1 Chapter 10 – Feedback Linearization Nonlinear System Linear System Control Input Transformation Linear Controller Big Picture: When does such a transformation.

1

Chapter 10 – Feedback Linearization

Nonlinear System

Linear System

Control InputTransformation

LinearController

Big Picture:

When does such a transformation exist?How do we find it?(not really control design at this point)

2

Given ( ) and system ( )

( ) ( ) f

V x x f x

V VV x x f x L V

x x

3

2 2 2 21 11 2 1 22 2

1 21 2 1 2

x x x xh h hx x

x x x x x

2 21 2 1 22 1x x x x

4

2 21 2 1 24 2 1x x x x

(continued)

from previous case

2 21 2

2 21 2 2 2

1 22 21 1 2 1 1 2

( )

2 2 ( )1 1

f g f

f

L L h x L x x

x x x xx x L h x

x x x x x xx

5

1 1

1 2

2 2

1 2

1 0

0 1

x x

x xg

x x x

x x

Other than here, square brackets still indicate a matrix

2 2

1 2

2 21 1 2 1 1 2

1 2

1 1

x x

x xf

x x x x x x x

x x

6

7

Condition to test if something is a diffeomorphism

8

1

1 1 1

1

Constant transformation in linear system :

then the transformed system is

Note that this is why we require that exists, we want to

create a new (tr

z Tx x Ax Bu

T z x

T z AT z Bu z TAT z TBu

T

ansformed) system.

9

10

11

12

Lie BracketsVectorsMatrix

13

Lie Bracket =

1

2

3

x

x x

x

1 2 3 1 2 3

1 1 12 11 2 1

1 2 3 1 2 321

2 2 21 1 1

1 2 3 1 2 3

0 0 0 1 1 1

1 00 0 0

0

11 1 1

0 0 0

1 0 0

0 0 0

x x x x x x

x x xf ff f x

x x x x x x x xx

x x x

x x x x x x

1

21 1

1 0 0 0 0 0

0 0 0 0 1

2 0 0 1 0

x

x x

14

1

2

3

x

x x

x

Only [f1,f2] here

15

Systemxu

Reminder from Chapter 2 – Linear approximation of a system

( )x f x

Systemxu

Control Law

linear

control

u (0)df

x x Axdx

Taylor series at origin

This is not what we will do in this section!

16

Transformation Complete: Transformed the nonlinear system into system that is linear from the input perspective.Control Design: Use linear control design techniques to design v.

( ) 1 vector

( ) 1 vector

nx

f x nx

g x nx

1;

( ) 1 1 vector

( ) 1 1 vector

nu x

w x x

x x

1B

nxn

nx

A

17

Transformation Complete: Transformed the nonlinear system into a linear system.Control Design: Use linear control design techniques to design v:

1

1v

B is 2x1, “directs u to a specific row”

Scalar

AA

1( ) 1 ( )w x w x

18

SystemLinearizing Controlxu

Control Law

linear

control

v

v

Example 7: Implementing the Result

Looks like a linear system

19

Transformation Complete: Transformed the nonlinear system into a linear system.Control Design: Use linear control design techniques to design v.

1

2

0 1 0

1

xv

x a b

A

20

Affine in uWe are restricted to this type of system

,

( ) is 1 vector

nz x

T x nx

21

T(x)

u(x)

( ) ( )z Az B z u z

Feedback linearization and transformation process:

v(z) Using standard linear control design techniques

Stable closed-loop system

22

( )z T x

Terms that multiple u

All other terms that

don't multiple u

23

( ) ( ) are scalar

for single input system

x x

24

1 2

2 3

3

0 1 0

0 0 1

0 0 0 0C

T T

A T T T

T

3Example: :x

0 0

( ) ( ) 0 ( ) ( ) 0

1 ( ) ( )CB x x x x

x x

0 0

( ) 0 ( ) 0

1 ( )CB x x

x

25

1

2

1 1 11

1 21

2 22 21

1

Note:

( )

( )( ) and

( )

is the Jacobian

( ) ( ) ( )( )( )

( ) ( )( ) ( ) =

( )( )

n

n

n

n

n

n n

T x

T xT x x

T x

T

x

T x T x T xT x x x xxT x

T x T xT x T xx xx

x

T xT x T

x x

2

( ) ( ) ( )n nn

x T x T xx x

26

Found the RHS, now do element-

wise comparison on each row

c c

Tf A T B

x

Found the RHS, now do element-

wise comparison on each row

c

Tg B

x

27These are the conditions we must satisfy to linearize the system.

