Unit 2: Fourier Series PDF

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B.M.S. College of Engineering

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Fourier series transform calculus mathematics numerical techniques

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These notes contain Fourier series calculations, examples of applying Dirichlet's conditions, and Fourier series expansions for specific periodic functions within the interval (0, 2π). Note, they seem to be part of a university course.

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B. M. S. COLLEGE OF ENGINEERING, BANGALORE – 560019 DEPARTMENT OF MATHEMATICS TRANSFORM CALCULUS, FOURIER SERIES AND NUMERICAL TECHNIQUES (Course Code: 23MA3BSTFN) UNIT-2...

B. M. S. COLLEGE OF ENGINEERING, BANGALORE – 560019 DEPARTMENT OF MATHEMATICS TRANSFORM CALCULUS, FOURIER SERIES AND NUMERICAL TECHNIQUES (Course Code: 23MA3BSTFN) UNIT-2: FOURIER SERIES Dirichlet’s conditions Any function f  x  can be developed as a FOURIER SERIES a0   f  x    an cos nx   bn sin nx where a0 , an , bn are constants provided: 2 n1 n 1 (i) f   is periodic, single-valued and finite; x (ii) f  x  has a finite number of discontinuities in any one period; (iii) f  x  has at the most a finite number of maxima and minima. Fourier series expansion of a function f ( x) over (a, a  2l ) is a   n    n  f ( x)  0   an cos  x    bn sin  x, 2 n1  l  n1  l  1 a  2l 1 a  2l  n  Where a0   f ( x)dx , an   f ( x) cos  x  dx n  1, 2,3 , l a l a  l  1 a  2l  n  and nb   f ( x)sin  x  dx n  1, 2,3. l a  l  I. Fourier series expansion for the following periodic functions in the interval : (0, 2 ) Ans: x    2  sin x   sin 2 x 1. f ( x)  x ..... .  2   .  4 4 sin nx 2 4 cos nx 2. f ( x)  x2 Ans: f ( x)   2  3 n n n 1  x  x sin nx  3. f ( x)  Ans: . 2 2 n 1 n 2𝜋 2 𝑐𝑜𝑠𝑛𝑥−𝑠𝑖𝑛𝑛𝑥 4. f ( x)  x(2  x) , 𝐀𝐧𝐬: + 4 ∑∞ 3 𝑖=1 2 𝑛   2 a 1 e 1  a cos nx  n sin nx 5. f ( x)  eax. Ans: e ax  .  n a 2 2 2a n 1   x    2 2 cos nx 6. If f ( x)    , show that f ( x)   2.  2  12 n1 n Fourier series expansion for the following periodic functions over the interval ( ,  ) : n 1  (1) 1. f ( x)  x. Ans : x  2 sin nx. n 1 n  2 (1) n  1  2. f ( x)  x , Ans: x    cos nx. 2  n 1 n 2 2 2  ( 1) n 1  2  x2  4 cos nx 3. f ( x)    x 2 2 Ans: 3 n 1 n2. BMSCE Page 1 TRANSFORM CALCULUS, FOURIER SERIES AND NUMERICAL TECHNIQUES Unit 2: Fourier Series  2  ( 1) cos nx n 4. f ( x)  x2 Ans: x 2  4. 3 n 1 n2  2n(1) n sin m 5. f ( x)  sin mx where m is neither zero nor an integer. Ans: sin mx   sin nx. n 1 m 2   n2  2a sin a  1  (1) n 1  6. f ( x)  cos ax where ‘a ‘is not an integer. Ans: cos ax   2a 2  cos nx .   n 1 n a 2 2  1  ( 1) n 7. f ( x)  x cos x and deduce that x cos x   sin x   2 sin nx. 2 n2 n  1 8. If f ( x)  x  x and f ( x)  x  x 2 2 Ans: 2  4 2  x  x 2     (1)n 1  2 cos nx  sin nx  3 n 1 n n   2  4 2  x  x2    (1)n  2 cos nx  sin nx  3 n 1 n n  sinh a 2 sinh a  ( 1) n e  ax     a cos nx  n sin nx . 9.  ax f ( x)  e Ans: a  n 1 a2  n2 II. Fourier series expansion for the piece-wise continuous / discontinuous periodic functions:  x   / 2   x  0  1  3( 1) n 1. f ( x)   Ans: f ( x)   sin nx. x   / 2 0  x   n 1 n 2. f ( x)    cos x   x  0. 