Assignment 3 Solutions 2009

1 1 2 1 2 note thatthecoefficientsaregivenas fo llows

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Unformatted text preview: Plot the output responses. What is the steady­state reached by the outputs? Solution: We start by linearizing the nonlinear equations, recognizing that we have two state variables and two inputs in the state equations. The first equation yields: dx 1 = f ( x , x , u , u ) 11212 dt @ f ( x s , x s , u s , u s ) + 1 1 2 1 2 ¶f 1 ¶x x 1 ( x - x s ) + 1 1 u s u 2 1s , x 2 s , 1 , s + ¶f 1 ¶u x 1 ( 1 - u s ) + u 1 u s 2 1 , x 2 s , 1 , u s s ¶f 1 ¶u x 2 ¶f 1 ¶x x 2 ( x - x s ) + 2 2 u s 2 1 , x 2 s , 1 , u s s ( 2 - u s ) u 2 u2 1 , x 2 s , u1s , s s It can be put into the form, dx1 = a 1 ( x1 - x1 s ) + a 2 ( x - x s ) + b 1 ( 1 - u s ) + b 2 ( 2 - u s ) 1 1 2 2 1 u 1 1 u 2 dt also recognizing the fact that f ( x s , x s , u s , u s ) = 0 from the steady­state condit ion. 1 1 2 1 2 Note that the coefficients are given as fo llows: a11 = ¶f 1 = [- 1 - 0 5 exp( x s )] . 1 ¶x s . s . 1 a 2 = 1 ¶f 1 = [- 1 5 1s ] . u ¶x s . s . 2 b 1 = 1 ¶f 1 = [- 1 5 2 s ] . x ¶u s . s . 1 b 2 = 1 ¶f 1 = 0 ¶u s . s . 2 One can derive the coefficients of the second equation dx 2 = f 2 ( x , x , u1 , u ) also as 12 2 d t fo llo ws: a 1 = 2 ¶f 2 = ¶x s . s . 1 é 2 ù ê1 + x ú 2 s û ë a 2 = 2 ¶f 2 = ¶x s . s . 2 é ( 2 x s ) - 2 1 + x s ) ù ( 1 2 ê - 1 + ú 2 (1 + x s ) ê ú ë 2 û b21 = ¶f 2 = 0 ¶u1 s . s . b22 = ¶f 2 = 4 ¶u s . s . 2 We can define the deviat ion variables: x1 = x - x s ; x = x - x s ; u = u - u s ; u...
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This note was uploaded on 01/12/2014 for the course CHE 4198 taught by Professor Hjortso,m during the Fall '08 term at LSU.

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