f x sin dx 2 0 2
x sin 2
0
n x dx 2 2
n x n x cos sin 2 2 x n n 2 2 2 0 4 4 n 1 n
Fourier Series
25
Hence the required half range sin e series is n x n 4 x 1 sin n 2 n 1 4 x 1 2 x 1 3 x sin sin sin ...... 2 2 2 3 2
(ii)
The half range Fourier cosine series be
f x
a0 n x an cos 2 n 1 2
…..(2)
where 2 2 2 f x dx x dx 2 0 2 0 n x 2 2 an f x cos dx 2 0 2 2 n x x cos dx 0 2
a0
2
n x n x cos sin 2 2 x n 2 2 n 2 4 0 4 n 2 2 1 1 n Hence the required half range cosine series is x 1
4
2
1 n 1 n 1
2
n
n x 1 cos 2
4 2 x 2 3 x 2 5 x 2 cos 2 cos 2 cos ...... 2 1 2 3 2 5 2 8 1 n 1 3 x 1 5 x 2 cos 2 cos ...... x 1 2 2 cos 1 2 3 2 5 2 1
Example 13. Find the half range Sine Fourier series for the function f x x in the interval 0 x .
(RGPV June 2007)
26
Engineering Mathematics - II
Sol: The half range Sine series be
f x bn sin n 1
n x l
For the interval 0, ,
f x bn sin nx
…..(1)
n 1
where
bn
2
f x sin nx dx 0
2
0
x sin nx dx
cos nx 2 cos nx x 0 dx n 0 n
2 x cos nx sin nx 2 n n
0
2 2 n cos n 1 n n
Hence, the half range Sine series be n 2 x 1 .sin nx n n 1 1 1 2 sin x sin 2 x sin 3 x ...... 2 3
Example 14. Find the half range cosine Fourier series of the function:
0 t 1 2t , f t 2 2 t , 1 t 2
(RGPV June 2003)
Sol: The half range cosine Fourier series in 0, 2 is
f t
a0 n t an cos 2 n 1 2
…..(1)
Fourier Series
27
where
a0
2 2 f t dt 2 0
f t dt f t dt 1
2
0
1
2t dt 2 2 t dt 1
2
0
1
2
t 2 t 2 2t 0 2 1 2 1
1 1 2 4 2 2 2 2 2 2 n t an f t cos dt 0 2 2 1 2 n t n t f t cos dt f t cos dt 0 1 2 2 1 2 n t n t 2t cos dt 2 2 t cos dt 0 1 2 2 1
2
n t n t n t n t cos sin cos sin 2 2 2 2 t 2 22 2 2 t 2 2 n n n n 2 2 4 0 4 1 2 n 4 n 4 4 4 n n 2 2 2 cos 2 sin 2 2 cos 2 2 2 2 cos n sin n 2 n 2 2 n 2 n n n n 4 4 8 2 2 2 cos 2 2 2 2 cos n 2 n n n 8 n 1 cos n an 2 2 2 cos n 2 For n 1 ,
a1
8 2 cos 1 cos 0 2 2 1
( cos 1, cos
2
For n 2,
a2
8 2 2
2
2cos 1 cos 2
32 22 2
( cos 2 1)
2
0)
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Engineering Mathematics - II
For n 3,
a3 For n 4,
a4 For n 5,
3 2 cos 1 cos 3 0 2 3 8
2
2
8 4 2 2
( cos
3 0, cos 3 1) 2
2 cos 2 1 cos 4 0
a5 0 For n 6 ,
a6
8 6 2
2
2 cos 3 1 cos 6
a7 0, a8 0, a9 0, a10
32 62 2
32 and so on. 102 2
Half range cosine series becomes 32 1 1 1 f x 1 2 2 cos t 2 cos 3 t 2 cos 5 t ...... 2 6 10 Example 15. Obtain the Fourier series for f x : 0, x 0 f x 2 x , 0 x
(RGPV Dec 2003)
n
Sol :- The Fourier series be a0 ..…(1) an cos nx bn sin nx 2 n 1 n 1 1 1 0 a0 f x f x dx f x dx 0 1 0 2 0 dx x 2 dx 0 3 1 1 0 an f x cos nx dx 0.cos nx dx x 2 cos nx dx 0
f x
where
1 2 sin nx sin nx x dx 0 2 x n 0 n
2 n
2 2 n cos nx sin nx x n n 2 n 2 cos n n 2 1 0
Fourier Series
bn
1
f x sin nx dx
1 0 0.sin nx dx x 2 sin nx dx 0
1 2 cos nx cos nx x dx 0 2 x n 0 n n 2 1 1 2 x sin nx cos nx n n n n 2 0
1
n
2 2 n3 n
Hence the Fourier series becomes
f x
2
1 1 2 cos x 2 cos 2 x 2 cos 3x ...... 6 2 3
1 2 2 13 1
2 2 sin x 3 2 2
sin 2 x ......
