372 IEEE TRANSACTIONS ON NEURAL NETWORKS AND LEARNING SYSTEMS, VOL. 24, NO. 3, MARCH 2013
Unlike available results, the controllers to be studied in this
paper are derived without using backstepping. To facilitate the
control design, a set of new state variables will be defined
and a coordinate transform will be proposed to transform
pure-feedback system (1) into a canonical system.
Define alternative state variables as
⎧
⎨
⎩
z
1
= x
1
z
i
=˙z
i−1
,
i = 2,...,n
y = z
1
= x
1
.
(2)
Then, according to the fact that z
2
=˙z
1
= f
1
(x
1
, x
2
),it
follows that:
˙z
2
=¨z
1
=
∂ f
1
(x
1
, x
2
)
∂x
1
˙x
1
+
∂ f
1
(x
1
, x
2
)
∂x
2
˙x
2
=
∂ f
1
(x
1
, x
2
)
∂x
1
f
1
(x
1
, x
2
) +
∂ f
1
(x
1
, x
2
)
∂x
2
f
2
(x
1
, x
2
, x
3
). (3)
Since the unknown function f
2
(x
1
, x
2
, x
3
) is continuously
differentiable with respect to x
3
, we apply the mean-value
theorem [36] for f
2
(x
1
, x
2
, x
3
) as
f
2
(x
1
, x
2
, x
3
) = f
2
(x
1
, x
2
, 0)+
∂ f
2
(x
1
, x
2
, x
3
)
∂x
3
|
x
3
=x
θ
3
x
3
(4)
where x
θ
3
= θ x
3
with 0 <θ<1 being a constant (not
necessarily known) [17], [20]. Then (2) can be rewritten as
˙z
2
= α
2
(x
1
, x
2
) + β
2
(x
1
, x
2
, x
θ
3
)x
3
(5)
where α
2
(x
1
, x
2
)=(∂ f
1
(x
1
, x
2
)/∂x
1
) f
1
(x
1
, x
2
)+(∂ f
1
(x
1
, x
2
)/
∂x
2
) f
2
(x
1
, x
2
, 0) and β
2
(x
1
, x
2
, x
θ
3
) = (∂ f
1
(x
1
, x
2
)/∂x
2
)
(∂ f
2
(x
1
, x
2
, x
3
)/∂x
3
)|
x
3
=x
θ
3
are unknown but smooth functions
of ¯x
2
=[x
1
, x
2
]
T
∈ R
2
and [¯x
2
, x
θ
3
]
T
∈ R
3
.
For i = 3, one can obtain
˙z
3
=¨z
2
=
∂α
2
( ¯x
2
)
∂ ¯x
2
˙
¯x
2
+
∂β
2
( ¯x
3
)
∂ ¯x
3
˙
¯x
3
x
3
+ β
2
( ¯x
3
) ˙x
3
=
2
j=1
∂α
2
( ¯x
2
)
∂x
j
f
j
(x
1
,...,x
j+1
)
+
2
j=1
∂β
2
( ¯x
3
)
∂x
j
f
j
(x
1
,...,x
j+1
)x
3
+
∂β
2
( ¯x
3
)
∂x
3
x
3
+ β
2
( ¯x
3
)
˙x
3
=
2
j=1
∂α
2
( ¯x
2
)
∂x
j
f
j
( ¯x
j+1
)+
2
j=1
∂β
2
( ¯x
3
)
∂x
j
f
j
( ¯x
j+1
)x
3
+
∂β
2
( ¯x
3
)
∂x
3
x
3
+ β
2
( ¯x
3
)
f
3
(x
1
,...,x
4
). (6)
Since f
3
(x
1
,...,x
4
) is continuously differentiable with
respect to x
4
, then based on the mean-value theorem, it can
be represented as
f
3
(x
1
,...,x
4
)= f
3
(x
1
,...,x
3
, 0)+
∂ f
3
(x
1
,...,x
4
)
∂x
4
x
4
=x
θ
4
x
4
(7)
where x
θ
4
= θx
4
with 0 <θ<1 being a constant [17], [20].
We then rewrite (6) as
˙z
3
= α
3
( ¯x
3
) + β
3
¯x
3
, x
θ
4
x
4
(8)
where
α
3
( ¯x
3
) =
2
j=1
(∂α
2
( ¯x
2
)/∂x
j
) f
j
( ¯x
j+1
)
+
2
j=1
(∂β
2
( ¯x
3
)/∂x
j
) f
j
( ¯x
j+1
)x
3
+(
(
∂β
2
( ¯x
3
)/∂x
3
)x
3
+ β
2
( ¯x
3
)
)
f
3
( ¯x
3
, 0)
and
β
3
( ¯x
3
, x
θ
4
) = (
(
∂β
2
( ¯x
3
)/∂x
3
)x
3
+β
2
( ¯x
3
)
)
(∂ f
3
( ¯x
4
)/∂x
4
)|
x
4
=x
θ
4
are smooth functions of ¯x
3
∈ R
3
and [¯x
3
, x
θ
4
]
T
∈ R
4
,
respectively.
Similar to above analysis, for i = 4,...,n − 1, we can
derive
˙z
i
=¨z
i−1
=
∂α
i−1
( ¯x
i−1
)
∂ ¯x
i−1
˙
¯x
i−1
+
∂β
i−1
( ¯x
i
)
∂ ¯x
i
˙
¯x
i
x
i
+ β
i−1
( ¯x
i
) ˙x
i
=
i−1
j=1
∂α
i−1
( ¯x
i−1
)
∂x
j
f
j
( ¯x
j+1
)
+
i−1
j=1
∂β
i−1
( ¯x
i
)
∂x
j
f
j
( ¯x
j+1
)x
i
+
∂β
i−1
( ¯x
i
)
∂x
i
x
i
+ β
i−1
( ¯x
i
)
f
i
(x
1
,...,x
i+1
). (9)
Applying the mean-value theorem for the smooth function
f
i
(x
1
,...,x
i+1
) with respect to x
i+1
, it follows:
f
i
(x
1
,...,x
i+1
) = f
i
(x
1
,...,x
i
, 0)
+
∂ f
i
(x
1
,...,x
i+1
)
∂x
i+1
x
i+1
=x
θ
i+1
x
i+1
(10)
where x
θ
i+1
= θ x
i+1
with 0 <θ <1 being a constant. Then
substituting (10) into (9) yields
˙z
i
= α
i
( ¯x
i
) + β
i
¯x
i
, x
θ
i+1
x
i+1
(11)
where
α
i
( ¯x
i
) =
i−1
j=1
(∂α
i−1
( ¯x
i−1
)/∂x
j
) f
j
( ¯x
j+1
)
+
i−1
j=1
(∂β
i−1
( ¯x
i
)/∂x
j
) f
j
( ¯x
j+1
)x
i
+
(
(∂β
i−1
( ¯x
i
)/∂x
i
)x
i
+ β
i−1
( ¯x
i
)
)
f
i
( ¯x
i
, 0)
and
β
i
( ¯x
i+1
, x
θ
i+1
) = ((∂β
i−1
( ¯x
i
)/∂x
i
)x
i
+β
i−1
( ¯x
i
))(∂ f
i
( ¯x
i+1
)/∂x
i+1
)|
x
i+1
=x
θ
i+1
are unknown functions of ¯x
i
∈ R
i
and [¯x
i
, x
θ
i+1
]
T
∈ R
i+1
.
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