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all the material for fall 2011

all the material for fall 2011 - EEL3105 Fall2011 FinalExam...

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EEL 3105 Fall 2011 Final Exam 1. Consider the differential equation 2 2 0. d y dy a by dt dt Suppose ( ) sin(3 ) t y t e t is a solution to this differential equation. Find a and b. [4 pts] Answer: Based on the form of y(t), 1 3 j   are roots of the polynomial corresponding to the differential equation: 2 . a b Thus, 2 2 ( 1 3)( 1 3) 2 10. a b j j Therefore, a=2 and b=10. An alternative approach would be to plug the expression for y(t) into the differential equation: 2 2 sin(3 ) 3 cos(3 ) sin(3 ) 6 cos(3 ) 9 sin(3 ) , [ 8sin3 6cos(3 ) sin(3 ) 3 cos(3 ) sin(3 )] 0 8 0 6 3 0 2, 10. t t t t t t dy e t e t dt d y e t e t e t dt Thus e t t a t a t b t a b a a b     2. Find Laplace transform of ( ) ( ) sin 5 , 0. 3 f t tU t t t Here U(t) is the unit step function. [4 pts] Answer: We will use superposition. Laplace transform of the first term is 2 1 ( ( )) . LT tU t s
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