FOURIER SERIES PERIODIC FUNCTIONS A function f(x) is said to be periodic with period T if f(x+T)=f(x) x , where T is a positive constant . The least value of T>0 is called the period of f(x). f(x+2T) =f ((x+T)+T) =f (x+T)=f(x) f(x+nT)=f(x) for all x Ex.1 f(x)=sin x has periods 2, 4, 6, …. and 2 is the period of f(x). Ex.2 The period of sin nx and cos nx is 2/n. FOURIER SERIES Let f (x) be defined in the interval (l , l ) and outside the interval by f ( x 2l ) f ( x) i.e assume that f (x) has the period 2l .The Fourierseries corresponding to f (x ) is given by a0 nx nx an cos bn sin 2 n1 l l where the Fourier coeffecients are l a0 1 f ( x)d x l l an 1 nx f ( x ) co s dx l l l bn 1 nx f ( x ) sin dx l l l l l n 1,2,3,.... If f (x) is defined in the interval (c,c+2l), the coefficients can be determined equivalently from a0 1 l an bn 1 l 1 l c 2l f ( x)d x c c 2l c c 2l c f ( x ) co s nx dx l nx f ( x ) sin dx l DIRICHLET CONDITIONS Suppose that 1. f(x) is defined and single valued except possibly at finite number of points in (-l,+l) 2. f(x) is periodic outside (-l,+l) with period 2l 3. f(x) and f ’(x) are piecewise continuous in( -l,+l) Then the Fourier series of f(x) converges to a)f(x) if x is a point of continuity b)[f(x+0)+f(x-0)]/2 if x is a point of discontinuity METHOD OF OBTAINING FOURIER SERIES OF f (x ) a0 nx nx 1. f (x) 2 an cos l bn sin l n 1 l 1 2. a0 f ( x)d x l l 3. an 1 nx f ( x ) co s dx l l l 4. bn 1 nx f ( x ) sin dx l l l l l n 1,2,3,.... SOLVED PROBLEMS 1. Expand f(x)=x2,0<x<2 in Fourier series if the period is 2 . Prove that 1 1 1 ... 2 12 22 32 6 SOLUTION Period = 2 l= 2 thus l= and choosing c=0 1 an l c 2 1 1 c nx f ( x) cos dx l 2 f ( x) cos nx dx 0 2 2 x cos nx dx 0 2 1 2 sin nx cos nx sin nx x 2 x 2 2 3 n n n 0 4 2 n0 n 2 2 8 2 a0 x dx 0 3 1 1 bn l 1 1 c 2 c nx f ( x ) sin dx l 2 f ( x ) sin nxdx 0 2 2 x sin nxdx 0 2 1 2 cos nx cos nx sin nx x 2 x 2 2 n n n 3 0 4 n a0 f ( x) an cosnx bn sin nx 2 n1 4 4 4 2 f ( x) x 2 cos nx sin nx 3 n1 n n 2 At x=0 the above Fourier series reduces to 4 2 4 3 n n 1 2 X=0 is the point of discontinuity By Dirichlet conditions, the series converges at x=0 to (0+4 2)/2 = 2 2 2 n 1 2 4 2 4 2 3 n 1 n 1 2 2 n 6 2. Find the Fourier series expansion for the following periodic function of period 4. Solution 2 x 2 x 0 f ( x) 2 x 0 x 2 f ( x 4) f ( x ) l a0 1 f ( x)d x l l 0 2 1 2 x d x 2 x d x 2 2 0 0 2 1 x2 x2 2x 2x 2 2 2 2 0 2 1 nx f ( x ) co s dx l l l l an 0 1 n xd x 2 x co s 2 2 2 n 2 x co s xd x 2 2 0 nx nx sin co s 1 2 2 x n2 (1) 2 ( n ) 2 2 4 0 2 nx nx sin co s 2 ( 1) 2 2 x 2 n ( n ) 2 4 4 1 (1) n n 2 2 0 fo r n even 8 fo r n o d d 2 2 n 2 0 1 nx bn f ( x ) sin dx l l l l 0 2 1 nx nx 2 x sin dx 2 x sin dx 2 2 l l 0 nx cos 1 2 2 x n 2 2 nx cos 2 2 x n 2 f ( x) 1 8 2 n 1 0 nx sin 2 (1) 2 2 n 4 2 nx sin 2 ( 1) 0 2 2 n 4 0 1 x cos 2 n 1 n2 2 2 EVEN AND ODD FUNCTIONS A function f(x) is called odd if f(-x)=-f(x) E x : x 3, s i n x , t a n x , x 5+ 2 x + 3 A function f(x) is called even if f(-x)=f(x) E x : x 4, c o s x , e x+ e -x, 2 x 6+ x 2+ 