H Optimal Control Problem Formulation

2010 Spring ME854 - GGZ Page 1HH∞∞ Optimal ControlOptimal Control

HH∞∞ Optimal ControlOptimal Control

Problem FormulationProblem Formulation

2

a) Two problems to be discussed: optimal and suboptimal control

b) Behavior of suboptimal controller as function of to be discussed

c) Integral control in and theory

d) filter de

H H

H

H H

H

γ

∞ ∞

∞

∞

∞ sign technique

e) Assumptions: and are rational proper with state space realizationsG K

Guy

K

z w

minimized is that such )( scontroller admissible all Find

controller gstabilizin

∞zwTsK�� Optimal H∞∞∞∞ Control

2a) Different from controller, the controllers are generally not unique

b) Finding an optimal controller is numerically and theoretically difficult

c) In practical, it is not necessary and pr

H H

H

∞

∞

actical to design an optimal controllerH∞

Suboptimal H∞∞∞∞ Control (feasible solution)

γγ <>∞zwTsK that such any, are thereif ),( scontroller admissible all find 0, Given

Main topic of this chapter



A Simplified A Simplified HH∞∞ Control ProblemControl Problem

Guy

K

z w

control & between prelatioshi some ly respective , and and , As b)

dom(Ric), that guarantee no definite signnot are blocks 2)-(1 matrices' Botha)

.: ,:

:matrices an Hamiltoni twofollowing theinvolves solution The

222

*

11

2

*

21

*

1

2*

*

1

*

1

*

22

*

11

2

∞∞∞

∞∞

−

∞

−

∞

∞

⇒→∞→

∈⇒

−−

−=

−−

−=

HHJHJH

JH

ABB

CCCCAJ

ACC

BBBBAH

H

γ

γγ

[ ] [ ]

=

=

=

ID

D

B

DCD

ACBA

ACBA

DC

DC

BBA

G

0 a4)

I0 a3)

detectable is ),( and ablestabiliz is ),( a2)

observable is ),( and ablecontroll is ),( a1)

sassumption following thewith

0

0

below nrealizatio the withdiagramleft by the described system heConsider t

*

21

21

1

121

*

12

22

11

212

121

21




∞Mu y

Q

2

There exist an admissible controller such that iff the following three conditions hold:

i) dom(Ric) and : Ric( ) 0;

ii) dom(Ric) and : Ric( ) 0;

iii) ( ) .

Moreover, when

zwT

H X H

J Y J

X Y

γ

ρ γ

∞

∞ ∞ ∞

∞ ∞ ∞

∞ ∞

<

∈ = >

∈ = >

<

2 * * * 2 1 1 1 2 2 2 2

ˆthese condition hold, one such controller is ( ) , where

0

ˆ : , : , : , : ( )

Futhermore, the set of all admissible contro

sun

A Z LK s

F

A A B B X B F Z L C F B X L Y C Z I Y Xγ γ

∞ ∞ ∞

∞

− − −

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

−=

= + + + = − = − = −

2

2

llers such that equals the set of all transfer

matrices from to is

ˆ

( ) 0 ,

0

where and .

zwT

y u

A Z L Z B

M s F I

C I

Q RH Q

γ

γ

∞

∞ ∞ ∞ ∞

∞ ∞

∞ ∞

<

−

= −

∈ <

Theorem 14.1

theorem theof portionfirst theproveonly togoing are We




rnYI

IX

YI

IX

Y

XX

XX

XX

XX

XXRXRX

rYYXXRYRX

n

n

n

n

rrrn

nnnn

+≤

≥

=

>

=∈∈

>=>=∈∈

−

××

××

rank and 0

ifony and if

**

* and 0

and that such , matricesexist thereThen

interger. positivea be Let .0 and ,0 with, , that Suppose

1

2

*

12

12

2

*

12

12

*

22212

**

Proof of Lemma 14.2

=

∑−=

∑

∑

>

∑

∑−

∑=

∑

∑

==∑=−

≤−

≥−⇒≥

−

=

⇐

−−

>

−

−

−

−

−−

−

**

*

**

*)(

00

0

00

have we,select and , Therefore,

)rank(

00

0

0

0

0

have weion,decompositschur Using)(

1*

11

1

*

1

*2/1

2/1

1

*

1

*2/1

0

*

11

2/1

1

*

1

*2/1

2/1

1

2

*

1212

*

11

1

1

1

1

11

1

YUUX

IU

UX

IU

I

I

UUXI

UI

IU

UX

IXXXUUYX

rYX

YX

IY

I

Y

YX

I

YI

YI

IX

r

rr

r

rr

n

r

Y

r

r

rn

rr

r

rr

n

n

n

n

n

n

��

Lemma 14.2



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemProof of Lemma 14.2 (cont’d)

rXXXYXXXXYX

XXXXYYXXXX

XX

XX

≤=−≥=−⇒

−=⇒

=

−=

⇒

−−−−

−−−−−

)rank()rank( and 0

**

*

**

*)(

have weion,decompositschur Using)(

*

12

1

212

1*

12

1

212

1

*

12

1

212

1

1*

12

1

212

1

2

*

12

12

Lemma 14.3

rnYI

IX

YI

IX

CCCCXBBXXAAX

X

BBBBYCCYAYAY

Y

Tt

n

n

n

n

zw

+≤

≥

<−+++

>

<−+++

>

<∞

γ

γ

γ

γ

γγ

γγ

γ

/

/rank and 0

/

/ iii)

