ˆ
ˆ
ˆ
) }
∞
F∞
F∞
F∞ F∞
Substituting (11) for the above formula,we get that for any (
u k) and (
w k) and x(0) = 0 ,
2
1
2
2
2
2
2
2
−
1
J < − z + γ w − γ U (
−
T
w − γ U B X∞ A x)
2
2
1
1
ˆ
ˆ
F∞
F∞
2
∞
Note that
2
z
= ∑ ˆ T
x ( k)Ωˆ
= − γ − −
0
(
x k) , and define that
2
1
r :
T
w
U B X∞ A x , we get
2
1
ˆ
ˆ
F∞
F∞
k=0
2
1
2
2
2
2
2
ˆ z − γ w < γ
−
U r
2
2
1
2
Suppose that Γ is the operator with realization
ˆ
(
x k + 1) = ( A +
+
2
B F∞) (
x k) Bˆ (
w k)
F∞
2
−
1
r( k)
−
T
= γ
−
U
+
1 Bˆ X∞ A ˆ
(
x k)
(
w k)
F∞
F∞
which maps w to r .
Since
1
−
Γ exists ( and is given by
ˆ
2
−
1
(
x k + 1) = (
−
T
A +
+ γ
+
2
B F∞
Bˆ U B X∞ A x k B r k ,
F
1
ˆ
ˆ ) ( )
ˆ
( )
∞
F∞
F∞
F∞
2
−
1
(
w k)
−
T
= γ U
+
1 Bˆ X∞ Aˆ
(
x k) r( k) ), we can write
F∞
F∞
2
1
2
2
2
2
2
2
2
2
ˆ z − γ w < γ
−
U r = γ
−
w
Γ
≤ κ w
2
2
1
2
2
2
166
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for some positive κ .This implies that there exists an admissible non-fragile controller such
that ˆ
<
γ −
>
zw
T
γ . Note that 2
~
I T T
is equivalent to
∞
ˆ
ˆ
0
zw zw
2
~
~
γ I − T T >
>
∈
∞
0
T 0
T
0
zw zw
for all w
2
L [0, )
so ˆ
<
<
zw
T
γ implies
, and we conclude that there exists an admissible non-fragile
∞
zw
T
γ
∞
controller such that
<
zw
T
γ . Q. E. D.
∞
3. State Feedback
In this section, we will consider the discrete time state feedback mixed LQR/ H∞ control
problem. This problem is defined as follows: Given the linear discrete-time systems (2)(3)
with w ∈
∞
=
2
L [0, ) and x(0) x 0 and the quadratic performance index (1), for a given
number γ 0,
>
determine an admissible controller K that achieves
sup inf{ J} subject to
( ) <
zw
T
z
γ .
K
∞
w∈ 2
L +
If this controller K exists, it is said to be a discrete time state feedback mixed LQR/ H∞
controller.
Here, we will discuss the simplified versions of the problem defined in the above. In order
to do this, the following assumptions are imposed on the system
Assumption 1 ( C 1, A) is detectable.
Assumption 2 ( A, 2
B ) is stabilizable.
Assumption 3 T
=
12
D [ C 1
1
D 2 ] [0 I] .
The solution to the problem defined in the above involves the discrete time Riccati equation
T
T
ˆ ˆ T
ˆ
ˆ 1
− ˆ
∞
− ∞ −
∞ (
∞ +
)
T
T
A X A X
A X B B X B R B X∞ A + C
+ =
1 C 1
Q 0 (13)
⎡− I
0 ⎤
where, ˆ
1
B = γ −
⎡
⎤
ˆ
⎣
=
1
B
2
B ⎦ , R ⎢
. If A is invertible, the stabilizing solution to the
0
R I⎥
+
⎣
⎦
discrete time Riccati equation (13) can be obtained through the following simplectic matrix
⎡
ˆ ˆ 1
− ˆ T − T
T
ˆ ˆ 1
− ˆ
A + BR B A (
T
− T
C
+
−
⎤
1 C 1
Q)
BR B A
S∞ := ⎢
⎥
− T
⎢
− A ( T
− T
C
+
1 C 1
Q)
A
⎥
⎣
⎦
In the following theorem, we provide the solution to discrete time state feedback mixed
LQR/ H∞ control problem.
Theorem 3.1 There exists a state feedback mixed LQR/ H∞ controller if the discrete time
Riccati equation (13) has a stabilizing solution X
T
∞ ≥ 0 and
2
U = − γ −
>
1
I
1
B X∞ 1
B
0 .
