ˆ
sup { J}
T
= x
<
0 X∞ x 0 subject to
zw
T
γ
∞
w∈ 2
L +
if for any admissible uncertainty F
Δ ( k) , there exists a stabilizing solution X∞ ≥ 0 to the
inequality (11) such that
2
−
T
U = − γ
>
1
I
Bˆ X∞ Bˆ
0 .
F∞
F∞
Proof: Suppose that for any admissible uncertainty F
Δ ( k) , there exists a stabilizing solution
X
−
T
∞ ≥ 0 to the inequality (11) such that
2
U = − γ
>
1
I
Bˆ X∞ Bˆ
0 . This implies that the
F∞
F∞
solution X
−
−
T
∞ ≥ 0 is such that
2
1
Aˆ + γ Bˆ U B X∞ A is stable. Then it follows from
F∞
F
1
ˆ
ˆ
∞
F∞
F∞
Lemma 2.3 that
<
zw
T
γ . Using the same argument as in the proof of Lemma 2.3, we get
∞
that Aîs stable and J can be rewritten as follows:
F∞
∞
2
1
2
2
2
2
2
2
−
1
J = ∑{−Δ V( (
x k))
−
T
− z + γ w − γ U
− γ
1 ( w
U 1 Bˆ X∞ Aˆ x)
F∞
F∞
k=0
(18)
T
T
2
−
T
1
−
T
T
ˆ T ˆ
+ x ( Aˆ X∞ Aˆ − X + γ
∞
Aˆ X∞ Bˆ U B X∞ A + C C + Q + F∞ RF∞ x
F∞
F∞
F∞
F
1
ˆ
ˆ
ˆ
ˆ
) }
∞
F∞
F∞
F∞ F∞
Substituting (11) for (18) to get
2
1
T
2
2
2
2
2
2
−
1
−
T
J < x
−
+ γ
− γ
− γ
0 X∞ x 0
z
w
U ( w
U B X∞ A x) (19a)
2
2
1
1
ˆ
ˆ
F∞
F∞
2
or
2
1
T
2
ˆ
2
2
2
−
1
−
T
J < x
−
− γ
− γ
0 X∞ x 0
z
U ( w
U B X∞ A x) (19b)
2
1
1
ˆ
ˆ
F∞
F∞
2
By letting
2
−
1
−
T
w = γ U
ˆ
=
1 Bˆ X∞ Aˆ x for all k ≥ 0 , we get that (
x k)
k
A x with
F
ˆ F 0
∞
F∞
∞
ˆ
2
−
1
−
T
A
ˆ
ˆ = A ˆ + γ
Bˆ U B X∞ A which belongs to L
+∞ since A is stable. It follows
F
2[0,
)
ˆ
∞
F∞
F
1
ˆ
ˆ
∞
F∞
F∞
F∞
Discrete Time Mixed LQR/H∞ Control Problems
171
from (19b) that
ˆ
sup{ J}
T
=
. Thus, we conclude that there exists an admissible
w∈
x X∞ x
2
L +
0
0
non-fragile controller such that
ˆ
sup{ J}
T
=
subject to
< γ . Q. E. D.
w∈
x X∞ x
zw
T
2
L +
0
0
∞
Remark 4.1 In the proof of Lemma 4.1, we let
2
−
1
−
T
w = γ U 1 Bˆ X∞ Aˆ x for all k ≥ 0 to get that
F∞
F∞
ˆ
(
x k)
k
= A
ˆ
−
−
T
ˆ
ˆ x with
2
1
A = A + γ B U B X∞ A which belongs to L
+∞ since A is
F
0
ˆ
ˆ
ˆ
F
2[0,
)
ˆ
∞
F∞
F
1
ˆ
ˆ
∞
∞
F∞
F∞
F∞
stable. Also, we have
2
4 T
w
γ −
=
x X x ,
2
T
z = x X x .
2
0
w 0
2
0
z 0
Then it follows from (19a) that
T
2
J < x
+ γ −
−
0 ( X∞
X
X )
w
z x 0 (20)
∞
∞
where,
ˆ k T T
2
−
T
ˆ
X = ∑{(
k
ˆ k T T
ˆ k
w
Aˆ ) Aˆ X∞ B U B X∞ A A
, and X = ∑{( A ) C C A } .
