contents 1.motivation for nonlinear controlmotivation for nonlinear control 2.the tracking...

Contents1. Motivation for Nonlinear Control2. The Tracking Problem

1. Feedback Linearization3. Adaptive Control4. Robust Control

1. Sliding mode2. High-gain3. High-frequency

5. Learning Control6. The Tracking Problem, Revisited Using the Desired Trajectory

1. Feedback Linearization2. Adaptive Control

7. Filtered tracking error r(t) for second-order systems)8. Introduction to Observers9. Observers + Controllers10. Filter Based Control

1. Filter + Adaptive Control11. Summary

12. Homework Problems1. A12. A23. A34. A4 – Design observer, observer + controller, control based on filter

2

Nonlinear Control

• Why do we use nonlinear control :– Tracking, regulate state setpoint– Ensure the desired stability properties– Ensure the appropriate transients– Reduce the sensitivity to plant parameters

find ( )

( )

u r x

u y

state feedbackoutput feedback

• Consider the following problem:

x is bounded & goes to 0 how it goes to setpoint

so that the closed loop system ( , ( )) or ( , ( )) exhibits

desired stability and performance characteristics.

x f x r x x f x x

n

n

Ruxhy

Rxuxfx

),(

),,(

Applications and Areas of Interest

Mobile Platforms

• UUV, UAV, and UGV• Satellites & Aircraft

Automotive Systems

• Steer-By-Wire• Thermal Management• Hydraulic Actuators• Spark Ignition• CVT

Mechanical Systems

• Textile and Paper Handling• Overhead Cranes• Flexible Beams and Cables• MEMS Gyros

Robotics

• Position/Force Control • Redundant and Dual Robots• Path Planning• Fault Detection• Teleoperation and Haptics

Electrical/Computer Systems

• Electric Motors• Magnetic Bearings• Visual Servoing• Structure from Motion

Nonlinear Control and Estimation Chemical Systems

• Bioreactors• Tumor Modeling

The Mathematical Problem

Typical Electromechanical System Model Classical Control Solution

Obstacles to Increased Performance– System Model often contains

Hard Nonlinearities– Parameters in the Model are

usually Unknown– Actuator Dynamics cannot

be Neglected– System States are Difficult

or Costly to Measure

x f x y·

( , )y g x y u·

( , , )u y x

Electrical Dynamics Mechanical Dynamics

x f x y·

( , )y g x y u·

( , , )u y x

LinearController

fLinear

f

x

gLinear

g

y

u y xy x y u· ?( , , ) x x y

· ?( , )

x f x y·

( , )?u y x

x f x y·

( , )y g x y u·

( , , )u ? ?

Nonlinear Lyapunov-Based Techniques Provide– Controllers Designed for the Full-Order

Nonlinear Models– Adaptive Update Laws for On-line

Estimation of Unknown Parameters– Observers or Filters for State

Measurement Replacement– Analysis that Predicts System

Performance by Providing Envelopes for the Transient Response

The Mathematical Solution or Approach

Mechatronics

Based Solution

AdvancedNonlinear Control

Design Techniques

RealtimeHardw are/Softw are+

NewControl

Solutions

u y x

NonlinearParameterEstimator

NonlinearController

y x y u· ?( , , ) x x y

· ?( , )

x f x y·

( , )y g x y u·

( , , )u ? x

NonlinearObserver

NonlinearController

t

Transient Performance Envelopes

6

Nonlinear Control Vs. Linear Control

• Why not always use a linear controller ?– It just may not work.

Ex: 3x x u x R

When 0, the equilibrium point 0 is unstable.u x

Choose3 3

.

.

u kx

x x k x

We see that the system can’t be made asymptotically stable at 0.x

On the other hand, a nonlinear feedback does exist :3( )u x kx

Then (1 )x x kx k x

Asymptotically stable if 1.k

Then

7

Example

• Even if a linear feedback exists, nonlinear one may be better.Ex: y ky v

y u

k

+_

v y

y u+_

vy

k

y

1 2y k y k y v

y ky v

for 0v

y

y

1x y

2x yfor 0v

8

Example (continued)

Let us use a nonlinear controller : To design it, consider the same system in the form:

1 2

2 1

x x

x kx

If 1k If 1k

1x

2x

1x

2x

2 1On the line: - exponentially stablex kx

Why is that especially interesting? If we could get onto that line then the system converges to the origin

Both systems have interesting properties, can we combine the best features of each into a single control?

9

Example (continued)

1x

2x

1 2 0x x

1k

1k

1k

1k

sliding line

1

1

if 01

if 01

x sk

x s

1 2where s=x x

Created a new trajectory: the system is insensitive to disturbance in thesliding regime Variable structure control

system. structure variableaobtain andely appropriat 1 to1 from Switch k

HW Simulate this system and control. Be sure to plot the evolution of the states

Example (continued)

10

x

x

x

x

1 2

2 1

x x

x kx

1

1

if 01

if 01

x sk

x s

1 2where s=x x

11

Consider the system:

( )

Need to accomplish two control objectives:

1) Control Objective-make ( is a desired trajectory), assuming , .

2) Hidden Control Objective-keep everythind d d d

x f x u

x x x x x L

g bounded (ie., , , ).

Need to make some assumptions first:

1) is measureable.

2) if , then ( ) .

3) has a solution.

4) (0) .

x x u L

x

x L f x L

x

x L

The Tracking Problem

12

The Tracking Problem (continued)

bounded! are signals All

)( ,,

bounded are signals all insure Finally,

)exp()0()(

equation aldifferenti thesolve Now,

)(

:get we, Letting

)(

:for substitutecan weNow

as defined be , error, trackingLet the

assumptionby

FeedbackForward Feed

LxLuLxfLxLeLxx

ktete

kexfxu

kee

uxfxe

x

xxe

xxe

e

dd

d

d

d

d

Feedback LinearizationExact Model Knowledge

Example Exact Model Knowledge

• Dynamics:

Mass

bx3

asin(t) bx3

u(t)Nonlinear Damper

Disturbance Velocity

Control Input

a,b are constants

• Tracking Control Objective:

• Open Loop Error System:

• Controller:

• Closed Loop Error System:

• Solution:

Feedforward Feedback Assume a,b are known

Drive e(t) to zero

Exponential Stability

3 sin( )x bx a t u

de x x 3 sin( )d de x x x bx a t u

3 sin( )du x bx a t ke

e ke( ) (0)exp( )e t e kt

Example Exact Model Knowledge

Mass

bx3

asin(t) bx3

u(t)Nonlinear Damper

Disturbance Velocity

Control Input

a,b are constants

• Open Loop Error System:

• Control Design:


• Solution:

Feedforward Feedback

Assume a,b are known

Exponential Stability

3 sin( )d de x x x bx a t u

3 sin( )du x bx a t ke

( ) (0)exp( )e t e kt

• Lyapunov Function: 2 31; ( sin( ) - )

2 dV e V ee e x bx a t u

2V ke

A different perspective on the control design

15

Adaptive Control

~)(

get we(2), and (1) combining Now,

ˆ~

as defined be ~

Let

(2) ˆ)(

where

controlour (1) )(

Let

constant.unknown an is and known, is )( where

)sin( )( examplefor )()(

function zableparameterilinearily aConsider

32

xWkee

kexWxu

uxWxe

xW

L

RxxxxfxWxf

d

d

By Assumption 2: both f(x) and W(x) are bounded.

