Adv Control & Robotic Lec 3

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METR4202/7202AAdvancedContro

l&Robotics,

Semester2,

2005:Page:1

METR4202 Advanced Control & Robotics

Lecture 3

Material from Various Sources,Mainly Nise Chapters 5.8 and 12

Similarity Transformations andIntroduction to State-Space Control

G. Hovland 2004-2006

METR4202/7202AAdvancedControl&Robotics,

Semester2,

2005:Page:2

The Simplest Controller: The P-Controller

K G(s)-

Goal: Relate the value of K to percentage overshoot

and settling/peak time in y(t) for a stepresponse in u(t)

u(t) y(t)

Before we start with state-space design, we will lookbriefly at frequency response design.


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2005:Page:3

Frequency Design

In Frequency Domain: Y(s)/U(s) =

++=

=

42

2

412

2tan

1

a

Tw

M

p

n

Example: 10% overshoot = 0.5912, M=58.6o

Peak time Tp = 1.0 sec n = 3.90 rad/sec

Assumption:

wn is the correct frequency for all systems.

We are in practice only placing the dominant

2nd order pole. After the design, we cross ourfingers and hope that higher-order poles do notaffect the step-response.

s

nT

w4

=

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cedControl&Robotics,

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2005:Page:4

Phase Margin vs Gain Adjustment

Bode plots.For a simple gain adjustment, desired phase margincan be achieved.

To increase cross-over frequency, a PD controller canbe used (topic for Pendulum prac)

n

You can alsouse the Nyquistplot to design thecontroller


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2005:Page:5

Limitations of Frequency Domain Techniques

We want to place all poles, not only the 2nd

order dominant

In frequency domain, we have threeparameters: gain, compensator pole andcompensator zero.

Three parameters are not sufficient to place allpoles for systems of high order

Example: For the 5th order system

design a controller that yields 10% overshoot and peak time 1.0 secfor a step-response.

01

2

2

3

3

4

4

5

1

)(

)(

asasasasassR

sY

+++++=

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2005:Page:6

Controller Design via State-Space (Ch. 12.1-4)

State-space formulation of the uncontrolled system:

State-space formulation of the controlled system:

CxyBuAxx =+=

CxyBrxBKAKxrBAxBuAxx =+=+=+= )()(


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l&Robotics,

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2005:Page:7

Figure 5.31State-space forms for

)(

)6)(4(3

)()(

tcy

sss

sRsC

=

++ +=

Summary State Space Forms

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2005:Page:8

Phase-Variable Form

u

x

x

x

x

aaax

x

x

x

n

n

nn

n

+

=

1

0

0

0

1000

0100

0010

1

2

1

110

1

2

1

A MatrixB Vector

+++

=

)()()(

1000

0100

0010

12110 nn kakaka

BKA

Majoradvantage ofthe phase-

variable andcontrollercanonicalforms


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2005:Page:9

Phase-Variable Form

Poles of uncontrolled system:

0011

1 =++++

asasasn

n

n

n system parameters to adjust

Poles of controlled system:

0)()()( 10211

1 =+++++++ kaskaskas nnnn

METR4202/7202AAdvan


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2005:Page:10

Example 12.1: State-Space Controller Design

Given the plant

)4)(1(

)5(20)(

+++

=sss

ssG

design the phase-variablefeedback gains to yield9.5% overshoot and asettling time Ts of 0.74 sec.

= 0.60 0.94==

sn

Tw

Desired poles:

))(2(22

pswsws nn +++

Should not

interfere withdesign requirements!


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l&Robotics,

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2005:Page:11

Example 12.1: Desired Poles

Bode plots

)2(

20)(

)2)(1.5(

)5(20)(

222

221

nn

nn

wwssG

wwss

ssG

++=

+++

+=

and

In general:Use extra poles to cancelout zeros. If no cancellations

required, place poles far away

from 2nd

order pole.

NB: Watch input saturation - fastpoles require high feedbackgains.

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2005:Page:12

Example 12.1: Signal-Flow Diagram

)4)(1(

)5(20)(

+++

=sss

ssG

]020100[

540

100

010

=

=

C

A

How?


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l&Robotics,

Semester2,

2005:Page:13

Example 12.1: Controller Gains

++

=

)5()4(

100

010

321 kkk

BKA

Characteristic equation:

0)4()5( 122

3

3 =+++++ ksksks

Must match design requirements:

1.41308.1369.15))(2(2322

+++=+++ ssspswsws nn

Controller gains by inspection:

k1 = 413.1, k2 = 132.08, k3 = 10.9

Large poles result in largefeedback gainsand possible saturation.

