1 chapter 10 – feedback linearization nonlinear system linear system control input transformation...
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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