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Fourier series

This document discusses Fourier series, including definitions, properties, and applications. It describes how to derive Fourier series for various functions, the concept of periodic functions, and conditions under which a function can be expressed as a Fourier series. Additionally, it presents methods for constructing half-range Fourier sine and cosine series, along with examples and problems for further understanding.

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Contents Euler’s Formula Functions having point of discontinuity Change of interval Even and Odd functions Half Range series Harmonic analysis
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 (1) 0 1 ( ) ( cos sin ). 2 n n n a f x a nx b nx ∞ = = + +∑ 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 [0,2π]. 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
( ) ( )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).  Any function that satisfies Periodic Function
Euler’s Formula The Fourier series for the function f(x) in the interval Is given by - These values of a0,an&bn are known as Euler’s Formula. παα 2+<< x { } ∫ ∫ ∫ ∑ + + + = ∞ = = = ++= πα α πα α πα α π π π 2 2 2 0 1 0 sin)( 1 cos)( 1 )( 1 sincos 2 )( nxdxxfb nxdxxfa dxxfa nxbnxa a xf n n n nn
Problems Q1.Obtain the Fourier series for f(x)=e-x in the interval 0<x<2 . Sol. We know that, π { }∑ ∞ = ++= 1 0 sincos 2 )( n nn nxbnxa a xf { } π π π π π x x x n nn x e ea ea nxbnxa a e 2 2 00 2 0 0 1 0 1 1 1 sincos 2 − − − ∞ = − − = −= = ++= ∫ ∑
( ) ( ) ( ) π π ππ π π π π π ππ π π π 2 02 2 0 2 2 2 1 2 2 2 02 2 0 cossin 1 1 sin 1 ,.... 5 11 & 2 11 1 1 . 1 sincos )1( 1 cos 1 nxnnxe n b nxdxeb e a e a n e a nxnnxe n a nxdxea x n x n n x n x n −− + = =       − =      − =∴ +      − = +− + = = − − −− − − − ∫ ∫
....... 5 2 . 1 & 2 1 . 1 1 . 1 2 2 2 1 2 2       − =      − =∴ +      − = −− − ππ π ππ π e b e b n ne bn Putting the values of a0,an&bn in the Fourier Series Answer             ++++      ++++ − = − − .....3sin 10 3 2sin 5 2 sin 2 1 .....3cos 10 1 2cos 5 1 cos 2 1 2 11 2 xxxxxx e e x π π
Q2. Find a Fourier series to represent x-x2 from x=- to x= Sol. We know that, π π { }∑ ∞ = ++= 1 0 sincos 2 )( n nn nxbnxa a xf { }∑ ∞ = ++=− 1 02 sincos 2 n nn nxbnxa a xx ∫ ∫ − − − −= − = −= −= π π π π π π π π π π nxdxxxa a xx a dxxxa n cos)( 1 3 2 32 1 )( 1 2 2 0 32 0 2 0
( ) ( ) ( ) ( ) ( ) ( ) n b n nx n nx x n nx xxb nxdxxxb aa n a n nx n nx x n nx xxa n n n n n n n )1(2 cos )2( sin 21 cos1 sin)( 1 ....... 2 4 ; 1 4 14 sin 2 cos 21 sin1 32 2 2 2221 2 32 2 −− =             −+      − ×−−      − −= −= − ==∴ −− =             − −+      − ×−−−= − − − ∫ π π π π π π π π π
