PTFE4761_Midterm

PTFE4761_Midterm - Question 1 (25 of 100 points) A process...

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Question 1 (25 of 100 points) A process is described by the differential equation: t e y dt dy 2 - = + . At t = 0, y = dy/dt = 0. a) Find Y(s) (function of s only) Laplace transform of both sides: sY(s) + Y(s) = ) 2 ( 1 + s (1+s) Y(s) = ) 2 ( 1 + s ) 1 )( 2 ( 1 ) ( + + = s s s Y b) . 0 ) ( lim = t y t Find dt t y t t 0 ) ( lim . Since the value of y goes to 0 at long time, we expect the integral to converge to value (the limit exists instead of blowing up to infinity). According to Final Value Theorem: Y(s) lim s Y(s) lim ) ( lim ) ( lim 0 0 0 0 0 = = = s s t s t t s dt t y L s dt t y So 2 1 2 1 1 2) 1)(s (s 1 lim ) ( lim 0 0 = = + + = s t t dt t y c) Determine the value of y(t) when t = 3. By partial fraction, ) 1 ( ) 2 ( ) 1 )( 2 ( 1 ) ( + + + = + + = s n s m s s s Y . Multiplying both sides with (s+2)(s+1), we get: 1 = m(s+1) + n(s+2) = ms+m+ns+2n. Then ms+ns = 0 m = -n. Also, m+2n = 1. –n+2n = 1 n = 1 and m= -1 ) 1 ( 1 ) 2 ( 1 ) 1 )( 2 ( 1 ) ( + + + - = + + = s s s s s Y By inverse Laplace, y(t) = -e -2t + e t . At t = 3, y(t) = -e
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PTFE4761_Midterm - Question 1 (25 of 100 points) A process...

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