of time invariant part,a model set of the nonlinear system can
be given.Multiple model iterative learning controller will be ob-
tained based on this model set.An index function with the form
of integral of output error between different models and the ac-
tual system will be given for the decision of the best model.The
ILC based on this best model will improve the transient re-
sponse of ILC system greatly.
In the first part of this paper, the ILC of nonlinear system and
its stability analysis will be given.In the second part,the struc-
ture and the training procedure of fuzzy neural network will be
proposed.Then the ILC based on multiple models can be ob-
tained in the third part. Finally, simulation cases will be given
to show the effectiveness of the proposed method.
2. System description and ILC
Consider a class of discrete-time nonlinear dynamic systems
described by the following equations:
x(t + 1) = f(x(t), t) + B(x(t), t)u(t) (1)
where 0 ≤ t ≤ T − 1, x(t) ∈ R
n
and u(t) ∈ R
r
(t ∈ 0, 1, · · · , T)
denotes the state and control vector of the system respectively.
Suppose that the system is all-state output,and also we know
that the estimated model of {f, B} is {
˜
f,
˜
B}.The desired trajectory
is given as x
d
(t)(0 ≤ t ≤ T),when B is column full rank ,the
desired control u
d
(t) can be obtained by Eqn.(2)
u
d
(t) = B
+
(x
d
(t), t)[x
d
(t + 1) − f(x
d
(t), t)] (2)
where B
+
(x
d
(t), t) = [B
T
(x
d
(t), t)B(x
d
(t), t)]
−1
B
T
(x
d
(t), t).By
imposing an arbitrarily selected continuous control u
0
(t)(0 ≤
t ≤ T) onto system (1),the error between the output trajectory
x
0
(t) and the desired trajectory x
d
(t) will be decided by the error
of the control sequences below:
∆u
0
(t) =u
d
(t) − u
0
(t)
=B
+
(x
d
(t), t)[x
d
(t + 1) − f(x
d
(t), t)]
− B
+
(x
0
(t), t)[x
0
(t + 1) − f(x
0
(t), t)]
(3)
The approximation of the control error ∆u
0
(t) can be obtained
at several levels by using the model {
˜
f,
˜
B}
∆u
0
(t) ≈
˜
B
+
(x
d
(t), t)[x
d
(t + 1) −
˜
f(x
d
(t), t)]
−
˜
B
+
(x
0
(t), t)[x
0
(t + 1) −
˜
f(x
0
(t), t)] (4)
∆u
0
(t) ≈
˜
B
+
(x
0
(t), t)[x
d
(t + 1) − x
0
(t + 1)−
˜
f(x
d
(t), t) +
˜
f(x
0
(t), t)] (5)
∆u
0
(t) ≈
˜
B
+
(e
0
(t), t)[e
0
(t + 1) −
˜
f(e
0
(t), t)] (6)
where e
0
(t) = x
d
(t)−x
0
(t).Apparently ,the input can be adjusted
by the approximation of control error.Here Eqn.(5) is used :
u
k+1
(t) = u
k
(t)+
˜
B
+
(e
k
(t), t)[e
k
(t+1)−
˜
f(x
d
(t), t)+
˜
f(x
k
(t), t)] (7)
where k indicates the iterative times.
Suppose that f,
˜
f,B are globally uniformly Lipschitz in the
sense of
kf(x
1
, k) − f(x
2
, k)k ≤ c
f
kx
1
− x
2
k
k
˜
f(x
1
(k)) −
˜
f(x
2
(k))k ≤ c
˜
f
kx
1
− x
2
k
and
kB(x
1
, k) − B(x
2
, k)k ≤ c
B
kx
1
− x
2
k
for positive constants c
f
,c
˜
f
and c
B
.Simultaneously B,
˜
B are
bounded.
For system (1),when control strategy (7) is applied,the fol-
lowing theorem can guarantee the eventual convergence.[5]
Theorem 1. For accessible desired trajectory x
d
(t)(0 ≤ t ≤
T),if system(1) satisfies the following conditions
1. kI −
˜
B
+
(e(t), t)B(x(t), t)k < 1 (0 ≤ t ≤ T)
2. x
k
(0) = x
d
(0)
Then by the effect of learning algorithm(7),ke
k
(t)k → 0(k →
∞)(0 ≤ t ≤ T).
P. From Eqn.(7)
u
k+1
(t) = u
k
(t) +
˜
B
+
(e
k
(t), t)[e
k
(t + 1) −
˜
f(x
d
(t), t) +
˜
f(x
k
(t), t)]
for
e
k
(t + 1) =B(x
k
(t), t)∆u
k
(t) + f(x
d
(t), t) − f(x
k
(t), t)+
[B(x
d
(t), t) − B(x
k
(t), t)]u
d
(t) (8)
so
∆u
k+1
(t) =∆u
k
(t) −
˜
B
+
(e
k
(t), t)B(x
k
(t), t)∆u
k
(t)−
˜
B
+
(e
k
(t), t)[B(x
d
(t), t) − B(x
k
(t), t)]u
d
(t)−
˜
B
+
(e
k
(t), t)[f(x
d
(t), t) − f(x
k
(t), t) −
˜
f(x
d
(t), t)+
˜
f(x
k
(t), t)]
(9)
Taking norms of both sides of Eqn.(8) gives
ke
k
(t + 1)k ≤kB(x
d
(t), t)kk∆u
k
(t)k + c
f
kx
d
(t) − x
k
(t)k
+ k
B
kx
d
(t) − x
k
(t)kku
d
(t)k
= k
1,t
k∆u
k
(t)k + k
2,t
ke
k
(t)k (10)
where k
1,t
= kB(x
k
(t), t)k,k
2,t
= c
f
+ k
B
ku
d
(t)k,and k
2,t
is irrele-
vant with k.
Taking norms of both sides of Eqn.(9) gives
k∆u
k+1
(t)k ≤kρ
k
(t)kk∆u
k
(t)k + k
˜
B
+
(e
k
(t), t)k(k
B
ke
k
(t)kku
d
(t)k
+ c
f
ke
k
(t)k + c
˜
f
ke
k
(t)k)
=kρ
k
(t)kk∆u
k
(t)k + k
t
ke
(
t)k (11)
where ρ
k
(t) = I −
˜
B
+
(e
k
(t), t)B(x
k
(t), t), k
t
=
k
˜
B
+
(e
k
(t), t)k(k
B
ku
d
(t)k + c
f
+ c
˜
f
).They are both relevant
with k.
Using Eqn.(10) and Eqn.(11),we obtain
k∆u
k+1
(t)k ≤kρ
k
(t)kk∆u
k
(t)k + k
t
(k
1,t−1
k∆u
k
(t − 1)k
+ k
2,t−1
ke
k
(t − 1)k)
≤kρ
k
(t)kk∆u
k
(t)k + K
t−1
k∆u
k
(t − 1)k
+ · · · + K
0
k∆u
k
(0)k (12)
2