;....... 2 2 ; 1 2 21 − =∴ bb Putting this value in Fourier Series, we get Answer       +−+      +−+ − =− ..... 2 2sin 1 sin 2..... 2 2cos 1 cos 4 3 22 2 2 xxxx xx π
Functions having point of discontinuity In the interval (α, α+2π) f(x)= Φ(x), α<x<c = Ψ(x),c<x< α+2π     +=     +=     += ∫ ∫ ∫ ∫ ∫ ∫ + + + c c n c c n c c nxdxxnxdxxb nxdxxnxdxxa dxxdxxa α πα α πα α πα ψφ π ψφ π ψφ π 2 2 2 0 sin)(sin)( 1 cos)(cos)( 1 )()( 1
Problems Q1. Find the Fourier Series expansion for f(x), if f(x)=-π<x<0 x, 0<x< π. Deduce that Sol. We know Then 2 222 85 1 3 1 1 1 π =⋅⋅⋅⋅⋅+++ ∑ ∑ ∞ = ∞ = ++= 1 1 0 sincos 2 )( n n n nxdxbnnxdxa a xf ]| 2 ||[| 1 ])([ 1 0 2 0 0 0 0 π π π π π π π π x xdxdxa +−= +−= − − ∫ ∫ ---A
)1(cos 1 ] 1 cos 1 0[ 1 ] cossinsin [ 1 ]coscos)([ 1 2 ) 2 ( 1 2 22 0 2 0 0 0 2 2 −= −+= ++−= −+−= −= +−= − − − ∫ ∫ π π π π π π π π π π π π π π π π n n n n n n nx n nxx n nx nxdxxnxdxa therefore n
,....... 4 1 ,1, 2 1 ,3 )cos21( 1 ]cos)cos1([ 1 ] sincoscos [ 1 ]sinsin)([ 1 , 5 2 ,0, 3 2 ,0, 1 2 4321 0 2 0 0 0 25423221 − == − ==∴ −= −−= +−+= +−= ⋅ −== ⋅ −== ⋅ − =∴ − −∫ ∫ bbbb n n n n n n n nx n nx x n nx nxdxxnxb therefore aaaaa n π π π π π π π π π π πππ π π π π
....... 4 4sin 3 3sin3 2 2sin sin3...... 5 5cos 3 3cos cos 2 4 )( 22 +−+−+      +++−−= xxx x xx xxf π π Hence putting the values of a’s and b’s in equation –A -----B Hence this is the required result. Putting x=0 in equation B We get, -------C is discontinuous at x=0,       ∞+++−−= ...... 5 1 3 1 1 2 4 )0( 22 π π f )(xf 0)00()00( =+−=−∴ andff π 2 )]00()00([ 2 1 )0( π− =++−=∴ fff
    +++−−=− ..... 5 1 3 1 1 12 42 222 π ππ Hence C equation takes the form
Q2. Find the Fourier Series to represent the function given by for and= for Deduce that Sol. We know that, -----A )(xf xxf =)( ,0 π≤≤ x x−π2 ππ 2≤≤ x 8 .... 5 1 3 1 1 1 2 222 π =∞+++ ∑ ∑ ∞ = ∞ = ++= 1 1 0 sincos 2 )( n n n nxdxbnnxdxa a xf       −−++=     −+=∴         −+=     −+=∴ ∫ ∫ ∫∫ π π π π π π π π π π π π π π π π π π π π 2 2 0 2 0 2 2 2 0 2 2 0 0 sin )2( cossin1 cos)2(cos 1 2 2 2 1 )2( 1 n nx n nx x n nx n nxx nxdxxnxdxxa x x x dxxxdxa n
( ) ( ) ( ) 0 sin)2(sin 1 112 11111 0 2 2 2222 =∴     −+=       −− =∴         − +−        − − ∫ ∫ n n n n nn b nxdxxnxdxxb n a nnnn π π π π π π π Required Fourier Series -------B Put x=a in equation B We get, ∑ ∞ =       −− += 1 2 cos 1)1(2 2 )( n n nx n xf π π
∞+++=⇒ ∞+ − + − + − = ×− =       −− += ∑ ∞ = ...... 5 1 3 1 1 1 8 ...... 5 2 3 2 2 2 4 1)1(2 2 )0( 222 2 22 1 2 π ππ π π n n n f
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. π ),( ππ−
∑ ∞ = += 1 0 cos 2 )( n l xnana xf π ∫= l dxxf l a 0 0 )( 2 ∫       = l n dx l xn xf l a 0 cos)( 2 π Where, 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. ∫ ∑       =       = ∞ = l n n n dx l xn xf l b l xn bxf 0 1 sin)( 2 sin)( π π
