chapter 14

# chapter 14 - Chapter 14 The Laplace Transform Exercises Ex...

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Chapter 14: The Laplace Transform Exercises Ex. 14.3-1 () 1 cos and 2 jt at ee ft t e sa ωω ω +− +  == =  L 22 11 1 [cos ] 2 s Fs t sj s ==+ =  ++ ±L Ex. 14.3-2 2 2 2 3 ( ) [ sin ] [sin ] 21(2 ) ( 1 tt ss e t e t −− =+= + =+ = ++ ++ LL L ) Ex. 14.4-1 44 23 ( ) [2 ( ) 3 ( )] 2 [ ( )] 3 [ ( )] 4 ut e ut =+ = + = + + L Ex. 14.4-2 2 1 ( ) [sin( 2) ( 2)] [sin ] 1 t e t e s  =− = =  +  Ex. 14.4-3 1 1 ( )[][ ] (1 ) t te t →+ == = = + Ex. 14.4-4 ( 55 54 . 2 33 t t =− + ) 4 . 2 4.2 4.2 15 5 1 5 3 s s se e s s + = 2 14-1

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Ex. 14.4-5 2 2 2 00 0 3 3(1 ) () 3 st s st st e e Fs fte d t e d t s s −− == = = ∫∫ Ex. 14.4-6 52 0 2 0o t h e r w i tt ft << = s e () ( ) ( ) () () 22 2 55 5 5 2 2 2 2 2 ( 2 2 2 51 2 12 ss tu t u t tut t u t t u t u t ee e s e s s =−   ) =  L Ex. 14.5-1 [] ( ) () ( 1 ,t a n cos sin cos cos jj at at at c jd c jd me me d w h e r em c d c s a js a a a j ) f t e c t d e c d t ut me t θθ θ ωω ω ωθ +− =+=+ = + = ++ =− = + + = + Ex. 14.5-2 (a) () ( ) () 2 2 1 2 28 3 83 1 41 32 29 38 2 2, 8, 3 & 3 6.33 3 6.33 tan 38.4 , 8 6.33 10.2 8 10.2 cos 3 38.40 t s s s ac c ad d m e t −° × == = −= −⇒ = =  = + =   ⇒= + 14-2
(b) () ( ) () () () () ( ) 1 2 22 1 11 1 23 33 1 Given , first consider . 21 7 7 2 6 Identify 1, 0, 4 3 3 4. Then | | 3 4, tan 3 4 0 90 So ( ) (3 4) sin 4 . Next, 1 . Finally s ts e Fs ss s a c and d d m d ft e tut eFs ωω θ −° −− == = × ++ = = = = ⇒ =− = = = = ( ) ( ) (1 ) (3 4) sin 4 1 1 t f te t u t   Ex. 14.5-3 (a) 2 5 1 sA B C s + + + + + where ()( ) 2 01 51 | 5 and 1 | 4 As C s 5 = −= + = = 6 B Multiply both sides by 2 1 + () () 2 2 551 14 B s s s − + + = Then 2 56 4 1 1 s =+ + + + Finally ( ) ( ) ( ) 4 tt f t e =−+ + u t (b) 2 32 4 3 sABC s ==++ + 3 3 s + Where ( ) ( ) 2 2 1 34 , 3 2 dd F sB s F s ds ds =+ = = + = 2 4 6 and ) 3 3 s Cs F s = += Then 42 4 3 6 3 s + Finally ( ) ( ) ( ) 4 1 8 t f t e =− + u t 14-3

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Ex. 14.6-1 (a) () 2 65 21 s Fs ss + = ++ () () ( ) 2 2 0l i m l i m lim lim 0 00 fs F s F s +  ==  →∞ →∞  + ∞= = = →→ 6 = (b) 2 6 = −+ 2 2 0 6 i m 0 6 undefined no final value s s s f s f →∞ = Ex. 14.7-1 KCL: 1 6 7 5 t v ie += KVL: 11 43 0 4 di di iv v i dt dt 3 + −= ⇒ = + Then 66 35 72 54 tt di i di dt i e dt + + = Taking the Laplace transform of the differential equation: 35 1 35 1 (0 ) 2() 46 4 (2 ) (6 sI s i I s = ⇒ = ) s + Where we have used . Next, we perform partial fraction expansion. (0) 0 i = 2 1 where and ) ) 2 6 6 4 2 4 AB s s s s =− = − =+ = = = = + + + + 6 1 1 Then 26 35 35 35 1 35 1 16 2 16 6 16 16 I si t e e 14-4
Ex. 14.7-2 Apply KCL at node a to get 12 1 1 1 22 48 24 dv v v vv dt dt =⇒ + = Apply KCL at node b to get 1 2 2 2 50cos2 1 03 6 0 20 24 30 24 vt v v v d v d v dt dt −− ++ + = + + = c o s 2 t Take the Laplace transforms of these equations, using (0) 10 V and (0) 25 V = = , to get () () 2 1 2 2 25 60 100 2 () 2 () 10 and () 3 4 ss sVs V s Vs sV s s +−= + += + Solve these equations using Cramer’s rule to get ( ) ( ) 2 2 2 2 32 2 25 60 100 21 0 2 25 60 100 10 4 4 2( 3 ) 2 414 25 120 220 240 s ss s s s s sss s   + ++ + + +  == +++ = Next, partial fraction expansion gives * 2 V 1 4 A ABC s sj s s =++ + + −+ + where 2 * 2 1 2 4 25 120 220 240 240 240 66 14 2 4 0 25 120 220 240 115 23 15 3 44 25 120 220 240 320 16 60 3 41 s s j Aj j B C =− ++ − = = = + Then 14-5

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() 2 66662 3 31 6 3 221 jj Vs sj s s 4 + =+++ + −+ + Finally 4 2 23 16 12cos2 12s in2 V 0 33 tt vt t t e e t
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## This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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chapter 14 - Chapter 14 The Laplace Transform Exercises Ex...

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