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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 Spring '08
 Perreira
 Radioactive Decay, Steady State, Exponential decay

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