ECE634S09_L8_DifferentialEquations

ECE634S09_L8_DifferentialEquations - ECE634 Signals and...

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ECE634 Signals and Systems II, Spring 2009 - Lecture 8, February 9 Daniel S. Brogan 1 4.3 Solution of Differential and Integro-Differential Equations: Example 4.10 (pp.371-372) (Linear differential equation with constant coefficients) Solve the second-order linear differential equation ( ) ( ) ( ) ( ) 2 5 6 1 D D y t D x t + + = + for the initial conditions ( ) 0 2 y = and ( ) 0 1 y = ± and the input ( ) ( ) 4 t x t e u t = . The equation is ( ) ( ) 2 2 5 6 d y dy dx y t x t dt dt dt + + = + If we let ( ) ( ) y t Y s , then using the time-differentiation property, ( ) ( ) ( ) 0 2 dy sY s y sY s dt = ( ) ( ) ( ) ( ) 2 2 2 2 0 0 2 1 d y s Y s sy y s Y s s dt = ± For the input ( ) ( ) 4 t x t e u t = , ( ) 1 4 X s s = + and ( ) ( ) 0 0 4 4 dx s s sX s x dt s s = = + + We could also take the derivative in the time domain and then take the Laplace transform… ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 4 4 4 4 4 4 4 t t t t t t d e u t d e d u t dx t u t e e u t e t e u t t dt dt dt dt δ δ = = + = − + = − + 1 4 1 4 4 dx s dt s s ⇔ − + = + + Note that this method required more work. Using the Laplace transform on the differential equation ( ) ( ) ( ) ( ) ( ) 2 2 2 1 5 6 2 1 5 2 6 4 4 d y dy dx s y t x t s Y s s sY s Y s dt dt dt s s + + = + + + = + + + Now for some algebra…
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