springps08_soln

springps08_soln - 18.03 PS8 SolutionsSpring 09 1a. L ( e at...

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Unformatted text preview: 18.03 PS8 SolutionsSpring 09 1a. L ( e at f ( t )) = integraldisplay e st e at f ( t ) dt = integraldisplay e ( s a ) t f ( t ) dt = F ( s- a ) 1b. L ( u ( t- a ) f ( t- a )) = integraldisplay e st u ( t- a ) f ( t- a ) dt = integraldisplay a e s ( + a ) u ( ) f ( ) d = e as integraldisplay e s f ( ) d = e as F ( s ) 1c. Let g ( t ) = t + 1, so L ( g ( t )) = L ( t ) + L (1) = 1 s 2 + 1 s = s + 1 s 2 . Define G ( s ) = s + 1 s 2 . By (a), F ( s ) = L ( f ( t )) = L (( t + 1) e t +1 ) = e L ( e t ( t + 1)) = eG ( s- 1) = es ( s- 1) 2 . By (b), L ( u ( t- 1) te t ) = L ( u ( t- 1) f ( t- 1)) = e s F ( s ) = e 1 s s ( s- 1) 2 . 1d. L ( f ( at )) = integraldisplay e st f ( at ) dt = integraldisplay e s ( /a ) f ( ) 1 a d = 1 a integraldisplay e ( s/a ) f ( ) d = 1 a F parenleftBig s a parenrightBig 2. Let Y ( s ) = L ( y ( t )). Then L ( y ) = sY ( s )- y (0) = sY...
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.

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springps08_soln - 18.03 PS8 SolutionsSpring 09 1a. L ( e at...

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