Review Problem 7 solution

Review Problem 7 solution - ( ) ( )( ) 1 2 9 3 2 5 X s s s...

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7. Utilize the “final value theorem” to determine the final value of ( ) x t given the conditions of problem 6. (The solution should be the same as that obtained by taking the limit of the solution in problem 6 as t goes to infinity.) The final value theorem states that ( ) 0 lim ( ) lim t s y t sY s →∞ = where ( ) Y s in our case is ( ) X s . From Problem 1, ( ) ( ) ( ) 0 0 2 2 2 2 5 2 5 U s s x x X s s s s s + + = + + + + + d . From superposition, like in Problem 6, ( ) ( ) ( ) 1 2 X s X s X s = + where ( ) ( ) 1 2 2 5 U s X s s s = + + and ( ) ( ) 0 0 2 2 2 2 5 s x x X s s s + + = + + d . From Problem 2, where the input is zero and the initial conditions are the same as those of Problem 7, ( ) ( ) 2 2 2 .2 3 1 .6 .2 2 5 2 5 s s X s s s s s + - - = = - + + + + . And from Problem 5, where the initial conditions are both zero and the input is like that of Problem 7,
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Unformatted text preview: ( ) ( )( ) 1 2 9 3 2 5 X s s s s s = + + + . Thus ( ) ( ) ( ) ( )( ) ( ) 1 2 2 2 .2 3 1 9 2 5 3 2 5 s X s X s X s s s s s s s + = + =-+ + + + + . Applying the theorem, ( ) ( )( ) ( ) 2 2 .2 3 1 9 3 lim ( ) lim lim 2 5 5 3 2 5 t s s s s x t sX s s s s s s + = =-= + + + + + . Examine the solutions of problems 2 and 5 and note that the steady state solution of problem 2 is 0 due to the exponential decay and that of problem 5 is 3/5 (again due to the exponential decays), confirming the solution obtained here....
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This note was uploaded on 02/29/2008 for the course ME 242 taught by Professor Perreira during the Spring '08 term at Lehigh University .

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