Ch_02 - Laplace Transform - Problems

# Ch_02 - Laplace Transform - Problems - Problems for Laplace...

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Problems for Laplace Transform

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Laplace Transform Fundamentals Fundamentals 1. Example of the definition of Laplace transform Solve following integral using Laplace transform. (A) 3 0 sin 2 t et d t (B) 2 0 cos 2 t te t dt (C) 2 0 sin t tdt Solution: (A) Use the definition of the Laplace transform: {} 3 3 0 sin 2 sin 2 t s d t t = = L Let () s in2 f tt = , then 2 2 ˆ () 4 fs s = + , therefore 3 2 3 0 3 22 sin 2 sin 2 13 4 t s s d t t s = = == = + L (B) Let 2 t τ = , then 2 dt d = , so 2 1 00 11 cos2 cos cos 44 t s e d ττ ∞∞ −− = ∫∫ L Let () c o s f = , then 2 ˆ 1 s s = + 2 1 0 1 1 1 1 2 ˆ ' cos 2 ( ) ( ) 0 4 ( 1 ) t s s s ss f f s s = = = +− = = + L (C) 2 2 0 sin sin t s te t dt t t = = L Let s in f = , then 2 1 1 s = + 2 2 0 2 2 24 ˆ ' sin ( ) ( ) 25 (1 ) t s s s s te t dt t f t f s s = = = = = + L
Laplace Transform Fundamentals 2. Examples of shifting theorem (A) Find the Laplace transform of the following function () f t . ( 2 ) ft tut = (B) Find the Laplace transform of the following function f t . 2 ( 2 ) = where () ut is the unit step function. Solution: (A) Use the property of {( ) ( ) } {( ) } as ftu t a e ft a −= + LL { } { } {} 2 2 2 2 ( 2 ) 2 2 12 s s s et e s s =− =+ ⎛⎞ ⎜⎟ ⎝⎠ L L (B) ) ( ) } ) } as e + { } 2 22 2 32 ( 2 ) (2 ) 44 244 s s s t ut t e s ss = ++ + L L

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Laplace Transform Fundamentals 3. Examples of change of scaling property (A) Given {} 2 cos 1 s t s = + L , determine { } cos bt L (B) Given 2 1 sin 1 t s = + L , determine { } sin bt L Solution: Use the formula of 1 () ( ) s s a fa t ft a = LL (A) () cos f tt = 2 2 2 22 1/ 1 cos (/) 1 s bb s s bt b s b bsb sb =⋅ = ⋅ = + ++ L (B) Since () s in f = 2 2 2 11 1 sin bt b s s b s b = + L 4. Examples of derivative and integral property (A) Given 2 cos 1 s t s = + L , determine { } sin t L (B) Given 2 1 sin 1 t s = + L , determine { } cos t L Solution: Use the formula of { } ' (0 ) f ts f tf =− (A) f = , then ' s f { } 2 2 sin cos cos(0) 1 1 1 sin 1 ss s t s s t s −− −= = ⋅− = = + = + L (B) f = , then ' f = 1 cos sin sin(0) 0 s t s = = + +
Laplace Transform Fundamentals 5. Examples of derivative and integral property Evaluate the following by using the derivatives of transform , { } sin tt ω L Solution: Use the formula {} 2 ˆ () () (0 ) ) ft s fs s f f ′′ =− L Let s in f t = , then 2 2 s in cos 2 cos s 2c o s ( ) t t t f t t tf t ωω ωωω = + and (0) (0) 0 ff == . { } { } { } 22 2 2 s o ( ) o s s i n 2 f ts f t t t t s s = + LL L L Hence 2 2 sin s s = + L General Method ˆ ( 1 ) n nn n d tft fs ds L 2 ˆ sin ( 1) ( ) 2 d f s ds ds ds ss ⎡⎤ = ⎢⎥ ++ ⎣⎦ L

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Laplace Transform Fundamentals 6. Examples multiplication and division property Find the Laplace transform of the following functions. (A) 3 () cos2 t zt te t = (B) 3 s in2 t t = Solution: (A) Let () cos2 g tt = , then, {} 2 co s2 4 s gs t s == + L By the formulas of {( ) } ( 1 ) ( ) n nn n d tft fs ds =− L , apply 1 n = , 222 22 2 2 2 42 4 cos 2 4( 4 ) ( 4 ) ds s s s ds s s s ⎛⎞ +− ⎡⎤ = ⎜⎟ ⎢⎥ ++ + ⎣⎦ ⎝⎠ L by the shifting theorem, ˆ ( ) at ef t f sa = + L ( 3 a = ), therefore 3 2 2 (3 )4 65 ˆ cos 2 ( ) [( 3) 4] ( 6 13) t ss s te t f s a s + + =+ = = + + L (B) Let () s g = , then, 2 2 s 4 t s + L By the formulas of ) } ( 1 ) ( ) n n d ds L , apply 1 n = , 2 24 sin 2 4 ) ds s s = L by the shifting theorem, ˆ ( ) at t f = + L ( 3 a = ), therefore 3 2 2 4( 3) 4 12 ˆ s [( 3) 4] ( 6 13) t t e t s === + + L
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## Ch_02 - Laplace Transform - Problems - Problems for Laplace...

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