Fourier Analysis of Waves

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Αγάπη Αλεβίζος
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1 Exercises for the Feynman Lectures on Physics by Richard Feynman, Et Al. Chapter 36 Fourier Analysis of Waves Detailed Work by James Pate Williams, Jr. BA, BS, MSwE, PhD From Exercises for the Feynman Lectures on Physics by Richard Feynman, Robert Leighton, Matthew Sands, et al. 36 Fourier Analysis of Waves. Refer to The Feynman Lectures on Physics Vol. I, Chapter 5. Fourier series over a period P where x is contained in the half-open interval [x, P). f(x) = a 2 + [a n cos ( 2nx P ) + b n sin ( 2nx P )] n= x +P a n = 2 P x x +P f(x) cos (2nx P ) dx b n = 2 P x f(x) sin (2nx P ) dx Euler s equation used to derive trigonometric identities. e i(θ±φ) = cos(θ ± φ) + i sin(θ ± φ) = e iθ e ±iφ = (cos θ + i sin θ)(cos φ ± i sin φ) = cos θ cos φ sin θ sin φ + i(cos θ sin φ ± sin θ cos φ) cos(θ ± φ) = cos θ cos φ sin θ sin φ sin(θ ± φ) = cos θ sin φ ± sin θ cos φ 2cos θ cos φ = cos(θ φ) + cos(θ + φ) 2sin θ cos φ = sin(θ + φ) sin(θ φ) e ix e ix = e = 1 = (cos x + i sin x)(cos x i sin x) = (cos x) 2 + (sin x) 2 e 2ix = cos 2x + i sin 2x = e ix e ix = (cos x + i sin x)(cos x + i sin x) = (cos x) 2 (sin x) 2 + i(cos x sin x + sin x cos x) cos 2x = (cos x) 2 (sin x) 2 = 1 (sin x) 2 (sin x) 2 = 1 2(sin x) 2 1

2 36.1 (a) y(x) = const. (b) y(x) = sin x <= x <= 2 * pi. (a) (sin x) 2 = 1 (1 cos 2x) 2 sin 2x = 2 cos x sin x y(x) = C x [, P] x +P a n = 2C P cos (2nx P ) dx x a n = 2C P P 2n x=x +P sin (2nx P )] x=x = C n sin [2n(x + P) ] C P n sin (2nx P ) sin [ 2n(x + P) ] = cos ( 2nx P P ) sin(2n) + sin (2nx P ) cos(2n) = sin (2nx P ) a n = C n [sin (2nx P ) sin (2nx P )] = x +P b n = 2C P sin (2nx P ) dx x b n = 2C P P 2n x=x +P cos (2nx P )] x=x = C n {cos [2n(x + P) ] cos ( 2nx P P )} = C n [cos (2nx P ) cos(2n) sin (2nx P ) sin(2n) cos (2nx P )] = C n [cos (2nx P ) cos (2nx P )] = a 2 = C a = 2C (b) y(x) = sin x x [,2] x =, P = 2 2 a = 1 sin x dx = 1 cos x] x= x=2 = 1 [cos(2) cos()] = 1 (1 1) = 2

3 2 a n = 1 sin x cos nx dx = [ sin(x + nx)dx sin(x nx)dx] = 1 { sin[(1 + n)x] dx sin[(1 n)x] dx 2 = 1 2 = 1 2 { cos[(1 + 2n)x] 1 + 2n { cos[(1 + 2n)x] 1 + 2n ] ] 2 x=2 x= x=2 x= + cos[(1 2n)x] ] 1 2n cos[(1 2n)x] 2n 1 ] } x=2 x= x=2 x= } = } = b n = 1 sin x sin nx dx = n > 1 2 b 1 = 1 (sin x)2 dx = 1 (1 cos 2x)dx = x [, ] f(x) = { 1 x (, 2) a n = 1 ( cos nx dx cos nx dx) = 1 n (sin nx] x= x= sin nx] x=2 x= ) = (a) b n = 1 ( sin nx dx sin nx dx) = 1 n (cos nx] x= x= cos nx] x= 2 = 1 [cos n + cos n 2] n b 2n+1 = 4( 1)2n+1 (2n + 1) = 4( 1)2n+2 (2n + 1) = 4 (2n + 1) f(x) = 4 sin[(2n + 1)x] (2n + 1) n= x=2 ) f ( 2 ) = 4 sin[(2n + 1) 2 ] (2n + 1) n= sin[(2n + 1) 2] = sin(n + 2) = cos n sin( 2) + sin n cos( 2) = ( 1) n 3

