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HW5SOL - HW#5 Chapter 6.1 3 f(t is a continuous function#4...

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Unformatted text preview: HW #5. Chapter 6.1 # 3. f(t) is a continuous function. #4. F(t) is a piecewise continuous function and have a jump discontinuity at t= 1. #7. f t f t F s 1 2 s e f t dt dt s b for |s| 1 2 1 s b e 1 s b e 1 1 2 s b 1 s b cosh bt 2 #10. f t sinh bt 2 f t F s 1 2 1 2 1 2 e f t dt dt s s 1 a 1 a b e b s 1 a s b 1 a s b e b a b for |s a| #18. f t f t F s e f t dt 1 2 dt Note Integratoin by part a t b t dt a 1 s n s a a t t a t b t , a t b t dt b n s a t dt dt n n 1 t s a n n s s n! a 1 a 2 t n! s a| a dt for |s #27. a) Show that t change of variables Then t By the definition of p dt x s dt p s 1 x st, dx sdt dx s 1 s x dx 1 from Prob. 26 t p s 1 b) Show that t , where n is a positive integer From Prob. 26 p and Therefore, t n! s s 0 1 p p 1 1 1 n s p 0 n s 1 n n s n n 1 n! 1 s n! s c) Show that t / 1 s 1 s dx s 0 t change of variables then t x 1 2 s x 2ydy 1 s dx y , dx dy 1 s dx d) Show that t / / 2s t 1 s / d 2ydy 2 s 0 change of variables then t x 2 s y , dx d d Note Integratoin by part a t b t dt a Then t 2 s d 2 s 1 s 1 2 d 1 2 2s d , a t b t b a t b t dt Chapter 6.2 # 8. F s 8s s s 4s 4 a Therefore F s 5s s F s #10. F s 12 as s 5, b 4 4 3 s b 4 c s 4, c 5s s 4 s a c s s s bs 4 4c 3 4 4 3 s 3 5 cos 2 2 sin 2 2s 3 s 2s 10 2 s s 2s 1 1 5 9 2 s s 1 5 e 3 1 3 sin 3 5 3 s 3 1 3 F s 2e cos 3 #23. y 2y y 4e , y 0 2, y 0 1 Take the Laplace transformation of the equation, y y 4e Then s Y s sy 0 s Y s s 1 s y 0 2s 4 2s 1 s 2t e s 1 1 22 s 1 Y s 2 s 1 s 1 y t s Y s sY s 4 s 1 sy 0 y 0 y 0 2y 0 4 s 1 2s 2s 2s 1 22 s 1 2e Y s 3 s 3 2s 2 s 1 s 1 1 1 4 s 1 2sY s 1 Y s te #28. F s e f t dt a) Show that F s dF ds b) Show that F s t f t d ds tf t e f t dt e f t dt s tf t e dt tf t d F ds d ds e f t dt e s f t dt t f t e dt t f t Chapter 6.3 # 3. g t f t g t u t , where f t t 0, , 0 t t #6. g t t 1 u t t t 1 t 1 2 t 2 2 t t 2 t 0, 1, 2 , 3 , 2 u t 0 1 2 t 1 2 3 3 t 3 u t t 3 0, 1, 2t, 0, 0 1 2 t 1 2 3 3 g t #15. f t f t t t 0, , 0, t 0 t t 2 2 t u t u t 2 u take the Laplace transformation of f t f t F s e s e s e s #23. F s F s set G s s s s s s 2 e 4s 4 s s s 2 2 1 2 2 1 e 2 e 4s 4 Take the inverse Laplace transformation, G s Therefore, F s G s e g t 1 u t e cosh t 1 u t g t e cosh t Chapter 6.4 # 6. y 3y 2y u t , y 0 0, y 0 1 Take the Laplace transformation of the equation, y y s Y s sY s sy 0 y 0 y 0 u t Then s Y s sy 0 s Y s e s s 3s 2 a Y s set F s e s s 3s 2 s s 1 , b 2 1 3s y 0 3s 1 3s e s 3y 0 1 a s e s b s 1, e 1 s 2 s 1 s 1 1 1 2 s 1 1 s 1 2 s c 2 s d 1 s e 2 2Y s e s 3sY s 2 Y s 2 1, c 2 e 1 ,d 2 e 2 1 s 2 s 1 s 1 2 Take the inverse Laplace transformation of F s , F s e 2 Therefore, Y s e 2 u t 1 2 F s 2e 1 s 1 e s 1 2 e e F s 1 u t 1 2 2e 2e e e #9. y y g t t , 2 0 3, t 6 6 , y 0 0, y 0 1 g t can be written as g t t 1 2 u t 3u t t 2 1 t 2 6 u t Take the Laplace transformation y g t s s Y s 11 2s 1 Y s sy 0 y 0 1e 2 s 11 2s 1 e 2s s e 1 2 1e 2 s 1 Y s 1 1 2s s 1 1 s 