2 4 n cos Ans: f ( x)    n 2 cos nx.    cos x 0  x     n  2 n 1 2 1   x  0 4 sin 3x sin 5 x  f ( x)  sin x  ......   3 5 . 3. f ( x)   0 x0 Ans: 1 0 x      x  0 4. f ( x)   and hence deduce that sum of the reciprocal squares of odd integers  x 0 x  2  1  1  (1)   1  2(1) n  n is equal to. Ans: f ( x)    cos nx     sin nx. 8 4  n1 n2 n 1  n  0    x  0  2 cos x cos3x 5. f ( x)  .Ans: f ( x)    2  2    sin x sin 2 x    1  2    x 0  x   4  1 3 k   x  0 4k  sin(2n  1) x 6. f ( x)   Ans: f ( x)  . k 0 x   n1 2n  1 (  x) / 2   x  0 sin 2 x sin 3x 7. f ( x)  . Ans: f ( x)  sin x   .  (  x) / 2 0 x  2 3  x 2   x  0   2    (1)n1  3 1  (1)n  sin nx. 2 2 8. f ( x)   2. Ans: f ( x)  x 0 x   n1  n n  1  2 x /    x  0 1  (1) n 4  f ( x)   f ( x)  2  cos nx. 9. 1  2 x /  0 x  Ans:  n 1 n 2. BMSCE Page 2 TRANSFORM CALCULUS, FOURIER SERIES AND NUMERICAL TECHNIQUES Unit 2: Fourier Series   x 0 x   4 1 10. f ( x)  . Ans: f ( x)    cos nx..  x     x  2 2  n1,3,5..... n2  x   x  0  2  1  (1) n 11. f ( x)   Ans: f ( x)    cos nx. x 0 x  2  n 1 n 2 0   x  0 13. f ( x)   2 which is assumed to be periodic with period 2. x 0 x  2 (1) n  1  n 2 2 Ans: f ( x)  2 cos nx   ( 1) (  ) sin nx. 6 n 1 n 2  n 1 n3 n  x  1  x  0 14. f ( x)  . Is the function even or odd?  x 1 0 x   4  cos(2n  1) x Ans: f ( x)   1  . 2  n 1 (2n  1) 2  x 0 x  15. Show that the Fourier series of f ( x)   is 2  x   x  2  4  cos x cos3x . f ( x)    2  2   2  1 3 III. Fourier series expansion for arbitrary period : l 2l  1  (1) n n f ( x)   2 cos x 1. f ( x)  x. Ans: 2  n 1 n 2 l. cos n x / l n sin n x / l (l , l ). Ans: f ( x)  sinh l   2l  (1)n 2 2 2  2  (1)n1 2 2 2  1   2. f ( x)  e x , l n 1 l n  n 1 l n  . 𝑛𝜋𝑥 𝑛𝜋𝑥 2𝑙2 4𝑙2 (cos −sin ) 3. 𝑓(𝑥) = 𝑥(2𝑙 − 𝑥), (0,2𝑙) 𝐀𝐧𝐬: + 2 ∑∞ 𝑖=1 𝑙 2 𝑙 3 𝜋 𝑛 (1) n 1 n x  4. f ( x)  sin mx in (l , l ). Ans: f ( x)  sin ml sin. n 1 n   m l 2 2 2 2 l 1 4  (1) n 2  (1) n 1 5. f ( x)  x  x2 (1,1). Ans: f ( x)   2  2 cos n x   sin n x. 3  n 1 n  n 1 n l 3  (1) n 4l 2 n x 6. f ( x)  x (l, l ). 2 Ans: f ( x)    cos. 3 n 1 n  2 2 l 7. f ( x)  e x in the interval 0  x  2. e2  1  e2  1  n (e 2  1) Ans: f ( x)    cos n  x   sin n x. 2e2 n1 e2 (1  n2 2 ) n  e (1  n  ) 2 2 2 8 16 (1) n 1 n t 8. f (t )  4  t 2 in (2, 2). Ans: f ( x)   2  cos. 3  n 2 2  9  2n x   3  2n x  f ( x)   2 2 cos    sin   9. f ( x)  2x  x2 (0,3) Ans: n 1 n   3  n 1 n  3 . sin n x 2  10. f ( x)   x (0, 2). Ans: f ( x)  .   n1 n  x   x   1  sin n x 11. f ( x)  in 0  x  2. Ans:  . 2 2 2 n 1 n BMSCE Page 3 TRANSFORM CALCULUS, FOURIER SERIES AND NUMERICAL TECHNIQUES Unit 2: Fourier Series  4x 3 1   x0  12. f  x    3 2 1  4 x 0  x  3   3 2  x 0  x 1 13. f ( x)   show that Fourier expansion of the function   2  x  1  x  2  4  cos  x cos3 x cos5 x  2  1 f ( x)    2   .................. and deduce that . 2  1 32 52  8 n 1  2n  12 2  x in 0 x4 14. If f ( x)   , express f ( x) as a Fourier series and hence deduce that  x  6 in 4 x8 2  1 8 1  (1)n n . Ans: f ( x)   cos x. 8 n 1  2n  12  n 1 n 2 2 4  0 2  x  1 1  x 1  x  0  1 4  1  n  n x 15. f ( x)  . Ans: f ( x)   2  2 1  cos  cos. 1  x 0  x 1 4  n1 n  2  2  0 1 x  2  x 1  x  0 2  1 16. f ( x)   Ans: f ( x)  1   [1  2(1)n ]sin n x. x  2 0  x 1  n1 n 𝑙−𝑥 0

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