Regional Solution of Constrained
LQ Optimal Control
José De Doná
September 2004
Centre of Complex Dynamic
Systems and Control
Outline
1
Recap on the Solution for N = 2
2
Regional Explicit Solution
Comparison with the Maximal Output Admissible Set
3
Example
4
Conclusions
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Motivation
In the previous part, we provided a global characterisation of
receding horizon constrained optimal control for horizon
N = 2.
Later in the course, we will extend the solution to the case of
arbitrary horizon. This gives practically valuable insights into
the form of the control law. Indeed, for many problems it is
feasible to compute, store and retrieve the function u = KN (x ),
thus eliminating the need to solve the associated QP on line.
In this part we address a different question: Given that we
only ever apply the first move from the optimal control
sequence of length N , is it possible that the first element of
the control law might not change as the horizon increases
beyond some modest size, at least locally in the state space?
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Recap on the Solution for N = 2
Theorem (RHC Characterisation for N = 2)
The RHC law has the form
⎧
⎪
⎪
−sat∆ (Gx + h ) if x ∈ Z− ,
⎪
⎪
⎪
⎨
K2 (x ) = ⎪
if x ∈ Z,
−sat∆ (Kx )
⎪
⎪
⎪
⎪
⎩−sat∆ (Gx − h ) if x ∈ Z+ ,
(1)
where the gains K , G ∈ R1×n and the constant h ∈ R are
K = R̄ −1 B t PA ,
G=
K + KBKA
,
1 + (KB )2
h=
KB
∆,
1 + (KB )2
and the sets (Z− , Z, Z+ ) are defined by
Z− = {x : K (A − BK )x < −∆} ,
Z = {x : |K (A − BK )x | ≤ ∆} ,
+
Z = {x : K (A − BK )x > ∆} .
◦
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Recap on the Solution for N = 2
Thus, the solution for N = 2 is given by:
⎧
⎪
⎪
−sat∆ (Gx + h ) if x ∈ Z− ,
⎪
⎪
⎪
⎨
K2 (x ) = ⎪
−sat∆ (Kx )
if x ∈ Z,
⎪
⎪
⎪
⎪
⎩−sat∆ (Gx − h ) if x ∈ Z+ ,
We now ask the following question:
Under what conditions is the control law
uk = −sat∆ (Kxk )
optimal for the receding horizon control problem,
and for arbitrary horizon N ?
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Regional Explicit Solution
Consider the fixed-horizon optimisation problem for arbitrary
horizon N :
PN (x ) :
VNopt (x ) = min VN ({xk }, {uk }),
subject to:
x0 = x ,
xk +1 = Axk + Buk
for k = 0, 1, . . . , N − 1,
uk ∈ U = [−∆, ∆] for k = 0, 1, . . . , N − 1,
with objective function:
N −1
1 t
1 t
VN ({xk }, {uk }) = xN PxN +
xk Qxk + ukt Ruk .
2
2
k =0
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Regional Explicit Solution
We have already shown, for N = 2, that
ukopt = −sat∆ (Kxk ),
k = 0, 1,
for all x ∈ Z, where
Z = {x : |K (A − BK )x | ≤ ∆} .
We can see that this characterisation is valid locally, i.e., inside a
region Z, defined by linear inequalities.
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Regional Explicit Solution
To extend the regional solution ukopt = −sat∆ (Kxk ) to larger
horizons, we need to define a number of additional sets in the state
space. This is no surprise, since as we increase the horizon we
would expect additional sets of inequalities to be required in order
for the solution ukopt = −sat∆ (Kxk ) to still be valid.
Define the gains
K̄i = KA i −1 (A − BK ),
i = 1, 2, . . . , N − 1.
Also, define the sets Xi by the linear inequalities:
Xi = x : K̄i x ≤ ∆i , −K̄i x ≤ ∆i ,
i = 1, 2, . . . , N − 1.
(2)
where the saturation bounds ∆i are defined as
⎛
⎞
i −2
⎜⎜⎜
⎟⎟⎟
k
⎜
∆i ⎜⎜⎝1 +
|KA B |⎟⎟⎟⎠ ∆,
i = 1, 2, . . . , N .
(3)
k =0
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Regional Explicit Solution
Also define the nonlinear mapping
φnl (x ) = Ax − B sat∆ (Kx ),
and the sets:
Yi =
i −1
j =1 Xj ,
i = 2, 3, . . . , N ,
(4)
k
Z = x : φnl
(x ) ∈ YN −k , k = 0, 1, . . . , N − 2 , for N ≥ 2.
(5)
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Regional Explicit Solution
Theorem
Consider the fixed horizon optimal control problem PN (x ). Then for
opt
all x ∈ Z the optimal control sequence {u0opt , . . . , uN
} is
−1
ukopt = −sat∆ (Kxk ),
k
for k = 0, 1, . . . , N − 1, where xk = φnl
(x ).
◦
The proof follows the same arguments as in the proof of the case
N = 2 and uses induction. (More involved.)
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Regional Explicit Solution
We now consider the receding horizon control (RHC)
implementation of the previous (fixed horizon) control law. That is,
opt
given the control sequence {u0opt , . . . , uN
} that achieves the
−1
minimum in PN (x ), the control applied to the plant is
KN (x ) = u0opt .