METR4202/7202AAdvan


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2005:Page:14

Example 12.1: Final Transfer Function

1.41308.1369.15

)5(20

)2)(1.5(

)5(20)(

2322 ++++

=+++

+=

sss

s

wswss

ssT

nn

RWhat happens tothe transfer functionif we increase thecontroller gains?


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State-Space Design: Summary so far

For systems on phase-variable state-space form, wecan easily place all poles by state feedback.

For systems of high order, you can design the poles asa series of 2nd order poles, which all meet the designrequirements on overshoot and settling time.

Alternatively, you can use additional poles to cancel outzeros.

If there are left-over poles, place them at frequencieswhere they have little influence on design requirements.

METR4202/7202AAdvan


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2005:Page:16

State-space control for other model forms

If we can transform the original model to phase-variableor controller canonical form, we can easily place allpoles for any controllable state-space model.

After the state feedback gains have been designed, theymust be transformed back to the original system.

In the following material, similarity transformations willbe introduced. Keep in mind the motivation for thesetransforms: easy state-feedback control design inphase-variable (or controller canonical) form.


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2005:Page:17

Similarity transformations

Similar systems: systems of different state

space representations, but have the same

transfer function, so the same poles /eigenvalues and the same response.

We will study how to transform between similar

systems without using signal flow graph andtransfer function.

This can be realised by transfer matrix.

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2005:Page:18

Expressing any vector in terms of Basic vectors

State variables form the axis of the state space.

The same points can be represented by differentrepresentations of state space in different coordinatesystems.

Coordinate frames may be rotated, but can not bedilated because the coordinate systems are based onunit vectors, and the origin is not allowed to be shifted.


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State-space transformations

=+=

2

1

2211x

xxx xx UUx

||,

|| 2

2

1

1

21 x

xU

x

xxx==U

O

xx

zz

=+=

2

1

2211z

zzz zz UUz

METR4202/7202AAdvan


Semester2,

2005:Page:20

Vector transformations

The basis vectors of z1z2 space canbe represented by basis vectors of x1x2 space:

2221122

2211111

xxz

xxz

pp

pp

UUU

UUU

+=

+=

=+=

2

1

2211z

zzz zz UUz

Seeing x=z, we have:

+

+=+++=

222211

122111

22222111122111 )()(pzpz

pzpzpzpzpzpz xx UUx

Pzx =

=

2

1

2221

1211

z

z

pp

ppxPz 1

=

Also:

=+=

2

1

2211x

xxx xx UUx , compare coefficients:


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2005:Page:21

Finding the transformation matrix, P=[Uz1, Uz2 ,,Uzn], for n-th system

2221122

2211111

xxz

xxz

pp

pp

UUU

UUU

+=

+=

=+=

2

1

2211

z

zzz zz UUz

The columns ofP are the coordinates of the basis vectors of the z1z2 space

expressed as linear combinations of the basis vectors of the x1x2 space.

1st column of P is Uz1, 2nd column of P is Uz2. We have P=[Uz1, Uz2], or :

Pzx =

=

2

1

2221

1211

z

z

pp

ppxPz 1

=

P=[Uz1, Uz2 ,,Uzn], for n-th order system

METR4202/7202AAdvan


Semester2,

2005:Page:22

Example 5.9 vector transformations

Transform vector x=[1 2 3]T expressed with its basis vectors, Ux1=[1 0 0]T, Ux2=[0 1 0]

T

and Ux3=[0 0 1]T into a vector expressed in Uz1=[0 1/2 1/2]T, Uz2=[0 -1/2

1/2]T,Uz3=[1 0 0]T.

SOLUTION:

332211 zzz zzz UUUz ++=

( ) ( )( ) ( )

+

=

+

+

=

21

21

3

321

2121

2121

0

0

1

2/1

2/1

0

2/1

2/1

0

zz

zz

z

zzzx

Pz=

=

3

2

1

02121

02121

100

z

z

z

=

==

1

0

83.2

3

2

1

001

707.0707.00

707.0707.001xPz

!x x1x2 z z1z2 "x z #


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2005:Page:23

Transforming the state equations

Vector transformation selection of different set of state variables torepresent the same system transfer function.