4 (b) (c) n= 1 = 4 ( 1)n (2n + 1) n= ( 1)n = (2n + 1) 4 n= ( 1)n ( 1)m (2n + 1) (2m + 1) = ( 4 )2 = 2 16 m= ( 1)n+m = 1 (2n + 1)(2m + 1) 2 1 (2n + 1) 2 = 2 16 n= m= n= 1 (2n + 1) 2 = 2 8 n= 4 n n= = = (2n + 1) m = (2n + 1) 2 4 m n= m= n= m= = = g(x) = { 1 + x x [, ) x x [, 2) a = 1 g(x) dx = 1 2 x dx + 1 dx + 1 dx 1 2 x dx 2 = (2 ) 2 2 [(2)2 2 ] = = 1 a n = 1 g(x) cos nx dx = 1 x cos nx dx + 1 cos nx dx + 1 cos nx dx 1 x cos nx dx d(uv) = udv + vdu 2 d(uv) = uv = udv + vdu

5 x cos nx dx = x n sin nx] 1 sin nx x= n 2 x= 2 = (2n + 1) 2 x cos nx dx = x n sin nx] x= 2 x=2 = 1 n 2 [1 ( 1)n ] = udv = uv vdu u(x) = x dv = cos nx dx cos nx dx = 1 sin nx n 2 1 sin nx n dx = 1 n 2 cos nx] x= x= = 1 n 2 (cos n 1) = 1 n 2 [( 1)n 1] 2 (2n + 1) 2 dx = 1 n 2 cos nx] x= x=2 = 1 (cos 2n cos n) n2 2 a 2n+1 = (2n + 1) (2n + 1) 2 2 = 4 (2n + 1) 2 2 b n = 1 g(x) sin nx dx = 1 x sin nx dx + 1 sin nx dx + 1 sin nx dx 1 x sin nx dx u(x) = x dv = sin nx dx sin nx dx = 1 cos nx n x sin nx dx = x n cos nx] 1 cos nx x= n x sin nx dx = x n cos nx] x= x=2 x= 2 1 cos nx n 2 dx = 2( 1)n n dx = 2 n 2( 1)n n b n = 1 n 2 [( 1)n + ( 1) n 2] 2 1 (cos 2n cos n) = n n 2 [4( 1)n 4] 2 n [1 ( 1)n ] 8 b 2n+1 = (2n + 1) (2n + 1) = 4 (2n + 1) (1 2 ) g(x) = (2n + 1) 2 cos[(2n + 1)x] + 2 (2n + 1) (1 2 ) sin[(2n + 1)x] n= n= 2 5

6 (a) (b) g() = g(2) = = (2n + 1) 2 2 n= n= 2 1 (2n + 1) 2 = 8 [ 1 (2n + 1) 2 ] n= 2 = 4 64 ( ) ( ) = ( ) ( ) ( ) ( ) = s + 1 (2n + 1) 4 1 (2n + 1) 4 n= = 4 64 s n= s = 2 [ ( ) + ] s = s = δ δ δ = 96 (1) 1 (2n + 1) 4 n= 6 = (2n + 1) (2n + 2) 4 = 1 n 4 n= n= n=

7 (2) 1 (2n + 2) 4 = (n + 1) 4 n= 1 16 dx (x + 1) 4 1 n 4 n= n= = 1 16 dy y 4 = = t t = (n + 1) n= t = 96 9 t = = = 4 9 t = (n + 1) 4 = 1 16 ( ) = m 4 = ζ(4) 16 n= m=1 = = = Evaluate the following integral: ζ(4) = 4 9 x3 dx e x = x3 e x dx 1 1 e x 1 1 e x = 1 + e x + e 2x + e 3x + = (e x ) n 7 n= x3 e x dx 1 e x = x 3 e (n+1)x dx n= u(x) = x 3 dv = e (n+1)x dx v(x) = e (n+1)x dx = e (n+1)x n + 1 = e nx n=

8 x 3 e (n+1)x dx = 3 n + 1 x2 e (n+1)x dx u(x) = x 2 dv = e (n+1)x dx v(x) = e (n+1)x dx = e (n+1)x n + 1 x 2 e (n+1)x dx = 2 n + 1 x e (n+1)x dx u(x) = x dv = e (n+1)x dx 36.5 x v(x) = e (n+1)x dx = e (n+1)x n + 1 e (n+1)x dx = 1 n e (n+1)x dx = (n + 1) 2 x3 e x dx 1 e x = 6 1 (n + 1) 4 n= = 6ζ(4) = 64 9 = 4 15 y(x, t) = 8h 2 ( 1)n+1 1)x (2n 1)at sin [(2n ] cos [ ] (2n 1) y (1, 2 a ) = 8h 2 ( 1)n+1 1) sin [(2n ] cos[(2n 1)] = 8h (2n 1) ( 1)n+1 ( 1) n+1 ( 1) n+1 (2n 1) = 8h 2 ( 1)3n+3 (2n 1) 2 = 8h 2 ( 1)n+1 (2n 1) 2 = 8h 2 ( ) y(1,) = 8h 2 ( 1)n+1 1) sin [(2n ] = ( 1)n+1 ( 1) n+1 (2n 1) 2 2 (2n 1) 2 = 8h 2 4 = 2h = 8h 2 ( 1)2n+2 (2n 1) 2 = 8h 2 1 (2n 1) 2 = 8h 2 ( ) = A 1 A = 1, A 2 A =, A 3 A = = 1 9 8