0/

thatsuch 0 an exits There ii)

0/

thatsuch 0a exists There i)

:hold conditions

threefollowing theifonly that such controller admissibleorder th- an exists There

1

1

1

1

2

*

2

2

1

*

1

2

1

*

1111

*

1

1

*

22

2*

11

2

11

*

11

*

11

1



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemProof of Lemma 14.3

Suppose that there exists a -th order controller ( ) such that . Let ( ) have a state space

ˆ ˆrealization ( ) . Then using star product formulea, we have

ˆ ˆ

( , )

zw

zw l

r K s T K s

A BK s

C D

A B

T F G K

γ∞

<

=

+

= =

2 2 2 1 2 21

2 21

1 12 2 12 12 21

2 * 2 *

1 12

*

12 2

ˆˆ ˆ

ˆˆ ˆ : .

ˆˆ ˆ

Let and . Since , by Corollary 12.3, there exists an

0 such that

c c

c c

c c c c zw

DC B C B B DDA B

BC A BDC D

C D DC D C D DD

R I D D R I D D T

X XX

X X

γ γ γ∞

+

=

+

= − = − <

= >

ɶ

ɶ

1 * 1 * * 1 * * 1

* * 2 * 2 *

1 1 1 1 1 1 1 1 2 2

( ) ( ) 0

After much algebraic manipulation, we have

/

c c c c c c c c c c c cX A B R D D A B R D D X XB R B X C R C

X A A X X B B X C C C Cγ γ

− − − −+ + + + + <

+ + + −

+

ɶ ɶ ɶ ɶ ɶ

2 * 2 * 1 2 * *

1 1 12 2 1 1 12 2

* 2 * 2 *

1 1 1 1 1 1 1 1 2 2

ˆ ˆ ˆ ˆ ˆ ˆ( )( ) ( ) 0

which implies

/ 0.

X B D X B C I D D X B D X B C

X A AX X B B X C C C C

γ γ γ

γ γ

−+ + − + + <

+ + + − <




rnYI

IX

YI

IXY

XX

XX

XY

XYXXYX

YCCYBBBBAYAY

BCYDCYDDIBCYDCY

YCCYBBBBAYAY

BRBYCRCYCDRBAYYCDRBA

YY

YYYYXY

n

n

n

n

cccccccccccc

+≤

≥

=

=

=>>

<+−++

<++−+++

+−++

<+++++

>

==

−

−

−

−

−−−−

−

γ

γ

γ

γγ

γγ

γγ

γγ

γγγγγ

γγ

γγγ

γγ

γ

/

/rank ,0

/

/ iff

**

*/

//

//

:))/~

(/~

(

~~

that such / and / exists there,0/ and 0/ given 14.2,By Lemma

.0/

implies which

0)ˆˆ()ˆˆ)(ˆˆ(

/

provides this,Similarily

0~~~~

)(~~

)(

Then .0~

as ~

partition and ~~

Let

1

1

1

11

1

2

*

12

121

1

12

21211

2

11

*

11

*

22

2*

11

*

11

*

2

2*

12

**

11

1*2

2

2*

12

**

11

2

11

*

11

*

22

2*

11

*

11

*11***1*1

2

*

12

12112



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemTheorem 14.4 *

*

Let 0 and suppose ( , ) is controllable and there is an such that

( ) : 0

Then there exists a solution to the Riccati equation

R A R X X

X XA A X XRX Q

X X+

≥ =

= + + + <

>

Q

* 0

such that is antistable.

X A A X X RX Q

A RX

+ + + ++ + + =

+

Proof of Theorem 14.4 *

0

0 0

Let for some . Note that ( , ) is controllable iff ( , ) is. Let be such that ( ) 0.

Since ( , ) is controllable, there is an such that

:

is

R BB B A R A B X X

A B F

A A BF

= <

= +

Q

*

0 0

* *

0 0 0 0 0 0

*

0 0

antistable. Now let be the unique solution to the Lyapunov equation

0

ˆDefine : and we have the following equation

X X

X A A X F F Q

F F B X

=

+ − + =

= +

* *

0 0 0 0 0 0

0 0

ˆ ˆ ( ) ( ) ( ) 0

Since is antistable, we have .

Now we are going to prove the Theorem by an inductive method.

