Nonlinear Systems and Control
Lecture # 32
Robust Stabilization
Sliding Mode Control
– p. 1/17
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Robust Stabilization of Nonlinear Systems: Sliding Mode Control, Slides of Nonlinear Control Systems
An overview of robust stabilization techniques for nonlinear systems using sliding mode control. The concept of a sliding manifold, the conditions for reaching and maintaining the manifold, and methods for reducing chattering. It also discusses the analysis of the system and the behavior of trajectories inside the positively invariant set.
Typology: Slides
2011/2012
1 / 17
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Nonlinear Systems and Control
Lecture # 32
Robust Stabilization
Sliding Mode Control
Docsity.com
Example
x
1
x
2
x
2
h
x
g
x
u,
g
x
g
0
Sliding Manifold (Surface):
s
a
1
x
1
x
2
s
t
x
1
a
1
x
1
a
1
lim t
→∞
x
1
t
How can we bring the trajectory to the manifold
s
How can we maintain it there?
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s <
u
β
x
V
g
x
s
x
g
x
su
g
x
s
x
g
x
β
x
s
V
g
x
s
x
g
x
̺^
x
β
0
s
g
x
β
0
s
sgn(
s
s >
s <
u
β
x
) sgn(
s
V
g
x
β
0
s
g
0
β
0
s
V
g
0
β
0
V
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V
g
0
β
0
V
dV √
V
g
0
β
0
dt
V
V
(
s
(
t
))
V
(
s
(0))
g
0
β
0
t
V
s
t
V
s
g
0
β
0
t
s
t
s
g
0
β
0
t
s
t
reaches zero in finite time
Once on the surface
s
, the trajectory cannot leave it
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x
1
x
2
x
2
h
x
g
x
β
x
)sgn(
s
x
1
a
1
x
1
s
s
a
1
x
2
h
x
g
x
β
x
)sgn(
s
s
s
g
0
β
0
s
if
β
x
x
β
0
V
1
1 2
x
2 1
V
1
x
1
x
1
a
1
x
2 1
x
1
s
a
1
x
2 1
x
1
c
s
c
and
x
1
c
a
1
x
1
c a
1
s
c
is positively invariant if
a
1
x
2
h
x
g
x
x
over
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x
1
c a
1
s
c
6
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
x
1
x
2
s
c/a
1
c
a
1
x
2
h
x
g
x
k
1
< k,
x
u
k
sgn(
s
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Reduce the amplitude of the signum function
s
a
1
x
2
h
x
g
x
u
u
[
a
1
x
2
ˆh
x
)]
ˆg
x
v
s
δ
x
g
x
v
δ
x
a
1
[
g
x
ˆg
x
]
x
2
h
x
g
x
ˆg
x
ˆh
x
δ
x
g
x
x
β
x
x
β
0
v
β
x
) sgn(
s
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Replace the signum function by a high-slope saturationfunction
u
β
x
) sat
s ε
sat(
y
y,
if
y
sgn(
y
if
y
6
y
sgn(
y
6
−
y
ε
sat
y^ ε
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Inside the boundary layer:
x
1
a
1
x
1
s
s
a
1
x
2
h
x
g
x
β
x
s ε
x
1
x
1
a
1
x
2 1
x
1
ε
< θ <
x
1
x
1
θ
a
1
x
2 1
x
1
ε
θa
1
The trajectories reach the positively invariant set
ε
x
1
ε
θa
1
s
ε
in finite time
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What happens inside
ε
Find the equilibrium points
a
1
x
1
s
x
2
a
1
x
2
h
x
g
x
β
x
s ε
φ
x
1
h
x
a
1
g
x
β
x
x
2
=
x
1
εφ
x
1
Suppose
x
1
εφ
x
1
has an isolated root
¯x
1
εk
1
h
¯x
1
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z
1
a
1
z
1
z
2
z
2
z
g
x
β
x
z
2 ε
z
1
z
1
2
z
2
g
x
β
x
g
0
β
0
V
1 2
z
(^21)
1 2
z
(^22)
V
z
1
a
1
z
1
z
2
z
2
[
z
g
x
β
x
z
2 ε
]
V
a
1
z
(^21)
1
z
1
z
2
2
z
(^22)
g
0
β
0
ε
z
(^22)
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V
a
1
z
(^21)
1
z
1
z
2
2
z
(^22)
g
0
β
0
ε
z
(^22)
V
[
z
1
z
2
]
T
[
a
1
1 2
1
1 2
1
g
0
β
0
ε
2
]
Q
[
z
1
z
2
]
det(
Q
a
1
g
0
β
0
ε
2
1 4
1
2
h
lim t
→∞
x
t
h
lim t
→∞
x
t
[
¯x
1
0
]
Read Section 14.1.
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