me375_sp11_hwk03 - m , c and k . The following initial...

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ME 375 – Spring 2011 Homework No. 3 Due: Friday, February 4 Problem No. 1 (30%) a) For u t ( ) shown below, determine the Laplace transform L u t ( ) . b) If Y s ( ) = s + 6 5 s 2 + 4 s is the Laplace transform of y t ( ) : i. what is y t = 0 ( ) ? ii. what is y t →∞ ( ) ? c) If Y s ( ) = s + 6 5 s 2 + 4 s is the Laplace transform of y t ( ) , what is the Laplace transform of e 2 t y t ( ) ? 2 2
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Problem No. 2 (40%) a) Determine the solution x t ( ) for the following differential equation with the initial conditions of x 0 ( ) = 2 : 10 x + 5 x = 20 h t ( ) , where h t ( ) is the unit step function. b) Determine the solution x t ( ) for the following differential equation with the initial conditions of x 0 ( ) = 0 and x 0 ( ) = 2 :  x + 12 x + 100 x = 0 c) The Laplace transform of a response x t ( ) is known to be: X s ( ) = s 2 + 4 s s 2 + 9 ( ) . Determine the response x t ( ) .
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Problem No. 3 (30%) The system has the following equations of motion: y 1 + 2 y 1 2 y 2 = 0  y 2 2 y 1 + 2 y 2 = 0 for selected values of
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Unformatted text preview: m , c and k . The following initial conditions are known for the system: y 1 ( ) = y 2 ( ) = and y 2 ( ) = 1 . a) Determine the expression for Y 2 s ( ) , where Y 2 s ( ) is the Laplace transform of the response y 2 t ( ) . Write your final answer for Y 2 s ( ) as a single fraction Y 2 s ( ) = N s ( ) D s ( ) . b) Determine the three poles p 1 , p 2 and p 3 of Y 2 s ( ) . c) Recall that if the poles of Y 2 s ( ) are distinct (not repeated), the partial fraction of Y 2 s ( ) can be written as: Y 2 s ( ) = A 1 s p 1 + A 2 s p 2 + A 3 s p 3 . Based on the poles found above in b), describe the qualitative nature of the time response y 2 t ( ) . You do NOT need to solve for the coefficients A 1 , A 2 and A 3 in order to provide this description....
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue University-West Lafayette.

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me375_sp11_hwk03 - m , c and k . The following initial...

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