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 + e 2 Therefore, L{ cosh } = 1 2 Z 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 = e e 2 Therefore, L{ sinh } = 1 2 Z 0 h e ( s β ) t e ( s + β ) t i 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 } = dF ( s ) ds = s 2 ω 2 ( s 2 + ω 2 ) 2 AP 12.3 F ( s 6 s 2 +26 s ( 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 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 + 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 = j 12
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Problems 12–3 F ( s )= K 1 s +5 + K 2 s 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 2 + K 2 s K 0 = 1 (1) 2 =1 ; K 1 = 4 7+1 1 =2 K 2 = d ds " 4 s 2 s s # s = 1 = s (8 s +7) (4 s 2 s 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 j 1) 2 + K 2 ( s j 1) + K 1 ( s j 1) 2 + K 2 ( s j 1) K 1 = 40 ( j 2) 2 = 10 = 10/180 and K 1 = 10 K 2 = d ds " 40 ( s 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 s s 2 +6 s +8 =5 s ( s + 2)( s s ( s + 2)( s = 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 s f ( t δ ( t )+[ 3 e 2 t e 4 t ] u ( t ) AP 12.9 F ( s 2 s 3 s 2 s 4 s 2 +5 s =2 s 2+ 4( s +1) ( s + 1)( s s 4 s 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 · ( )=0 lim s →∞ sF ( s ) = lim s →∞ " s 3 [4+(7 /s )+(1 /s 2 )] s 3 [1+(1 /s )] 2 # =4 · (0 + )=4
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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 · (0 + )=0 lim s 0 sF ( s ) = lim s 0 " 40 s ( s 2 +4 s +5) 2 # · (
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12–6 CHAPTER 12. Introduction to the Laplace Transform Problems P 12.1 [a] f ( t )=5 t [ u ( t ) u ( t 2)] + 10[ u ( t 2) u ( t 6)]+ (
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ch12_ism - 12 Introduction to the Laplace Transform...

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