Problems Q1. Find the Fourier cosine series for the function f(x)=x2 in the range . Sol. The given function f(x)=x2 is a even function. So, we apply case 1 i.e. π≤≤ x0 ∑ ∞ = += 1 0 cos 2 )( n l xnana xf π ∫       = l n dx l xn xf l a 0 cos)( 2 π [ ] 3 2 3 2 2 1 2 0 0 3 0 0 2 0 2 0 π π π π π π π π = = = = ∫ ∫− a xa dxxa dxxa
[ ] 3 2 3 2 2 1 2 0 0 3 0 0 2 0 2 0 π π π π π π π π = = = = ∫ ∫− a xa dxxa dxxa ( ) ( ) π π π π π π π 0 32 2 0 2 2 sin2cos2sin2 cos 2 cos 1       − + − −= = = ∫ ∫− n nx n nxx n nxx a nxdxxa nxdxxa n n n
∑ ∞ = − += − =     = 1 2 2 2 2 2 cos)1( 4 3 )1(4 cos22 n n n n n n nx x n a n n a π ππ π       −+−−= .......3cos 3 1 2cos 2 1 cos4 3 22 2 2 xxxx π Hence the required result is ,
Q2. Obtain the Fourier expansion of xsinx as a cosine series in (0, ). Hence deduce Sol. Let 4 2 ...... 75 1 53 1 31 1 − =∞− ⋅ + ⋅ − ⋅ π π ( ) ( ) [ ]∫ ∫ ∫ −−+= = = −−−= = π π π π π π π π 0 0 0 00 0 0 )1sin()1sin( 1 cossin 2 2 sin1cos 2 sin 2 dxxnxnxa nxdxxxa a xxxa xdxxa n n
)1( 1 )1cos( 1 )1cos( )1( )1sin( )1( )1sin( 1 1 )1cos( 1 )1cos(1 0 22 ≠       + + − − − =             − − − + +− −       − − + + +− = n n n n n a n xn n xn n xn n xn xa n n ππ π π 2 1 2 2cos1 2 2sin 1 2 2cos1 2sin 1 cossin 2 ,1 1 0 1 0 1 0 1 − =     − =      − −     − = = == ∫ ∫ ππ π π π π π π π a xx xa xdxxa xdxxxan When,
      ∞− ⋅ + ⋅ − ⋅ +=       ∞− ⋅ + ⋅ − ⋅ −−= ...... 75 1 53 1 31 1 21 2 ..... 75 4cos 53 3cos 31 2cos 2cos 2 1 1sin π xxx xxx Putting x= ,then 2 π 4 2 ...... 75 1 53 1 31 1 − =∞− ⋅ + ⋅ − ⋅ π Hence,
Even and Odd Functions:-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
• 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. How To Determine Whether The Function IsHow To Determine Whether The Function Is Even Or OddEven Or Odd f(x)=sin x -3π -5π/2 -2π -3π/2 -π -π/2 π/2 π 3π/2 2π 5π/2 3π -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 x y 2 5 2:sin π π <<− xx f(x)=sin x -3π -5π/2 -2π -3π/2 -π -π/2 π/2 π 3π/2 2π 5π/2 3π -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 x y
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 Or then the function is even function and then the function is odd function If Function is in this form
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. π ),( ππ−
∑ ∞ = += 1 0 cos 2 )( n l xnana xf π ∫= l dxxf l a 0 0 )( 2 ∫       = l n dx l xn xf l a 0 cos)( 2 π Where, 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. ∫ ∑       =       = ∞ = l n n n dx l xn xf l b l xn bxf 0 1 sin)( 2 sin)( π π
Problems Q1. Find the Fourier cosine series for the function f(x)=x2 in the range . Sol. The given function f(x)=x2 is a even function. So, we apply case 1 i.e. π≤≤ x0 ∑ ∞ = += 1 0 cos 2 )( n l xnana xf π 3 22 0 0 3 3 2 0 0 22 0 21 0 π a πx π a π dxx π a π π dxx π a =    = ∫= ∫−=