Example 16. Obtain the Fourier series for the function:
0 x 1 x, f x 2 x , 1 x 2
(RGPV June 2002, Dec 2004)
Sol: The Fourier series be
f x
a0 an cos nx bn sin nx 2 n 1 n 1
where
a0 f x dx f x dx 1
2
0
1
x dx 2 x dx 1
2
0
1
1
2
x2 x2 2 x 2 1 2 0
…..(1)
29
30
Engineering Mathematics - II
an x cos nx dx 2 x cos nx dx 1
2
0
1
1
2
sin n x sin n x cos n x cos n x x 2 2 2 x 2 2 n n n 0 n 1 1 cos 2n cos n cos n 2 2 2 2 n n 2 n n 2 n 2 1 1 n bn x sin n x dx 2 x sin n x dx 1
2
0
1
1
2
cos n x sin n x cos n x sin n x x 2 2 2 x 2 2 n n n 0 n 1 cos n cos n 0 n n Hence the required Fourier series becomes
f x
2
4 cos x cos 3 x cos 5 x ...... 2 2 2 1 3 5
Example 17. Obtain a half range cosine series for
l for 0 x kx, 2 f x k l x , for l x l. 2
(RGPV Dec 2004, Feb 2006)
Sol: Let the cosine series be
f x
a0 n x an cos 2 n 1 l
where
2 l f x dx l 0 l 2 l 2 kx dx l k l x dx l 0 2
a0
…..(1)
Fourier Series
31
l l 2 kx 2 2 x 2 k lx l 2 0 2 l 2
l 2 l 2 2 kl 2 kl 2 kl 2 l2 k l 2 k l 8 2 2 8 l 4 2 2 l n x an f x cos dx l 0 l l n x n x 2 l 2 kx.cos dx l k l x cos dx l 0 l l 2
l l 2 l n x l2 n x 2 l n x l2 n x kx k 2 2 cos sin sin k 2 2 cos k l x l n l n l 0 n l n l l 2
2 2 kl 2 n kl 2 n kl 2 n kl 2 n kl sin sin 2 2 cos 1 2 2 cos n 2 2 cos 2 n 2 2n 2 l 2n l n n
2 2kl 2 n kl 2 kl 2 cos 2 2 2 2 cos n 2 2 2 n l n n
an
2kl n 2 2
n 2 cos 2 1 cos n
For n 1,
a1
2kl 12 2
2 cos 2 1 cos 0
cos 1, cos 0 2
For n 2,
2kl 8kl 2cos 1 cos 2 2 2 2 2 2 2 For n 3, a2
a3
2kl 32 2
3 2 cos 2 1 cos 3 0
For n 4,
a4
2kl 2 cos 2 1 cos 4 0 42 2
cos 2
1
3 0, cos 3 1 cos 2
32
Engineering Mathematics - II
For n 5,
8kl 8kl , a7 0, a8 0, a9 0, a10 2 2 and so on. 2 2 6 10 Fourier cosine series becomes a5 0, a6
f x
kl 8kl 1 2 x 1 6 x 1 10 x 2 2 .cos 2 .cos 2 .cos ...... 4 2 l 6 l 10 l
f x 1 cos x , 0 x 2 in a Fourier series. Hence
Example 18. Expand evaluate
1 1 1 ...... 1.3 3.5 5.7
(RGPV Sept 2009)
Sol: Let f x 1 cos x
2 sin
x 2
The Fourier series be
f x
a0 an cos nx bn sin nx 2 n 1 n 1
…..(1)
where
a0
1
2
0
x 2 sin dx 2
x 2 2 cos 2
0
4 2
2