Constant that can be factored out

Yet to be designed, feed-forward term based on an estimate of the parameters

16

Adaptive Control (continued)

0)(limthen

)( 3)

0)( where),( 2)

0 1)

if :Lemma

.in definite positive

and ,in unboundedradially zero),by y necessaril(not boundedlower isIt :A

candidate? good a thisis Why :Q

~ where,21~~

21

21

candidate Lyapunov theChoose

2

tg

Ltg

tgtgV

V

z

z

ezzzeV

t

TT

Lyapunov-like lemma

if ( ) is bounded the ( ) is uniformly continuousg t g t

Note: detailed in deQueirozwill use this lemma by getting get and into and satisfying the conditions on ge g

"explodes" as and "explode"V e

17

eWWkee

Leketgketg

LuLxeLV

keV

eW

eWkeV

WkeeV

eeV

eV

T

T

TT

T

T

T

~ and

~

is system loop closedour So

2)( and )(

problem For this

bounded! are signals all,ˆ,,~

Therefore

getfinally we,ˆ Letting

)ˆ(~

ˆ~)

~(

~~

gives derivative theTaking

~~2

12

1

function Lyapunov candidateour With

2

2

2

2

y!necessarilNot :A

?0~

does So, :Q

Adaptive Control (continued)

closed-loop

error system

since 0

ˆdesign to help Lyapunov function

can't identify the parameters

2didn't get here analysis is more complicated

e 0 dx x

Now have a dynamic control

(control has dynamics) compared

to state-feedback which ia s static control

Example Unknown Model Parameters

• Open Loop Error System:• Control Design:

a,b are unknownconstants

Same controller as before, but and are functions of time

How do we adjust and ?

Use the Lyapunov Stability Analysis to develop an adaptive control design tool for compensation of parametric uncertainty


At this point, we have not fully developed the controller since and are yet to be determined.

parameter error

3 sin( )d de x x x bx a t u 3ˆ ˆ( ) ( )sin( )du x b t x a t t ke

ˆ( )a t ˆ( )b t

ˆ( )a t ˆ( )b t

3( ) ( )sin( )e ke b t x a t t ˆ( ) ( )

ˆ( ) ( )

a t a a t

b t b b t

ˆ( )a t ˆ( )b t

( is UC)


Fundamental Theorem

V (t) ¸ 0

V (t) ¸ 0

effects of conditions i) and ii)

i) If

ii) IfV (t) ¸ 0is bounded

iii) If is bounded

satisfies condition i)

V (t) ¸ 0

finally becomes a constant

V (t) ¸ 0

• Non-Negative Function:

• Time Derivative of V(t):

is bounded

examine condition ii)

design and

substitute the dynamics for

constant

effects of condition iii)

l imt! 1

e(t) = 0

( ) 0V t ( ) 0V t

( )V t ( )V t

lim ( ) 0t

V t

( )V t

2 2 21 1 1( )

2 2 2V t e a b

ˆ( )a t ˆ( )b t3( ) ( )sin( )e ke b t x a t t

ˆˆ( )V t ee aa bb


• Substitute Error System:

How do we select and such that ?

• Update Law Design:

• Substitute in Update Laws:

and

Fundamental Theorem

is boundedV (t) ¸ 0 all signals are bounded

Fundamental Theorem

Feedforward Feedback

control structurederived fromstability analysis

control objective achieved

is bounded

2( ) 0V t ke

ˆ( )b tˆ( )a t

( ) 0V t 3ˆˆ sin( ) a e t b ex

2 3 ˆ ˆ( ) ( )( ) ( )( sin( ) )V t ke b t ex b a t e t a

( )V t

( ) 0V t

lim ( ) 0t

V t

( ) 0V t

lim 0t

e

3 3

0 0sin( ) sin( )

t t

du x x ex dt t e t dt ke

How Can We Use the Adaptive Controller?

Design adaptive control to track a desired trajectory while compensating for unknown, constant parameters (parametric uncertainty)

,x f x u xu

1

2

, ˆ

ˆ

u h x

h x

2

Backstepping - intermediate controller is adaptive

, ˆ ˆ,dy h x h x

Adaptive control with backstepping in cascaded subsytems to track a desired trajectory while compensating for unknown, constant parameters (parametric uncertainty)

,x f x y xy 2 2,y f y u u


2 3 2 42

Backstepping - intermediate and input controller are adaptive

, , and ˆ ˆ ˆ , ˆ, , ,dy h x h x u h x y h x y

(continued) Adaptive control with backstepping in cascaded subsytems to track a desired trajectory while compensating for unknown, constant parameters (parametric uncertainty)

,x f x y xy

22 ,y f y u u

2 23 4

Backstepping - input controller i

ˆ

s adaptive

, , , ˆ ,u h x y h x y

,x f x y xy 22 ,y f y u u


What about the case where input multiplied by an unknown parameter, can we design adaptive control to track a desired trajectory while compensating for unknown, constant parameters (parametric uncertainty)

2,x f x u xu?u

Homework A.2-2

24

Robust Control

Recall the system defined by the following:

( )

( )

We can try to make several assumptions about the system:

1) ,

2) if , then ( ) .

3) and all si

d

d

d d

d

x f x u

e x x

e x f x u

x x L

x L f x L

x x

knownunknownboundingdynamicsfunction

gnals are bounded

4) ( ) is linearily parameterizable (ie., ( ) ( ) )

Adaptive control ONLY

5) ( ) ( )

We use this assumption for Robust (S

f x f x W x

f x x

liding Mode) control ONLY!

Restriction on the structure but not the uncertainty

25

Robust Control (continued)

1

22

2

3

Now, let the control be

where is a function that we can choose. Consider the three following functions

Sliding mode

1 Robust, high gain

Robus

d R

R

R

R

R

u ke x V

V

eV

e

V e

eV

e

t, high frequency

where 0. We will consider each separately.RV

sgn( )e

2 3

In reality can't implement this

control because it is not really defined at 0.

Practically, as in MATLAB, use:

and are a response

to fix this mathematicallyR RV V

auxiliary control

if is small then it looks

like the high frequency

controller ( frequency)

frequency

Feedback and feedforward motivated by: ( )de x f x u


( ) ( )f x x

26

1

2

1

21

22

2 2

Let's try the first function

( )

Now, take a Lyapunov candidate

1 2

( ( ) )

( ) ( )

( ) ( )

( )

R

R

Re

e ke f x V

V e

V ee e ke f x V

V ke eV

eV ke e x x

e

V ke e x e x k

f

e

x

e

2 2

2 2 0

2 , where ( 0( ) )

V

V kV V kV

V kV s t s t

Robust (Sliding Mode) Control

Use assumption 5 here: ( ) ( )

more positive less negative bound on

f x x

V

differential inequlaity differential equality solve

this differential equation will not have a solution

1Substitute proposed control

Note it doesn't exist at 0

where we are trying to drive the error system

R

eV

e

e

27

Robust (Sliding Mode) Control (continued)

0

2 2

Solving the differential equation, we get

( ) (0)exp( 2 ) exp( 2 ) exp(2 ) ( )

( ) (0)exp( 2 )

1 1 ( ) (0)exp( 2 )2 2 ( ) (0) exp( )

So, the system is globally exponent

t

V t V kt kt k s d

V t V kt

e t e kt

e t e kt

ially stable, and all signals are bounded!

always negative

upper bound

Base of the natural

logarithm.

28

2

2

22

22 2 2 2

2

Now, let's try it with and the same Lyapunov function

( )

( )

1 1 ( ) ( ) ( ) ( )

1 ( ) 1 ( )

R

R

R

V

e ke f x V

V ke e x eV

V ke e x e x e ke e x x e

V ke e x e x

Robust (High-Gain) Control

new robust control term

1Same basic proof as VR


Reminder of assumption:

5) ( ) ( ) ( ) 0f x x x

2Recall, we proposed d Ru ke x V

High-gain controller

29

2

2

Lyapunov analysis continued. As a reminder we started with:

1 2Now we have:

1 ( ) 1 ( )

V e

V ke e x e x

Robust (High-Gain) Control (continued)

This is what the new robust control term

accomplished. Is it useful?

