1635 Inverse Laplace Transforms Special Functions Heaviside or Step function

# 1635 inverse laplace transforms special functions

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Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Laplace Transform of Step Function Theorem If F ( s ) = L [ f ( t )] exists for s > a 0 , and if c is a nonnegative constant, then L [ u c ( t ) f ( t - c )] = e - cs L [ f ( t )] = e - cs F ( s ) , s > a. Conversely, if f ( t ) = L - 1 [ F ( s )] , then u c ( t ) f ( t - c ) = L - 1 [ e - cs F ( s )] . This theorem states that the translation of f ( t ) a distance c in the positive t direction corresponds to the multiplication of F ( s ) by e - cs Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (17/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Example with Step Function 1 Example with Step Function: Consider the following initial value problem : y 00 + 2 y 0 + 5 y = u 2 ( t ) - u 5 ( t ) , y (0) = 0 , y 0 (0) = 0 . Take Laplace transforms and obtain ( s 2 + 2 s + 5) Y ( s ) = e - 2 s s - e - 5 s s This rearranges to Y ( s ) = e - 2 s - e - 5 s s ( s 2 + 2 s + 5) Partial fraction decomposition gives 1 s ( s 2 + 2 s + 5) = A s + B ( s + 1) + 2 C ( s + 1) 2 + 4 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (18/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Example with Step Function 2 Example with Step Function: Partial fraction decomposition gives 1 = A ( ( s + 1) 2 + 4 ) + ( B ( s + 1) + 2 C ) s With s = 0, 1 = 5 A or A = 1 5 The s 2 coefficient gives 0 = A + B , so B = - 1 5 The s 1 coefficient gives 0 = 2 A + B + 2 C , so C = - 1 10 Thus, Y ( s ) = 1 5 s - 1 5 ( s + 1) + 2 10 ( s + 1) 2 + 4 ( e - 2 s - e - 5 s ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (19/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Example with Step Function 3 Example with Step Function: With the Laplace transform Y ( s ) = 1 5 s - 1 5 ( s + 1) + 2 10 ( s + 1) 2 + 4 ! ( e - 2 s - e - 5 s ) The theorem for step functions allows the inverse Laplace transform yielding y ( t ) = u 2 ( t ) 10 2 - 2 e - ( t - 2) cos(2( t - 2)) - e - ( t - 2) sin(2( t - 2)) - u 5 ( t ) 10 2 - 2 e - ( t - 5) cos(2( t - 5)) - e - ( t - 5) sin(2( t - 5)) 0 1 2 3 4 5 6 7 8 9 10 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 t y ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (20/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Periodic Functions Definition A function f is said to be periodic with period T > 0 if f ( t + T ) = f ( t ) for all t in the domain of f . 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 t y ( t ) A sawtooth waveform f ( t ) = t, 0 t < 1 , 0 , 1 t < 2 . and f ( t ) has period 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (21/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Periodic and Window functions Consider a periodic function f ( t ). Define the window function , f T ( t ), as follows: f T ( t ) = f ( t ) [1 - u T ( t )] = f ( t ) , 0 t T 0 , otherwise The Laplace transform F T ( s ) satisfies: F T ( s ) = Z 0 e - st f T ( t ) dt = Z T 0 e - st f T ( t ) dt. Subscribe to view the full document. • Fall '08
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