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
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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
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