Solve differnetial equation (as before): 2 ( )V kV s t

2

Case 1: if ( ) :

1 1 1 ( ) 1 1 ( ) 0 ( ) 1 ( ) 0

Case 2: if ( ) :

1 1 ( ) 1 0 1 ( ) 1

e x

e x e x e x e x

V kee x

e x e x

2

2

1( ) 1 ( ) ( )

2

2 ( )

e x e x e x

V ke

V ke

V kV

V kV s t

30

Robust (High-Gain) Control (continued)

0 0

1exp( 2 ) 1

2

2 2

Solving the differential equation yields

( ) (0)exp( 2 ) exp( 2 ) exp(2 ) ( ) exp( 2 ) exp(2 )

( ) (0)exp( 2 ) 1 exp( 2 )2

1 1 ( ) (0)exp( 22 2

t t

ktk

V t V kt kt k s d kt k d

V t V kt ktk

e t e k

2

) 1 exp( 2 )2

( ) (0)exp( 2 ) 1 exp( 2 )2

The system is Globally Uniformly Ultimately Bounded (GUUB),

and all signals are bounded.

t ktk

e t e kt ktk

discard this negative term

( ) is less than a constant.V t

( ) will go to a ball of size 2

e tk

22

Can make small to reduce the size of the ball, 2

1but the trade-off is that the control term becomes large.R

k

V e

dSignal chasing: e,x are bounded x bounded u is bounded.

31

3

2 22 2 22 2

2 2 2 22

2

2

Using the third function, , we obtain similar results

As you can see, the sol

RV

e ee eV ke e e ke

e e e

e e eV ke

e

eV ke

e

V ke

2ution to this equation will be the same as for RV

Robust (High-Frequency) ControlHigh frequency control chattering

2Same basic proof as VR

2Same basic result as V

Note Global doesn't depend on intial timeR

Upper bounded by 1

This is what the new robust control term

accomplished. Is it useful?

=

32

Let's take another look at the system from the previous control

( ) , , and ( )

For each control type, we attempt to make different assumptions. Those

assumptions eventually he

d dx f x u e x x e x f x u

lp us in the proof of stability and boundness of

the system. For instance, we made the assumption that ( ) was linearily

parameterizable ( ( ) ( ) ). For the Robust (Sliding Mode)

control, we made th

f x

f x W x

e assumption that ( ) was unknown, but that it could

be bounded by some known function ( ) ( ) . For learning control,

we make the assumption that ( ) is periodic:

( ( )) ( ( ))

Let ( ) ( (

f x

f x x

f x

f x t f x t T

d t f x

)), that leaves us with

( ) , where means "disturbance"

We also know, via our assumption that

( ) ( )

d

t

e x d t u d

d t d t T

Learning ControlNeed to compensate for an unknown periodic disturbance.

33

Learning Control (continued)

Now, take the control to be

ˆ

ˆ ( )

where

ˆ

ˆOur task is to design . So, let's try

for ˆ ˆ ( ) ( ( )) , where ( )sgn( ) for

d

d

u x ke d

e ke d d ke d

d d d

d

x xd t sat d t T k e sat x

x x

We make the assumption that the magnitude of the disturbance, ( ) is bounded:

( ) , where is a constant

So, then we can say

( ) ( ( )) ( ( ))

ˆ( ) ( ( )) ( ( ))

d t

d t

d t sat d t sat d t T

d t sat d t T sat d t T

use this in the stability proof!dk e

ˆWithout this we can't prove that is bounded.d

() sat

Don't actually use in the control

since is an upper bound

ˆsubstitute proposed d

ˆcloseness of to d d

34


22

2

We choose the following Lyapunov candidate to investigate stability:

1 ˆ1 ( ( )) ( ( ))2 2

0 Can you prove this?

1 ˆ ( ( )) ( ( )) ( ( ))2

t

d t T

d

V e sat d sat d dk

V

V e ke d t sat d t sat d tk

2

2 2

1 ˆ ( ( )) ( ( ))2

1 1ˆ ( ( )) ( ( )) ( ( ))2 2

( ( )

d

dd k e

dd d

sat d t T sat d t Tk

V e ke d t sat d t sat d t d k ek k

V e ke d t

221 1ˆ) ( ( )) ( ( )) 2

2 2 dd d

sat d t sat d t d k dek k

2

2 22 21 1ˆ ˆ ( ( )) ( ( )) ( ) ( )

2 2

d

dd d

k e

V ke sat d t sat d t d t d t k ek k

dx

duxuf

dx

dvxvfdttf

dx

dxv

xu

)()()(

:rule sLeibniz')(

)(

Derivative of an integral

derived on the previous slided

Cancels the cross term

From definition d

35


22 2

2 1 ˆ ˆ ( ( )) ( ( )) ( ) ( )2 2

d

d

k eV ke sat d t sat d t d t d t

k

22

2

2

Math Note: ( ) ( ) ( )

So, ( ), for ( ) 0 and

( ) ( ) 2 ( )

So ( ) 0 and that means

ˆ ( ), , , 0

The system is Globally Asymto

x y sat x sat y

V ke

V g t g t

g t ke g t kee g t L

g t

V L e L x L d t e x u L e

tically Stable (GAS),


36

The Tracking Problem, RevisitedUsing the Desired Trajectory

We want to build a tracking controller for the following system:

( )

where our controls are

( )

which yields

( )

For this pro

d d

d

aux d

aux

x f x u

e x x x x e

e x f x u

u ke V x

e ke f x V

1 1blem, we assume ( ) (where C means once differentiable)

( ) , where ( ) ( )

We also assume ( , ), where is non-decreasing, and ( ) 0. We

also assume that ( ) is

d aux d

d

f x C

e ke f x f V f f x f x

f e x e x

f x

known. So, we can say

( )

aux dV f x

e ke f

Differentiable assumption needed in analysis but not required to implement control.

if 0 then 0e f

Feedback Linearization

( )f x

xdx

( )df x

( ) ( )( )

( ) ,

d

d

d

f x f x ff c

x x e

f e f c e e x

Mean Value Theorem for scalar function

if 0 then 0e f

http://en.wikipedia.org/wiki/File:Mvt2.svg

38

The Tracking Problem, Revisited (continued)

decreasing-non is because true))(2())0(2( So

))0(2( if )2exp()0()(

))(2( if )2exp()0()(

)2( if 2

)( if

)(

have then we,1Let

)(

~

~

221

be candidate Lyapunovour Let

2

22

22

2

2

22

tVVk

VktVtV

tVktVtV

VkVV

ekeV

eekeV

kk

eekeV

fekeV

fekeeeV

eVeV

n

n

n

n

n

n

n

Let's see what does to the system.f

Dropped the notation just to

clarify next steps (it is still there).dx

39

2 2 2

Now we can write

1 1 ( ) (0)exp( 2 ) if (0)2 2

( ) (0) exp( ) if (0)

So, we have semi-global exponential tracking! It is semi-global (instead of just

local) because we can, in the

n

n

e t e t k e

e t e t k e

ory, set as high as we want. Also, as long as

assumptions are met, all signals will remain bounded.nk


Region is adjustable(not a fixed local region)

Design alternatives:

( )

( ) (this may work better in an experiment if there is noise on x,

could pre-compute ( ))

aux

aux d

d

V f x GES

V f x SGES

f x

40

Adaptive Control

40

What if we assumed ( ) was linearily parameterizable (ie., ( ) ( ) )? Then

we get

( ) ( )

( )

ˆLetting ( ) , would make ( ) , and we have

ˆ

d d

d aux

aux d

T

f x f x W x

f x W x

e ke W x f V

V W x e ke W f

W

( )

( )

d

Td

x e

W x e


41


2

22

0 (design of the adaptation law)

2

If we let our Lyapunov function be

1 1 2 2we get

ˆ ˆ ˆ ( ) ( ) recall that and

if

T

Tn d

n

V e

V e e k e eW x

V e k e

V

2 2

dependson !

if 2 (0) Be careful! We can't plug in 2 for . Why?

