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2
x
W
−
1
[
R
T
(2)
B
2
j
R
R
(
B
3
+
D
(
e
))
B
1
YW
−
1
]
x
+ 2
x
W
−
1
A
1
j
+
+
R
+
B
1
φ
(
K
x
)
+2
x
W
−
1
R
(
B
4
j
+
D
(
e
)
B
5
j
)
x
)
M
(
x
)+
ZW
−
1
x
)
−
ω
ω
<
ω
−
2
φ
(
K
φ
(
K
0
.
Thus, by using (15.12) one can upper-bound the term containing
T
(2)
as [17]
2
x
W
−
1
R
T
(2)
B
2
j
R
x
(
W
−
1
R
R
1
R
W
−
1
+
u
1(2)
R
B
2
j
R
−
1
x
≤
B
2
j
R
)
x
1
with
R
1
=
R
1
>
0. In the same way, by using both (15.12) and (15.19), one can
upper-bound the term containing
T
(3)
and
D
(
e
) as
2
x
W
−
1
R
(
T
(3)
(
B
3
+
D
(
e
))
R
x
+
D
(
e
)
B
5
j
ω
)=
⎡
⎤
x
ω
B
3
R
0
R
T
(3)
I
3
T
(3)
D
(
e
)
D
(
e
)
⎣
⎦
2
x
W
−
1
R
0
0
B
5
j
(
u
1(3)
(1 +
2
)+
2
)
x
W
−
1
R
R
W
−
1
x
≤
ε
β
β
⎡
⎤
⎡
⎣
⎤
x
ω
B
3
R
0
B
3
R
0
ε
−
1
x
ω
⎣
⎦
⎦
+
R
0
R
0
.
0
B
5
j
0
B
5
j
Thus, the satisfaction of relations (15.36) to (15.39) with
M
=
S
−
1
,
=
YW
−
1
K
and
=
ZW
−
1
G
allows to verify that
V
(
x
)
<
ω
ω
,
(15.42)
for all
x
(0)
satisfying (15.11). Hence the trajectories of the closed-
loop system (15.18) remain bounded in
∈E
0
and any
ω
E
1
(
W
,
ζ
,
δ
1
) for all
x
(0)
∈E
0
(
W
,
ζ
) and any
ω
satisfying (15.11). This completes the proof of statement (i). To prove statement
(ii), assume that
= 0. Then, from (15.42), we have
V
(
x
)
ω
<
0,
∀
x
∈ E
1
,which
means that
Remark 15.1.
The study of system (15.18) subject to constraints (15.19) means that
the constraints on the error (C2) and on the velocity (part 1 of C3) are linearly
respected (saturation avoidance case). On the contrary, saturation of the acceleration
(part 2 of C3) is allowed. Nevertheless, if one wants to consider that saturation on
the velocity is also allowed then one can modify the closed-loop system as
E
0
=
E
1
is a set of asymptotic stability for system (15.21).
e
=
L
(
z
ω
,
u
=
sat
u
0
(
K
1
e
+
K
2
sat
u
1
(
u
))
,
e
)
sat
u
1
(
u
)+
B
(
z
,
e
)
,
(15.43)
T
=
sat
u
1
(
u
)
.
Thus, by considering
x
n
=
e
u
∈
ℜ
6
and the same type of matrices as in (15.15),
the closed-loop system reads
x
n
=(
A
(
z
,
x
n
)+
B
1
K
)
x
n
+
B
1
φ
0
+(
A
(
z
,
x
n
)+
B
1
K
)
B
1
φ
1
+
B
2
ω
,
(15.44)
where
φ
0
=
sat
u
0
(
K
1
e
+
K
2
sat
u
1
(
u
))
−
(
K
1
e
+
K
2
sat
u
1
(
u
)) =
sat
u
0
(
K
x
n
+
K
2
φ
1
)
−
(
K
x
n
+
K
2
φ
1
) and
φ
1
=
sat
u
1
(
u
)
−
u
=
sat
u
1
(
C
x
n
)
−
C
x
n
. In this case, Problem 15.1
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