1 1 1 s s 1 s 1 s 1 1 1 1 1 2 s Take the inverse Laplace transformation Y s y t 1 t 2 sin t u t 2 t 6 sin t 6 sin t #16. u 1 u 4 u g t u / t u / t , u 0 0, u 0 0 a) b) Take the Laplace transformation u u g t s 1 s 4 1 U s s U s sU s e su 0 / u 0 e / / u 0 s e s s e / s 1 1/4 s s s s s U s Set e e e / e e e / s s 1 s 1 s 1 1 1/8 63/64 / / 1/4 1/4 s / / 1/8 1/8 H s Then H s 1 s 1 s 1 s s s 1/8 1/8 1/8 63/64 s s 1/8 1/8 s 1/8 63/64 1/8 37/8 1 37 s 37/8 1/8 37/8 s 1/8 Take the inverse Laplace transformation of H s H s Therefore, U s ,where e u t h t e 3 2 H s u t h t u t 5 2 h t 1 e cos 37t 8 1 37 sin 37t 8 h t 1 e cos 37t 8 1 37 sin 37t 8 c) d) From part(c) as k gets larger, the maximum value of y(t) gets larger. Also the smallest value of k, kmin, for max(y(t)) = 2 is larger than 2. Hence, to get the kmin the maximum values of y(t) are observed while k is increasing by 0.01 usin Matlab. e) From the figure 25.68 Chapter 6.5 # 4. y y 20 t y 20 t s Y s 1 Y s 3 , y 0 1, sy 0 20e 20e s s 1 cosh t s y 0 y 0 0 Take the Laplace transformation s Y s 3 Substitute these into the equation and rewrite it 20e 1 s 20 sinh t Take the inverse Laplace Transformation Y s y t 3 u t #9. y y u / t 3 t 3 2 u t , y 0 0, y 0 0 Take the Laplace transformation u / t 3 2 e s / 3e / 3 t u t e s e s e s s s 1 / s Y s e s s e Set 1 Y s 3e s 1 s 3e e / e s e 1 s s 1 3e s 1 1 e 1 s 1 3e s 1 H s 1 s s s 1 u 2 1 cos t Take the inverse Laplace transformation of H s H s Therefore, e e H s h t 1 / h t t 2 h t u t 2 u 1 t cos t 2 u t cos t Finally take the inverse Laplace transformation of Y s Y s e e 2 1 s u t s s 1 1 cos t 3e s y t 3 u 2 1 t 1 cos t 2 u 3 sin t t #14. y a) 1/2 y Take the Laplace transform t 1 s 2 s e 1/4 1 e 1 y 2 y t 1 y y t 1 , y 0 0, y 0 0 s Y s Set H S e s/2 1 Y s e 4 15 s 15/4 e 15/16 1/4 s 1 15/16 s 15/4 15/16 1/4 Take the inverse Laplace transformation of H S H s Therefore, Y s 4 15 H s e e sin 15 t 4 y t 4 15 e sin 15 t 4 1 u t b) y t reaches the maximum when dy dt 1 15 e sin 0 15 t 4 15 t 4 1 e cos 15 t 4 1 e 1 15 sin 1 cos 15 t 4 1 e 1 1 15 cos t 15 4 1 , where atan 1 15 0.2527 In order to get dy/dt = 0 15 t 4 Therefore, t c) 1/4 y Take the Laplace transform t 1 s 4 1 e 1 y 2 y t 1 2.3613 and y t 0.711531 1 2 s 1 Y s e Y s Set s e s/4 1 s e 1/8 63/64 8 37 s 37/8 e 63/64 1/4 H S s 37/8 63/64 1/8 Take the inverse Laplace transformation of H S H s Therefore, Y s 8 37 H s e y t 8 37 e sin 37 t 8 1 u t e sin 37 t 8 For the maximum value of y t dy dt 1 37 e 1 37 sin 37 t 8 37 t 8 1 e cos 37 t 8 1 1 e sin 1 cos 37 t 8 e 1 37 1 cos t 63 8 1 atan 1 317 0.1253 0, where Therefore, 37 t 8 and t d) y Take the Laplace transform t s Y s s e s e 1 s /2 1 /4 s 1 1 Y s e e y y t 2.4569 and y t 1 2 0.833508 1 1 1 /4 s 1 /2 /4 e 1 /4 Therefore, Y s y t 1 1 /4 e sin 1 /4 t 1 u t For the maximum value of y t dy dt e 2 1 /4 2 1 e 1 /4 4 1 /4 sin sin 1 1 /4 t /4 t 1 1 e cos cos 1 1 /4 t 1 1 /4 t e , where cos 1 /4 t 1 0 atan 2 1 /4 For 0, 0 then t 1 2 which gives us t 1 2 2.5708 and y t 1.00000 As decreases t1 and ymax get larger. ...
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