State-space feedback 3 transformation to get a canonical...

State-space feedback 3 transformation to get a canonical

form

J A Rossiter

1

Slides by Anthony Rossiter

Introduction

• The previous video showed how state feedback can place poles precisely when the system is in control canonical form.

• More generally, the system is not in canonical form, but we may still wish to place the closed-loop poles.

• This video shows how the a transformation can create the canonical form, and thus implicitly allows explicit and straightforward pole placement.


2

xBKAxKxu

BuAxx

)(

Control canonical form


3

zbbby

uz

aaaa

zdt

d

C

nn

BA

nn

021

0121

;

0

01

0100

0001

Remark: A transformation to control canonical form only exists if the system is

fully controllable.

Observation

In control canonical form, the eigenvalues are determined solely by the parameters along the top row of the A matrix.


4

Corollary: One can place the poles precisely if one can select the top row of A precisely without changing any other coefficients.

0

1

1

0121

0100

0001

aaAI

aaaa

A

n

n

n

nn

Similarity transformation (chapter 1)

Consider how the state feedback varies for two alternative representations linked by a similarity transform.


5

BuzATzTBuAxx 11

uBzAz

uTBzTATzTTBA

ˆˆ

ˆˆ

11

zCTy

Cxy

C

1

zTxTxz 1;

Similarity transform continued

Let the state feedback for system A, B be K.

In the transformed state, an equivalent control law is given by:


6

xBKAxKxu

BuAxx

)(

zKBAzzKTKxu

uBzAz

K

ˆˆ

1 )ˆˆˆ(

ˆˆ

zTxTxz 1;

Compare controllability matrices

For the original system.

For the transformed system.


7

],,,,[ 12 BABAABBM n

cx

]ˆˆ,,ˆˆ,ˆˆ,ˆ[ 12 BABABABM n

cz

])(,,)(,,[

ˆ,ˆ

11211

1

TBTATTBTATTBTATTBM

TATATBB

n

cz

cx

n

cz TMBABAABBTM ],,,,[ 12

Controllability matrices are linked by the

transformation matrix!

REMARK

We can get the control canonical form directly from the transfer function so this is known for any A,B,C,D.


8

zbbby

uz

aaaa

zdt

d

C

nn

BA

nn

ˆ

021

ˆˆ

0121

;

0

01

0100

0001

)()(01

2

2

1

1

01

1

1 sUasasasas

bsbsbsY

n

n

n

n

n

UDBAsICY ])([ 1

INSIGHT

If we know the system in control canonical form, then we can find the transformation that takes us from an arbitrary form to the control canonical form!


9

BuAxx

Given form Control canonical form

uBzAz ˆˆ zTxTxz 1;

cxcz M

n

M

n BABAABBTBABABAB ],,,,[]ˆˆ,,ˆˆ,ˆˆ,ˆ[ 1212

INSIGHT CONTINUED If we know the TRANSFORMATION that takes us from control canonical form to another form, we can do pole placement in control canonical form, and then find the equivalent state feedback for an alternative form.


10

KxuBuAxx ;

Given form Control canonical form

zKuuBzAz ˆ,ˆˆ

zTxTxz 1; TMM cxcz 1][

TKKKTK ˆˆ 1

Pole placement algorithm

1. Find transfer function representation.

2. Find control canonical form.

3. Find pole placement state feedback for control canonical form.

4. Find transformation matrix using controllability matrices.

5. Find state feedback for original state space system.


11

uBzAz ˆˆ

zKu ˆ

TMM cxcz 1][

TKK ˆ

BAsIC 1)(

EXAMPLES


12

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


13

First find the control canonical form.

43;2

1;

4.01

21

CBA

Next find the similarity transform relating the two.

2.95ˆ;0

1ˆ;

01

4.24.1ˆ

CBA

4.24.1

2.95)(

2

1

ss

sBAsIC

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


14

Find the controllability matrices.

8.12

31

],[

;2

1;

4.01

21

cx

cx

M

ABBM

BA


10

4.11

]ˆˆ,ˆ[

;0

1ˆ;

01

4.24.1ˆ

cz

cz

M

BABM

BA

TMMTxz cxcz 1][

Example 1


15

Define feedback for control canonical form, Desired pole polynomial is s2+2s+1.

Find feedback for original system.

01

12

01

4.24.1ˆˆˆ

;0

1ˆ;

01

4.24.1ˆ

21 kkKBA

BA 4.16.0ˆ K

52.044.0

13.026.0

56.013.0][;ˆ 1

K

TMMTKK cxcz

Confirm results using MATLAB


16

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


17

First find the control canonical form.


39.018.0

21.02.12)(

23

21

sss

ssBAsIC

431;

0

1

1

;

4.01.00

6.04.02.0

5.021

CBA

21.02.12ˆ;

0

0

1

ˆ;

010

001

39.018.01

ˆ

CBA

Example 2


18

Find the controllability matrices.

02.01.00

1.06.01

25.211

],,[ 2

cx

cx

M

BAABBM


100

110

82.011

]ˆˆ,ˆˆ,ˆ[ 2

cz

cz

M

BABABM

TMMTxz cxcz 1][

Example 2


19

Define feedback for control canonical form, Desired pole polynomial is s3+3s2+2.75s+0.75.

Find feedback for original system.

14.157.22ˆ K

57.2018.218.0

][;ˆ 1

K

TMMTKK cxcz

010

001

75.075.23

010

001

39.018.01

ˆˆˆ321 kkk

KBA

Confirm results using MATLAB


20

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

1. Show that, assuming full controllability, there exists a transformation matrix to generate the equivalent control canonical form.

2. Pole placement design can be done using the canonical form.

3. Feedback parameters for the original states are obtained using the corresponding similarity transformation which is defined from the controllability matrices (see algorithm).

4. Not paper/pen exercise in general.


21

Kxu

© 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 3 transformation to get a canonical...

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