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Bài giải phần giải mạch P15

Chapter 15, Solution 1. e at + e - at 2 1 1 1 s L [ cosh(at ) ] = + = s2 − a2 2 s − a s + a (a) cosh(at ) = (b) sinh(at ) = e at − e - at 2 1 1 1 a L [ sinh(at ) ] = − = s2 − a2 2 s − a s + a Chapter 15, Solution 2. (a) f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ) F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ] F(s). | Chapter 15 Solution 1. a eat e-at cosh at ------ L cosh at 1 2 L s - a s a J s s2 - a2 b eat - e-at sinh at ------------ L sinh at 1 2 L s - a s a J a s2 - a2 Chapter 15 Solution 2. a f t cos ot cos 0 - sin ot sin 0 F s cos 0 L cos ot - sin 0 L sin ot s cos 0 - m sin 0 F s -------2----2----- s2 m2 b f t sin ot cos 0 cos ot sin 0 F s sin 0 L cos ot cos 0 L sin ot z x ssin 0 - cos 0 F s -------2----2----- s2 m 2 Chapter 15 Solution 3. a L e-2tcos 3t u t s s2 22 9 b L e2t sin 4t u t s 24 16 s c Since L cosh at s2 a2 d L e-3t cosh 2t u t s 3 s 3 2 - 4 Since L sinh at _2 _ 2 s2 - a2 a L e-4t sinh t u t 4 2 - 1 e L e-s n 2t sll .4 If f t ------------ F s - d tf t ---- V-F s ds Thus L te-t sin 2t 2 s 1 2 4 -1 ds s 1 2 4 2 2 s 1 L1 1 S4 S 2 l 2 Chapter 15 Solution 4. a G s .e - s 6se-s s2 16 b 2 e -2s F s 73 Chapter 15 Solution 5. a L cos 2t 30 s cos 30 - 2 sin 30 s2 4 r id2 L t2 cos 2t 30 d scos 30 -1 . s2 4 . _d_d_ ds ds r A s -1 s2 4 -1 2 k 2 7 d ds i-1 2 k 2 -1 4 2 7 _4 -2s s2 4 2 r A 2 2 -1 k 7 2s r 4 . 3 A 8s2 -s -1 k 2 7 s 4 s h s - I . 03 A r r r 8s2 s -1 _-V3s-V3s 2-V3s k J s 0 s 4 3 _ -O3s 2 s2 4 03 s3 - 8 s2 s 0 s 0 .r z i 8 - 1O3s - 6s2 V3s3 L t cos 2t 30 -----7 2 ------------ s2 4 b c L 30 4e 30 720 s 2 5 L 2tu t - 4 0 t dt 2 2 - 4 s 1 - 0 - 4s s2 .