# K 1 2 3 k 3k hence the fourier expansion

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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....
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## This document was uploaded on 09/29/2013.

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