x 2 sin cos nx dx 2 2 2 x 2 cos nx sin dx 0 2 2 1 2 1 1 sin n x sin n x dx 2 2 2 0
an
1
2
0
2
1 2 2 2n 1 2n 1 2n 1 cos 2 x 2n 1 cos 2 x 2 0
Fourier Series
33
2 1 1 cos 2n 1 1 cos 2n 1 1 2n 1 2 2n 1
2 2 2 4 2 2n 1 2n 1 4n 2 1
cos 2n 1 cos 2n 1 1 2
x 2 sin sin n x dx 2 2 2 x 2sin n x sin dx 2 0 2 1 2 1 1 cos n x cos n x dx 0 2 2 2
bn
1
0
1 2
2
2 2 2n 1 2n 1 2n 1 sin 2 x 2n 1 sin 2 x 0
2 1 1 sin 2 1 0 sin 2n 1 0 n 2n 1 2n 1
0 Hence the required Fourier series becomes
f x
2 2
4 2 cos nx 2 n 1 4n 1
when x 0, we have
0
2 2
4n n 1
1 2
1
1 2 n 1 4n 1 1 1 1 1 'or' ...... 1.3 3.5 5.7 2 'or'
1
4 2
2
2 Example 19. Find the Fourier series expansion of f x 2 x x in 0,3 and hence
deduce that
34
Engineering Mathematics - II
1 1 1 1 2 ...... . 12 22 32 42 12
(RGPV June 2008)
Sol: The required Fourier series be
f x
a0 n x n x where 2l 3 an cos bn sin 2 n 1 l l n 1
..…(1)
where
a0
1 2l 2 3 f x dx 2 x x 2 dx 0 3 0 l 3
2 x3 x2 0 3 3 0 1 2l n x 2 x x 2 cos dx l 0 l 2 3 2n x dx 2 x x 2 cos 3 0 3
an
3
cos 2n x sin 2n x 2n x cos 2 2 3 3 3 2x x 2 2x 2 2 3 n 2 3 2n 2n 3 3 3 0 2 9 9 . 2 2 4 cos 2n 2 2 2 3 4n n bn
1 2l n x 2 3 2n x 2 x x 2 sin dx 2 x x 2 sin dx l 0 l 3 0 3 3
sin 2n x cos 2n x cos 2n x 2 2 3 3 3 2x x 2 2 2x 2 3 2n 3 2n 2n 3 3 0 3 2 6 27 3 2 2 cos 2n 3 3 cos 2n 1 3 n 4n n
Fourier Series
35
Hence, the Fourier series becomes
2 x x 2 n 1
Putting x
3
9 n 2 2
cos
2n x 3 2n x sin 3 3 n 1 n
3 , we have 2
9 9 2 2 cos n 4 n 1 n
1 1 1 1 2 or 2 2 2 2 ...... 1 2 3 4 12 Example 20. Expand
1 4 x, f x x 3 , 4
if 0 x if
1 2
1 x 1 2
as a Fourier series of sin e terms.
(RGPV Sept 2009)
Sol: The Fourier sin e series be
f x bn sin n x
…..(1)
n 1
where
2 1 f x sin n x dx 1 0 1 11 3 2 2 x sin n x dx 1 x sin n x dx 0 4 2 4
bn
1
1
1 3 cos n x sin n x cos n x sin n x 2 2 x 2 2 2 x 2 2 4 n n 0 n 1 n 4
2
36
Engineering Mathematics - II
n n sin sin 1 1 1 1 n n cos cos n cos 2 2 22 2 22 4 2 4 4 4 2 n n n n n n n 4sin 1 n 1 1 2 22 2n n
1
For
n 1,
b1
For
n 2,
b2 0,
For
n 3,
b3
For
n 4,
b4 0,
For
n 5,
b5
For
n 6,
b6 0
4
2
,
1 4 2 2, 3 3
1 4 2 2, 5 5 and so on.