9 2 a n = 1 h(x) cos nx = dx = h(x) = x x [,2) 2 x cos nx dx = x 2 n sin nx] x= x=2 2 1 sin nx n dx = 1 n 2 cos nx] x=2 x= Conjecture: 2 2 b n = 1 h(x) sin nx x sin nx dx = x n cos nx] x= 2 dx = 1 2 x=2 b n = 1 n a = 1 h(x) dx = x 2 2 x sin nx dx 2 2 h(x) = sin nx n 1 cos nx dx = 2 n n dx = x2 ] x=2 x= = h ( 2 ) = sin (n 2 ) = 1 n 2 1 ( 1)n+1 = 1 (2n 1) 2 1 ( ) =.25 = = ( 1)n+1 = (2n 1) 4 = tan 1 1 S 2 y x 2 σ 2 y t 2 = S tension and σ mass density x A x [, x x p ] p y(x, ) = A (1 x x p ) x (x { L x p, L] p y (x, ) = A amplitude, L the length of the string 9

10 y(x, t) = sin ( nx L ) [a n sin ( nct L ) + b n cos ( nct L )] y (x, t) = nc sin (nx L L ) [a n cos ( nct L ) b n sin ( nct L )] c = S σ y(x, ) = 8A 2 a n = 2L 2 sin ( nx p b n = A L ) n 2 2 x p (L x p ) y(x, t) = 2A L2 sin ( nx p L ) sin (nx n 2 2 x p (L x p ) L ) cos ( nct L ) y(x, ) = 2A L2 sin ( nx p L ) sin (nx n 2 2 x p (L x p ) L ) T = 2L c = 2L σ S y(x, t) = 8A 1 n 2 sin (nx 2 L ) cos ( nct L ) 1 sin (nx n2 L ) = 8A 2 [sin (x L ) + 1 sin (2x 4 L ) + 1 sin (3x 9 L ) + ] T 2 = L c y (x, L c ) = 8A 1 n 2 sin (nx 2 L ) cos(n) = 8A 2 [ sin (x L ) + 1 sin (2x 4 L ) 1 sin (3x 9 L ) + ] 36.8 y ( L 2, L c ) = 8A 2 ( 1)n+1 n 2 f(x) = sin x x [, ) f(x) = a 2 + [a n cos(2nx) + b n sin(2nx)] 1

11 a = 2 sin x dx = 2 cos x} x= x= = 2 ( 1 1) = a n = 2 sin x cos(2nx) dx = 2 { sin[(1 + 2n)x] dx sin[(1 n)x] dx} = 1 = 1 cos[(1 + 2n)x] { ] 1 + 2n { cos[(1 + 2n)x] 1 + 2n ] x= x= x= x= + cos[(1 2n)x] ] 1 2n cos[(1 2n)x] 2n 1 ] x= x= x= x= } = a 1 = 1 cos(3) [ cos 1] = 1 (2 3 2) = 1 ( ) = a 2 = 1 cos(5) [ cos(3) ] = 2 ( ) = 2 ( ) = a 2 = 1 cos(7) [ cos(5) ] = 2 ( ) = 2 ( ) = b n = 2 sin x sin(2nx) dx sin x = ( 1)n x 2n+1 (2n + 1)! n= = x x3 3! + x5 5! sin x sin(2nx) dx = x sin(2nx) dx 1 3! x3 sin(2nx) dx + 1 5! x5 sin(2nx) dx u(x) = x dv = sin(2nx) dx v(x) = 1 2n cos(2nx) x sin(2nx) dx = x x= 2n cos(2nx)] + 1 x= 2n cos(2nx) dx = 1 n 2 sin(2nx)] = x= u(x) = x 3 dv = sin(2nx) dx } x= 11

12 v(x) = 1 2n cos(2nx) x 3 sin(2nx) dx = x3 x= 2n cos(2nx)] + 3 2n x2 cos(2nx) dx x= u(x) = x 2 dv = cos(2nx) dx Conjecture: (a) x 2 cos(2nx) v(x) = 1 2n sin(2nx) dx = x2 x= 2n sin(2nx)] x= x 2n+1 sin(2nx) dx = b n = 1 x sin(2nx) dx = n 1 f(x) 2 dx = a 2 + a n 2 (b) 2 = ( ) = 2a + a 2 ( ) a 2 = 4 15 = a 15 Fourier coefficients and graph for Exercise 36.1 (a) for f(x) = 1 for all x in the half-open interval [x, P). 12

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15 Fourier coefficients and graph for Exercise 36.1 (b) f(x) = sin x for all x in the half-open interval [, 2). 15

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17 Fourier coefficients and graph for Exercise 36.2, the square wave. 17

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22 Fourier coefficients and graph for Exercise 36.3, the triangle wave. 22

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27 Fourier coefficients and graph for Exercise 36.6, the sawtooth wave. 27

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29 Fourier coefficients and graph for Exercise 36.8, the rectified sine wave. 29

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