Moreover, this state feedback mixed LQR/ H∞ controller is given by
1
−
T
K = U
− 2 2
B U 3 A
where,
T
U = + +
T
=
+ γ −
−
2
R I
2
B U 3 2
B , and
2
1
U 3 X∞
X∞ 1
B U 1 1
B X∞ .
Discrete Time Mixed LQR/H∞ Control Problems
167
In this case, the state feedback mixed LQR/ H∞ controller will achieve
T
2
sup inf{ J} = x
+ γ −
−
subject to
< .
0 ( X∞
X
X )
w
z x 0
zw
T
γ
∞
K
w∈ 2
L +
∞
where,
ˆ
2
1 T
=
+ γ −
−
ˆ k T T
−
T
ˆ k
K
A
K
A
K
B U 1 K
B X∞ K
A ,
2
X = ∑{( A )
, and
w
K
K
A X∞ K
B U 1 B X∞ A A }
K
K K
k=0
∞
ˆ k T T
ˆ
X = ∑ {( A )
k
C C A } .
z
K
K K K
k=0
Before proving Theorem 3.1, we will give the following lemma.
Lemma 3.1 Suppose that the discrete time Riccati equation (13) has a stabilizing solution
X
T
−
T
∞ ≥ 0 and
2
U = − γ −
>
= +
= −
1
I
1
B X∞ 1
B
0 , and let K
A
A
2
B K and
1
K
U 2 2
B U 3 A ; then K
A is
stable.
Proof: Suppose that the discrete time Riccati equation (13) has a stabilizing solution X∞ ≥ 0
and
2 T
U = − γ −
>
1
I
1
B X∞ 1
B
0 . Observe that
1
−
T
1
−
T
γ
⎡
⎤
⎡−
⎤ ⎡
−
γ
⎤
T
1
ˆ
ˆ
ˆ
B
1
I
0
U
−
1
1
B X∞ 2
B
B X
+ = ⎢
⎥
⎡
⎤
∞ B
R
X γ
∞
+
=
1
B
2
B
⎣
⎦ ⎢
⎥ ⎢
⎥
T
1
⎢⎣ B ⎥⎦
⎣ 0
−
T
T
R + I⎦
γ
⎢
+ +
2
⎣
2
B X∞ 1
B
2
B X∞ 2
B
R I⎥⎦
Also, note that
2 T
U = − γ −
>
T
=
+ γ −
−
= +
1
I
1
B X∞ 1
B
0 ,
2
1
U 3 X∞
X∞ 1
B U 1 1
B X∞ , and U 2 R I
T
+ 2
B U 3 2
B ; then it can be easily shown by using the similar standard matrix manipulations as
in the proof of Theorem 3.1 in Souza & Xie (1992) that
1
−
1
−
⎡
ˆ
1
− ˆ T
1
−
1
− ˆ
1
U
−
+ U B U B U
U B U− ⎤
ˆ T
ˆ
ˆ 1
−
1
1
1 2
1
1
1
1 2
( B X∞ B + R) = ⎢
⎥
1
− ˆ T
1
−
1
⎢
U
−
2
1
B U 1
U 2
⎥
⎣
⎦
where, ˆ
1 T
= γ −
1
B
1
B X∞ 2
B .
Thus, we have
T
ˆ ˆ T
ˆ
ˆ 1
− ˆ T
2
−
T
1
−
T
T
1
∞ (
∞ +
)
−
T
A X B B X B R B X∞ A = γ
−
A X∞
+
1
B U 1 1
B X∞ A A U 3 2
B U 2 2
B U 3 A
Rearraging the discrete time Riccati equation (13), we get
T
2
−
T
1
−
T
T
1
−
T
T
X∞ = A X∞ A + γ A X∞
−
+
+
1
B U 1 1
B X∞ A A U 3 2
B U 2 2
B U 3 A C 1 C 1 Q
T
2
−
T
1
−
T
T
T
1
−
T
2
−
1
−
T
= A X∞ A + γ A X
+
+ −
+ γ
∞ 1
B U 1 1
B X∞ A C 1 C 1 Q A U 3 2
B U 2 2
B ( X∞
X∞ 1
B U 1 1
B X∞) A
T
2
−
1
−
T
1
− A (
−
T
X + γ
∞
X∞ 1
B U 1 1
B X∞ ) 2
B U 2 2
B U 3 A
T
1
−
T
2
−
1