Discrete Time Systems by Mario A. Jordan and Jorge L. Bustamante - HTML preview

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