State-space feedback 4 Ackermann’s...

State-space feedback 4 Ackermann’s approach

J A Rossiter

1

Slides by Anthony Rossiter

Introduction

• The previous videos showed how state feedback can place poles precisely as long as the system us fully controllable.

• Both relied on control canonical forms.

• This video shows how one can find a suitable state feedback without resorting to canonical forms within the design procedure.


2

xBKAxKxu

BuAxx

)(

Remark

Much of this resource derives and explains the result.

If you only want to know the result, go directly towards the end.


3

Pole polynomials

The approach given here relies on clear definitions of two polynomials:

• the open-loop pole polynomial.

• the desired closed-loop pole polynomial.


4

o

n

n

n

o asasasp

1

1

1

o

n

n

n

c sssp

1

1

1

Cayley-Hamilton Theorem

A matrix satisfies its own characteristic equation.

This implies that.


5

0

1

10 aaAI n

n

n

00

0

1

1

AaAaA n

n

n

The proof is omitted but relatively straightforward and available in most text

books.

Corollary

The following 2 identities are known to be true.


6

001

1

1

IaAaAaA n

n

n

o

n

n

n

o asasasp

1

1

1

o

n

n

n

c sssp

1

1

1

0)()()( 01

1

1

IBKABKABKA n

n

n

Finding K such that this last

identify is satisfied implies achieving target

poles.

Ackermann’s approach

The intention is to exploit the latter of these identities and full rank nature of the controllability matrix.

First define:

The first concept is to show that pc(A) is a component of pc(A-BK) and exploit this in the algebra.

NOTE, by definition: Slides by Anthony Rossiter

7

IBKABKABKABKAp n

n

n

c 01

1

1 )()()()(

IAAAp n

c 01 )()()(

0)(

0)(

Ap

BKAp

c

c


The first concept is to expand pc(A-BK) and unpack the underlying structure. First consider each term.


8

BKABKA

BKBKAABKABKA

BKABfBKABKAABKA

BKABfBKABKAABKA

BKABfBKABKAABKA

n

nnnnnn

n

nnnnnn

n

nnnnnn

222

2

233322

1

122211

111

)()(

),()1()()1()(

),()1()()1()(

),()1()()1()(

Deliberate separation of terms with a leading ‘A’

and a leading ‘B’.

Ackermann’s approach The next step is to show that one can express each term using the controllability matrix.


9

n

c

v

nn

n

n

M

nnnn

n

nnnnnn

K

BKK

BKAf

BABAABBABKA

BKABfBKABKAABKA

21

12

111

)()1(

),()1(

],,,,[)(

),()1()()1()(

3

0

)(

),(

],,,,[)(

),()()(

3

1233

3

2233

v

M

nK

BKK

BKAf

BABAABBABKA

BKABfBKABKAABKA

c

The ZERO is key


Next expand pc(A-BK) and unpack the underlying structure. First consider each term.


10

1

2

22

1

11

)(

)(

)(

vMABKA

vMABKA

vMABKA

vMABKA

c

c

nc

nn

nc

nn

Only vn is non-zero in the bottom row!

IBKABKABKABKAp n

n

n

c 01

1

1 )()()()(

][

)(

1111

01

1

1

vvvM

IAAABKAp

nnnc

n

n

n

c


Show that pc(A) is a component of pc(A-BK) and exploit this in the algebra.


11

][

)(

1111

01

1

1

vvvM

IAAABKAp

nnnc

n

n

n

c

0)(

0)(

Ap

BKAp

c

c

][)(0 1111 vvvMAp nnncc

][)( 1111

1 vvvApM nnncc


Exploit the structure of vn by extracting only the last row of each side of the equation and thus extracting the feedback gain K.


12

nv

nn

n

n

K

BKK

BKAf

21 )()1(

),()1(

][)( 1111

1 vvvApM nnncc

KApM cc )(100 1

,0100,0100 21 nn vv

As pc is by definition, one can find the required state feedback using this formulae.

Summary of Ackermann’s approach

Define the desired closed-loop pole polynomial.

The required feedback is given from the formula where Mc is the controllability matrix.


13

KApM cc )(100 1

o

n

n

n

c sssp

1

1

1

EXAMPLES


14

Example 1: Choose K to set the closed-loop poles at -1 and -2.


15

First find the required pole polynomial

43;2

1;

4.01

21

CBA

232 sspc

Find pc(A)

04.16.1

2.32

20

02

4.01

213

4.01

21)(

2

Apc



16

Define the controllability matrix

;2

1;

4.01

21

BA

8.12

31],[ ABBM c

Use Ackermann’s formulae

95.031.0)(100 1 KApM cc

04.16.1

2.32)(Apc


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Example 2: Choose K to set the closed-loop poles at -0.5, -1 and -1.5.


18

First find the required pole polynomial

431;

0

1

1

;

4.01.00

6.04.02.0

5.021

CBA

75.075.23 23 ssspc

Find pc(A) [Use MATLAB as tedious]

608.2257.004.0

742.1248.0046.0

715.156.023.0

75.075.23)( 23 IAAAApc



19

Define the controllability matrix

02.01.00

1.06.01

25.211

],,[ 2BAABBM c

Use Ackermann’s formulae

57.2018.218.0)(100 1 KApM cc

;

0

1

1

;

4.01.00

6.04.02.0

5.021

BA

608.2257.004.0

742.1248.0046.0

715.156.023.0

)(Apc

Summary

Introduced concepts of pole placement state feedback without a control canonical form.

1. Show that, assuming full controllability, one can use Ackermann’s formulae to define the required feedback to achieve any desired pole polynomial.

2. Not paper/pen exercise in general as involves substantial matrix multiplication and inversion.

3. Numerically poorly conditioned for weakly controllable and large dimension systems.


20

Kxu

KApM cc )(100 1

© 2016 University of Sheffield This work is licensed under the Creative Commons Attribution 2.0 UK: England & Wales Licence. To view a copy of this licence, visit http://creativecommons.org/licenses/by/2.0/uk/ or send a letter to: Creative Commons, 171 Second Street, Suite 300, San Francisco, California 94105, USA. It should be noted that some of the materials contained within this resource are subject to third party rights and any copyright notices must remain with these materials in the event of reuse or repurposing. If there are third party images within the resource please do not remove or alter any of the copyright notices or website details shown below the image. (Please list details of the third party rights contained within this work. If you include your institutions logo on the cover please include reference to the fact that it is a trade mark and all copyright in that image is reserved.)

Anthony Rossiter Department of Automatic Control and

Systems Engineering University of Sheffield www.shef.ac.uk/acse

http://creativecommons.org/licenses/by/2.0/uk/

http://engsc.ac.uk/an/oer-project/oer-project.asp

State-space feedback 4 Ackermann’s...

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