K 1 2 3 k 3k hence the fourier expansion

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 0 2π = 0 2 sin kx. 3k dx = 1 π 1 ( 2 )2 dx = −1 8 . π 1π 11 4 Now a0 = f (x) dx = ( 2 ) dx = , so the energy of the constant term π −π π −1 π 16 1 8 A2 8/π 2 1 a2 = 2 . Hence 0 = = ≈ 0.3183. About 31.83 % is A2 = 0 = 0 2 2 2 π π E 8/π π of the energy is in the constant term. 1π (b) Since f (x) is an even function, bk = 0. Now ak = f (x) cos kx dx = π −π 1 4 sin k 2 sin kx 11 = 2 cos kx dx = , k = 1, 2, · · · . The energy of the π −1 kπ kπ −1 16 sin 2 k k th harmonic is A2 = a2 + b2 = , k = 1, 2 , · · · . k k k k2 π2 16 sin 2 1 A2 16 sin 2 1 1 2 π2 ≈ 0.4508 so = (c) The energy in the first harmonic is A1 = 8 π2 E π A2 + A2 1 giving 0 ≈ 0.3183 + 0.4508 = 0.7691. Hence the constant term and the E first harmonic contain about 77 % of the energy of f . Since we are still less than 4 sin 2 2 16 sin 2 2 A2 2 2 π2 ≈ 0.1316. 80 % of the energy, we try A2 = which gives = 8 4π 2 E π A2 + A2 + A2 1 2 Now 0 ≈ 0.3183 + 0.4508 + 0.1316 = 0.9007 or about 90 % of the E MATB42H page 4 Solutions # 1 energy. We need the first 3 terms of the Fourier series to capture 80 % of the energy of f . (d) The N th 1 Fourier polynomial FN (x) is given by FN (x) = a0 + 2 N sin k cos kx k k =1 sin N sin 3 cos 3x + · · · + cos Nx . 3 N bk sin kx 4 2 + = π π 2 = 2 4 + π π N ak cos kx + k =1 sin 1 cos x + sin 2 cos 2x + 2 fx F3 3Π 2Π Π -1 1 2Π Π 3Π 5. f (x) = x2 − 16 restricted to [−4, 4] and extended to all of R with period 8. -4 4 -16 We first note that, since the periodic extension of f (x) is even, bk = 0, k = 1, 2, · · · . 4 1 64 1 42 x3 242 even (x − 16) dx = (x − 16) dx = − 16x = − 64 = Now a0 = 8 −4 20 3 23 0 −64 3 1 42 2πkx k πx 242 even dx = dx = (x − 160 cos (x − 16) cos ak = 8 −4 8 20 4 4 4 4 1 4(x2 − 16) sin kπx 4 kπx kπx 16 4 − − x sin dx = 0 + 2 2 x cos 2 kπ kπ 0 4 kπ 4 0 0 X $ =0 4 $ kπx $ 16 $$$ dx cos $$ k$2 4 $2 π$ 0 64 , k = 1, 2 , · · · . k2 π2 ∞ 32 64 k πx (−1)k The Fourier series is − + 2 . cos 2 3 π k 4 = (−1)k k =1 MATB42H page 5 Solutions # 1 -4 4 F2 x2 y 16 -16 1π′ f (x) cos kx dx 6. (a) Let the Fourier coefficients of f be ak and bk . Now ak = π −π π 1 kπ f (π ) − f (−π ) parts + f (x) cos kx f (x) sin kx dx = ± + k bk = k bk , =π π −π π −π 1π′ ′ f (x) sin kx dx parts since f is periodic of period 2π , and bk = = π −π π π 1 k − f (x) sin kx f (x) coskx dx = 0 − k ak = −k ak . π π −π −π ′ ′ ′ ′ (b) The amplitude of the k th harmonic of f is Ak = a2 + b2 , k ≥ 1. The amk k plitude of the k th harmonic of f ′ is (ak ′ )2 + (bk ′ )2 = (k bk )2 + (−k ak )2 = k 2 (a2 + b2 ) = k k k a2 + b2 = k Ak , k = 1, 2, 3, · · · . k k 7. To do this question we need to regard sin x, x ∈ [0, π ], as the restriction of an even function and as the restriction of an odd function. To obtain an even function we simply extend sin x over R with period π . Thus bk = 0, π 4 2π 2 1 a0 = sin x dx = − cos x = and π0 π π 0 2π ak = sin x cos(2kx) dx = π0 1π Π Π 2Π (sin((2k + 1)x) + sin((1 − 2k )x)) dx = π0 π cos((2k + 1)x) cos((1 − 2k )x) 1 = − − π 2k + 1 1 − 2k 0 1 1 1 1 1 1 = + + + π 2k + 1 1 − 2k 2k + 1 1 − 2k 1 4 1 2 =− + . The 2 − 1) π π 2k + 1 2k − 1 (4k Π Π 2Π ∞ 2 cos 2kx 4 Fourier series is F (x) = − . π π k=1 4k 2 − 1 -1 To obtain an odd function we need only use sin x itself with period 2π . The Fourier series is F (x) = sin x....
View Full Document

This document was uploaded on 09/29/2013.

Ask a homework question - tutors are online