Nonlinear Systems and Control
Lecture # 27
Stabilization
Partial Feedback Linearization
– p. 1/11
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Partial Feedback Linearization-Feedback in Non Linear Control Systems-Lecture Slides, Slides of Nonlinear Control Systems
Dr. Javed Iftikhar delivered this lecture at A.P. University of Law for Non Linear Control Systems course. It includes: Stabilization, Partial, Feedback, Linearization, Change, Variables, Controllable, Nonsingular, Hurwitz
Typology: Slides
2011/2012
1 / 11
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Nonlinear Systems and Control
Lecture # 27Stabilization
Partial Feedback Linearization
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Consider the nonlinear system
x
f
x
G
x
u
[
f
(0) = 0]
Suppose there is a change of variables
z
[
η ξ
]
T
x
[
T
1
x
T
2
x
]
defined for all
x
D
R
n
, that transforms the system into
η
f
0
η, ξ
ξ
Aξ
Bγ
x
)[
u
α
x
)]
A, B
is controllable and
γ
x
is nonsingular for all
x
D
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If the origin of
η
f
0
η,
is globally asymptotically stable,
will the origin of
η
f
0
η, ξ
ξ
A
BK
ξ
be globally asymptotically stable?
In general
No
Example
η
η
η
2
ξ,
ξ
v
The origin of
η
η
is globally exponentially stable, but
the origin of
η
η
η
2
ξ,
ξ
kξ,
k >
is not globally asymptotically stable.
The region of
attraction is
ηξ <
k
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Example
η
1 2
ξ
2
η
3
ξ
1
ξ
2
ξ
2
v
The origin of
η
1 2
η
3
is globally asymptotically stable
v
k
2
ξ
1
kξ
2
def
Kξ
A
BK
[
k
2
k
]
The eigenvalues of
A
BK
are
k
and
k
e
(
A
−
BK
)
t
kt
e
−
kt
te
−
kt
k
2
te
−
kt
kt
e
−
kt
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Lemma 13.2:
The origin of
η
f
0
η, ξ
ξ
A
BK
ξ
is globally asymptotically stable if the system
η
f
0
η, ξ
is input-to-state stable Proof:
Use
Lemma 4.7:
If
x
1
f
1
x
1
, x
2
is ISS and the origin of
x
2
f
2
x
2
is globally asymptotically stable, then the
origin of
x
1
f
1
x
1
, x
2
x
2
f
2
x
2
is globally asymptotically stable
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u
α
x
γ
−
1
x
KT
2
x
What is the effect of uncertainty in
α
γ
, and
T
2
Let
α
x
ˆγ
x
, and
T
2
x
be nominal models of
α
x
γ
x
, and
T
2
x
u
α
x
ˆγ
−
1
x
K
T
2
x
η
f
0
η, ξ
ξ
A
BK
ξ
Bδ
z
δ
γ
[ ˆ
α
α
γ
−
1
KT
2
ˆγ
−
1
K
T
2
]
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Proof–First Part:
As in Lemma 13.
ξ
t
≤
cε,
t
t
0
η
t
β
0
η
t
0
, t
t
0
γ
0
(sup
t
≥
t
0
ξ
t
η
t
β
0
η
t
0
, t
t
0
γ
0
cε
Proof–Second Part:
c
1
η
2
V
1
η
c
2
η
2
∂V
1
∂η
f
0
η,
c
3
η
2
∂V
1
∂η
c
4
η
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V
z
bV
1
η
ξ
T
P ξ
V
[
η
ξ
]
T
Q
[
η
ξ
]
Q
[
bc
3
k
P B
bc
4
L/
k
P B
bc
4
L/
k
P B
]
b
k
Q
is positive definite for sufficiently small
k
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