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...
View
Full Document
 Fall '08
 Hjortso,M
 pH, Alternat ive Linear, ive Linear model, transfer funct ions

Click to edit the document details