X X A A X X F F X

A X X

− + − = − >

>

Q



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemProof of Theorem 14.4 (cont’d-1)

0Starting with , define a sequence of Hermitian matrices { }. Associated with { }, define a

sequense of antistable matrices { } and a sequense of matrices { }. Assume inductively that

we have alre

i i

i i

X X X

A F

0 1 1

*

1

ady defined matrices { }, { }, { } for up to 1 such that is Hermitian and

, is antistable for 0,1, , 1

(*) ,

i i i i

n i i

i i

X A F i n X

X X X X A A BF i n

F B X

−

−

−

≥ ≥ ≥ > = − = −

=

⋯ ⋯

* *

*

1

1, 2, , 1;

Now let

and

First we show that is antistable. Using (*) with 1, we have

i i i i i i

n n n n

n

i n X A A X F F Q

F B X A A BF

A i n

X

−

= − + = −

= − = −

= −

⋯

1

*

*1 1

* *

1 1 1 1 1 1

* * * * * *

1 1 1 1 1 1 1 1

( ) ( )

( ) ( ) 0

( ) ( ) 0

or

n

n n

n n n n

A

n n n n n n

A F

n n n n n n n n n n n n n n

F F F F

A BF A BF X F F Q

X A BF A BF X F F F F X B F F B X F F Q

−

− −

− − − − − −

− − − − − − − −

− − −

− + − − + =

− + − − − − − − + =

��

��

��

* * *

1 1 1 1

*

* * *

1 1 1 1

( ) ( ) 0 (**)

ˆLet : , then

ˆ ˆ ( ) ( ) ( ) ( ) ( ) 0

which implies that

n n n n n n n n n n

n n

n n n n n n n n n n

n

X A A X F F F F F F Q

F F B X

X X A A X X F F F F F F X

A

− − − −

− − − −

+ − − − − + =

= +

− + − = + − − − >Q

1is antistable by Lyapunov theorem since 0.

nX X− − >



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemProof of Theorem 14.4 (cont’d-2)

1

* *

Now we prove . Let be the unique solution of the following Lyapunov equation

.

Then is Hermitian. Next we hav

n n n

n n n n n n

n

X X X X

X A A X F F Q

X

− ≥ >

+ = −

* *

* *

1 1 1 1

e

ˆ ˆ ( ) ( ) ( ) 0

Using equation (**), we have

( ) ( ) ( ) ( ) 0

Since is antistable, we h

n n n n n n

n n n n n n n n n n

n

X X A A X X F F X

X X A A X X F F F F

A

− − − −

− + − = − >

− + − = − − ≥

Q

1

ave

Since we have a lower bounded (by ) nonincreasing sequence { }, the limit exists

n n

i

X X X

X X

− ≥ >

*

*

: lim

is Hermitian and . As , we get ( ) 0. Therefore is the solution of

0

Note that

( ) ( )

nn

X X

X X n X X

X A A X X RX Q

X X A A X X

+→∞

+ + +

+ + + +

+ + + +

=

≥ → ∞ =

+ + + =

− + −

Q

*( ) ( ) ( ) 0

Hence 0 and is antistable.

X X X R X X

X X A A RX

+ +

+ + +

= − + − − >

− > = +

Q



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemLemma 14.5

2

1

1

*

112

*

2

2

1

*

1

*

1

*

1

*

22

2*

11

*

)(or 0 iii)

0)/(

to0 solution gstabilizina exists There ii)

0)/(

to0 solution gstabilizina exists There i)

:hold conditions threefollowing theifonly that such controller admissible an exists There

γρ

γ

γ

γ

<>

=+−++

>

=+−++

>

<

∞∞−

∞

−

∞

∞∞∞∞

∞

∞∞∞∞

∞

∞

YXγXI

IγY

BBYCCCCYAYAY

Y

CCXBBBBXXAAX

X

T

n

n

zw

Proof of Lemma 14.5

1

* * 2 * 2 *

1 1 1 1 2 2

* 2

1 1

Apply Theorem 14.4 to part i) of Lemma 14.3, it can be concluded that there exists a 0 such

that

/ 0

and / is antistable. Let :

Y Y

AY YA YC C Y B B B B

A C C Y X

γ γ

γ ∞

> >

+ + + − =

+ 2 1

* * 2 * *

1 1 2 2 1 1

* 2 * 1 * 1 1 * 2

1 1 2 2 1 1 1 1

, we have

( / ) 0

and

( / ) ( ) ( / )

is stable.

Y

X A A X X B B B B X C C

A B B B B X X A C C X X X A C C Y X

γ

γ

γ γ

−

∞ ∞ ∞ ∞

− − −

∞ ∞ ∞ ∞ ∞ ∞

=

+ + − + =

+ − = − + = − +




1

* * 2 * 2 *

1 1 1 1 2 2

* 2

1 1

Similarily, apply Theorem 14.4 to part ii) of Lemma 14.3, we concluded that there exists an

0 such that

/ 0

and / is antistable.

X X

XA A X XB B X C C C C

A B B X

γ γ

γ

> >

+ + + − =

+ 2 1

* * 2 * *

1 1 2 2 1 1

* 2 *

1 1 2 2

Let : , we have

( / ) 0

and ( / ) is stable.

