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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
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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