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Unformatted text preview: Review Given a signal f : [0 , ) R that is piece wise continuous and of polynomial growth, its Laplace Transform F ( s ) = L{ f ( t ) } ( s ) is defined as F ( s ) def = Z e st f ( t ) dt , s > . The Laplace transform is linear and one toone. Some Laplace Transforms: f ( t ) F ( s ) t n n ! /s n +1 cos t s/ ( s 2 + 1) sin t 1 / ( s 2 + 1) Cosh t s/ ( s 2 1) Sinh t 1 / ( s 2 1) e ct f ( t ) F ( s c ) f ( ct ) F ( s/c ) c f ( t ) sF ( s ) f (0) f 00 ( t ) s 2 F ( s ) sf (0) f (0) u c ( t ) f ( t c ) e cs F ( s ) c ( t ) e cs ( f * g ) ( t ) F ( s ) G ( s ) 2 If y ( t ) is the unique solution of the CCSOLIVP ay 00 + by + cy = g ( t ) , y (0) = y ,y (0) = y , then its Laplace transform Y ( s ) satisfies ( as 2 + bs + c ) Y ( s ) = ( asy + ay + by ) + G ( s ) = Y ( s ) = asy + ay + by as 2 + bs + c + G ( s ) as 2 + bs + c Hence y ( t ) = L 1 asy + ay + by as 2 + bs + c  {z } y h ( t ) + L 1 G ( s ) as 2 + bs + c  {z } y p ( t ) From this it is obvious that y h is the solu tion of the homogeneous CCIVP ay 00 + by + cy = 0 , y (0) = y ,y (0) = y . ( H ) Instead of looking up the first Laplace in verse above in the table, we might pre 3 fer to compute it as the solution of the CCIVP ( H ) . Example: The Laplace inverse of 1 as 2 + bs + c is the solution of the IVP ay 00 + by + cy = 0 , y (0) = 0 ,y (0) = 1 /a . 4 Step Functions Definition: The Unit Step Function at c is the function u c ( t ) = for t < c 1 for c t < . Given a signal f ( t ) , its version shifted by c is the function u c ( t ) f ( t c ) for t < c f ( t c ) for c t < . u c ( t ) 1 f ( t ) u c ( t ) f ( t c ) c t 5 Theorem: L{ u c ( t ) f ( t c ) } ( s ) = e cs L{ f ( t ) } ( s ) and thus L 1 { e cs F ( s ) } ( t ) = u c ( t ) L 1 { F ( s ) } (...
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This note was uploaded on 09/07/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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