Finally, we can show

( ), where ( ) 0

( ) 2 ( ) , so lim ( ) 0

We have semi-global asymptotic trac

n

V

t

e k V V e

V g t g t

g t ee g t L e t

king.

from f

Design alternatives:

ˆ ( )

ˆ ( ) (this may work better in an experiment if there is noise on x,

could pre-compute ( ))

T

Td

d

W x e GAS

W x e SGAS

f x

42

Continuous Asymptotic Tracking

2 2 2

2

2

2

If we let

ˆ ˆ1 1 2 2ˆhaving made and .

Then let's us say

ˆ if

Now try the new approach:

where , then

V x V xx V x x

x x

V x x

V x P

V x x P

P x P x

0

( ) ( )t

a

t

d

Scalar System:

ˆLet so that

x u

u x x x

Don't know P V is unknown

43

If you knew that , then

21 ( )2 2 21 1 ( ) ( ) ( ) ( )2 2 00 02 21 , so ( )2 0

This solution is not unique even though we found it two different ways.

x

dt tP d d t t

a a adt t

V x ta

Continuous Asymptotic Tracking (continued)Constant of integration

44

Continuous Asymptotic Tracking (continued)

Lx

xf

x

xfxfL

x

xm

x

xmxm

Lx

Lx

xf

x

xfxfL

x

xm

x

xmxm

Cx

xm

xfxgxfxgxmuxfxxm

txtx

xg

gfuxgxfx

d

d

d

dd

d

d

d

dd

d

d

2

2

2

2

2

2

2

2

3

11

)(,

)(),( and

)(,

)(),( A3)

as long as

)(,

)(),( and

)(,

)(),( A2)

A1)

:sassumption following themake We

0)( here,

)()()( and )()( where,)()(

as system therewritecan we),()( want We

0)( here,

unknown are and wheresystemscalar a)()(

system heConsider t

if ( ) goes through zero then we can't control the systemg x

General problem: control is multiplied by an

unknown function that we can't invert

45


true?). thisis(Why then , if and ,0 then ,0 if shown that becan It

as variablenew a define sLet'

)(sgn)()1()()1()(

gives of derivative theTaking

.)continuous wise-piece is controller This :(Note

as defined is ableerror vari theHere

.0)( that see Weconstants. positive are and , , where

)(sgn)()1()()1()()1(

be controlour Let

0

0

0

LeLrer

eer

r

tetektektu

u

xxe

tuk

deektekteku

ss

d

s

t

t

sss

Proportional Integral

From linear systems:

e+ e=r

1E(s)= R(s) if r(t) 0 then ( ) 0, then ( ), r(t) 0 e( ) 0 e t e t t

s

Note that we can write

( 1) ( ) sgn ( )s

r e e

k r t e t

46

Now, from our original system, we can write

( ) ( ) ( ) (where )

We also know

and we can then proceed as

( ) ( ) ( ) ( ) (

d d

d d

m x e m x x f x u e x x

r e e

m x e m x e m x x m x x f

)

( ) ( )( ) ( )

1 ( ) ( ) ( , , )2where

1 ( , , ) ( )( ) ( )( ) ( )2Substituting for gives

1 ( ) ( ) ( ) ( 1)2

d

d

s

x u

m x r m x x e m x x f u

m x r m x r e N x x t u

N x x t m x x e m x r x e f x

u

m x r m x r e N k r

sgn( )e


Derivative

1 ( )2 m x eWhy?, Motivated by the analysis

Analyze these

closed-loop dynamics

47

2 2

2

2 2

Let's study the stability of our control using the following Lyapunov candidate:

1 1 ( )2 21 1 ( ) ( ) ( 1) ( ) sgn( ) ( )2 2

( ) sgn(

new

s new

V e m x r V

V e e r r m x r e k r r N e m x r V

V e r r N e

2 2

)

( ) sgn( ) ( ( ) ( ) )

Let us define a new variable , as follows:

( ) sgn( ) ( , , ) | ,

We assume that

s new

d d s new

d d d d

k r V

V e r r N e r N N k r V

L

L t r N e N N x x t x x x x

can be bounded as follows

( ) ,

where, ( ), is a non-decreasing, positive, scalar function

So, due to the above assumptions , .

d

d d

N N N

eN z z z

r

N N L

Continuous Asymptotic Tracking (continued)The second term that results from the derivative of the Lyapunov function

is canceled by the term introduced in previous slide.

Solve

for

r e e

e

Cancelation by

term introduced

in previous slide.

Crucial step: ( , , )d d dN x x t

always boundeddN

small if dx x

48

0

2 2

Let ( ) ( is a positive constant) then, ( )

where we still have to show that 0.

Substituting these definitions into the equaton for we get

( ( )

t

new b b new

t

new

s

V L d V L t

V

V

V e r r N k r

222

2 2 2

4 2

222

3

22

3 3

)

Now use the bound for :

4

, where min{ ,1}4

ss s

s

z zz k r z

k k

s

s

N

V e r r z z k r

z zV z

k

zV z

k


challenging

2 22

2 2

2

2

Complete the square

2

222

add/subtract b then write as a squared term

then find an upper bound by throwing away the

negative term

s

ss

a ab ba b

k r r z z b

z z z zab r z z b r

kk r

49

2 2

1 2

1 11 22 2

22

3

1 01We can also write as:

0 ( )2

where, ( ) and

Now, let min{1, } and ( ) max ( ),1

We then have

if , where4

new

new

s

eV V e r V

m x r

zy V x y y

V

m x m x

zV z

k

22

3

1

2

2 1

03

0

if 4

( )Knowing that we can write

( )

if Here, we can replace with .4

s

s

yV z k

V ty

V t

V z k t t


eigenvalues of the diagonal matrix

(continuing from previous slide)

50


2 2

So, we have Semi-Global Asymptotic tracking!

How do you know?

Remember our Lemma involving V g(t)?

Recall our Lyapunov candidate

1 1 ( )2 2

and

new

new

V e m x r V

V negative terms L V

0

( ) ( ) ( ) sgn ( )

( )

So, this gave us ( ) ( ) Asymptotic stability

Why not follow this procedure all the time?

Difficult to show t

new

d

new

V

L t r t N t e t

V L t

V negative terms L t L t negative terms

hat is lower bounded by zero (i.e. the integral is always 0).newV

51


0

0 0

So our result is only valid if 0.

( ) ( ) ( sgn ( )

( ) ( ) sgn( ( )) ( ) ( ) sgn( ( )) Expanded

Remember sgn . We now show th

new

t

new b d

t

t t

new b d d

t t

d

V

V e e N e d

V e N e d e N e d

L r N e

0

0 0 0

at if is selected as

1 ( ) ( ) ,

then ( ) .

So ( ) ( ) ( ) 0

d d

t

b

t

b d

N t N t

L d

e t e t N t

This condition is actually

developed in the next slide.

