Lecture Notes Laplace Transforms Part B 2735 Inverse Laplace Transforms Special

# Lecture notes laplace transforms part b 2735 inverse

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Lecture Notes – Laplace Transforms: Part B — (27/35)

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Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function IVP with Periodic Forcing Function 3 Example: So Y ( s ) = 1 4 1 s - s s 2 + 4 X k =0 ( - 1) k e - ks Taking the inverse Laplace transform gives: y ( t ) = 1 4 (1 - cos(2 t )) + 1 4 X k =1 ( - 1) k u k ( t ) (1 - cos(2( t - k ))) 0 1 2 3 4 5 6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 f ( t ) y ( t ) t y ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (28/35)
Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Impulse Function Impulse Function: Some applications have phenomena of an impulsive nature , e.g. , a large magnitude force over a very short time ay 00 + by 0 + cy = g ( t ) , where g ( t ) is very large for t [ t 0 , t 0 + ε ) and is otherwise zero Example: Let t 0 = 0 be a real number and ε be a small positive constant Suppose t 0 = 0 and g ( t ) = I 0 δ ε ( t ), where δ ε ( t ) = u 0 ( t ) - u ε ( t ) ε = 1 ε , 0 t < ε, 0 , t < 0 or t ε. Consider the mass-spring system m = 1, γ = 0, and k = 1 y 00 + y = I 0 δ ε ( t ) , y (0) = 0 , y 0 (0) = 0 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (29/35)

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Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Impulse Function - Example Example: With δ ε ( t ) = u 0 ( t ) - u ε ε , the Laplace transform is easy for y 00 + y = I 0 δ ε ( t ) , y (0) = 0 , y 0 (0) = 0 . It satisfies ( s 2 + 1) Y ( s ) = I 0 ε 1 - e - εs s , so Y ( s ) = I 0 ε 1 s - s s 2 + 1 ( 1 - e - εs ) The inverse Laplace transform gives y ε ( t ) = I 0 ε ( u 0 ( t )(1 - cos( t )) - u ε ( t )(1 - cos( t - ε ))) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (30/35)
Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Impulse Function - Example Example: Since y ε ( t ) = I 0 ε ( u 0 ( t )(1 - cos( t )) - u ε ( t )(1 - cos( t - ε ))) , equivalently: y ε ( t ) = 0 , t < 0 , I 0 ε (1 - cos( t )) 0 t < ε, I 0 ε (cos( t - ε ) - cos( t )) t ε. The limiting case is y 0 ( t ) = lim ε 0 y ε ( t ) = u 0 ( t ) I 0 sin( t ) = 0 , t < 0 , I 0 sin( t ) , t 0 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (31/35)

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Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Unit Impulse Function Unit Impulse Function: Rather than using the definition of δ ε ( t - t 0 ) to model an impulse, then take the limit as ε 0, we define an idealized unit Impulse Function, δ The “function” δ imparts an impulse of magnitude 1 at t = t 0 , but is zero for all other values of t Properties of δ ( t - t 0 ) 1 Limiting behavior: δ ( t - t 0 ) = lim ε 0 δ ε ( t - t 0 ) = 0 2 If f is continuous for t [ a, b ) and t 0 [ a, b ), then Z b a f ( t ) δ ( t - t 0 ) dt = lim ε 0 Z b a f ( t ) δ ε ( t - t 0 ) dt = f ( t 0 ) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (32/35)
Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function δ ( t - t 0 ) Dirac delta function, δ ( t - t 0 ): This is not an ordinary function in elementary calculus, and it satisfies: Z b a

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