1 1 1 11 2

Remember: ( ) ( ) ( ) ( )n

T x T x T x T xx x x x

28

Not matched to u

Matched to u

Note that n=2 in the above procedure.

29

1

(continued)

First constraint required:

30

Will only depend on T2

Moved the nonlinearities to the bottom equation where they are matched with u

( )z T x

(continued)

31

2

2

will ever go to -1? no

1so never goes to

1

z

z

22 1xz e

2x

2 2

Note:

For to be bounded requires 1x z

(continued)

32

SystemLinearizing Controlxu

Control Law

linear control

designed

for z system

v Kz

v

Example 9: Implementing the Result

( )z T x

z

2

1

2 xu x e v

33

Always good to try this approach but may not be able to find a suitable transformation.

Con

diti

ons

on

the

dete

rmin

ant

u “lives” behind g, so g must possess certain properties so that u has “enough access” to the system

Questions:• What is adf g(x)? Review Lie Bracket• What is a span? set of vectors that is the set of all linear combinations of the

elements of that set.• What is a distribution? Review Distributions sections• What is an involutive distribution?

Lie BracketsMatrix

34

=

n=2 (size of x)

35

0 0( ) ( )

1 1rank rank

36

2 21 2

Maybe can't linearize from to ,but perhaps we could make up an interesting ( )e.g., (to represent the the norm of )

u xh x

y x x x

Systemyu

37

Differentiated until

the control appears

Remove nonlinearities

Linearized from input to output

Now design v

38

Lie Derivatives

39

40

41

can't say input has appeared can't say input has appeared

42

Differentiate until

the control appears

(r=2)

43

Input-output

linearization

44

System?

u y

45

46

Linearizing control

0

0

u

y y y

47

=

48

Nonlinear System

Linear System

Control InputTransformation Linear

Controller

Main IdeaSummary

Input to state linearization

Input to output linearization

1. Conditions to know if linearization is possible2. Procedure to find x Ax Bu

( )

is the relative degree

ry Ay Bur

Homework 10.1Problems 10.4, 10.5, 10.6

-

+++

--

-

-

--

21 2

233 3

1 2 3

1 1

2

2 2

2

3 3

2

0

( ) ; ( ) 1

0

( ) 1 0

( ) 1 0

( ) 1 0

x x

f x x g x

x x x

T Tg x

x x

T Tg x

x x

T Tg x

x x

2 2 3 31 1 1 11 2 3 1 2 3 2

1 2 3

2 2 3 32 2 2 21 2 3 1 2 3 3

1 2 3

( )

( )

T T T Tf x x x x x x x T

x x x x

T T T Tf x x x x x x x T

x x x x

-

2 2 3 31 1 1 11 2 3 1 2 3 2

1 2 3

2 3 31 11 2 1 2 3 2

1 3

1 2

( )

0

T T T Tf x x x x x x x T

x x x x

T Tx x x x x T

x x

T c T

1 11 2

2

2 22 2

2

3 3

2

( ) 1 0 is independent of

( ) 1 0 is independent of

( ) 1 0

T Tg x T x

x x

T Tg x T x

x x

T Tg x

x x

Homework 10.2Problems 10.6, 10.7, 10.8

10.7 (needed)

Homework 10.3

3

22 3

3 2 3

2

3

11

tan( ) 0

tan( ) 1( ) ; ( ) ; requires and

cos( ) cos( )cos( ) 2 2

tan( ) 0

cos( )

let tan ( ) which means must limit ; now use stadard tools to design2

x

xf x g x x x

a x b x x

x

a x

u u u

1

1 1

2 2 3

2 2

2 2 3

3 3

2 2 3

1( ) 0

cos( )cos( )

1( ) 0

cos( )cos( )

1( ) 0

cos( )cos( )

u

T Tg x

x x b x x

T Tg x

x x b x x

T Tg x

x x b x x

1 1 1 2 1 23 2

1 2 3 3 3

2 2 2 2 2 23 3

1 2 3 3 3

tan( ) tan( )( ) tan( )

cos( ) cos( )

tan( ) tan( )( ) tan( )

cos( ) cos( )

T T T x T xf x x T

x x x a x x a x

T T T x T xf x x T

x x x a x x a x

Find the transformation for the input-state linearization.

