1

Chapter 9

Controller Design Two Independent Steps:

1) Feedback Design – Control Law u=-Kx – assumes all states are accessible (a lot of sensors are necessary)

2) Design of Estimator – (also called an Observer) which estimates the entire state vector given the outputs and inputs

Control Law Design Assumed system for control law design u=-Kx

for an nth order system there are n feedback gains K1,…,Kn. By choice of K the roots can generally be placed anywhere

PLANT

BuAxx

x C y

-

+

ˆ x

r

ESTIMATOR

CONTROL LAW -K

matrix of constants

state vector estimate

COMPENSATION

BuAxx

C y

u=-Kx

x

2

xBKAx

Kxu

BuAxx

)(

characteristic equation is

I (ABK) 0

Placing Roots

Example: Undamped oscillator with freq. 0

y(s)

u(s)

1

s2 02

state space description

ux

x

x

x

1

0

0

10

2

1

2

02

1

let's place the roots both at -20

we want to double the natural frequency and increase

damping from =0 to =1

desired characteristic equation is

212

0

00

22

0

1

0

0

10

0

0det)(det

44)2()(

KKs

sBKAsI

ssssd

or

s2 K2s02 K1 0

equating same coefficients:

K2 40

02 K1 40

2

K1 302

K2 40 02021 43 KKK

n

jn 12

jn

1 2

-20 -j0

j0

j

cos

REVIEW

3

Use of Canonical Forms A simple way of calculating the gains when order is greater than three is to use special

“canonical” forms of the state equations. The special structure of the system matrix is referred to as companion form. Example: Third order case: The characteristic equation is

s3 a1s2 a2s a3

Recall the phase variable form (a lower companion form)

G(s) b1s

n1 .... bn1s bn

sn a1sn1 .... an

11 ....

1

:

0...100

0...010

aaa

A

nn

1

0

:

0

B 11 ... bbbC nn 0D

The closed loop system matrix is

ABK Third order case:

312213

000

000

321

123

100

010

1

0

0

100

010

321

KaKaKa

KKK

aaa

BKA

KKK

Characteristic equation

sI (ABK) 0

s3 (a1 K3)s2 (a2 K2)s (a3 K1) 0 (1)

4

if the desired pole locations result in the characteristic equation

d (s) s3 1s

2 2s3 0 (2) equating like coefficients of (1) and (2)

K3 a1 1

K2 a2 2

K1 a3 3

Example: Drill problem D9.1 page 635 of text

u

x

x

x

x

x

x

1

0

0

763

100

010

3

2

1

3

2

1

rx

x

x

kkku

3

2

1

321

1

2

3

2 0 1

x

y x

x

Find ki to place the closed-loop system poles at s= -3, -4, -5 ANS desired characteristic polynomial

d (s) (s 3)(s 4)(s 5)

s3 12s2 47s 60

1 12, 2 47, 3 60 we have

k3 a1 1

7 12 5

k2 a2 2

6 47 41

k1 a3 3

3 60 57

5

Design Procedure

Given (A,B) and desired d(s), transform to (Ac, Bc) and solve for gains We then need to transform gain back to original state space note: the poles can only be placed arbitrarily if the system is fully controllable This procedure is encapsulated in Ackermann’s formula Ackermann’s Formula

)(10...0 1 AMk dC

where

BABAABBM nC 12 ... (controllability matrix) where n is the order of the system or the number of states and

d (A) is defined as

IAAAA nnnn

d ...)( 22

11

where the si ' are the coefficients of the desired characteristic polynomial

d (s) sn 1s

n1 ...n Example Apply the formula for the undamped oscillator

2

00

2

2

0

44

)2()(

ss

ssd

01 4 , 2

02 4

recall

0

102

0A

1

0B

2

0

3

0

0

2

0

2

02

0

02

0

2

0

34

43

10

014

0

104

0

0)(

Ad

control canonical (companion) form i.e. phase variable form

6

also

01

10ABBM c

01

101cM

2

0

3

0

0

2

021

34

43

01

1010

KKK

020 43 K which is the same as the result previously obtained.

Tracking Problems

For step input: Will find

N to ensure zero steady-state error to step inputs

rNKxu

rNBxBKA

BuAxx

)(

now

rNBBKACy

rNBBKAx

rNBxBKAx

ss

ss

ssss

1

1

)(

)(

)(00

A-BK is stable inverse exists now

rNBBKAC

rNBBKACryre ssss1

1

)(1

)(

to get zero steady state error

N 1

C(A BK)1B

steady-state error

7

Integral Control (used to get zero steady state error) Integrator increases the system order by one, i.e. augment the plant model by an added state variable xi

Cxrx

dtCxrdtyredtx

i

i

)()(

Augmented systems becomes

ruB

x

x

C

A

x

x

ii

1

0

00

0

Control law is

i

iiix

xKKxKKxu

The design now proceeds as before. Example

Double Integrator

G(s) 1

s2

xy

uxx

01

1

0

00

10

Augment the plant

zero matrices (compatible dimensions)

integrator in the forward path

R(s) E(s)

