Bài giải phần giải mạch P17

Chapter 17, Solution 1. (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = [sin(A + B) + sin(A – B)], g(t) = sin 3t cos 4t = [sin 7t + sin(–t)] = – sin t + sin7t which is harmonic or periodic with the fundamental frequency ω = 1 or T = 2π/ω = 2π. (d) h(t) = cos 2 t = (1 + cos 2t). Since the sum of a periodic. | Chapter 17 Solution 1. a This is periodic with n which leads to T 2n 2. b y t is not periodic although sin t and 4 cos 2nt are independently periodic. c Since sin A cos B sin A B sin A - B g t sin 3t cos 4t sin 7t sin -t sin t sin7t which is harmonic or periodic with the fundamental frequency 1 or T 2n 2n. d h t cos 2 t 1 cos 2t . Since the sum of a periodic function and a constant is also periodic h t is periodic. 2 or T 2n n. e The frequency ratio makes z t periodic. or T 2n 10. f p t 10 is not periodic. g g t is not periodic. Chapter 17 Solution 2. a The frequency ratio is 6 5 . The highest common factor is 1. a 1 2n T or T 2n b d 2 or T 2nD n c f3 t 4 sin2 600n t 4 2 1 - cos 1200n t d 1200n or T 2n a 2n 1200n 1 600. d f4 t e10t cos 10t y sin 10t. d 10 or T 2n a . Chapter 17 Solution 3. T 4 o0 2n T n 2 g t 5 0 t 1 10 1 t 2 0 2 t 4 ao 1 T g t dt J dt 10dt an 2 T J0g t cos noot dt 2 4 5 cos -nn t dt J Wcos t dt 1 5 sin t nn 2 0 .2 . nn 10 sin t nn 2 1 -1 nn 5 sin nn 2 an 5 nn -1 n 1 2 n odd 0 n even bn 2 T J0g t sin noot dt 2 4 j05sin 22n t dt J Wsin t dt 1 2x5 nn ------cos t nn 2 0 2x10 nn nn cos t 21 5 nn 3 - 2 cos nn cos nn 2 2 Chapter 17 Solution 4. f t 10 - 5t 0 t 2 T 2 Oo 2n T n ao 1 T J0f t dt 1 2 J2 10 - 5t dt 10t - 5t2 2 0 5 an 2 T Jf t cos noot dt 2 2 10 - 5t cos nnt dt J0 10 cos nnt dt- 5t cos nnt dt 2 -5 n2n2 cos 2nn - 1 0 0 5 2 cos nnt n n 5t . _ sinnnt nn bn 2 2 10 - 5t sin nnt dt 10 sin nnt dt- 5t sin nnt dt - 5 2_2 n n 2 sin nnt 5t 2 cos nnt nn 0 0 10 nn cos 2nn 10 nn - 10 1 . z x Hence f t 5 ---------V sin nnt n Zin Chapter 17 Solution 5. ao T 2n 2n T 1 T 1 T _ i _ t J z t dt 1xn - 2xn an bn 2T 1 n 1 2n 1 n Jz t cosn odt J1cosntdt---J 2cosntdt n T n n J nn 0 0 0 n 2t _ 1 n z t cosn . dt T J0 o nJ 2 . 2n -----sin nt 0 nn---n 1 sin ntdt - I 2 . 1 n 2 2n 2sinntdt -----cosnt L ---cosnt n 1 nn 0 nn n n 6 n odd i nn 0 n even Thus œ 6 z t - V sin nt n .