1 1 1 2 1 23 2

1 2 3 3 3

tan( ) tan( )( ) tan( )

cos( ) cos( )

T T T x T xf x x T

x x x a x x a x

1 11 2

2 2 3

2 22 2

2 2 3

3 3

2 2 3

1( ) 0 is independent of

cos( )cos( )

1( ) 0 is independent of

cos( )cos( )

1( ) 0

cos( )cos( )

T Tg x T x

x x b x x

T Tg x T x

x x b x x

T Tg x

x x b x x

1 1 1 23 2

1 3 3

1

3

1 13 2

1

11 1 2 3

1

Have some freedom to start:

tan( )( ) tan( )

cos( )

0

( ) tan( )

now 1 tan( )

T T T xf x x T

x x x a x

Tselect

x

T Tf x x T

x x

Tselect T x T x

x

2 2 2 2 2 23 3

1 2 3 3 3

2 3

2 22

3 3 1

2 2 23 3

3 3

tan( ) tan( )( ) tan( )

cos( ) cos( )

Based on previous selection of tan( )

1 and 0

cos ( )

tan( ) 1( )

cos( ) co

T T T x T xf x x T

x x x a x x a x

T x

T T

x x x

T T xf x T T

x x a x

2 22 3

3 3 3

tan( ) tan( )

s ( ) cos( ) cos ( )

x x

x a x a x

3 3

2 2 3

3 23 23 3 2

2 2 3 3 2

1

3

23

3

Final constraint

1( ) 0

cos( )cos( )

tan( ) 10 (given the range of and )

cos ( ) cos ( ) cos ( )

Transformation:

( ) tan( )

tan( )

cos ( )

T Tg x

x x b x x

T xx x

x x a x a x x

x

z T x x

x

a x

Homework 10.4

2

2

2

actuator position

* velocity,

* acceleration

input torque

q

q

q

u

1. Find the transformation for the input-state linearization.2. Place the eigenvalues at -1,-2,-3,-4 and simulate

1 1 1 2

2 1 2

Flexible-joint robotic link

sin( ) ( ) 0

( )

Iq MgL q K q q

Jq K q q u

1

1

1

link position

* velocity,

* acceleration

q

q

q

spring connecting

actuator and link

K

Homework 10.4 (sol)

1 1

2 1

3 2

4 2

2

1 1 3

4

1 3

Let then ( ) ( )

0

0sin( ) ( )( ) and ( ) 0

1( )

x q

x qx x f x g x u

x q

x q

x

MgL Kx x x

I If x g xx

Kx x J

J

1 1 1 2

2 1 2

Flexible-joint robotic link

sin( ) ( ) 0

( )

Iq MgL q K q q

Jq K q q u

Arrange in state-space form:

2

1 1 3

4

1 3

1 11 4

4

2 22

4

0

0sin( ) ( )( ) and ( ) 0

1( )

Transformation:

1( ) 0 is independent of

1( ) 0 is independent o

x

MgL Kx x x

I If x g xx

Kx x J

J

T Tg x T x

x x J

T Tg x T

x x J

4

3 33 4

4

4 44 4

4

f

1( ) 0 is independent of

1( ) 0 depends on

x

T Tg x T x

x x J

T Tg x T x

x x J

1 1 1 1 12 1 1 3 4 1 3 2

1 2 3 4

2 2 2 2 22 1 1 3 4 1 3 3

1 2 3 4

3 3 32

1 2

( ) sin( ) ( ) ( )

( ) sin( ) ( ) ( )

( )

T T T T TMgL K Kf x x x x x x x x T

x x x I I x x J

T T T T TMgL K Kf x x x x x x x x T

x x x I I x x J

T T Tf x x

x x x

3 31 1 3 4 1 3 4

3 4

sin( ) ( ) ( )T TMgL K K

x x x x x x TI I x x J

Homework 10.4 (sol)

Find T1-T4

1 1

2 3

1 12 2

1

11 1 2 2

1

Have some freedom to start:

0 and 0

( )

now 1

T Tselect

x x

T Tf x x T

x x

Tselect T x T x

x

1 1 1 1 12 1 1 3 4 1 3 2

1 2 3 4

( ) sin( ) ( ) ( )T T T T TMgL K K

f x x x x x x x x Tx x x I I x x J

2 2 2 2 22 1 1 3 4 1 3 3

1 2 3 4

( ) sin( ) ( ) ( )T T T T TMgL K K

f x x x x x x x x Tx x x I I x x J

21 1 3 3( ) sin( ) ( )

T MgL Kf x x x x T

x I I

Homework 10.4 (sol)

31 2 4 4

4 1 2 2 4

4 4

4

( ) cos( )

cos( )

1 1( ) 0

T MgL K Kf x x x x T

x I I I

MgL K KT x x x x

I I I

T T Kg x

x x J I J

3 3 3 3 32 1 1 3 4 1 3 4

1 2 3 4

( ) sin( ) ( ) ( )T T T T TMgL K K

f x x x x x x x x Tx x x I I x x J

Homework 10.4 (sol)