This would ensure that is positive.NEWV

52


0 0 0 0

0 0

0 0

Integrate by parts

Working with just the integral:

( ) ( ) ( ) ( ) sgn ( ) ( ) ( ) sgn ( )

( ) ( ) ( ) ( ) | ( ) ( ) (

t t t t

d d

t t t t

t ttt d

d t tt t

de deL d N d e d e N e d

d d

dNL d e N e d e e

d

0

1

0

2

2

0 0 0 and are positive since > ( )1

) ( ) sgn ( )

12

2Note: sgn( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

b

t

d

t

t

d d

t

N tb d

N e d

xxd x x

x x xdt xx

L d e t N t e t e t N t e t

0

0

This term is always negative( )1

( ) ( ) sgn ( )

So, we have ( )

td

d

t

t

b

t

dNe N e d

d

L d

Because of the condition on .

Thus 0V Done !

53

Feedback Linearization for Second-Order Systems

dmmpv

mpvd

pv

mpvd

md

m

d

mT

m

qMMNNqVVekekMeM

NqVekekqM

ekeke

NqVekekqM

NqVqMeM

qFqGqqNqqNqqqVqqM

qqe

xqqVqMx

qMqFqGqqqVqqM

)ˆ(ˆ)ˆ()(ˆ

ˆˆ)(ˆ

try weifWhat

0

)(

can write wemodel), (the system about the everything know weIf

))()(),(( ),(),()(

as system therewrite could We

0),()(21

symmetric. definite, positive is )( where,)()(),()(

system heConsider t

General Dynamic Equation for an n-link Robot

54

Feedback Linearization Problem (continued)

1 1

Continuing from previous slide:

ˆ ( )( ) ( )

ˆ ˆ ˆwhere , and

( , , , ) ( , , , , , , ) Not good. Why?

v p v p d m

m m m

v p m d d d

e k e k e I M M k e k e M Mq V q N

M M M V V V N N N

e k e k e f M M V N f e e q q q q q

Let's try something else. Define

Multiplying through by gives

( )

( ) ( )

( , ,

d m

m d m d

m

r e e

r e e

M

Mr Me M e

Mr M q e V q N

Mr V r M q e V q e N

Mr V r Y q q

, , )

ˆ ˆDesign your control, letting and . Now, we can write

ˆ ˆ , where , , and .

d d d

T

Tm

q q q

Y kr Y r

Mr V r kr Y Y r

Filtered Tracking ErrorFrom linear systems:

e+ e=r

1E(s)= R(s) if r(t) 0 then ( ) 0,

then ( ), r(t) 0 e( ) 0

e ts

e t t

55


Our Lyapunov candidate can be selected to be

1 1 2 2which gives

1 1 1 1 1 2 2 2 2 2

ˆ1 21 ( ) 2

T T

T T T T T

T T T

T T T Tm

V r Mr

V r Mr r Mr r Mr

V r Mr r Mr

V r V r kr Y Y r r Mr

V

2( ) recall that

So, all signals are bounded, and 0 (due to our stability lemma). Notice that this

way did not feedback linearize the system like the previous one.

T Tr kr g t M r r Mr

r

56


212

Example - Simple case : scalar state, exact model knowledge

( , )

, , ( , )

( , )

Our Lyapunov candidate can be selected to be

d d d

d

x f x x u

e x x e x x e x f x x u

r e e

r e e x f x x u e

V r

2

which gives

( , )

design

( , )

then

is , is ND 0,

since r(t) 0 then ( ) 0,

since ( ), r(t) 0 e( ) 0

d

d

V rr r x f x x u e

u x f x x e r

V r

V PD V r

e t

e t t

From linear systems:

e+ e=r

1E(s)= R(s) if r(t) 0 then ( ) 0,

then ( ), r(t) 0 e( ) 0

e ts

e t t

Converting the 2nd-order problem into a 1st-order problem

Opportunity to design the control ( )u t

57

For the previous system, we want to apply a robust control:

( ) ( ) ( , )

1We made the assumption that ( ) was p.d. symmetric and ( ) 0.2Let our control

m

d m d

Tm

Mr V r W

W M q e V q e N q q

M q x M V x

2 2

1 2 3

be

, where we choose from , , or

So, our system can be written

Where

Choose the Lyapunov candidate to be

1 2Taking the derivative

R R R R

m R

T

r r rkr V V V V

r r

Mr kr V r W V

W

V r Mr

gives

( )T TRV r kr r W V

Previous Problem Using a Robust Approach

58

Previous Problem Using a Robust Approach

222

min

1

2

min

2 2

1 2 1 2

Continuing from the previous slide:

{ }

{ }

Since is p.d. symmetric, we can write

1 1 ( , are constants)2 2

Where

rr

rV k r r

V k r

M

m r V m r m m

2 2

1 2

min

2

min

2

the assumption ( ) was used.

2 { }Let , which leads to

2 { }

( ) (0)exp( ) 1 exp( )

Therefore, the system is GUUB.

On a practical note, high ga

Tm x x M q x m x

k

m

kV V V V

m

V t V t t

ins cause noise to corrupt actual experiments.

Nonlinear Lyapunov-Based Techniques Provide– Observers or Filters for State

Measurement Replacement

Observers

Mechatronics

Based Solution

AdvancedNonlinear Control

Design Techniques

RealtimeHardw are/Softw are+

NewControl

Solutions

x f x y·

( , )y g x y u·

( , , )u ? x

NonlinearObserver

NonlinearController

Estimate of the State

Observers Alone

x f x y· ( , )y g x yu

·

= ( , , )u ? x

Nonlinear

Observer

Ex: Motor with robotic load: sin( )=

Standard approach: measure and to control .

Could we reduce cost or improve reliability if we didn't need to measure ?

61

ObserversGiven system ( , ) ( ) we have assumed that all states could be measured

and used in feedback (full-state feedback (fsfb)).

x f x x g x u

Example: If angle is measured with an encoder then the velocity must be estimated, e.g. using backwards difference.

Encoder Measured Position

Position

Velocity Estimate

Backwards difference may yield noisy estimate of actual velocity

62

Observers (continued)

Consider the linear system with plant

A full-state feedback control would look like

Specify a Luenberger Observer as

ˆ ˆ

ˆ ˆ

fsfb

x Ax Bu

y Cx

u kx

x Ax Bu Ly

y Cx

ˆ where

Modifying the above control, an observer-based feedback control

ˆ ˆwould use state estimate and look like ( or the plant the observer)

The separation principle (linear sys

o

y y y

x u kx f

tems ONLY) says that for the

plant works just like for the plant and the observer.o

fsfb

u

u

Solution for linear systems was to design an observer for unmeasurable states.

In a linear system, can design the Observer and the Controller Separately

use a formula to find L based on plant parameters

use a formula to find k

based on plant parameters

63


11 2

2

What about a nonlinear system? Consider the system

( ) ( ) (nonlinear)

( ) not all appear in so you will want an observer!

ˆ ˆ( , , ) ( ) and ( ) are

ˆ( )

x f x g x u

y h x x y

x x u y

u x

designed

Then, you could try

ˆ ˆ ˆ( ) ( )

ˆ ˆ ( ) Difficult to prove stability result

ˆ

x f x g x u Ly

y h x

u kx

(Using the Linear Systems Approach)

Note what this means:

ˆ ˆif then x x x u kx kx kx This estimation error term could

destabilize the system (Kokotovic peaking)

In a nonlinear system, may not be able to design the Observer and the Controller Separately

Can't assume the Separation Principle holds for nonlinear systems.

64

Observers (continued)1Let's try to develop an observer for the scalar ( ), second-order nonlinear system of the form

( , )

The nonlinear system above can be represented by two cases:

Case 1) ( ) is known, bu

x

x f x x u

f

2 4

2 4

t unmeasurable e.g. ( , )

Case 2) ( ) is uncertain and unmeasurable, e.g. ( , ) { is unknown}.