2 EXPANSIONS OF EVEN AND ODD PERIODIC FUNCTIONS If f (x) is a periodic function defined in the interval (l , l ) , it can be represented by the Fourier series Case1. If f (x) is an even function l a0 1 f ( x )d x l l 2 l l 0 f ( x )d x 2 nx an f ( x) cos dx l 0 l l nx is also even function f ( x) cos l 1 nx bn f ( x) sin dx l l l l nx 0 f ( x) sin is odd function l If a periodic function f (x) is even in (l , l ) , its Fourier series expansion contains only cosine terms a0 nx f ( x) an co s 2 l n 1 a0 an l 2 l 2 l l f ( x)d x 0 0 nx f ( x ) co s dx l Case 2. When f (x) is an odd function l 1 a0 f ( x ) dx 0 l l 1 nx nx an f ( x) cos dx 0 f ( x) cos is odd l l l l l 2 nx nx bn f ( x) sin dx f ( x) sin is even l0 l l l If a periodic function f (x) is odd in (l , l ) ,its Fourier expansion contains only sine terms nx f ( x) b n 1 2 bn l l 0 n sin l nx f ( x ) sin dx l SOLVED PROBLEMS 1.For a function defined by f ( x) x , x obtain a Fourier series. Deduce that 1 1 1 2 2 2 .... 2 1 3 5 8 Solution f ( x) x is an even function a0 nx f ( x) an cos 2 l n 1 SOLUTION a0 an 2 2 2 x dx 0 0 2x xd x 2 2 x co s n xd x 0 0 2 x co s n xd x 0 2 sin n x co s n x x 2 n n 0 2 n 1 1 2 n f ( x) 2 2 1 n 1 n n cosnx 1 2 At x=0 the above series reduces to n 2 1 1 2 n 1 n2 x=0 is a point of continuity, by Dirichlet condition the Fourier series converges to f(0) and f(0)=0 0 0 2 2 2 2 1n 1 2 n n 1 2 2 2 2 2 2 .... 3 5 1 1 1 1 2 2 .... 2 1 3 5 k PROBLEM 2 f ( x) k 2 8 when 3 x 0 when 0 x 3 Is the function even or odd. Find the Fourier series of f(x) SOLUTION f (x) is odd function a0 0 bn an 0 l 2 l 2 3 3 0 0 nx f ( x ) sin dx l nx k sin dx 3 nx k co s 2 3 n 3 3 2k [1 ( 1) n ] n 3 0 [1 (1) ] nx f ( x) sin n 1 n 3 2k n nx k co s 2 3 n 3 3 2k [1 ( 1) n ] n 3 0 [1 (1) ] nx f ( x) sin n 1 n 3 2k n nx k co s 2 3 n 3 3 2k [1 ( 1) n ] n 3 0 [1 (1) ] nx f ( x) sin n 1 n 3 2k n HALF RANGE SERIES COSINE SERIES A function f (x) defined in (0, l ) can be expanded as a Fourier series of period 2l containing only cosine terms by extending f (x) suitably in (l ,0) . (As an even function) a0 nx f ( x) an cos 2 l n 1 where 2 an l l 0 nx f ( x ) cos d x, n 0 l SINE SERIES A function f (x) defined in (0, l ) can be expanded as a Fourier series of period 2l containing only sine terms by extending f (x) suitably in (l ,0). [As an odd function] nx f ( x ) bn sin l u 1 wh ere bn 2 l l 0 nx f ( x ) sin d x, n 1 l SOLVED PROBLEMS Obtain the Fourier expansion of (x sinx )as a cosine series in (0, ) .Hence find the value of 1 2 2 2 1.3 3.5 5.7 SOLUTION Given function represents an even function in ( , ) a0 f ( x) an cos nx 2 n1 2 an f ( x) cos nxdx 0 2 a f ( x)dx 0 0 2 2 a x sin xdx x( cos x) 1( sin x) 0 0 0 2 an 2 0 x sin x cos nxdx 1 x sin(1 n) x sin(1 n) x dx 0 2 2 cos(1 n) x sin(1 n) x x 1. 2 1 n (1 n) 0 1 cos(1 n) x sin(1 n) x x 1. 2 1 n (1 n) 0 n 1 1(1) n 1 1 1 n (1) n 1 1 2 1 1 (1) 2 n 1 n 1 n 1 n 1 n 1 n 1 if n 1 2 1 a1 x sin x cos xdx x sin 2 xdx 0 0 1 cos 2 x sin 2 x x 1. 