1s

U(s) X(s) + +

+ -

Ki +

+ B

1s C

Y(s) Xi(s)

A

-K

8

xCy

x

xCy

Cxy

i

0

i

ii

x

x

y

u

x

x

x

x

001

0

1

0

001

000

010

Select poles of the closed loop system to be at

1 j,5 N.B. 3 poles because of extra state

10712 K

Steady state output to a unit step input can be derived as follows

rCxx

BuAxx

i

ruB

x

x

C

A

x

x

dt

d

ru BB

ii

1

0

00

0

i

ix

xKKu

rBxKBAx

rBxKBxAx

xKu

rBuBxAx

ru

ru

ru

)(

in steady state 0ssx

9

For the example

011

0

00

10

CBA

0011

0

0

0

1

0

001

000

010

CBBA ru

r

r

ry ss

1

1

0

0

001

10712

010

001

1

0

0

10712

0

1

0

001

000

010

001

1

1

det 112 10

1 0

10

adj(A)13 1 0

7 10

10

only term of interest

10

Observer Design

We will estimate states rather than measure them Observer simulates the original system Original System

Cxy

BuAxx

x(0) x0

Observer

)ˆ(ˆˆ xCyLBuxAx 0ˆ)0(ˆ xx

(A,B)

x(t) C y(t)

-

+

ˆ x (t)

(A,B)

C

ˆ y (t)

L

u(t)

+

+ + -

+

+ B C

A

B

L

A

C

1s

1s

U(s)

Y(s)

ˆ X (s)

PLANT

11

Error between states and their estimates

˜ x x ˆ x

0~)0(~~)(

)ˆ(ˆ

ˆ~

xxxLCA

xCyLBuxABuAx

xxx

Observer error will go to zero asymptotically A-LC is stable (i.e. eigenvalues are in the LHP)

note: the eigenvalues of A-LC are the observer poles, which can be placed arbitrarily if the system is observable

this can be done by choice of the observer gains L (a column vector for single output

systems)

Definition: 1) a system is detectable if the unstable modes are observable

2) a system is stablizable if the unstable modes are controllable Duality

Control Estimation Control Estimation

A AT MC Mo

T B CT K LT C BT

Example

MC B AB .... An1B

duality

11

1

1

::

)(....)(

....

n

o

T

n

TnTT

TnTTTTTo

CA

CA

C

M

CA

CA

C

CACAC

CACACM

12

Ackermann’s formula to find observer gains

1

:

0

0

)( 1oe MAL

DERIVATION USING DUALITY Ackermann’s Formula for control problem

)(1.....0 1 AMK CC

Duality

T

oe

T

e

T

o

T

MA

AML

1

:

0

)(

)(1.....0

1

1

AT BTCT (CBA)T

1

:

0

)(1

oe MAL

Example Design an observer for

xy

uxx

ssG

01

1

0

00

10

1)(

2

Note the system is completely observable

Design the observer with poles at

2 j2 Actual characteristic equation:

21

2

2

101

00

10

10

01)(

lsls

l

lsLCAsI

13

Desired characteristic equation:

(s 2 j2)(s 2 j2) s2 4s 8

Equating coefficients

l1 4

l2 8

8

4L

The observer equations are

uxyx

xyxx

)ˆ(8ˆ

)ˆ(4ˆˆ

12

121

STRUCTURE OF OBSERVER

+ - +

+ 4

1s

1s

u y

ˆ x 1

PLANT

1s

+

+ 8

1s

ˆ x 2

2x 12 xx

x1

14

Control Using Observers

Plant: Cxy

xxBuAxx

0)0(

Observer: 0ˆ)0(ˆ)ˆ(ˆˆ xxxCyLBuxAx

Estimated state feedback:

uKˆ x closing the loop

x

x

LCBKALC

BKA

x

x

LCxxBKLCA

LyxBKxLCAx

xBKAxx

ˆˆ

ˆ)(

ˆˆ)(ˆ

ˆ

15

Separation Principle Introduce transformation

1

0

ˆ

0

ˆ ˆ

N N

N N

N N

N N

Ix zP where P

I Ix w

note P P

Iz x x x

I Iw x x x x

Therefore using this transformation the old augmented state vector comprising

x (the plant states) and

ˆ x (the estimator sates) now becomes

x and

˜ x (the estimator error). The new

system matrix

˜ A P1AP

x

x

LCA

BKBKA

x

x~0~

note

˜ A is block-triangular Eigenvalues of a block-triangular matrix are equal to the eigenvalues of the diagonal blocks. So

the eigenvalues of the full system comprise the eigenvalues of the plant (i.e. eigenvalues of A-BK) and the observer (i.e. eigenvalues of A-LC).

