Solution_to_problem_3.8 - Solution to problem 3.8 a The...

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3.8 – 1 Solution to problem 3.8 a) The Laplace transform of - = + + t d e x x x 0 2 2 3 τ can be obtained as follows: For the right hand side of the above expression, we know that ( ) ( ) s F s dt t f L t 1 0 * * = And thus we can write that: ( ) ( ) 1 1 1 0 + = = - - s s e L s d e L t For the left hand side of the above expression, we apply the superposition principle (and we assume that we have a system in steady-state, i.e. x(0)=x’(0)=0): ( ) ( ) ( ) ) ( 2 ) ( 3 2 3 2 3 2 2 2 s X s X s s X s x L dt dx L dt x d L x x x L + + = + + = + + Combining the Laplace transforms for left and right hand side we now obtain: ( ) ( ) 1 2 ) ( 2 ) ( 3 2 + = + + s s s X s X s s X s ( ) ( ) 2 1 2 ) ( 2 + + = s s s s X This expression can now be expanded in partial fractions: ( ) ( ) ( ) 2 1 1 2 1 2 ) ( 4 2 3 2 1 2 + + + + + + = + + = s s s s s s s s X α Applying Heaviside expansion we obtain values for α 1 , α 3 and α 4 . Multiplication by s and putting s equal to 0 gives: ( ) ( ) 1 2 1 2 0 2 1 = + + = = s s s Multiplication by (s+1) 2 and putting s equal to -1 gives: ( ) 2 1 2 2 2 1 3 - = - = + = - = s s s Multiplication by (s+2) and putting s equal to -2 gives:
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Solution_to_problem_3.8 - Solution to problem 3.8 a The...

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