Note that the rank condition in part iii) is automatically satisfied by choose

Y X

AY Y A Y C C C C Y B B

A C C C C Y

γ

γ

γ

−

∞

∞ ∞ ∞ ∞

∞

=

+ + − + =

+ −

11

11

2

, and

/ / 0

/ /

or ( )

n nn

n nn

r n

X I X IY I

I Y I YI X

X Y

γ γγ

γ γγ

ρ γ

−

∞

−

∞

∞ ∞

≥

= > ≥

<



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemProof of Theorem 14.1 To complete the proof, we need to show that the controller given in Theorem 14 renders

. Note that the closed loop transfer function with is given by

sub

zw sub

zw

K

T K

A

T

γ∞

<

=

2 1

2 21

1 12

2 1 2 1 1

2 1 * 1 2 1 1

ˆ :0

0

Note that

0( )

and satisfies

c c

c

B F BA B

Z L C A Z L DC

C D F

Y Y ZP

Z Y Y Z

γ γ

γ γ

∞

∞ ∞ ∞ ∞ ∞

∞

− − −

∞ ∞ ∞

− − − −

∞ ∞ ∞ ∞

− − =

−= >

−

* * 2 *

* 1 * 1 1

* 2 1 1 2 1 1

* 2

1 1 2

* 1

1 1 1

/ 0

Moreover

/0 /

has no eigenvalues on the imaginary axis since is antistable and

c c c c c c

c c c

PA A P PB B P C C

A B B Y B F B B Y ZA B B P

A B B X B F

A B B Y A B

γ

γγ

− − −

∞ ∞ ∞ ∞

∞ ∞

−

∞

+ + + =

+ −+ =

+ +

+ + * 2

1 2/

is stable. Thus, by Corollary 12.3 (v), .zw

B X B F

T

γ

γ

∞ ∞

∞

+

<



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemRemark 14.1

1

a) It worth to mention that these conditions stated in Lemma 14.3 are necessary and sufficient.

b) When applying the necessity, one need to be suitably interpreted. For example, one finds 0

a

X >

1

1 1

nd 0 satisfying conditions (i) and (ii), but not (iii), this does not means there is no admissible

controller since there might be other 0 and 0 that satisfy all three conditions.

c)

Y

H X Y∞

>

> >

2

1 1

1 12

1 1

For instance, consider 1 and

1 1 0 1

1 0 0 0 ( ) , ( 0.7321)

0 0 0 1

1 0 1 0

i) 2 0 when select 0.5

ii) 2 0

optG s

Y YY X

X X

γ

γ

=

− = =

⇒ − + < = =

⇒ − + <

1 1

both conditions i) and ii) are satisfied

0.5 1 iii) is not positive semidefinite

1 0.5

But if selecting both 1 , 2, conditions i), ii), and iii) are satisfied.Y X

⇒

⇒

< <



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemExample 14.1

++ +

+

-eW

eyd

uW

id

pu

u

u~

K P

:)()(

)()(~

thatNoteminimized.is~to

fromnormthat thesocontrollera designshallWe

10

1,

2.0

2,

)2)(1(

)4.1(50

right withonshownsystemfeedback heConsider t

11

11

=

+−+−

++=

=

=

+

+=

+=

++

+=

−−

−−

∞

i

zw

iuu

ee

i

ue

d

dT

d

d

PPKIKWPKIKW

PPKIWPKIW

u

e

u

ez

d

dw

H

s

sW

sW

ss

sP −

−

−−

−−

−

=

−

−

−−

−−

−

=

+=

+=

−+−=

−+−=

−=

=

−−−=

++−=

u

d

d

x

x

x

x

y

u

y

x

x

x

x

u

d

d

x

x

x

x

y

u

e

x

x

x

x

dyy

xxy

udxx

udxx

uxu

xe

uxx

ydxx

WWP

i

u

p

p

e

u

p

p

e

i

u

p

p

e

u

p

p

e

p

ppp

ipp

ipp

u

e

uu

pee

ue

2

1

2

1

2

1

2

1

21

22

11

0010110

1000000

0000110

90010000

303000200

202000010

0020222.0

0010110

1001000

0000001

90010000

303000200

202000010

0020222.0

~

,)(30

)(20 ~),(910

),(22.0

are ,,for equations aldifferenti The

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemExample 14.1 (cont’d)