2 2 2 0 1 1 2 2 1 (1) n1 x sin x 1 cos x 2 2 cos nx 2 n2 n 1 in (0, ) x At 2 1 2 n2 x 2 the above series reduces to 1 n cos 1 2 n 1 n2 is a point of continuity The given series converges to f( ) 2 2 2 1 2 n2 1 n co s 1 2 n 1 n 2 1 1 1 2 ........... 1.3 3.5 5.7 4 2) Expand f ( x) x, 0 x 2 in half range (a) sine Series (b) Cosine series. SOLUTION (a) Extend the definition of given function to that of an odd function of period 4 i.e x; 2 x 0 f ( x) x; 0 x 2 an 0 Here 2 nx bn f ( x ) sin dx l 0 l l 2 nx bn f ( x ) sin dx 20 2 2 2 nx nx sin n cos 2 4( 1) 2 x( ) 1( ) 2 2 n n n 0 2 22 (1) n nx f ( x) sin n 1 n 2 4 (b) Extend the definition of given function to that of an even function of period 4 x;2 x 0 f ( x) x ; 0 x 2 bn 0 2 nx an f ( x) cos dx l 0 l nx l 2 2 an f ( x) cos dx 20 2 2 nx nx cos n sin 2 4 ( 1 ) 1 2 x( ) 1( ) ; 2 2 2 2 n n n 0 2 22 n0 2 a0 xdx 2 0 f ( x) 1 4 2 n 1 (1) n n 2 cos nx 1 2 Exercise problems 1. 0; x o f ( x) sin x;0 x f (x ) Find Fourier series of 2. f (x ) e x in (0,2 ) Find Fourier series of f (x ) 3.Find the Fourier series of in ( , ) 4.Find the Fourier series of in f ( x) x 3 (-2 ,2) f ( x) 4 x 2 5.Represent function x f ( x) sin L In (0,L) by a Fourier cosine series 6.Determine the half range sine series for x f ( x) 8 x 0 x4 4 x8 PARSEVAL’S IDENTITY • To prove that 2 1 2 2 2 f ( x ) dx l a ( a b n ) 2 0 n l n 1 l Provided the Fourier series for f(x) converges uniformly in (-l, I). The Fourier Series for f(x) in (-l,l) is a0 nx nx f ( x) an cos bn sin ................(1) 2 l l n 1 n 1 Multiplying both sides of (1) by f(x)and integrating term from – l to l ( which is justified because f(x) is uniformly convergent) a0 nx nx 2 l f ( x) dx 2 n1 an n1 f ( x) cos l dx n1 bn n1 f ( x) sin l dx a0 la0 an (lan ) bn (lbn ) 2 n 1 n 1 l 2 a 2 2 2 0 f ( x) dx l (an bn ).....................(2) l 2 n1 l CASE-I If f(x) is defined in (0,2l) then Parseval’s Identity is given by 2 1 2 2 2 (an bn )...............(3) 0 f ( x) dx l 2 a0 n 1 2l CASE-II If half range cosine series in (o,l) for f(x) is a0 nx f ( x) 2 an cos n 1 l . Then Parseval’s Identity is given by 2 l 1 2 2 an ..............(4) 0 f ( x) dx 2 2 a0 n 1 . l CASE-III If the half range Sine sereies in (0,l) for f(x) is nx f ( x) bn sin n 1 l Then Parseval,s Identity is given by 2 l 2 f ( x ) dx b .......... ....( 5 ) n 0 2 n 1 l RMS VALUE OF FUNCTION If a function y=f(x) is defined in ( c , c+2l ),then 1 2l c 2l y 2 dx c is called the root mean square value (RMS value) of y in ( c , c+2l ).It is denoted by y 1 y 2l 2 c 2l c 2 f ( x ) dx . Equation(2) becomes 1 2 2 2 y 2l l a0 (an bn ). n 1 2 2 y 2 1 2 1 2 2 a0 (an bn ). 2 n1 4 Equation(3) becomes 1 2 2 2 y 2l l a0 (an bn ). n 1 2 2 y 2 1 2 1 2 2 a0 (an bn ). 2 n1 4 Equation(4) becomes l 1 2 2 a0 an . 2 2 n 1 2 y l y 1 2 1 2 a0 an . 2 n1 4 2 Equation(5)becomes y 2 1 2 bn . 2 n1 SOLVED PROBLEMS 1) Find the Fourier series of periodic function f ( x) x x 2 in ( , ) Hence deduce the sum of series 1 1 1 1 4 4 4 ....... 4 1 2 3 4 1 2 2 6 n 1 n Assuming that SOLUTION a0 nx nx f ( x) an cos bn sin . 