Now, convert state space representation with state vector x into a statespace representation with a state vector z.

u

u

DCxy

BAxx

+=

+= Let x=Pzu

u

DCPzy

BAPzzP

+=

+=

u

u

DCPzy

BPAPzPz

+=

+= 11

Multiply P-1

$%

#$%

#$&%

METR4202/7202AAdvan


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2005:Page:24

Similarity transformations on state equations

[ ]x001

1

0

0

752

100

010

3

2

1

3

2

1

=

+

=

y

u

x

x

x

x

x

x

'%(%(%

$)*+),

xPxz 1

541

023

002=

=

u

u

DCPzy

BPAPzPz

+=

+= 11

=

=

2.64.05.2

4.07.025.1

015.1

541

023

002

752

100

010

541

023

0021

1APP

=

=

5

0

0

1

0

0

541

023

0021BP [ ] [ ]005.0

541

023

002

001

1

=

=

CP,

[ ]z

zz

005.0

5

0

0

2.64.05.2

4.07.025.1

015.1

=

+

=

y

u


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2005:Page:25

Similarity transformations on state equations

[ ]x001

1

0

0

752

100

010

3

2

1

3

2

1

=

+

=

y

u

x

x

x

x

x

x

xPxz 1

541

023

002=

=

( )

=

=

=

=

==

2.04.05.0

05.075.0

005.0

4810

01015

0010

0*32*2)1*04*2(2*14*3

)0*30*2(1*05*2)1*05*3(

2*00*0)4*05*0(4*05*2

)det(

)(

541

023

002

201

201

1

1

1

11

P

PPP

adj

-

METR4202/7202AAdvan


Semester2,

2005:Page:26

Diagonalising a system matrix decoupled stateequations

1. Eigenvalues and eigenvectors:

Axi=ixi

All vectors xi0 are the eigenvectors ofA corresponding to(constant) eigenvalue i.

Solve for xi and i :

ii xAI0 )( =

0AI

AI0AIx )det(

)()(

1

==

i

iii

adj

0)det( =AIi i


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To be an eigenvector, the transformation Ax

must be co-linear with x; thus in (a), x is not

an eigenvector; in (b),it is.

Axi=ixi

Verify eigenvector solution

METR4202/7202AAdvan


Semester2,

2005:Page:28

Diagonalising a system matrix

# ./%%&%

=diag{i} &x=[x1,x2,xn].

# * P = x=[x1,x2,xn].

0# . , Axi=ixi (: AP=P,

( =diag{i}.

1# ( =P-1AP

2# $&%%.

u

uu

DCPzy

BPzBPAPzPz

+=

+

=+= 1

2

111

0

0

[ ]

== 2,22,1

1,21,1

21 xx

xx

P xx


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2005:Page:29

Diagonalising a system in state space

[ ]x

xx

32

2

1

31

13

=

+

=

y

u

8631

13

31

13

0

0)det( 2 ++=

+

+=

=

AI

.det(I-A)=0&%A 2 4#&Axi=ixi (

=

2

1

2

12

31

13

x

x

x

x $%

=

=

1

11

choosec

cx.3

.31

=

2

1

2

1

431

13

x

x

x

x $%3

=

= 1

12

choosec

cx

[ ]

==

11

1121 xxP

*

[ ]zDCPzy

zBPAPzPz

15

2/1

2/3

40

0211

=+=

+

=+=

u

uu

METR4202/7202AAdvan


Semester2,

2005:Page:30

Example: Similarity Transformation

[ ]x

xx

41

3

1

64

31

=

+

=

y

u

xz

=

41

23


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2005:Page:31

Class Question

[ ]x

xx

41

3

1

64

31

=

+

=

y

u

Find the transformation matrix P such that

[ ]zy

uzz

6.2121.2

20

39.18

30

02

=

+

=

What advantage does system (2) have that (1) does not?

Use:P-1APP-1B

CP

METR4202/7202AAdvan


Semester2,

2005:Page:32

Summary Similarity Transformations

Given the base vectors of the two systems x and z, wecan easily form the transformation matrix P. Thecolumns of P contain the basis vectors of z.

The transformation P formed by the eigenvectors alongthe coloumns of P, will diagonalise the system.

For two system descriptions, the transformation matrix Pcan be cumbersome to find. We need a better procedureof finding P, than solving A2*P = P*A1

We will develop such a procedure, but first we need to

define controllability of a system.