We need to introduce the concept of an invariant set for the
closed loop system xk +1 = φnl (xk ) = Axk − B sat∆ (Kxk ).
To this end, define the set
k
(x ) ∈ YN , k = 0, 1, 2, . . . ,
Z̄ = x : φnl
N ≥ 2.
The set Z̄ is the maximal positively invariant set contained in Z and
YN for the closed loop system xk +1 = φnl (xk ) = Axk − B sat∆ (Kxk ).
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Regional Explicit Solution
We then have the following result:
Theorem
For all x ∈ Z̄ the RHC law KN is given by
KN (x ) = −sat∆ (Kx ).
◦
The proof follows from the fact that Z̄ ⊆ Z and that Z̄ is positively
invariant under the control KN (x ) = −sat∆ (Kx ).
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Comparison with the Maximal Output Admissible Set
Consider the maximal output admissible set O∞ , defined as the
set where constraints are always satisfied with the linear controller
u = −Kx :
O∞ x : |K (A − BK )i x | ≤ ∆ for i = 0, 1, . . . .
(6)
It can be shown that the set Z̄, where the control law
KN (x ) = −sat∆ (Kx ) is optimal, is bigger than the maximal output
admissible set O∞ , as we will prove next.
In fact, Z̄ can be considerably bigger than O∞ , as we will illustrate
with an example.
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Comparison with the Maximal Output Admissible Set
Proposition
O∞ ⊆ Z̄.
Proof.
Since Z̄ is the maximal positively invariant set in YN , it suffices to
−1
show that O∞ ⊆ YN N
i =1 Xi (see (4)). Assume, therefore, that
x ∈ O∞ , so that (see (6))
|KAKj x | ≤ ∆,
where AK = A − BK .
j = 0, 1, . . . ,
(7)
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Comparison with the Maximal Output Admissible Set
Proof (Ctd.)
For any i ∈ {1, 2, . . . , N − 1},
AKi
= (A − BK )AKi −1 = AAKi −1 − BKAKi −1
= A (A − BK )AKi −2 − BKAKi −1 = A 2 AKi −2 − ABKAKi −2 − BKAKi −1
= A 2 (A − BK )AKi −3 − ABKAKi −2 − BKAKi −1
= A 3 AKi −3 − A 2 BKAKi −3 − ABKAKi −2 − BKAKi −1
..
.
i −2
i −1−j
i −1
A j BKAK ,
= A AK −
j =0
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Comparison with the Maximal Output Admissible Set
Proof (Ctd.)
From the previous expression, we obtain
KA i −1 AK x = KAKi x +
i −2
i −1−j
x.
(8)
|KA j B ||KAKi −1−j x |
(9)
j =0
KA j BKAK
From (7) and (8), we obtain the inequality
|KA i −1 AK x | ≤ |KAKi x | +
i −2
j =0
⎛
⎞
i −2
⎜⎜⎜
⎟⎟⎟
≤ ⎜⎜⎜⎜⎝1 +
|KA j B |⎟⎟⎟⎟⎠ ∆ = ∆i
(see (3)).
(10)
j =0
This implies x ∈ Xi for all i ∈ {1, 2, . . . , N − 1} (see (2)), yielding the
desired result.
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Example
Consider the system xk +1 = Axk + Buk with
1
A=
0.4
0.4
B=
,
0.08
0
,
1
∆ = 1, N = 10, Q = I and R = 0.25.
8
YN =
6
YN
4
i =1
Xi
O∞ = maximal output
O∞
xk2
2
N −1
admissible set
E
0
−2
E = positively invariant
−4
ellipsoid.
−6
−8
−2.5
O∞ ∪ E ⊆ Z̄ ⊆ YN .
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
xk1
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Example
Simulation with initial condition x0 = [1.27 − 0.1]t . Notice that the
control remains saturated during the initial three steps.
1.5
1
1
0.8
0.6
0.5
0.4
uk
xk2
0.2
0
0
−0.2
−0.4
−0.5
−0.6
−0.8
−1
−1
−1.5
−1
−0.5
0
xk1
0.5
1
1.5
−1.5
0
5
10
k
15
20
25
Shown in the figures are the sequences obtained with numerical
(QP) RHC and with the control law uk = −sat∆ (Kxk ). (Both
sequences coincide.)
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Example
Simulation with initial condition x0 = [0.25 4.7]t . Notice that the
control remains saturated during the initial five steps.
8
1.5
6
1
4
0.5
uk
xk2
2
0
0
−2
−0.5
−4
−1
−6
−8
−2.5
−1.5
−2
−1.5
−1
−0.5
0
xk1
0.5
1
1.5
2
2.5
0
5
10
k
15
20
25
Shown in the figures are the sequences obtained with numerical
(QP) RHC and with the control law uk = −sat∆ (Kxk ). (Both
sequences coincide.)
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Conclusions
State Space
MPC = - sat(Kx)
Unconstrained
Locally, the tactical MPC design has a closed-form solution
KN (x ) = −sat∆ (Kx ).
The control law KN (x ) = −sat∆ (Kx ) can be thought of as a
particular case of an Anti-windup design.
Simple implementation of optimisation based strategies can
lead to new ideas in robustness, stability, etc.
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