For Case 1, we can estimate :

ˆ ˆ a) Open-loop observer : ( , )

b) Cl

f x x x x

f f x x ax x a

x

x f x x u

ˆ ˆosed-loop observer : ( , ) ( , ), where x f x x f x x x x x

No feedback. Other possible approaches include

a Kalman or particle filter as an estimator.

if we knew and

then we know ( , )

x x

f x x

We will address Case 1 with an observer (Case 2 is more difficult)

(this will not be a general result)

We now seek to design a closed-loop observer.

65


A filtered tracking error (a change of variables),

that transforms a second-order into a first-order problem can be defines as:

s x x

ˆ

ˆ ˆ ˆStart with the estimation error, , then and

Substituting the system dynamics ( , ) x

x x x x x x x x x

x f x x u

We need , described by the dynamics in , to go to zero, this seems similar to our previous use of the Lyapunov functions to design the controllers. We can see a hint of what the obse

x x

ˆrver should do (via ) to make the estimationerror dynamics go to zero:

1) Cancel ( , )

2) Add feedback (stabilizing) terms

x

f x x u

This linear system can be transformed (Laplace transform) into

1( )= S( ) if s(t) 0 then ( ) 0,

then ( ),s(t) 0 ( ) 0

X x t

x t x

s

t

ss

66


Mathematically, this may make go to zero but it

includes - the quantity we are trying to estimate!

There is a solution that we will see later.

x

x

Motivated by the use of the filtered tracking error (and a lot of trial and error),

let's apply the change of variables. Substitution from the system dynamics yields:

ˆ s x x x x x

01 02

ˆ

ˆ

( , )

Anticipating the Lyapuniov analysis, propose an observer

ˆ ( , )

x

x

f x x u x

f x x u k x k x

67

01 02 01 02

01 02

01 02

ˆ

Substitute observer

ˆ( , ) ( , )

Now substitute into the s-dynamics

Make and 1, then

x xx k x k x f x x f x x k x k x f

x

s x x

k x k x f x

k k k k

s k

1

=

x k x f x

kx kx x f

ks x f


01 02

01 02

Note this can be arranged as a linear system: should be able to pick and to make go to zero (if =0)

x k x k x fk k x f

Just substitute filter

x x s

68

2 2

2 2

1 2

22 21 2

Consider the Lyapunov candidate:

1 1 1 (Here , )2 2 2where

( ) ( )

Assume , then

Note:

TTV x s z z z x s

V x x s s ks x f

V x ks sf

f x s

V x ks x s s

22

2 211 2

Where is a positive constant

We used

We can use the property

which allows us to write

( )

All signals bounded! (Can you show th

x s x

xx y y

V x k s

is?) Here we assume that , x x L


If and k are selected large enough,

negative definite, so and 0!x s

2 22

22

Note:

2 0

2

x xy x y y

xy x y x y

Done if 0 f

69

01 02

01

02

ˆ

ˆ

Start with the orginal

ˆ( , )

and introduce a new variable with derivative . Rewrite as two first-order equations

ˆimplementable, closed

ˆ( , )

x

x f x x u k x k x

p p

x p k x

p f x x u k x

loop observer

Observers (continued)ˆClean-up: remember we introduced the to make go to zero but it

included - the quantity we are trying to estimate. We need to fix that now!

x x

x

01

01

This is a trick to make the observer implementable i.e. can be applied using measurable quantities.

ˆTo see how it works, differentiate

ˆ

x p k x

x p k x

Not measurable

All signals are measurable!

Term we needed to stabilize the observation error dynamics (not measurable)

Terms that we don't want to differentiate go in and appear here.p

0

Example: Design an observer to estimate in the open-loop system:

( is measureable but is not).

Define:

ˆ

(similar to the filtered tracking error ) then

u

x

x x x u

x x

x x x

s x x r s x

2 21 12 2

2

ˆ

propose

ˆ

rearrange definition of s:

ˆ

ˆ

substitute the open-loop system ( with

x x x x

V x s

V xx ss xx s x x x

x s x

V x s x s x x x

x sx s x x x

x

2

2 2

2 2

u=0):

ˆ

ˆwe would like to have only and - in , design to make this happen:

ˆstabilize cancel cross termcancel

V x sx s x x x x

x s V x

x x x x s x

V x s


Implement the closed-loop observer

ˆ 1 1

ˆ 1

1

x x x x s x x x x x

x p x x

p x x

71

ˆWhat kind of terms can we put in ( , ) and cancel directly with ?f x x x

2

For open-loop observer, analysis leads to

ˆ( , )

Two-part implementation of the filter:

ˆ ˆ(terms to get differentiated to make )

terms that don't get differentiated to

V x sx s f x x x x

x p x

p

1 2

ˆ

2

1

ˆmake

ˆ ( ) 1 ( , ) 1

Implementable observer:

ˆ ( , ) 1

( ) 1

put in p put in x

x

x f x x f x x x

x p f x x dt x

p f x x


2Basically we need to be able to find ( , )f x x dt

212 2 2

22 2

Examples of favorable terms:

( , ) , ( , ) ,

Examples of unfavorable terms:

( , ) ( , ) ?

f x x x xx f x x dt x x

f x x x f x x dt

0

2

Example: Design an observer to estimate in the open-loop system:


Define:

ˆ


u

x

x x u

x x

x x x

s x x r s x

2 21 12 2

2

ˆ

propose

ˆ


ˆ

ˆ

substitute the open-loop system ( with u=

x x x x

V x s

V xx ss xx s x x x

x s x

V x s x s x x x

x sx s x x x

x

2 2

2 2

2

2 2 2 2

0):

ˆ


ˆ ˆ

ˆ

stabilize cancel cross termcancel

V x sx s x x x

x s V x

x x x s x

V x s x x


Implement the closed-loop observer

ˆ 1 1

ˆ 1

1

x x x x s x x x x x

x p x x

p x x

2Can't cancel the term with x

73

0

2 2 2 2

1 2

22 21 2

22

Example (cont): Design an observer to estimate in the open-loop system:

ˆ

Assume , then

We can use the property

u

x

V x s x x

f x s

V x ks x s s

xx y y

2 211 2

0

which allows us to write

( ) V x k s


2 2

Definition of derivative ( ) and the Mean Value Theorem:

ˆ( ) note ( ) (y)

ˆ

Apply norm and rearrange:

( )

Since , is a known function: ( ) 2

2

Triangle

f x

x x f df c f c f

x dyx x

f f c x

f x x f c c

f c x

1 2

Inequality

2 2

2

f c x c s x

c s x

x s

74

Combining Observers & Controllers (continued)

2

22

2

2

2

? (assume is a positive gain)

if then 0

if then / and

=

Thus we have the greatest upper bound

zy kz k

zy kz z y k z z y k z

k z y z y k z

k z y z y k y k z y

y yz y k z y

k k

yzy kz

k

Tool for Lyapunov analysis - "nonlinear damping"

75


Modification to previous observer design

Use estimate in place of .x

Motivated by the use of the filtered tracking error (and a lot of trial and error),

let's apply the change of variables. Substitution from the system dynamics yields:

ˆ s x x x x x

01 02

ˆ

ˆ

( , )

Anticipating the Lyapunov analysis, propose an observer

ˆ ˆ ˆ ( , )

x

x

f x x u x

f x x u k x k x

x

76

1

ˆSuppose we redefine :

ˆ ˆ ˆ ˆ ˆ ( ) ( , ) so now it depends on instead of

If , then we can use the Mean Value Theorem to state

ˆ ˆ ( , , , ) ( , , , )

ˆ ˆwhere ( , ) (

f

f f x x x x

f c

xf x x x x x x x x

x

f f x x f

1

ˆ, ).