2

2

1 1 34

1 2 14

1 3

22

1 1 1 1 3 1 3

sin( ) ( )( ) sin( ) cos( ) 0

( )

sin( ) cos( ) sin( ) ( ) ( )

x

MgL Kx x x

T MgL MgL K K I If x x x xxx I I I I

Kx x

J

MgL MgL K MgL K Kx x x x x x x x

I I I I I JI

2

4

42

21 1 1 1 3 1 3

4

1( ) ( )

( )( ) sin( ) cos( ) sin( ) ( ) ( )

( )

T K Kx g x

x I J IJ

Tf x IJ MgL MgL K MgL K Kxx x x x x x x x x

T K I I I I I JIg xx

Homework 10.4 (sol)

1

12

2

1 1 33

41 2 2 4

Transformation:

( ) sin( ) ( )

cos( )

xz xz MgL Kz T x x x xz I Iz MgL K K

x x x xI I I

Homework 10.4 (sol)

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

z z v

v Gz

A =

0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0

B =

0 0 0 1

where

>> A=[0 1 0 0;0 0 1 0;0 0 0 1;0 0 0 0]>> B=[0;0;0;1]>> P=[-1,-2,-3,-4]>>G=place(A,B,P)G = 24.0000 50.0000 35.0000 10.0000

In MATLAB

Homework 10.4 (sol)

1

2

1 1 3

1 2 2 4

24.0000  50.0000  35.0000 1  0.0000

24.0000  50.0000  35.0000 1  0.0000 sin( ) ( )

cos( )

v Gz z

x

x

MgL Kx x x

I IMgL K K

x x x xI I I

Homework 10.4 (sol)Expression:-(M*g*L/I)*sin(u(1))-(K/I)*(u(1)-u(3))

Expression:(K/J)*(u(1)-u(3))

Expression:-(24*(u(1))+50*(u(2))+35*(-(M*g*L/I)*sin(u(1))-(K/I)*(u(1)-u(3)))+10*(-(M*g*L/I)*u(2)*cos(u(1))-(K/I)*u(2)+(K/I)*u(4)))

Expression:(-(I*J/K)*((M*g*L/I)*sin(u(1))*u(2)*u(2)+(-(M*g*L/I)*cos(u(1))-(K/I))*(-(M*g*L/I)*sin(u(1))-(K/I)*(u(1)-u(3)))+(K*K/(J*I))*(u(1)-u(3

Homework 10.4 (sol)

Legend:X1-YellowX2-MagentaX3-CyanX4-Red

Response to initial conditions x1=2, x3=-2

1

2

22 1 1

21 1

Let then ( ) ( )

2 0( ) and ( )

12

xx x f x g x u

x

bx x xf x g x

x x

1 1

1 2

2 2

1 2

1

2

2

2

0, since is a constant vector.

1

,0

0( , , ) 2 provided b 0.

1 0

f f

x xg f ff g f g g g

f fx x x

x x

f

x bf g

f

x

brank g f g rank

21 2 1 1

22 1 1

1 2

Given the system:

2

2

1. Use Simulink to plot the response of the system to initial conditions 0 and 1 . Assume b=5

Show both states on the same plot, label the state

x bx x x

x x x u

x x

1

s.

2. What are the conditions on b such that the system is linearizable?

3. Find the transformation for the input-state linearization.

4. Place the eigenvalues at -1,-2 and simulate response to 0 andx 2 1 with b=5.

Show both states on the same plot, label the states.

x

01

1

0 0( , , ) 1 Involutive

1 0

The system is linearizable if 0.

g

rank g rank

rank g g g rank

b

1 1 1

2

2 2 2

2

21 1 1 1 11 2 1 2 1 1 2

1 2 1 1

211 1 2 2 1 1

1

1

0( ) 0

1

0( ) 0

1

( ) 2

propose 1 and 2 .

Controller ( ) ( )

T T Tg x

x x x

T T Tg x

x x x

T T T T Tf x f f f bx x x T

x x x x x

TT x T bx x x

x

u x x v

2 2

2

2

2 2 2

2 1 2

22 1 12

1 1 21 1

2 2 21 1 2 1 1 1 1

( ) ( )

( ) 1 1( ) ( ) ( )

( )

214 2 2

2

14 2 2 2 2

T Tx g x b

x x

Tf x T T Txx f x f x

T b x b x xg xx

bx x xx x b

b x x

x x bx x x b x xb