Hence Fourier sine series becomes
4 4 1 4 1 1 f x 2 sin x 2 2 sin 3 x 2 2 sin 3 x ...... 3 3 5 5 Example 21 If f x cos x , expand f x as a Fourier series in the interval , . Sol: As f x cos x cos x f x . Thus cos x is an even function. Therefore, the Fourier series be
f x
a0 an cos nx 2 n 1
..…(1)
Fourier Series
where
a0
2
0
cos x dx
2 2 0 cos x dx cos x dx 2 2 4 sin x02 sin x 2
an
2
0
cos x cos nx dx
2 2 0 cos x cos nx dx cos x cos nx dx 2 1 2 cos n 1 x cos n 1 x dx cos n 1 x cos n 1 x dx 0 2
n x n 1 si n x 2 sin n 1 x sin n 1 x 1 sin 1 n 1 n 1 0 n 1 n 1 2 sin n 1 sin n 1 sin n 1 sin n 1 1 2 2 2 2 n 1 n 1 n 1 n 1 n n cos cos 2 2 2 n 1 n 1 n 4 cos 2 , n 1 2 n 1
For
2 2 n 1, a1 cos 2 x dx cos 2 x dx 0 0 2
37
38
Engineering Mathematics - II
Hence Fourier series becomes
cos x
2
4 1 1 cos 2 x cos 4 x ...... 15 3
Example 22. Prove that in 0 x l : x deduce that
1 4l 2 2
x 1 3 x cos l 32 cos l ...... and
1 1 1 4 . ...... 14 34 54 96
(RGPV Dec 2003)
Sol: Let f x x, then cosine series in 0,l be
f x
a0 n x an cos 2 n 1 l
…..(1)
where
2 l 2 l2 x dx . l l 0 l 2 2 l n x an x cos dx l 0 l l l l 2 l n x n x sin dx x. sin 0 l n l 0 n l a0
n x 2 l l cos . l n n l 0 2l 2l n 2 2 cos n 1 2 2 1 1 n n l
Hence Fourier cosine series becomes
x
l 2l n x n 2 2 1 1 cos 2 n 1 n l
l 4l 2 2
n 1 3 x 1 5 x cos 2 32 cos l 52 cos l ......
…..(2)
Fourier Series
39
Now, using Parsevel’s theorem:
b
a
f x dx 2
b a a0 2 an 2 bn 2 2 2 n 1
l 0 l2 4l x dx 0 2 2 n odd n 2 2 l
2
2
l 3 l l 2 16l 2 1 1 1 4 4 4 4 ...... 3 2 2 1 3 5
2l 2 l 2 16l 2 1 1 1 4 4 4 4 ...... 3 2 1 3 5
l 2 16l 2 1 1 1 4 4 4 ...... 4 6 1 3 5
4
96
1 1 1 ...... 14 34 54
2 Example 23. Obtain the Fourier series for the function f x x , x . Hence
deduce that
1 1 1 2 ...... RGPV June 2009 6 12 22 32 2 1 1 1 (ii) 2 2 2 ...... 12 1 2 3 1 1 1 2 iii 2 2 2 ...... 8 1 3 5 1 1 1 4 iv ...... 4 4 4 RGPV June 2008, Dec 2010 90 1 2 3 2 Sol: Let f x x . i.e. an even function then Fourier series be
i
f x
a0 an cos nx 2 n 1
…..(1)
where
a0
2
0
2 x3 2 x dx 2 3 0 3 2
40
Engineering Mathematics - II
an
2
0
x 2 cos nx dx
2 sin nx cos nx sin nx x2 2x 2 2 3 n n n 0 2 cos n 4 n 2 2 1 2 n n
Hence Fourier series becomes
x 2
x 2
(i)
2 3
4 n 1
1 n cos nx n2
2
cos x cos 2 x cos 3x cos 4 x 4 2 ...... 2 2 2 1 3 2 3 4
…..(2)
Putting x , we get
2
1 1 1 1 4 2 2 2 2 ...... 3 2 3 4 1 2 2 1 1 1 1 4 2 2 2 2 ...... 3 3 4 1 2
2
(ii)
1 1 1 1 2 ...... 6 12 22 32 42
Putting x 0 , we get
2
1 1 1 1 4 2 2 2 2 ...... 3 3 4 1 2 2 1 1 1 1 2 2 2 2 ...... 12 1 2 3 4 0
(iii)
..…(3)
…..(4)
Adding (3) and (4), we get 2 1 1 1 2 2 2 2 ...... 5 1 3 4 2 1 1 1 2 2 2 ...... 8 1 3 5
…..(5)
Fourier Series
(iv)
41
Using Parseval’s theorem:
a0 2 2 f x dx an 2 bn 2 2 n 1 4 4 16 x 4 dx 4 18 n 1 n
x5 2 16 5 4 n 1 n 5 9
2 5 2 5 1 1 1 16 4 4 4 ...... 5 9 1 2 3
8 5 1 1 1 16 4 4 4 ...... 45 1 2 3
4 1 1 1 ...... 14 24 34 90
Practice Problems 1. Find the Fourier series representation of f x x cos x, x . n 2n 1 1 Ans : sin nx 2 sin nx 2 n 2 n 1
2. Obtain Fourier series expansion for f x x sin x in the interval x . Hence deduce that
4
1 1 1 1 ...... 2 1.3 3.5 5.7 1 2 2 Ans: 1 2 cos x 1.3 cos 2 x 2.4 cos 3 x ......