10-4

10-2

100

102

104

106

108

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

10-4

10-2

100

102

104

106

108

0

0.5

1

1.5

2

2.5

7875.0 where

)12.0/)(14.1/)(197.21/)(110335.2/(

)14.1/)(127.7/)(110/(78.12ˆ

0.001 tol withController

7849.0 where

)12.0/)(14.1/)(119.22/)(110245.3/(

)14.1/)(127.7/)(110/(82.12

0.0001 tol withobtained function transfer Controller

subopt

3

subopt

7

=

++++×

+++=

=

=

++++×

+++=

=

γ

γ

ssss

sssK

ssss

sssK

Weighting with)( Responses Frequency LoopClosed zwTσ

Weighting w/o)( Responses Frequency LoopClosed zwTσ

K with

K̂ with

K with

K̂ with

⇒ weightingincluding function transfer loop Closed

⇒ weightingexcluding function transfer loop Closed



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemExample 14.2

+

=

+

−−

−−=

======

+

=

+

=

+−+−

−−

2

1

2

1

2

1

4

3

2

1

4

3

2

1

212121

2

1

2

1

2

1

1

1

4

3

2

1

)()(

4

3

2

1

00

00

0010

0001

5.00

01

00

00

15.01.05.25.0

2.02.011

1000

0100

1.0,2.0,2,1,4,1 where

00

00

0010

0001

0

0

00

00

1000

0100

:equations aldifferenti system following by the described be can system

dynamical The right. on systemdamper -spring-massa Consider

2

1

2

21

2

1

2

21

2

1

1

1

1

1

1

1

1

1

F

Fx

x

x

F

F

x

x

x

x

x

x

x

x

bbmmkk

F

Fx

x

x

F

F

x

x

x

x

x

x

x

x

m

m

m

bb

m

b

m

kk

m

k

m

b

m

b

m

k

m

k

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

ɺ

1x

2x2F

1F

1m

2m

2k

1k1b

2b



A Simplified A Simplified HH∞∞ Control ProblemControl ProblemExample 14.2 (cont’d-1)

0.01( 10)

1100

0.01( 10)

2100

1 1 1

1

2 2 2

The system diagram is shown on right, where

00 5 , ,

0 500

As shown in the book, let

, ,

s

s

n e us

s

n

W sW W W

W s

y x ny W u F

y x n

+

+

+

+

+= = =

+

= = + = [ ]

1212

21

2 2 1

0

0 0 , , ( )

But this interpretation is difficult to obtain lowest order realization, we consider another arrangeme

ee

e

u

nu

W PW PxFW

Wxw n z G s

n P W PW u

= = =

[ ]

1

1 1 1

2 11 2 1

2 2 2

2 2 2 1

nt

0 0

0 , , , , ( )

We can have a 9-th order state space realization (

u u

e en

e

n

W u Wny x n

W P W Py W u F w n z G sxWy x n

F x W P P

= = + = = = =

4 , 2 , 2 , 1 ) using tranfer

matrices augmentation.

G22 mmult(abv(We,eye(2))); G12 sbs(0,Wu);G21 abv(zeros(2,2),Wn),G11 zeros(1,2);

G1 sbs(G11,G12);G2 sbs(G21,G22);

G abv(G1,G2)

th nd nd st

n e uP W W W→ → → →

>> = = = =

>> = =

>> =

+

+

eW1z

y

uW

1Fu =

2z

K P

nW

2

1

x

x

=

2

1

2n

nw

21 Fw =

y




:systems for three designed werescontroller and 2HH∞

+

+

eW1z

y

uW

1Fu =

2z

K P

nW

2

1

x

x

=

2

1

2n

nw

21 Fw =

y

10-4

10-3

10-2

10-1

100

101

102

0

1

2

3

4

5

6

7

scontroller and of responsesfrequency loop Closed 2HH∞

2H

∞H

4.7949 6.6068

8.6758 3.9247

designH ∞designH 2

normH ∞

normH 2

=

2

1

x

x

uW

z

u

0.6744 0.6764

0.1855 0.1849


normH ∞

normH 2

=

2x

uWz

u

0.9663 1.5994

29.5785 1.6320


normH ∞

normH 2

=

1x

uWz

u

=

2

1

x

x

uW

z

u



10-4

10-2

100

102

104

106

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6


+

+

eW1

zy

uW

1Fu =

2z

K P

nW

2

1

x

x

=

2

1

2n

nw

21 Fw =

y

2Closed loop frequency responses of and controllers H H∞

10-4

10-3

10-2

10-1

100

101

102

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

=

2x

uWz

u

∞H

2H

2H

∞H

=

1x

uWz

u



Optimality and Limiting BehaviorOptimality and Limiting Behavior

a) We are going to discuss the behavior of the suboptimal solution as reduces

to . Since Theorem 14.1 provides necessary and sufficient conditions for existance

of an admissible contr

opt

H

γ

γ∞

2 2 2

oller such that , is an infimum over all such

that conditions i), ii) and iii) are satisfied.

b) It's clear: as , , , is the minimum entropy

solution and also

zw opt

sub sub

T γ γ γ

γ H H X X K K K

∞

∞ ∞

<

→ ∞ → → → ⇒2 22

2 2

2 1 1 2 1 2

a minimax controller for

c) ( ) ( ) and ( ) ( ) and are decreasing

function of .

d) The formulea in Theorem 14.1 are not well defined in the optimal case since

opt

z w

X X Y Y X Y

γ

γ γ γ γ γ γ γ

γ

∞ ∞ ∞ ∞ ∞ ∞

−

≥ > ⇒ ≥ ≥ ⇒

2

1 1 1

( ) is not invertible. It is possible but far less likely that conditions i) and ii)

would fail before iii)

e) If complementary fail for at , then ( ) as . For ,

optI X Y

H X H

γ

γ γ ρ γ γ γ γ

−

∞ ∞

∞ ∞ ∞

−

= → ∞ → < may

again belond to dom(Ric) but may become indefinite. For such , the corresponding

controller may make but not stabilizingzw

X

T γ

γ∞

∞<




[ ]