[ ] 3 2 3 2 2 1 2 0 0 3 0 0 2 0 2 0 π π π π π π π π = = = = ∫ ∫− a xa dxxa dxxa ( ) ( ) π π π π π π π 0 32 2 0 2 2 sin2cos2sin2 cos 2 cos 1       − + − −= = = ∫ ∫− n nx n nxx n nxx a nxdxxa nxdxxa n n n and, ∫       = l n dx l xn xf l a 0 cos)( 2 π
∑ ∞ = − += − =     = 1 2 2 2 2 2 cos)1( 4 3 )1(4 cos22 n n n n n n nx x n a n n a π ππ π       −+−−= .......3cos 3 1 2cos 2 1 cos4 3 22 2 2 xxxx π Hence the required result is ,
Q2. Obtain the Fourier expansion of xsinx as a cosine series in (0, ). Hence deduce Sol. Let 4 2 ...... 75 1 53 1 31 1 − =∞− ⋅ + ⋅ − ⋅ π π ( ) ( ) [ ]∫ ∫ ∫ −−+= = = −−−= = π π π π π π π π 0 0 0 00 0 0 )1sin()1sin( 1 cossin 2 2 sin1cos 2 sin 2 dxxnxnxa nxdxxxa a xxxa xdxxa n n
)1( 1 )1cos( 1 )1cos( )1( )1sin( )1( )1sin( 1 1 )1cos( 1 )1cos(1 0 22 ≠       + + − − − =             − − − + +− −       − − + + +− = n n n n n a n xn n xn n xn n xn xa n n ππ π π 2 1 2 2cos1 2 2sin 1 2 2cos1 2sin 1 cossin 2 ,1 1 0 1 0 1 0 1 − =     − =      − −     − = = == ∫ ∫ ππ π π π π π π π a xx xa xdxxa xdxxxan When,
      ∞− ⋅ + ⋅ − ⋅ +=       ∞− ⋅ + ⋅ − ⋅ −−= ...... 75 1 53 1 31 1 21 2 ..... 75 4cos 53 3cos 31 2cos 2cos 2 1 1sin π xxx xxx Putting x= ,then 2 π 4 2 ...... 75 1 53 1 31 1 − =∞− ⋅ + ⋅ − ⋅ π Hence,
Harmonic Analysis The process of finding the Fourier series corresponding to the function when the function by numerical values is known as harmonic analysis. The Fourier series for the function f(x) in the interval is given by- 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 the following relations. ∫ ∫ ∫ ∑ + + + ∞ =       =       = =             +      += c n c n c n nn dx c xn xf c b dx c xn xf c a dxxf c a c xn b c xn a a xf 2 2 2 0 1 0 sin)( 1 cos)( 1 )( 1 sincos 2 )( α α α α α α π π ππ Where, cx 2+<< αα
a0 = 2x[Mean value of f(x) in the interval ( )] an = 2x[Mean value of in the interval( )] bn = 2x[Mean value of in the interval ( )] In formula 1 the first term of expansion is known as first or fundamental harmonic. The second term is known as second harmonic and the term is known as third harmonic and so on….. c2, +αα       ⋅ c xn xf π cos)( c2, +αα       ⋅ c xn xf π sin)( c2, +αα c xb c xa n ππ sincos1 + c xb c xa ππ 2sin2cos 22 + c xb c xa ππ 3sin3cos 33 +
Problems Q1. In a machine the displacement y of a given point is given for a certain angle as follows. Find the coefficient of in the Fourier series representing the above variations. Sol. The Fourier series for the function y in the given interval 0-360o or ( ) is given by θ 0 30 60 90 120 150 180 210 240 270 300 350 7.9 8 7.2 5.6 3.6 1.7 0.5 0.2 0.9 2.5 4.7 6.8 °θ y ( ).sincos 2 1 0 ∑ ∞ = ++= n nn xnbxna a y ππ π2,0 θ2sin
Presented By :- Shiv Prasad Gupta Naveen Kumar Avinash

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Fourier series