We can then write

( , , , )

( , )

x x

xf x x x s

s

xf x s

s

Observers (continued)Still considering that () is known function but

want to distiguish the fact that we are using an estimate

of a measurable quantity.

f

( )f x

xx

ˆ( )f x

assume ( )

ˆ( ) ( )( )

ˆ

( ) ,

f x x

f x f x ff c

x x x

f x f c x x x

Mean Value Theorem (in one variable)

if 0 then 0x f

http://en.wikipedia.org/wiki/File:Mvt2.svg

77


2 2 2 21

22 21

For the observer problem:

( , ) remember we found that

let 1 and 1. Then,

,

221

General approach to find

n

n

xV x ks s x s V x ks sf

s

k k

xV x s z z s k s z

s

z z

kn

2

2

1

2

22 2

2

upper bound for :

( ) and

There are two different cases:

Case 1) 0

Case 2) so,

then

n

n

n nn n

n

N x k x

N z L N s k s

k s N s L

N Nk s N s s L k s

k k

NL

k

Nonlinear Damping

78

222 1

221

2 21

2

21

1

So,

1

if , where is a positive constant.

1Recall that . So, we can write2

2 if 2

( ) (0)exp( 2 ) if

n

n

n

n

n

zV z

k

V zk

V z k z

V z

V V k V

V t V V k

2

2 2 2 21 1

1

2 ( )

1 1 ( ) (0) exp( 2 ) for 2 (0) 2 ( )2 2

( ) (0) exp( ) for (0)

This gives us a semi-global exponential result! Why not global?

n

n

V t

z t z t k V V t

z t z t k z


Observer + Controller

x f x y· ( , )y g x yu

·

= ( , , )u ? x

NonlinearController

Nonlinear

Observer

80

Combining Observers & Controllers

Can we develop a combined observer/controller for the previous system?

( , )

In the observer alone, we assumed ( , ) if , but we couldn't measure .

Our control objective is to

x f x x u

f x x L x x L x

1 1

force when only is measureable.

The observer/controller is more complex since all signals must be shown

to be bounded.

We can choose from two different error systems

ˆ ˆCase 1)

C

d

d d

x x x

e x x e x x

1

1 01 01

ase 2)

Let's use Case 1 since is measureable.

This gives us

ˆ

where

d d

d d d d

d

e x x e x x

e

e x x x p k x x k x p p

p p p

01

02

Closed-loop observer

ˆ

ˆ ( )

x p x k x

p f k x u

Inject a new term by

...d

d d

p

p p p

81

Combining Observers & Controllers

1

2 21 1 1 1 1

We are using to facilitate the stability analysis (seen later).

Here we see it makes "nice":

if we had ... ... ...

d

c

p

e

V e V e e V k e

1

1

In this step we have enhanced the role for the observer.

Recall that is our tracking error. We will see that the

observer will act to promote the stability of .

e

e

1 01

01 1 1

1 1 1

Now, given

Letting gives

d d

d d c

c

e x k x p p

p x k x k e

e k e p

"Interconnection term"

82


1

01 1 1 02

1 1 1 02 01

measureable

Recognizing that , we can write

ˆ ( )

ˆ ( )

We can design the control as follows

d

d c

d c c

W

p p p

p x k x k e f k x u

p x k k e p f k x k x u

1 2 1

2 1 01

2 21

21 1 1

is a control input designed during the stability proof

Taking the following Lyapunov functions

1 1 2 2

c aux

c aux

c

c c

Vauxu W k p e V

p k p e k x V

V e p

V k e e p

1

2 2 22 1 01 1 1 2 01

2 2

2 2

2 2 2 21

( ) ( )

1 1 2 2

where the combined Lyapunov function can be written

1 1 1 1 2 2 2 2

c aux c c aux

e

o

o

o c

k p e p p k x V k e k p p k x V

V x s

V x ks sf

V V V x s e p

Feedback

"Interconnection Buster"

Observer Lyapunov function

01

02

Closed-loop observer

ˆ

ˆ ( )

x p x k x

p f k x u

Controller Lyapunov function

83


2 2 2 21 1 2 01

good terms (Why?) bad terms (Why?) injected term

Now, we can write the derivative of the combined Lyapunov function as

Using

L

c c auxV x ks k e k p sf pk x pV

01 01

21 01 1 01

201 1 01 1 01 1

will be designed later1

the definition of : ( )

and letting lets us write

( ) ( )

aux aux n

aux n aux

Vaux

x pk x pk x s

V V k x k k p

L sf pk x s pV p k x k k p sf pV pk

2 201 1 01

2201 1 01

2

1

2 2 2 21 1 2 1

1

Now, we can say

( )

( )

So, we can write

1

n

n

n

c c auxn

s k k p

L k p s k k p

sL

k

V x k s k e k p sf pVk

Unmeasurable mismatch

01 1 01

1 01 01 1 01

1 01 01 1

2

01 1 011 1

Nonlinear damping on one term

if then 0

if then /

=

n

n n

n n

nn n

k p s k k p

k k p s k p s k k p

k k p s k p s k

s sk p s k k p s

k k

Done if 0

Why not use ND on ?

f

f

84


1

1 2 3 1

Recall that

ˆ ˆ ˆ ( , ) ( , )

Let's assume that , then

ˆ ˆ ( , , , ) , where and

It can be shown that

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( , , , ) ( , , , ) ( , , , ) ( , ,d

f f x x f x x

f c

x xf x x x x z z z

s s

x x x x x x x x x x x s x e

4 1 1

5 1 6 1

2 2 2 21 1 2 6 1 1

1

, , ) ( , , , , , )...

... ( , , , , , ) ( , , , )

Then we can write

1 ( , , , ) we let 0

d d

d d

c c auxn

x x s x x e e x s

x x e p x s e p x s

V x k s k e k p s e p x s z Vk

Variables in Lyapunov function

Turns out that

we don't need it

85


3

2 2 2 21 1 2 6

1

1

1 21

2 2

6

6

From the previous slide:

1 ( ) ,

where

1If we let , , 1 and 1, we can write

( )

If ( ) , then

c cn

c c Fn

F

N

F

V x k s k e k p s z zk

z

z e

p

k k k kk

V z s z z k s

k s z z

3 6

226

6 3

222 6

( ) 0

If ( ) , then

( )which gives,

F

FF

F

N s z z k s

z zk s z z N

k

z zV z

k

(See the nonlinear damping argument)

1

x

s

z e

p

Combined all 4 states into

86


222 6

226

2 26

2

26

26

From previous slide:

( )

( )So, 1

1Remembering that , we can say2

2 if 2

( ) (0)exp( 2 ) if 2 (0)

Now, we

F

F

F

F

F

z zV z

k

zV z

k

V z if k z

V z

V V k V

V t V t k V

2 2 26

26

can write

1 1 ( ) (0) exp( 2 ) if (0)2 2

( ) (0) exp( ) if (0)

F

F

z t z t k z

z t z t k z

Semi-global exponential

tracking

87


1

01 1 1 01 01 1 01

Remember that

ˆ

ˆ ˆ( ) ˆˆ

ˆ ˆ ˆ ˆ ˆ( ) ( )

Finally, we have semi-global exponent

dd

d c d c d

x xxxx x x xx xs

zx xx xe

p x k x k e x k x x k x k x x x k x x

so ,

16

ial stability, and we can say

ˆ ˆ ˆ ˆ , , , Occurs exponentially fast!

Recall that you can't measure ( ), which came from knowing ( , ) . Using t

d d

d d

x x x x

x x x x x x x x

f x x c

1

2 2

he

Mean Value Theorem and the fact that something is tells us that

ˆ ˆ ( ) ( ) ,

For example,

ˆ ˆ ˆ ( )

c

h x h x x x x

x x x x x x x x

goes to zero

What is bad about the observer approach?

Need to know the function f( )

88

Filter Based Control

( ) is a positive scalar functio

Assume we have the same system:

( , )

where only is measureable, and the structure of ( , ) is uncertain.

