Joseph M Mahaffy h mahaffymathsdsuedu i Lecture Notes Laplace Transforms Part A

Joseph m mahaffy h mahaffymathsdsuedu i lecture notes

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Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (15/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Short Table of Laplace Transforms Short Table of Laplace Transforms: Below is a short table of Laplace transforms for some elementary functions f ( t ) = L - 1 [ F ( s )] F ( s ) = L [ f ( t )] 1 1 s , s > 0 e at 1 s - a , s > a t n , integer n > 0 n ! s n +1 , s > 0 t p , p > - 1 Γ( p +1) s p +1 , s > 0 sin( at ) a s 2 + a 2 , s > 0 cos( at ) s s 2 + a 2 , s > 0 sinh( at ) a s 2 - a 2 , s > | a | cosh( at ) s s 2 - a 2 , s > | a | Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (16/26)
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Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform - e ct f ( t ) Laplace Transform - e ct f ( t ) : Previously found Laplace transforms of several basic functions Theorem (Exponential Shift Theorem) If F ( s ) = L [ f ( t )] exists for s > a , and if c is a constant, then L [ e ct f ( t )] = F ( s - c ) , s > a + c. Proof: This result immediately follows from the definition: L [ e ct f ( t )] = Z 0 e - st e ct f ( t ) dt = Z 0 e - ( s - c ) t f ( t ) dt = F ( s - c ) , which holds for s - c > a . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (17/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Example Example: Consider the function g ( t ) = e - 2 t cos(3 t ) . From our Table of Laplace Transforms , if f ( t ) = cos(3 t ), then F ( s ) = s s 2 + 9 , s > 0 . From our previous theorem, the Laplace transform of g ( t ) satisfies: G ( s ) = L [ e - 2 t f ( t )] = F ( s + 2) = s + 2 ( s + 2) 2 + 9 , s > - 2 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (18/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives 1 Theorem (Laplace Transform of Derivatives) Suppose that f is continuous and f 0 is piecewise continuous on any interval 0 t A . Suppose that f and f 0 are of exponential order with | f ( i ) ( t ) | ≤ Ke at | for some constants K and a and i = 0 , 1 . Then L [ f 0 ( t )] exists for s > a , and moreover L [ f 0 ( t )] = s L [ f ( t )] - f (0) . Sketch of Proof: If f 0 ( t ) was continuous, then examine Z A 0 e - st f 0 ( t ) dt = e - st f ( t ) A 0 + s Z A 0 e - st f ( t ) dt = e - sA f ( A ) - f (0) + s Z A 0 e - st f ( t ) dt, which simply uses integration by parts. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (19/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives 2 Sketch of Proof (cont): From before we have Z A 0 e - st f 0 ( t ) dt = e - sA f ( A ) - f (0) + s Z A 0 e - st f ( t ) dt. As A → ∞ and using the exponential order of f and f 0 , this expression gives L [ f 0 ( t )] = s L [ f ( t )] - f (0) . To complete the general proof with f 0 ( t ) being piecewise continuous, we divide the integral into subintervals where f 0 ( t ) is continuous.
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