Laplace transforms gives sY 1 1 4 Y 1 Y 2 2 s 1 sY 2 2 Y 1 2 Y 2 2 s 2 4

# Laplace transforms gives sy 1 1 4 y 1 y 2 2 s 1 sy 2

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Laplace transforms gives sY 1 - 1 = - 4 Y 1 - Y 2 + 2 s - 1 sY 2 - 2 = Y 1 - 2 Y 2 + 2 s 2 + 4 Equivalently, s + 4 1 - 1 s + 2 Y 1 Y 2 = 1 + 2 s - 1 2 + 2 s 2 +4 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (11/35)

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Inverse Laplace Transforms Special Functions Solving Differential Equations Laplace for Systems of DEs Laplace Transforms and Maple Example: Laplace for System 2 Example: If ( s I - A ) = s + 4 1 - 1 s + 2 , y 0 + F ( s ) = 1 + 2 s - 1 2 + 2 s 2 +4 , then the solution is Y 1 ( s ) Y 2 ( s ) = ( s I - A ) - 1 ( y 0 + F ( s )) , where ( s I - A ) - 1 = 1 ( s + 3) 2 s + 2 - 1 1 s + 4 The expressions for Y 1 ( s ) and Y 2 ( s ) are fairly complex, so we show how Maple can help solve these expressions into a form, which readily has an inverse Laplace transform Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (12/35)
Inverse Laplace Transforms Special Functions Solving Differential Equations Laplace for Systems of DEs Laplace Transforms and Maple Example: Laplace for System 3 Example: From the previous page, it is easy to see that Y 1 ( s ) = s ( s + 3) 2 + 2( s + 2) ( s - 1)( s + 3) 2 - 2 ( s 2 + 4)( s + 3) 2 Y 2 ( s ) = 2 s + 9 ( s + 3) 2 + 2 ( s - 1)( s + 3) 2 + 2( s + 4) ( s 2 + 4)( s + 3) 2 One can perform Partial Fractions Decomposition on these expressions, which is a very messy process. We demonstrate how this can be done with Maple , which can readily perform both Partial Fractions Decompositions and inverse Laplace transforms A Special Maple Sheet is provided along with a complete solution using Maple Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (13/35)

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Inverse Laplace Transforms Special Functions Solving Differential Equations Laplace for Systems of DEs Laplace Transforms and Maple Example: Laplace for System 4 Example: Maple gives the Laplace transform Y 1 ( s ) = 3 8 s - 1 + 12 s - 10 169 s 2 + 4 - 69 26 ( s + 3) 2 + 749 1352 s + 3 Y 2 ( s ) = 1 8 s - 1 + 88 - 38 s 169 s 2 + 4 + 69 26 ( s + 3) 2 + 2839 1352 s + 3 The inverse Laplace transform gives the solution y 1 ( t ) = 3 8 e t + 12 169 cos(2 t ) - 5 169 sin(2 t ) - 69 26 te - 3 t + 749 1352 e - 3 t y 2 ( t ) = 1 8 e t + 44 169 sin(2 t ) - 38 169 cos(2 t ) + 69 26 te - 3 t + 2839 1352 e - 3 t Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (14/35)
Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Discontinuous Functions Unit Step function or Heaviside function satisfies u ( t ) = 0 , t < 0 , 1 , t 0 . The translated version of the Unit Step function by c units is u c ( t ) = 0 , t < c, 1 , t c, which represents a switch turning on at t = c An indicator function , which is on for c t < d , satisfies u cd ( t ) = u c ( t ) - u d ( t ) = 0 , t < c or t d, 1 , c t < d, Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (15/35)

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Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Laplace Transform of Step Function 0 c 0 0.5 1 1.5 u c ( t ) t u c ( t ) Step Function Laplace Transform of Step Function; L [ u c ( t )] = Z 0 e - st u c ( t ) dt = Z c e - st dt = lim A →∞ Z A c e - st dt = lim A →∞ e - cs s - e - sA s = e - cs s , s > 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (16/35)
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