20095ee141_1_EE141_hw6_sol - EE141 Principles of Feedback...

EE141 Principles of Feedback Control (Fall 2009)
Solutions to Homework 6
Problem 9.2
(a)
The equilibrium pairs for u = 1 is
di
dt
= - i + v = 0
dv
dt
= - i + g (1 - v ) = - i + (1 - v )(1 - v - 1)(1 - v - 4) = 0
which gives
v = - 1 ±
√
3
Thus the other two equilibrium pairs are v
2
= i
2
= 0 . 732 and v
3
= i
3
= - 2 . 732.
(b)
x =
i
v
=
x
1
x
2
˙ x =
˙
i
˙ v
=
- i + v
- i + g ( u - v )
=
- x
1
+ x
2
- x
1
+ g ( u - x
2
)
=
f
1
( x, u )
f
2
( x, u )
F =



∂f
1
∂x
1
∂f
1
∂x
2
∂f
2
∂x
1
∂f
2
∂x
2



x
1
= x
2
=0 ,u =1
=
- 1 1
- 1 3
G =



∂f
1
∂u
∂f
2
∂u



x
1
= x
2
=0 ,u =1
=
0
- 3
Thus the linearized model here is
˙ x =
- 1 1
- 1 3
x +
0
- 3
u
(c)
(1) For v
2
= i
2
= 0 . 732 , u = 1
F =



∂f
1
∂x
1
∂f
1
∂x
2
∂f
2
∂x
1
∂f
2
∂x
2



x
1
= x
2
=0 . 732 ,u =1
=
- 1 1
- 1 - 1 . 54
G =



∂f
1
∂u
∂f
2
∂u



x
1
= x
2
=0 . 732 ,u =1
=
0
1 . 54
1

Thus the linearized model here is
˙ x =
- 1 1
- 1 - 1 . 54
x +
0
1 . 54
u
(2) For v
2
= i
2
= - 2 . 732 , u = 1
F =



∂f
1
∂x
1
∂f
1
∂x
2
∂f
2
∂x
1
∂f
2
∂x
2



x
1
= x
2
= - 2 . 732 ,u =1
=
- 1 1
- 1 - 8 . 46
G =



∂f
1
∂u
∂f
2
∂u



x
1
= x
2
= - 2 . 732 ,u =1
=
0
8 . 46
Thus the linearized model here is
˙ x =
- 1 1
- 1 - 8 . 46
x +
0
- 8 . 46
u
Problem 9.4
(a)
˙ x = - x
2
e
-
1
x
+ sin 0 = 0 ⇒ x = 0
Note: the x = 0 here is actually x = 0
+
.
(b)
Let ˙ x = - x
2