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# L17 - Review Given a signal f[0 R that is piecewise...

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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 0 e - st f ( t ) dt , s > 0 . The Laplace transform is linear and one– to–one.

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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 0 ( t ) sF ( s ) - f (0) f 00 ( t ) s 2 F ( s ) - sf (0) - f 0 (0) u c ( t ) f ( t - c ) e - cs F ( s ) δ 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 0 + cy = g ( t ) , y (0) = y 0 , y 0 (0) = y 0 0 , then its Laplace transform Y ( s ) satisfies ( as 2 + bs + c ) Y ( s ) = ( asy 0 + ay 0 0 + by 0 ) + G ( s ) = Y ( s ) = asy 0 + ay 0 0 + by 0 as 2 + bs + c + G ( s ) as 2 + bs + c Hence y ( t ) = L - 1 asy 0 + ay 0 0 + by 0 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 0 + cy = 0 , y (0) = y 0 , y 0 (0) = y 0 0 . ( H ) Instead of looking up the first Laplace in- verse above in the table, we might pre- 3

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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 0 + cy = 0 , y (0) = 0 , y 0 (0) = 1 /a . 4
Step Functions Definition: The Unit Step Function at c is the function u c ( t ) = 0 for 0 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 ) 0 for 0 t < c f ( t - c ) for c t < . u c ( t ) 1 f ( t ) u c ( t ) f ( t - c ) c t 5

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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 ) } ( t - c ) . Proof: L{ u c ( t ) f ( t - c ) } ( s ) = R 0 e - st u c ( t ) f ( t - c ) dt = R c e - st u c ( t ) f ( t - c ) dt = R c e - st f ( t - c ) dt with τ = t - c : = R 0 e - s ( τ + c ) f ( τ ) = e - sc R 0 e - f ( τ ) = e - sc L{ f ( t ) } ( s ) .
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L17 - Review Given a signal f[0 R that is piecewise...

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