[ ]

0.a and 0a :cases woconsider t togoing are We

1Im)_()_(

and exists )_( and )_( then, If

1)1(a

are and both of seigenvalue The

1

1 ,

0101

1

0

00

00

0

1

101

)(

by given nrealizatio system dynamical ctedinterconne heConsider t

1)1(a

1a

1

22

1

*

11

2

*

21

*

1

2*

1

*

1

*

1

*

22

*

11

2

22

2

2

2

2

2

<>

==

>

−+

±

−−=

−−

−=

−−=

−−

−=

=

−−+

∞∞

∞∞+

∞∞

−−

∞

−−

∞

γ

γγ

γ

γ

γ

γ

χχ

χχγ

γ

γ

γ

γ

a

JH

JH

JH

a

a

ABB

CCCCAJ

a

a

ACC

BBBBAHa

sG

Example 14.3




fail ii) and i) conditions before fails iii) condition

1 as )()1)1(a(

)(

but

satisfied ii) and i) conditions 01)1(a

and 01)1(a

and dom(Ric) and dom(Ric) 1, c)

1 if ;0

1 if ;0

1)1(a and dom(Ric)1 and

1a

1 b)

a eigenvaluebut ,invertiblenot 01)1(a1 a)

sinceproperty stability thebefore fail willdom(Ric) ofproperty ary complement thecase, thisIn

:0a Case

222

2

2222

1a

1222

1

22

2

⇒

→∞→⇒−−+

=

⇒>−−+

=>−−+

=

∈∈>

<<

>>=

−−+=∈⇒≠

+>

−=⇒=−−+⇒=

>

∞∞∞∞

∞∞

∞∞

+

∞∞

γργγ

γρ

γγ

γ

γγ

γ

γ

γ

γ

γγ

γγγ

γγγ

YXa

YX

aY

aX

JH

aXH

Xa

Example 14.3 (cont’d-1)




{ }

fail ii) and i) conditions before fails iii) condition

1a

12a ifonly and if

)1)1(a()(

thatshown be canIt .0)11(/ and

012)/(

equation Riccati the tosolutiona is 1

1

Im )_(

seigenvalueimaginary with include todom(Ric) extend can we,1a

1 if case, thisin Evenc)

imaginary)1(a failsproperty stability since dom(Ric) , ,1a

1for b)

1a

1for 0

1)1(a and 0

1)1(a

and dom(Ric) , re, Furthermosatisfied. always isproperty ary complement thecase, thisIn a)

:0a Case

2

22

222

2

21*

22

2

0

*

11

22

1

*

1

*

22

2

0

*

11

*

20

12

2

22222

2

⇒

+++><

−−+=

=−+−=−+

=++=+−++

−=⇒

−=

+=

⇐−+±⇐∉+

≤

+>>

−−+=>

−−+=

∈

<

∞∞

∞∞

∞∞∞∞∞∞

∞∞

∞

∞∞

∞∞

∞∞

aa

YX

aaXBBXBBA

aXaXCCXBBBBXAXXA

aX

aH

H

JH

aY

aX

JH

a

γγγγ

γρ

γ

γ

χ

γ

γ

γγγ

γ

γγ

γ

γ





[ ]

17321.0310

00

)31(31

311

3

1

310

00311

31

1

3

1 d)

7321.007321.0

001

7321.017321.1

re Furthermo.7321.0 and 7321.013 have we,1a wheninstance, For c)

.1)1(a

isThat constant.a is

ˆ ,ˆˆ)(

equations descriptor by the given controller optimal thefact, In b)

a2a

by given isfeedback output for the optimal the0,a and 0a cases both In a)

2

22

2

2

opt

<=

−+

−−

−=

−+−

−+=

−−

−−

=

−==−=−=

−−+−=

=−=−

++=

><

∞

∞∞∞∞

−

σ

γ

γγ

γ

γ

γ

γ

zw

zw

zw

optopt

optopt

opt

sopt

T

sT

T

yu

ya

u

xFuyLxAxXYI ɺ




Minimum Entropy ControllerMinimum Entropy Controller

( )

( ) ωωωγπ

γγ

γ

γ

γ

ωωσπ

γ

ωωσωσγγ

ωωσγπ

γγ

ωωωγπ

γγ

γ

γ

djTjTITI

T

H

TH

TdjTTI

jTijTjTTI

djTTI

djTjTITI

sTTT

zw

zw

i

i

ii

i

i

∫

∫ ∑

∫ ∑

∫

∞

∞−

−

∞

∞∞

∞

∞−∞→

−

∞

∞−

−

∞

∞−

−

∞

−−=

<

∞→

→

==

<−≥

−−=

−−=

<

)()(detln2

),(

entropy following theminimizes and

satisfies controller heactually t is 14.1 Theorem in controller suboptimal that theshown been hasIt b)