3. Find the Fourier series to represent the function f x sin x , x .
2 4 cos 2 x cos 4 x Ans: 3 15 ......
42
Engineering Mathematics - II
4. Find the half range cosine series for the function f x sin
x l
, 0 x l.
(RGPV Dec 2005)
2 x 4 x 6 x cos cos cos 2 4 l l l ...... Ans: 3.5 5.7 1.3
.
x 5. Obtain the Fourier series for f x e in the interval 0 x 2 .
1 e2 1 1 1 1 2 3 1 cos x cos 2 x cos 3x ...... sin x sin 2 x sin 3x ...... . Ans: 5 10 5 10 2 2 2
6. Find the Fourier series to represent the function
x 0 0, f x sin x, 0 x Hence show that
1 1 1 ...... 1.3 3.5 2 1 2 1 1 1 Ans: 3 cos 2 x 15 cos 4 x ...... 2 sin x
.
7. Find the Fourier series expansion of the periodic function of period 2 , defined
if x x, 2 2 by f x x, if x 3 . 2 2
4 sin x sin 3x sin 5 x Ans: 12 32 52 ...... . 8. Expand the function f x in Fourier series in the interval , :
2 x, f x x,
0 x x 0.
Fourier Series
43
21 1 1 1 Ans: 4 12 cos x 32 cos 3 x ...... 3 sin x 2 sin 2 x 3 sin 3 x ...... 9. If f x is a function defined by
0 x x, 2 f x x, x 2 Express f x by a sine Fourier series and also by a cosine series.
4 sin x sin 3 x sin 5 x 8 Ans: 12 32 52 ...... , 4
cos 2 x cos 6 x cos10 x ...... 2 2 2 2 . 6 10
10. Find the Fourier expansion of the periodic function
k , x 0 f x 0 x k , and f x 2 f x . Sketch the graph of f x .
4k sin x sin 3 x sin 5 x Ans: 1 3 5 ...... . 11. Expand f x in a Fourier series in the interval 0, 2 if
x, f x 0, 1 2 Ans: 4 2
0 x 1 1 x 2
1 1 1 cos x 32 cos 3 x 52 cos 5 x ......
12. Find the Fourier cosine series for Parseval’s theorem to prove that
1 4 4 90 n 1 n
sin 2 x sin 3 x sin x 2 3 ......
f x x x in 0 x and use
44
Engineering Mathematics - II
2 13. Find the Fourier series expansion for the function f x x x in 1 x 1.
1 4 cos x cos 2 x cos 3 x 2 sin x sin 2 x sin 3 x ...... Ans: 3 2 12 2 2 32 ...... 12 2 3 14. Find the Fourier series for the function
0, f x 1, 0,
x 0 0 x
2
2
x
1 2 cos x cos 3 x cos 5 x 1 sin x sin 2 x sin 3 x Ans: 4 1 3 5 ...... 1 2 3 ...... 15. Express coshx in Fourier series in the interval x .
1 1n 1 2 Ans: sinh 2 cos nx 2 n 1 n 1 16. Obtain Fourier series for the function f x given by
2x 1 , f x 1 2 x , Hence, deduce that
x 0 0 x
1 1 1 2 . ...... 12 32 52 8
.
.