Alternative Approach

xLCAx

xLCAxLCA

xLCBKALCxxBKAxxx

xLCBKALCxx

xBKAxx

~)(~

ˆ)()(

ˆ)(ˆˆ

ˆ)(ˆ

ˆ

xBKAxx ˆ

now

˜ x x ˆ x ˆ x x ˜ x

xBKxBKA

xxBKAxx~)(

)~(

16

Compensator Transfer Function

H(s) U(s)

Y(s)

we have

)()()(

)()()(ˆ

ˆ)(

ˆˆˆ)ˆ(ˆˆ

ˆ

)(

1

1

sYLLCBKAsIKsU

sLYLCBKAsIsX

LyxLCBKA

xLCLyxBKxAxCyLBuxAx

xKu

sH

Design Issues

1) Problem with pole placement is that there is no control over compenstor poles and zeros 2) Optimum choice for observer initial conditions is

3) Choice of observer poles:

i. Choose them to be faster than controller poles ii. Alternatively, choose them to be at plant zeros (if the sytem has RHP zeros,

use their LHP images).

Compensator output, which is the plant input

Compensator input, which is the plant output

u y PLANT

OBSERVER -K

ˆ x

Compensator Input

COMPENSATOR

Compensator Output

17

Reduced-Order Observer Design If system has n states and m measurements, then an observer of order (n-m) will be sufficient. If

y Cx we will assume C has the structure:

( )

11 12 1

21 22 2

22 21 2

0 , 0

0

( )

mxm mx n m

m

m

u

m m

u u

u u m

known input

C I I R R

xy I x

x

x xA A Bu

x xA A B

x A x A x B u

also u

tmeasuremenknown

um

xAuByAy

uBxAyAyx

12111

11211

Summarizing:

)1(

)1(

)1()(

12111

22122

bxAuByAy

auBxAxAx

u

tmeasuremenknown

inputknown

muu

Recall for full-order observer:

Plant: )2()0( 0

Cxy

xxBuAxx

Observer: )3(ˆ)0(ˆ)ˆ(ˆˆ 0xxxCyLBuxAx comparing (1) and (2) shows the correspondence

measured states

unmeasured states

dynamics of the unmeasured state

output relationship

18

22

21 2

11 1

12

(4)

u

m

x x

A A

Bu A x B u

y y A y B u

C A

substituting (4) into (3) we get the reduced order equation

12

12

22 21 2 11 1 12ˆ ˆ ˆ( ) (5)

u

u

A x

u u m u

A x

x A x A x B u L y A y B u A x

let us now define the estimator error

˜ x u xu ˆ x u

therefore subtracting (5) from (1a) and using (1b) (i.e. uxAuByAy 12111 ) we get

)6(~)(~ 1222 uu xLAAx Design proceeds by, given an A22 and A12 we choose an L to place estimator poles. Rewriting (5) we have

yLuLBByLAAxLAAx uu )()(ˆ)(ˆ 1211211222 (7)

The presence of the derivative of the measurement (i.e. y ) is not good since this amplifies the noise. To get around this we introduce a new state z where

Lyxz u ˆ (8)

ˆ x u z Ly (9)

substituting (9) into (7) leads to the final form of the reduced-order observer

Lyzx

GuFyDzz

u

ˆ

where

D A22 LA12

F DL A21 LA11

G B2 LB1

the error dynamics are given by this

equation

z is the state of the estimator

19

The block diagram of the reduced order observer is shown below

Example

Double integrator

G(s) 1

s2

Lyzx

GuFyDzz

u

ˆ

where

D A22 LA12

F DL A21 LA11

G B2 LB1

Dynamics of reduced order observer uu xLAAx~)(~ 1222

^

1

s

1

s

u

x2

x1 y

1

s2 ^

20

Require observer pole at

s2

I A22 LA12 0 where 2

I L 0

L 2

D 2

F 4

G 1

Ackerman’s formula for reduced order observer gains

1

:

0

0

0

)(1

022 MAL e

for example:

2

)1)(1)(2(

1

20)(

2)(

120

22

L

AM

A

ss

e

e

21

Reduced-Order Transfer Function

Lyz

u

y

m

u

mxKxK

x

xKKu

ˆ

ˆ2121

now

yLKKzKu

LyzKyKu

GuFyDzz

)(

)(

212

21

also

yLGKGKFzGKD

yLKKzKGFyDzz

)()(

)(

212

212

Transfer function:

U(s)

Y(s)C'(sI A')1B'D'

where

A' DGK2

B' F GK1 GK2L

C' K2

D' (K1 K2L)

When C is not of the form 0I ?

BuQAQzQzBuAQzzQQzx

DuCxy

BuAxx

11

y CQzDu

so find a transformation Q so that CQ is of form 0I

let

Q Q1 Q2

so 1 2 0CQ CQ CQ I

22

let

T

CP be nonsingular

arbitrary matrix

if

PQ I

1 21 21 2

1

0

0

CQ CQC IPQ Q Q

TQ TQT I

Q P