)( optimality and

)( optimality between off- trademeasuringindex eperformanc an isentropy Therefore

))((2

1),(lim

thatshown be canIt

).( of aluesingular v theis ))(( where),0))((1ln (since 0),( and

))((1ln2

),(

thatsee easy to sIt'

)()(detln2

),(

as defined is )( ofentrpy theThen . hmatrix wit a transfer be Let a)

**22

2

2

2

2

th22

222

*22



An Optimal ControllerAn Optimal Controller

KKYKXKKK

KK

KKKKopt

YiY

nn

XiX

nn

zw

CBYETXCBTEAXBC

CYBXYXYE

BAsECsK

YYXY

YXXX

YYYY

iTTY

Y

Y

YJR

Y

Y

XXXX

iTTX

X

X

XHR

X

X

T

2

*

1

*

122

*

2

*

2

*

22

*

2

2

1

*

1

1

*

22

*

2

1

2

*

2

1

1

*

2

1

*

22

*

1

2

1

2

12

2

1

1

*

22

*

1

2

1

2

12

2

1

: ,:

: ,:

where

)(:)(

is controller such one hold, conditions these whenMoreover,

.0 iii)

; and

0)(Re , that such

matrixrank column fulla exists There ii)

; and

0)(Re , that such

matrixrank column fulla exists There i)

:hold conditions threefollowing theiff that such controller admissible an exists There

∞∞∞

∞∞∞

−

∞∞

+

∞∞∞∞−

∞∞

−

∞∞

∞∞∞∞

∞

∞

∞

∞

∞×

∞

∞

∞∞∞∞

∞

∞

∞

∞

∞

×

∞

∞

∞

+=−=−=

=−=

−=

≥

=

∀≤

=

∈

=

∀≤

=

∈

≤

γ

γ

γ

λ

λ

γTheorem 14.6



An Optimal ControllerAn Optimal Controller

1 1

1 1 2 1 2 1

a) The three conditions are the generalization of these in Theorem 14.1

b) If both and are invertible, and , then condition iii) becomes

0 and 0, and ( )

X Y X X X Y Y Y

X Y X Yρ

− −

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

∞ ∞ ∞ ∞

= =

≥ ≥ 2

1 1

c) If and are singular, the controller is in a form of descriptor system (singular system)

d) The proof is too complicated to be provided in class

e) The general solution of will not b

X Y

H

γ

∞ ∞

∞

≤

e discussed in class neither

Remark 14.2



HH22 and and HH∞∞ Integral ControlIntegral Control

2a) It is interesting that either or control

design does not produce integral control.

b) In order to let output to track the reference

signal , we require to contain an integrator

H H

y

r K

∞

1

1

1

(i.e., ( ) has a pole at 0)

c) One way is to introduce an integral in output weighting . The transfer function between and is

( )

e

e d

K s s

W w z

z W I PK W−

=

= +

2

If the resulting controller stabilizings and the resulting CL system has finite 2-norm or -norm,

it shall contain a pole at 0. But the and control theories presented DO NOT allo

w

K P

s H H∞

∞

=

1

w

uncontrollable pole(s) on imaginary axis.

d) Factorize as ( ) ( ), where is proper

contains all imaginary poles of , and ,

and is stable and minimal phase.

e) Su

e e e

e

e

W W W s M s M

W M RH

W

−

∞

=

∈

ɶ

ɶ

ppose there exist a controller ( ) can be factorized

as such that there is no unstable pole/zero cancellations.

The problem can be reformulated on right.

K s

+ +

+

-eW

1zy

w

uW

u

2z

K P

dW

r

−

+ +

+

-eW

~ 1zy

w

uW

u

2z

K̂ M

dW

r

−

P

=

MPMW

W

MPWMWW

sG

d

u

ede

~

0

~

)(




−

−

−−

−−

=

−=

=

+=

+=

=

−−=

+

+=

=

−

=−+

−=

u

w

x

x

x

x

x

y

z

z

x

x

x

x

x

WsM

sW

s

sM

sW

s

sW

Wss

ssP

m

u

e

m

u

e

e

e

eu

d

2

1

1

2

1

2

1

~

0112100

1000010

0000001

1023000

0010000

0448000

9000001000

0112104~

:nrealizatio following thehave We .01

14~ and

11

40)( that Note

4

1~ ,

4 that generality of loss without choose We

1 ,

11

90100

100

10

1 ,

012

123

010

)3)(1(

2)(Let

ɺ

ɺ

ɺ

ɺ

ɺ

Example 14.4

eW~1z w

+

-

1y

uW

u

2z

K̂

M

dW

P




)883.19)(181.32(

)15564.0)(100)(1(85.7

)883.19)(107403.2)(181.32(

)15564.0)(100)(1(101518.2

have we

),(ˆ Since 7.8547. norm loop closed with

)883.19)(107403.2)(181.32()4(

)15564.0)(100)(4)(1(101518.2ˆ

is ˆ controller suboptimal The

4

5

42

5

−+

−++≈

−×++

−++×=

−=

−×+++

−+++×−=

∞

∞∞∞

∞

∞∞

sss

sss

ssss

sssK

sMKKH

ssss

ssssK

KH

Example 14.4 (cont’d)

