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ch12_ism

# ch12_ism - 12 Introduction to the Laplace Transform...

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12 Introduction to the Laplace Transform Assessment Problems AP 12.1 [a] cosh βt = e βt + e βt 2 Therefore, L{ cosh βt } = 1 2 0 [ e ( s β ) t + e ( s + β ) t ] dt = 1 2 e ( s β ) t ( s β ) 0 + e ( s + β ) t ( s + β ) 0 = 1 2 1 s β + 1 s + β = s s 2 β 2 [b] sinh βt = e βt e βt 2 Therefore, L{ sinh βt } = 1 2 0 e ( s β ) t e ( s + β ) t dt = 1 2 e ( s β ) t ( s β ) 0 1 2 e ( s + β ) t ( s + β ) 0 = 1 2 1 s β 1 s + β = β ( s 2 β 2 ) AP 12.2 [a] Let f ( t ) = te at : F ( s ) = L{ te at } = 1 ( s + a ) 2 Now, L{ tf ( t ) } = dF ( s ) ds 12–1

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12–2 CHAPTER 12. Introduction to the Laplace Transform So, L{ t · te at } = d ds 1 ( s + a ) 2 = 2 ( s + a ) 2 [b] Let f ( t ) = e at sinh βt , then L{ f ( t ) } = F ( s ) = β ( s + a ) 2 β 2 L df ( t ) dt = sF ( s ) f (0 ) = s ( β ) ( s + a ) 2 β 2 0 = βs ( s + a ) 2 β 2 [c] Let f ( t ) = cos ωt . Then F ( s ) = s ( s 2 + ω 2 ) and dF ( s ) ds = ( s 2 ω 2 ) ( s 2 + ω 2 ) 2 Therefore L{ t cos ωt } = dF ( s ) ds = s 2 ω 2 ( s 2 + ω 2 ) 2 AP 12.3 F ( s ) = 6 s 2 + 26 s + 26 ( s + 1)( s + 2)( s + 3) = K 1 s + 1 + K 2 s + 2 + K 3 s + 3 K 1 = 6 26 + 26 (1)(2) = 3; K 2 = 24 52 + 26 ( 1)(1) = 2 K 3 = 54 78 + 26 ( 2)( 1) = 1 Therefore f ( t ) = [3 e t + 2 e 2 t + e 3 t ] u ( t ) AP 12.4 F ( s ) = 7 s 2 + 63 s + 134 ( s + 3)( s + 4)( s + 5) = K 1 s + 3 + K 2 s + 4 + K 3 s + 5 K 1 = 63 189 + 134 1(2) = 4; K 2 = 112 252 + 134 ( 1)(1) = 6 K 3 = 175 315 + 134 ( 2)( 1) = 3 f ( t ) = [4 e 3 t + 6 e 4 t 3 e 5 t ] u ( t ) AP 12.5 F ( s ) = 10( s 2 + 119) ( s + 5)( s 2 + 10 s + 169) s 1 , 2 = 5 + 25 169 = 5 + j 12
Problems 12–3 F ( s ) = K 1 s + 5 + K 2 s + 5 j 12 + K 2 s + 5 + j 12 K 1 = 10(25 + 119) 25 50 + 169 = 10 K 2 = 10[( 5 + j 12) 2 + 119] ( j 12)( j 24) = j 4 . 167 = 4 . 167/90 Therefore f ( t ) = [10 e 5 t + 8 . 33 e 5 t cos(12 t + 90 )] u ( t ) = [10 e 5 t 8 . 33 e 5 t sin 12 t ] u ( t ) AP 12.6 F ( s ) = 4 s 2 + 7 s + 1 s ( s + 1) 2 = K 0 s + K 1 ( s + 1) 2 + K 2 s + 1 K 0 = 1 (1) 2 = 1; K 1 = 4 7 + 1 1 = 2 K 2 = d ds 4 s 2 + 7 s + 1 s s = 1 = s (8 s + 7) (4 s 2 + 7 s + 1) s 2 s = 1 = 1 + 2 1 = 3 Therefore f ( t ) = [1 + 2 te t + 3 e t ] u ( t ) AP 12.7 F ( s ) = 40 ( s 2 + 4 s + 5) 2 = 40 ( s + 2 j 1) 2 ( s + 2 + j 1) 2 = K 1 ( s + 2 j 1) 2 + K 2 ( s + 2 j 1) + K 1 ( s + 2 + j 1) 2 + K 2 ( s + 2 + j 1) K 1 = 40 ( j 2) 2 = 10 = 10/180 and K 1 = 10 K 2 = d ds 40 ( s + 2 + j 1) 2 s = 2+ j 1 = 80 ( j 2) 3 = j 10 = 10/ 90 K 2 = j 10

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12–4 CHAPTER 12. Introduction to the Laplace Transform Therefore f ( t ) = [20 te 2 t cos( t + 180 ) + 20 e 2 t cos( t 90 )] u ( t ) = 20 e 2 t [sin t t cos t ] u ( t ) AP 12.8 F ( s ) = 5 s 2 + 29 s + 32 ( s + 2)( s + 4) = 5 s 2 + 29 s + 32 s 2 + 6 s + 8 = 5 s + 8 ( s + 2)( s + 4) s + 8 ( s + 2)( s + 4) = K 1 s + 2 + K 2 s + 4 K 1 = 2 + 8 2 = 3; K 2 = 4 + 8 2 = 2 Therefore, F ( s ) = 5 3 s + 2 + 2 s + 4 f ( t ) = 5 δ ( t ) + [ 3 e 2 t + 2 e 4 t ] u ( t ) AP 12.9 F ( s ) = 2 s 3 + 8 s 2 + 2 s 4 s 2 + 5 s + 4 = 2 s 2 + 4( s + 1) ( s + 1)( s + 4) = 2 s 2 + 4 s + 4 f ( t ) = 2 ( t ) dt 2 δ ( t ) + 4 e 4 t u ( t ) AP 12.10 lim s →∞ sF ( s ) = lim s →∞ 7 s 3 [1 + (9 /s ) + (134 / 7 s 2 )] s 3 [1 + (3 /s )][1 + (4 /s )][1 + (5 /s )] = 7 · . . f (0 + ) = 7 lim s 0 sF ( s ) = lim s 0 7 s 3 + 63 s 2 + 134 s ( s + 3)( s + 4)( s + 5) = 0 · . . f ( ) = 0 lim s →∞ sF ( s ) = lim s →∞ s 3 [4 + (7 /s ) + (1 /s 2 )] s 3 [1 + (1 /s )] 2 = 4 · . . f (0 + ) = 4
Problems 12–5 lim s 0 sF ( s ) = lim s 0 4 s 2 + 7 s + 1 ( s + 1) 2 = 1 · . . f ( ) = 1 lim s →∞ sF ( s ) = lim s →∞ 40 s s 4 [1 + (4 /s ) + (5 /s 2 )] 2 = 0 · . . f (0 + ) = 0 lim s 0 sF ( s ) = lim s 0 40 s ( s 2 + 4 s + 5) 2 = 0 · . . f ( ) = 0

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12–6 CHAPTER 12. Introduction to the Laplace Transform Problems P 12.1
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