F
z
ˆ
ˆ
ˆ
ˆ
∞
F
1 1
1
ˆ
ˆ }
∞
F∞ F∞
F∞
F∞ F∞ F∞
k=0
k=0
Note that ˆ Aˆ depends on the controller uncertainty F
Δ ( k) , thus it is difficult to find an
F∞
upper bound of either of Xw and Xz . This implies that the existence of controller
uncertainty Δ F( k) makes it difficult to find sup ∈ { }
w
by using (20). Thus, it is clear that
2
L
J
+
the existence of the controller uncertainty makes the performance of the designed system
become bad.
In order to give necessary and sufficient conditions for the existence of an admissible non-
fragile controller for solving the non-fragile discrete-time state feedback mixed LQR/ H∞
control problem, we define the following parameter-dependent discrete time Riccati
equation:
T
T
ˆ ˆ T
ˆ
ˆ 1
− ˆ T
2
A X∞ A − X∞ − A X∞ (
B B X∞ B + R)
T
T
B X∞ A + ρ
+
+
=
K
E K
E
C 1 C 1 Qδ 0 (21)
⎡− I
0 ⎤
where, ˆ
1
B = γ −
⎡
⎤ ˆ
⎣
=
= + δ
δ >
1
B
2
B ⎦ , R ⎢
, Qδ Q
I with
0 being a sufficiently small
0
I R⎥
+
⎣
⎦
constant, ρ is a given number satisfying 2
T
ρ I − H
>
T
= − γ −
>
KU 2 H
0
K
,
2
U 1 I
1
B X∞ 1
B
0 ,
T
U =
+ +
T
=
+ γ −
−
2
2
B U 3 2
B
I R and
2
1
U 3 X∞
X∞ 1
B U 1 1
B X∞ . If A is invertible, the parameter-
dependent discrete time Riccati equation (21) can be solved by using the following
symplectic matrix
⎡
ˆ ˆ 1
− ˆ T − T
2 T
T
ˆ ˆ 1
− ˆ
A + BR B A (
T
− T
ρ
+
+
−
⎤
K
E K
E
C 1 C 1 Qδ )
ˆ
BR B A
S∞ := ⎢
⎥
− T
2
⎢
− A (
T
T
− T
ρ
+
+
K
E K
E
C 1 C 1 Qδ )
A
⎥
⎣
⎦
The following theorem gives the solution to non-fragile discrete time state feedback mixed
LQR/ H∞ control problem.
Theorem 4.1 There exists a non-fragile discrete time state feedback mixed LQR/ H∞
controller iff for a given number ρ and a sufficiently small constant δ > 0 , there exists a
stabilizing solution X∞ ≥ 0 to the parameter-dependent discrete time Riccati equation (21)
such that
2 T
U = − γ −
>
T
ρ −
>
1
I
1
B X∞ 1
B
0 and 2 I HKU 2 H
0
K
.
172
New Trends in Technologies
Moreover, this non-fragile discrete time state feedback mixed LQR/ H∞ controller is
1
−
T
F∞ = U
− 2 2
B U 3 A
and achieves
ˆ
sup{ J}
T
=
<
w∈
x X∞ x subject to
.
2
L +
0
0
zw
T
γ
∞
Proof: Sufficiency: Suppose that for a given number ρ and a sufficiently small constant
δ > 0 , there exists a stabilizing solution X∞ ≥ 0 to the parameter-dependent Riccati
equation (21) such that
2 T
U = − γ −
>
T
ρ −
>
1
I
1
B X∞ 1
B
0 and 2 I HKU 2 H
0
K
. This implies that the
solution X
ˆ ˆ T
ˆ
ˆ − ˆ T
∞ ≥ 0 is such that
1
A − (
B B X∞ B + R) B X∞ A is stable. Define respectively the
state matrix and controlled output matrix of closed-loop system
1
−
T
Aˆ = A + B U
−
B U A + H F k E
F
2 (
2
2
3
( ) )
K
K
∞
1
−
T
C ˆ = C + D
U
−
B U A + H F k E
F
1
12 (
2
2
3
( ) )
K
K
∞
and let
1
−
T
= −
−
T
−
+
F
A
A
2
B U 2 2
B U 3 A and
1
F∞ = U B U A H F k E , then it follows from the
∞
2
2
3
( )
K
K
square completion that
T
2
−
T
1
−
T
T
T
Aˆ X∞ Aˆ − X + γ
∞
Aˆ X∞ Bˆ U B X∞ A + C C + Q + F∞ RF
F
∞
∞
F∞
F∞
F
1
ˆ
ˆ
ˆ
ˆ
∞
F∞
F∞
F∞ F∞
T
2
−
T
1
−
T
T