2 n1 l l n 1 ( , ) in 1 1 2 2 a0 f ( x)dx x dx 3 2 1 sin nx cos nx an f ( x) cos nxdx ( x 2 x) cos nxdx x 2 ( ) 2x n n 2 1 1 4 2 (1) n n bn 1 if n0 xcosnx is odd function f ( x) sin nxdx 1 2 ( x x) sin nxdx 0 1 x sin nxdx 1 cos nx sin nx 2 n x( ) 1( ) ( 1 ) n n2 n 2 x sin nx is odd function (1) (1) f ( x) 4 2 cos nx 2 sin nx. 3 n n 1 n n 1 in ( , ) 2 n n Using the Parseval’s Identity 1 1 2 2 2 2 a0 (an bn ). y 2 n1 4 y 2 1 2 2 2 ( x x) d x 4 5 2 3 2 4 2 n n a0 ; an 2 (1) ; bn (1) 3 n n 4 2 4 1 4 1 16 1 4 4 2 5 3 4 9 2 n 1 n 2 n 1 n 2 n 1 1 4 4 n 90 n 1 1 2 2 n 6 i 2)By using sine series for Show that 2 8 1 f ( x) 1 in 0 x 1 1 1 ....... 2 2 2 3 5 7 SOLUTION nx f ( x) bn sin f ( x) f ( x) sin nx l n 1 n 1 bn 2 0 2 cos nx f ( x) sin nxdx sin nxdx ( ) 0 n 0 2 1 (1) n n 2 for n0 1 (1) n f ( x) 1 sin nx n 1 n 2 By Parseval’s Identity y 2 1 2 bn . 2 n1 1 4 2 n y 1 dx 1 1 ( 1 ) 0 2 n 1 n 2 2 1 2 2 1 1 1 1 2 2 2 ....... 8 3 5 7 2 0 xl 3)Prove that in l 4l x 1 3x 1 5x x 2 (cos 2 cos 2 cos ..........) 2 n l 3 l 5 l 1 1 1 1 4 and deduce that 14 34 54 7 4 ....... 96 SOLUTION In Half range cosine series l l 2 2 a0 f ( x)dx xdx l l 0 l 0 2 nx 2 nx an f ( x) cos dx x cos dx l 0 l l 0 l l l l nx nx sin cos 2 2l n l l x( ) 1.( ) ( 1 ) 1 2 2 2 2 n n l n 0 l l2 a0 nx f ( x) an cos 2 l n 1 l 1 x 2 2 4 nx cos 2 n l n 1, 3, 5. . . By Parseval’s Identity y 2 1 2 1 2 a0 an . 2 n1 4 2 l 1 l 1 l y f ( x) 2 dx x 2 dx l 0 l 0 3 2 l2 l2 1 4l 2 n 4 4 ( 1) 1 3 4 2 n 1 n 2 1 1 1 1 4 4 4 4 4 ....... 1 3 5 7 96 COMPLEX FORM OF FOURIER SERIES The Fourier series of function of period 2l is a periodic a0 nx nx f ( x) an cos bn sin 2 n 1 l l e i e i e i e i cos sin 2 2i nx nx nx nx i i i i a0 e l e l e l e l f ( x) an bn 2 n 1 2 2i nx i i nl x f ( x) c0 cn e c n e l n 1 1 1 1 c0 a0 , cn an ibn , c n an ibn 2 2 2 l inx 1 cn f ( x)e l dx for n 0,1,2,.... 2l l The Fourier series can be represented in the following way f ( x) n c n n e 1 where cn 2l inx l dx l f ( x )e inx l dx l SOLVED PROBLEM 1.Find the complex form of the Fourier series of the periodic function 0 when 0 x l f ( x) when l x 2l SOLUTION inx f ( x ) cn e l n cn 1 2l 2l f ( x )e inx l dx 0 l inx 1 cn 0 e l 2l 0 l e 2l in 2l inx dx e l dx l inx l 2l l e 2in e ni 2ni n 1 1 2ni i n 1 1 2n i inx n f ( x) (1 ( 1) )e l n 2 n 2.Find the complex form of Fourier seriesof f(x)=sinx in (0,) SOLUTION inx c f ( x) n n e 2 l 2 2 inx c e n n cn 1 1 f ( x )e 2 inx d x 0 2 inx sin xe dx 0 1 e 2in sin x co s x 2 1 4n 0 2 inx 1 2 in cn e 1 2 4n 1 2 2 4n 1 f ( x) 2 4n n 1 2 1 e 2 inx in (0, ) HARMONIC ANALYSIS a0 f ( x) an cos nx bn sin nx 2 n 1 a0 an bn 1 1 1 2 f ( x)dx 0 2 f ( x) cos nxdx 0 2 f ( x) sin nxdx 0 2 1 a0 2 f ( x ) dx 2 m ean of f ( x ) in 0 , 2 2 0 2 1 an 2 f ( x ) cos nxdx 2 0 2m eanof f ( x) cos nx in 0,2 1 2 bn 2 f ( x ) sin nxdx 2 0 2m eanof f ( x) sin nx in 0,2 T h e t e r m a 1c o s x + b 1s i n x i s c a l l e d t h e fundamental or first harmonic, t h e t e r m a 2c o s x + b 2s i n x i s c a l l e d t h e second harmonic and so on. Solved Problem 1.Find first two harmonics of Fourier Series from the following table