10-4

10-2

100

102

104

106

0

2

4

6

8

10

12

14

16

18

2H

∞H

7.8547 18.205

919.3 16.412


normH ∞

normH 2

)964.7)(81.41194.30(

)069.0)(100)(1(487.43

have we),(ˆ Since 16.412. norm with

)964.7)(81.41194.30()4(

)069.0)(100)(4)(1(487.43ˆ

:follows as

obtained be can controller optimal an ,Similarily

22

222

222

2

−++

−++=

−=

−+++

−+++−=

ssss

sssK

sMKKH

ssss

ssssK

H




−−

−−

−−

=

−=

−−==

−

=

=+

=

u

w

x

x

x

x

y

z

z

x

x

x

x

WWWsP

sW

u

e

u

e

eud

e

2

1

1

2

1

2

1

011200

100010

000001

102300

001000

900001000

01120001.0

:nrealizatio following thehave We

.01

1001.0,

11

90100,1,

012

123

010

)(

0.001 example, For this. enough smallfor 1

lettingby achieved be can control integral eapproximat An

ɺ

ɺ

ɺ

ɺ

εεε

Example 14.5

+ +

+

-eW

1zy

w

uW

u

2z

K P

dW

r

−



10-4

10-2

100

102

104

106

0

2

4

6

8

10

12

14

16

18


)883.19)(181.32(

)15447.0)(100)(1(85.7

have we7.8512, norm loop closed with

)875.19)(107161.2)(177.32)(001.0(

)15447.0)(100)(1(101318.2

is controller suboptimal The

4

5

−+

−++≈

−×+++

−++×=

∞

∞

∞

∞∞

sss

sssK

H

ssss

sssK

KH

Example 14.5 (cont’d)

2H

∞H

7.8512 18.197

914.82 16.407


normH ∞

normH 2

)962.7)(81.41194.30(

)069.0)(100)(1(487.43

and 16.407 norm with

)962.7)(81.41194.30)(001.0(

)069.0)(100)(1(487.43

:follows as

obtained be can controller optimal an ,Similarily

22

2

22

2

−++

−++≈

−+++

−++=

ssss

sssK

H

ssss

sssK

H



HH∞∞ FilteringFiltering

n.realizatiofor causala be tohasit that is problem filtering theof nrestrictio

The . oft measuremen theusing sense some in of ˆ estimation an find tois problem filtering The

0)0( ,

:equation following by the described is system dynamica Suppose

111

212

1

yzz

wDxCz

wDxCy

xwBAxx

+=

+=

=+=ɺ

.)(ˆ with

ˆsup:

that such existsit if )(filter causala find 0,a Given

2

2

2

2

2

),0[2

ysFz

w

zzJ

RHsF

Lw

=

<−

=

∈>

∞∈

∞

γ

γH∞∞∞∞ Filtering

212

111

1

DC

DC

BA

)(sF

w

∆z

z

z~

y

⇔

−=

0

0

)(

212

111

1

DC

IDC

BA

sG

)(sG

)(sF

problem Filtering∞H

2

2

2

2

2

),0[

ˆsup that such )(filter a Find

2

γ<−

∈∞∈

∞w

zzRHsF

Lw

∆z w

z~

y



HH∞∞ FilteringFiltering

[ ] [ ] 1

2

*

211

*

1

*

11121

11221121

112122112122

2

2

*

121

1

*

1111

*

111

*

111

*

2

*

1

*

11

*2*

21

11

21

11

111

2112111

21

11

11

21

212

1

2

~:

where

)(ˆ

by given is satisfying )(filter causal rationala thensatisfied, are conditions above theif Moreover,

~0: ,

00

0:

~

where,0)Ric( withdom(Ric) and )( ifonly and if

that such )( causala exists thereThen .00

asy conformabl dpartitione

and normalized be Let . allfor rank row full has and detectable is ),( Suppose

−∞∞∞∞

∞∞∞∞

−∞

∞∞∞

∞

++−=

−

−−++==

<

−−−

−−=

−

=

≥=∈<

<∈

=

−

RCYDBCYDBLL

yDCDC

DLLCDLCLAysFz

JsF

CBD

CBDR

DBDB

CC

ABB

AJ

I

D

D

D

DR

JYJD

JRHsFDD

D

DD

DDC

BIjAAC

γ

γ

γσ

γ

ωω

Theorem 14.8

dropped ist requiremenstability internal design,filter For b)

0)(

tosolution gstabilizin theis and 0

ˆ becomesfilter the,0 and 0 When a)

*

112

*

21

*

1

2*

1

*

22

*

2*

21111

=+−++

−===

∞−

∞∞∞

∞∞∞

BBYCCCCYAYAY

YyC

CYCCYAzDBD

γ

Documents

H Optimal Control Problem Formulation