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ECE108HW%233Sol_06

# ECE108HW%233Sol_06 - ECE 108 HW#3 Due 1 For a LTI system...

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ECE 108 HW#3 --- Due 2/13/06 1. For a LTI system described by the differential equation ( ) ( ) ( ) ( ) 2 2 3 5 2 4 32 d y t dy t y t x t dt dt + + = where ( ) ( ) x t tu t = (a) Determine the Characteristic Equation 2 3 5 0 4 32 p p + + = (b) Determine the Homogeneous Solution Roots of Characteristic Equation 1.2 3 1 9 20 3 1 8 2 16 32 8 8 p j =− ± =− ± Then the most general solution to the Homogeneous Equation is ( ) 1 2 3 1 3 1 8 8 8 8 1 2 1 2 j t j t p t b t y t Ae A e Ae A e + = + = + (c) Using the Method Of Undetermined Coefficients, determine the Particular Solution Assume we want to find the solution for t > 0, then over that range of values ( ) ( ) and 1 dx t x t t dt = = Further derivatives produce no new independent functions and so the assumed solution should be ( ) 0 1 p y t B Bt = + Substituting this assumed value for y ( t ) in the differential equation along with the value for x ( t ) we obtain the equation ( ) 1 0 1 3 5 2 4 32 B B Bt t + + = Then 1 0 1 3 5 5 2 0 4 32 32 B B B t + + = This is only possible for all values of t if 1 0 1 3 5 0 4 32 5 2 0 32 B B B + = =

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2 Solving these two equations for the constants 1 2 64 5 1536 25 B B = =−
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ECE108HW%233Sol_06 - ECE 108 HW#3 Due 1 For a LTI system...

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