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. dy d y ab y dt dt  Suppose () s in (3) t y te t is a solution to this differential equation. Find a and b. [4 pts] Answer: Based on the form of y(t), 13 j  are roots of the polynomial corresponding to the differential equation: 2 .   Thus, 22 (13 ) ) 21 0 . 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 ) , [ 8sin 3 6cos(3 ) sin(3 ) 3 cos(3 ) sin(3 )] 0 80 63 0 2, 10. tt t t dy et e t dt e t dt Thus t a t a t b t a         2. Find Laplace transform of () s in 5 , 0 . 3 ft tUt 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 We can use basic trigonometry to write
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22 2 sin 5 sin(5 )cos( / 3) cos(5 )sin( / 3) 3 31 sin(5 ) cos(5 ).
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all the material for fall 2011 - EEL3105 Fall2011 FinalExam...

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