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Controllable and Uncontrollable system

Controllable: the input u influencesall the states x1, x2 and x3.

Uncontrollable: the input u does notinfluence the state x1.

Parallel Form

METR4202/7202AAdvan


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Controllability by inspection

Form the diagonalised system by the transformationmatrix P given by the eigenvectors.

The system is controllable if the B-vector of thetransformed system (P-1B) has no rows that are zero(when the states are decoupled, the input u mustinfluence all the states as in the previous example.)


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An nth-order plant whose state equation is:

is completely controllable if the matrix

is of rank n.

Controllability: General Test

BuAxx +=

][ 12 BABAABBC nM=

METR4202/7202AAdvan


Semester2,

2005:Page:36

Controllability: Example 12.2

Convert the signal-flow diagram above to the form

Cxy

BAxx

=

+= u

The above form has a special name


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2005:Page:37

Controllability: Example 12.2

Controllability matrix CM:

The system is controllable if the rows of CM are linearlyindependent. One way to check this is to find the

determinant. If det(CM)0, then the system iscontrollable.

In Matlab: rank(ctrb(A,B)) must be 3

==

421

111

210

][ 2BAABBCM

123))1*1(2*1(*2)1*14*1(*1))2*1(4*1(*0)det( =+==MC

METR4202/7202AAdvan


Semester2,

2005:Page:38

Summary so far: state-space control

First check if the system is controllable, either bychecking the rank of CM or by transforming the system toparallel form (whichever method is easier).

Second, if the system is controllable, transform thesystem to phase-variable form or controller canonicalform and place the poles by state-feedback gains.

Third, transform the controller gains back to the originalsystem, ie. Korig = K*P

-1


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P from controllability matrix CM

For the original system

For the transformed system

Hence, P = CMZ * CMX-1

In Matlab: P = ctrb(Az,Bz) * inv(ctrb(Ax,Bx))

][ 12 BABAABBC nMZ=

][

])()([

121

11121111

BABAABBP

BAPPBPAPPBAPPPBPC

n

n

MX

=

=

PP-1

termsdisappear

METR4202/7202AAdvan


Semester2,

2005:Page:40

P-matrix that gives phase-variable form

If we have the original system

and the phase-variable form

P = ctrb(Az,Bz) * inv(ctrb(Ax,Bx)) will transform anysystem z to the phase-variable form x!!!!

uBzAz zz +=

uBxAu

x

x

x

x

aaax

x

x

x

xx

n

n

nn

n

+=

+

=

1

0

0

0

1000

0100

0010

1

2

1

110

1

2

1


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Controller design by transformation: Ex 12.4

Convert the signal-flow diagram above to the form

CxyBAxx

=+= u

METR4202/7202AAdvan


Semester2,

2005:Page:42

Ex 12.4: Check Controllability

==

111

310

100

][ 2BAABBCM

Is this system state controllable? Check by inspection


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Semester2,

2005:Page:43

Ex 12.4: Transform to Phase-Variables

10178

1

)5)(2)(1(

123 +++

=+++ ssssss

==

==

+

=

1710

015

001

*

4781

810

100

][

1

0

0

81710

100

010

1

2

3

2

1

3

2

1

MXMZ

M

CCP

BAABBC

u

x

x

x

x

x

x

Phase-variableform

METR4202/7202AAdvan


Semester2,

2005:Page:44

Ex 12.4: Desired Response

20.8% overshoot and settling time Ts=4.0

= 0.447 24.24==

sn

Tw

Desired poles:

20136)4)(52(

))(2(

232

22

+++=+++=

+++

ssssss

pswsws nn

We use the extra pole

to cancel the zero


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Semester2,

2005:Page:45

State-Feedback Controller

+++

=

)8()17()10(

100

010

321 kkk

BKA

20136)4)(52(

))(2(

232

22

+++=+++=

+++

ssssss

pswsws nn

Desired:

By inspection: k1=10, k2=-4, k3=-2

METR4202/7202AAdvan


Semester2,

2005:Page:46

State-Feedback Controller: Original

Korig = K * P-1 = [ -20 10 -2]

DBAIC +== 1)()(

)()( s

sU

sYsT

Verify design withdesired poles


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l&Robotics,

Semester2,

2005:Page:47

Class Exercise: DC Motor Control

Design a state-feedbackcontroller to yield20.8% overshoot andsettling time of Ts=4.0

seconds for a step-response

Shaft

position

Follow the procedure in Example 12.4

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