Page 314 - The Mechatronics Handbook

0066_frame_C14.fm Page 28 Wednesday, January 9, 2002 1:50 PM

The controller should be derived by finding the control value that attains the minimum to nonquadratic
functional. The first-order necessary condition (n1) leads us to an admissible bounded control law. In
particular,

T ∂Vx()


u = – Φ GB x() --------------- , u ∈U
∂x 

−1
The second-order necessary condition for optimality (n2) is met because the matrix G is positive-
definite. Hence, a unique, bounded, real-analytic, and continuous control candidate is designed.
If there exists a proper function V(x) which satisfies the Hamilton–Jacobi equation, the resulting closed-
loop system is robustly stable in the specified state X and control U sets, and robust tracking is ensured
in the convex and compact set XY(X 0 ,U,R,E 0 ). That is, there exists an invariant domain of stability
S = { x ∈ , e ∈ : xt() ≤ x x 0 , t( ) + u ( u ), et() ≤ e ( e 0 , t) + r ( r ) + y ( y ),
b
c
b
c
∀∈ ( ( × ,
x XX 0 ,U), t ∀∈ [t 0 , ∞), e ∀∈ EE 0 , R, Y)} ⊂
and control u(·), u ∈U steers the tracking error to the set
S E δ() = { e ∈ : e 0 ∈ E 0 , x ∈ XX 0 ,U), r ∈ R, y ∈ Y, t ∈ t 0 , ∞ )
[
(
b
et() ≤ ( e 0 ,t) + δ, δ ≥ 0, e ∈ EE 0 ,R,Y), t ∈ t 0 ,∞) } ⊂ b
(
∀
[
∀
e
Here x and e are the KL-functions; and u , r , and y are the K-functions.
The solution of the functional equation should be found using nonquadratic return functions. To
obtain V(·), the performance cost must be evaluated at the allowed values of the states and control. Linear
and nonlinear functionals admit the final values, and the minimum value of the nonquadratic cost is
given by power-series forms [9]. That is,
(
η 2 i+γ +1)
-----------------------
J min ∑ vx 0 2γ +1 , η = 0, 1, 2, …, γ = 0, 1, 2,…
=
()
i=0
The solution of the partial differential equation is satisfied by a continuously differentiable positive-
definite return function
η i+γ +1 T i+γ +1
2γ +

---------------
---------------
1
Vx() = ∑ ---------------------------- x 2γ +1   K i x 2γ +1
1) 
γ
2 i ++(
i=0
where matrices K i are found by solving the Hamilton–Jacobi equation.
1 T
The quadratic return function in V(x) = x K 0 x is found by letting η = γ = 0. This quadratic candidate
--
2
may be employed only if the designer enables to neglect the high-order terms in Taylor’s series expansion.
Using η = 1 and γ = 0, one obtains
1 T
Vx() = --x K 0 x + 1 2 T 2
-- x() K 1 x
2 4
while for η = 4 and γ = 1, we have the following function:
3
-- x (
Vx() = 3 2/3 T 2/3 + --x K 1 x + 3 4/3 T 4/3 + ----- x ( 5/3 T 5/3 + 1 2 T 2
1 T
-- x(
) K 0 x
-- x () K 4 x
) K 2 x
) K 3 x
4 2 8 10 4
©2002 CRC Press LLC