ch3soln - Chapter 3 Dynamic Response Problems and Solutions...

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Chapter 3 Dynamic Response Problems and Solutions for Section 3.1 1. Show that, in a partial-fraction expansion, complex conjugate poles have coe cients that are also complex conjugates. (The result of this relation- ship is that whenever complex conjugate pairs of poles are present, only oneo fthecoe cients needs to be computed.) Solution: Consider the second-order system with poles at α j β , H ( s )= 1 ( s + α + j β )( s + α j β ) P erf orm Partial Fraction Expansion: H ( s C 1 s + α + j β + C 2 s + α j β C 1 = 1 s + α j β | s = α j β = 1 2 β j C 2 = 1 s + α + j β | s = α + j β = 1 2 β j C 1 = C 2 2. Find the Laplace transform of the following time functions: (a) f ( t )=1+2 t (b) f ( t )=3+7 t + t 2 + δ ( t ) (c) f ( t e t +2 e 2 t + te 3 t (d) f ( t )=( t +1) 2 (e) f ( t )=sinh t 75
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76 CHAPTER 3. DYNAMIC RESPONSE Solution: (a) f ( t )=1+ 2 t L{ f ( t ) } = L{ 1( t ) } + L{ 2 t } = 1 s + 2 s 2 = s +2 s 2 (b) f ( t )=3+ 7 t + t 2 + δ ( t ) L{ f ( t ) } = L{ 3 } + L{ 7 t } + L{ t 2 } + L{ δ ( t ) } = 3 s + 7 s 2 + 2! s 3 +1 = s 3 +3 s 2 +7 s s 3 (c) f ( t )= e t e 2 t + te 3 t L{ f ( t ) } = L{ e t } + L{ 2 e 2 t } + L{ te 3 t } = 1 s + 2 s + 1 ( s +3) 2 (d) f ( t )=( t +1) 2 = t 2 t L{ f ( t ) } = L{ t 2 } + L{ 2 t } + L{ 1 } = 2! s 3 + 2 s 2 + 1 s = s 2 s s 3 (e) Using the trigonometric identity, f ( t )=s i n h t = e t e t 2 L{ f ( t ) } = L{ e t 2 } L{ e t 2 } = 1 2 ( 1 s 1 ) 1 2 ( 1 s ) = 1 s 2 1
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77 3. Find the Laplace transform of the following time functions: (a) f ( t )=3cos6 t (b) f ( t )=sin2 t +2cos2 t + e t sin 2 t (c) f ( t )= t 2 + e 2 t sin 3 t Solution: (a) f ( t )=3 c o s 6 t L{ f ( t ) } = L{ 3cos6 t } =3 s s 2 +36 (b) f ( t )=s i n 2 t t + e t sin 2 t = L{ f ( t ) } = L{ sin 2 t } + L{ 2cos2 t } + L{ e t sin 2 t } = 2 s 2 +4 + 2 s s 2 + 2 ( s +1) 2 (c) f ( t t 2 + e 2 t sin 3 t = L{ f ( t ) } = L{ t 2 } + L{ e 2 t sin 3 t } = 2! s 3 + 3 ( s +2) 2 +9 = 2 s 3 + 3 ( s 2 4. Find the Laplace transform of the following time functions: (a) f ( t t sin t (b) f ( t t cos 3 t (c) f ( t te t +2 t cos t (d) f ( t t sin 3 t 2 t cos t (e) f ( t )=1( t )+2 t cos 2 t Solution: (a) f ( t t sin t L{ f ( t ) } = L{ t sin t }
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78 CHAPTER 3. DYNAMIC RESPONSE Use multiplication by time Laplace transform property (Table A.1, entry #11), L{ tg ( t ) } = d ds G ( s ) Let g ( t )=s i n t and use L{ sin at } = a s 2 + a 2 L{ t sin t } = d ds ( 1 s 2 +1 2 ) = 2 s ( s 2 +1) 2 = 2 s s 4 +2 s 2 (b) f ( t )= t cos 3 t Use multiplication by time Laplace transform property (Table A.1, entry #11), L{ tg ( t ) } = d ds G ( s ) Let g ( t )=c o s 3 t and use L{ cos at } = s s 2 + a 2 L{ t cos 3 t } = d ds ( s s 2 +9 ) = [( s 2 +9) (2 s ) s ] ( s 2 2 = s 2 9 s 4 +18 s 2 +81 (c) f ( t te t t cos t Use the following Laplace transforms and properties (Table A.1, en-
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79 tries 4,11, and 3), L{ te at } = 1 ( s + a ) 2 L{ tg ( t ) } = d ds G ( s ) L{ cos at } = s s 2 + a 2 L{ f ( t ) } = L{ te t } +2 L{ t cos t } = 1 ( s +1) 2 +2( d ds s s 2 +1 ) = 1 ( s 2 2 ( s 2 (2 s ) s ( s 2 2 = 2 s 2 1 s 4 s 2 (d) f ( t )= t sin 3 t 2 t cos t Use the following Laplace transforms and properties (Table A.1, en- tries 11, 3), L{ tg ( t ) } = d ds G ( s ) L{ sin at } = a s 2 + a 2 L{ cos at } = s s 2 + a 2 L{ f ( t ) } = L{ t sin 3 t } 2 L{ t cos t } = d ds 3 s 2 +9 2( d ds s s 2 ) = (2 s 3) ( s 2 +9) 2 2 (( s 2 (2 s ) s ) ( s 2 2 = 6 s ( s 2 2 + 2( s 2 1) ( s 2 2
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80 CHAPTER 3. DYNAMIC RESPONSE (e) f ( t )=1 ( t )+2 t cos 2 t L{ 1( t ) } = 1 s L{ tg ( t ) } = d ds G ( s ) L{ cos at } = s s 2 + a 2 L{ f ( t ) } = L{ 1( t ) } +2 L{ t cos 2 t } = 1
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This note was uploaded on 03/17/2009 for the course MEEM 4700 taught by Professor Staff during the Spring '08 term at Michigan Technological University.

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ch3soln - Chapter 3 Dynamic Response Problems and Solutions...

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