solution-422home7-2009

solution-422home7-2009 - K G Y Y Y K K G K K G G s K = + +...

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1 155: 422 PROCESS SIMULATION AND CONTROL Spring 2009 Homework 7 Assigned: 04/09/09 Due date: 04/15/09 Problem : The process transfer function is: 6 ( ) 2 1 p G s s = - + (a) What is the value of y at the new steady state given a unit step input of the set point ( 1 sp Y s = ) for the following system: (b) For the following closed loop system derive the transfer function sp Y Y . For what values of K c the systems is stable if any? (c) Determine the dynamic behavior of the system under closed loop (y(t)) for K c =-0.2, and K c =-0.3. (d) What are the values of the offset for the above two cases K c =-0.2, and K c =-0.3. (e) Draw the dynamic responses for these two cases. G p (s) Y Y sp + - + + Y sp E D Y c K ( ) d G s ( ) p G s m K
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2 Solution: a) /2 6 1 ( ) ( ) 2 1 ( ) 6(1 ) , ( ) system is unstable. s sp t Y s G Y s s s y t e t y = = - + = - → ∞ ∞ → -∞ b) Fo r th e closed loop system , th e o utp ut Y (s): ( ) 1 1 T he transfer fu nctio n Y /Y : 1 th e characteristic equ ation of closed loop: 6 1 0 w here 2 1 1 c p sp d c M p c M p sp c p sp sp c M p c M p p K G Y G D Y s K K G K K G
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Unformatted text preview: K G Y Y Y K K G K K G G s K = + + + = + + = =-+ + 6 2 1 2 1 6 1 6 0 to get a stable system . 2 1 w here w e can assum e that K o f sen so r is unit. 6 1 6 c M c M c M c M M c K s s K K K K s K K K =-+-+ + = + = -- 3 c) ( ) 2 (1 6 ) 10 6 1 ( ) where , , 1 1 2 1 6 6 1 1 6 2 1 ( ) 6 2 1 1 2 1 1 6 6 ( ) 1 1 6 if 0.2 then ( ) 6 1 , t th c c p sp p sp M c M p c c c c c t K c c t c K G Y Y s G Y K K K G s s K K K s s Y s K s s s K K y t e K K y t e--+-= = = = +- + +- + = = -+ + -+ + =- + = -=- 2.5 en y( ) 6 (steady state value) if 0.3 then ( ) 2.25 1 , t then y( ) 2.25 (steady state value) t c K y t e- = -=- d, e) Kc= -0.2 Kc=-0.3...
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This note was uploaded on 09/09/2009 for the course 155 422 taught by Professor Pedersen during the Spring '08 term at Rutgers.

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solution-422home7-2009 - K G Y Y Y K K G K K G G s K = + +...

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