We will assume that

( , ) ( , )

x f x x u

x f x x

x x f x x

2 2

( , )( , )

n

In the analysis we use the fact that

Why couldn't we use this function in the control (if we know )?

It depends on (which we don't know)!

Example:

cos( )

x xf x x

x

x x a a x

( ) exists

( )The inequality is true, but depends on x

Assuming velocity is not measureable in second-order system,

build a "stunt double" for velocity.

Make e go to zero

without knowing ( ) or f x

Solves the same general problem as the estimator but in a different way.

89

Filtering Control (continued)

Let's define the following:

( , ) Why the 2nd derivative? We need it for the control.

We will need and , but is not measureable. So, we come up with another variable:

de x f x x u

e e e

a filter to help us with the problem

We now have three error systems:

error system 1 ) ( , )

error system 2 ) This is the

f

f d f

f f

e e e e

e e e x f x x e e u

e e k e

2 2 2

From the definition of ( )

filter design

error system 3 )

Our next step is to develop a Lyapunov candidate.

1 1 1 2 2 2Since is not measureable (due to the fact th

f

f

te e e

V e e

at is not measureable), we cannot use

it in the control. Later we show that is measurable.f

e

e

is similar to the filtered

tracking error that we used earlier

but is not measureable

r

Can't implement in this

form since it contains

90


2 2 2

Taking the derivative of our Lyapunov candidate gives

( ) ( ) ( ( , ) )

( ) ( )

( 1) ( , ) ( 2)

Is measurea

f f f d

f f

f d f

f

V e e e e e k e x f x x u

e e e k e

V e e k x f x x u k e e

e

1

2

1 2

ble?

1 1

Let's develop a new variable , where

We will need to find and . Differentiate to get

f f f f f

f

f

e e k e e k e e e e k e k e ke

p

p function

e p function

function function e

2 1 2

1

2

So, that means that

( 1) ( 1)

f

f

e p derivative function function derivative function

function k e k e

function ke

This is the only part contributed by the system,

everything else is from the filter design and error definition.

fe e

into fe into e

can we use

in the

control?

fe

Can't implement in this

form since it contains e

91


In previous slide:

1 1

1 1

Laplace tansform:

s 1 1

1

s+ 1

f f

f f

f f

f

e k e k e ke

e k e k e ke

e k e k e kse

ks ke e

k

High-pass filter:

( )s

H ss a

frequency

2

2 2 ( )H s

a

( ) ( )

which means the filter acts as a differentiator

over a certain range of frequencies

H s H s s

High-pass filter:

( )s

H ss a

frequency

( )H s

The filter acts as a differentiator of

over a certain range of frequencies

fe e

92


2 2 2

Now, we know

( 1) ( 1) and

So, is measureable, which leaves ( , ) and as the unmeasureable variables in

( 1) ( , ) ( 2)

We design the contro

f f

f

f d f

p k e k e e p ke

e f x x

V e e k x f x x u k e e

2 2 2

11

Note

By the Mean Value Theo

l:

( 2) , let 2

Now,

( , )

Also, we define

( , ) ( , )

If , then ( , , , )

d

d f n

f n

d d

f

d d

fd d

u x k e e k k

V e e f x x k

f f x x f x x

ef C f x x e e

e

2

1rem, since

,

f C

ef e e

e

Use this feedback term to deal with the

unknown, unmeasureable function ()f

New approach

to define f

93


2

3


,

Let's come up with a new variable , where

So, we know (because )

1Our Lyapunov function becomes

f

f

ef e e

e

z

e

z e

e e e

f z z

V

2 2 2

2

3

1 2

2 2

3 1 2

1 1 1 z with derivative2 2 2 2

Letting and allows us to write

Tf

T Td n d n

n n n d d

Tn d n

e e z

V z z f f k z z z z f k

k k k f

V z z z z k k

Add/sub = ( , ) and use defininition of d d df f x x f

94

2 2

3 1 2

2 2 23

1 2

2 223

1 2

2 223

1 2

As seen on the previous slide,

Now, we can say

1

1 , where

So, we can

Tn d n

T d

n n

d

n n

d

n n

V z z z z k k

z zV z z

k k

zV z

k k

zV z

k k

2 21 3

21 3

write

, if

2 , if 2 ( )

2 ( ) ( ) 0

n

n

V z k z

V V k V t

V V s t s t


Nonlinear Damping

1 Note:

2TV z z

Proof to show

Semi-global Uniformly Ultimately Bounded Tracking (SGUUB)

2

2

Nonlinear damping on one term

Given with a,b>0

if then 0

if then y

b

a x y ab x

ab x y a x y ab x

yab x y a x y ab x a x y

b

ya x y ab x

b

95


0 0

1exp(2 1

2


2 ( ) ( ) 0

Solving the differential equation gives

( ) exp( 2 ) (0) exp( 2 ) exp( 2 ) exp( 2 ) ( ) exp(2 )

(

t t

t

V V s t s t

V t t V t d t s d

V

) exp( 2 ) (0) 1 exp( 2 )2

t t V t

0 since is neagtives

96


21 3

21 3


( ) exp( 2 ) (0) 1 exp( 2 ) if 2 ( )2

So, ( ) is bounded such that ( ) (0) choose 2 (0)2 2

We can then write

( ) e

n

n

V t t V t k V t

V t V t V k V

V t

21 3

2 221 3

xp( 2 ) (0) 1 exp( 2 if 2 (0)2 2

which means

( ) (0) exp( 2 ) 1 exp( 2 ) if (0)

So, we have semi-global ultimate uniform boundness. We can easily

n

n

t V t k V

z t z t t k z

show that all

signals are bounded.

2

22

We don't show that e goes to zero, only show

that it can be made smaller by choice of gains.

Specifically, decrease by increasing dn

n

kk

97

Adaptive Approach

Reconsider the previous system:

( , )

( , )

( 1) ( , ) 2

Let

2

where

d

f

f f

f

f d f

d f ff f

x f x x u

e f x x u x

e e e

e e k e

e e e

e e e k x f x x u e

u x e e u ke

u

is a feed forward term, which was not included in our previous control.

This gives

( 1) ( , )

ff

f ff fk e f x x u ke

98

Adaptive Approach (continued)

2

2 2 2

Consider the Lyapunov candidate

1 (where [ ] )2which gives

( 1) ( , )

Assume ( , ) ( , ) Assume LP

We now write

( , ) (

Tf

f ff

ff

V z z e e

V e e k f x x u

f x x W x x

L f x x u f x

0

( , )

, ) ( , ) ( , )

( , ) Recall that ( , ) ( , )

ˆIf we can show that , and we let ( , ) , then

d d

d d d d ff

d d ff d d

f x x

ff d d

x f x x f x x u

L f W x x u f f x x f x x

f z z u W x x

ˆ ( , ) , where d dL f W x x

99


2 2 2

Now, consider the Lyapunov candidate:

ˆ1 1 , 2 2

where

ˆ ( 1) ( , )

Our system can now be written

( , ) , whe

T Tf

f d d

e

V z z z e

V e e k f W x x

x W x x u

1

1

re we assume that ( , )

ˆ 2 ( , )

We know that , , is true since

( , ) ( , ) and ( , )

Let's create a variable, , where

( 1)

d f f d d

d d

d d

f

W x x c

u x e e ke W x x

f x x z z

f W x x W x x W x x c

p

p k e

( 1) and

Let

f

f f

k e e p ke

e e k e

100


2

222

2

We address this below

If we let 2, then

ˆ ( , )

ˆWhere we let ( , ) is NOT measureable!

We can know say

( )

1

n

Tn d d

d d

n

n

k k

V z z z k W x x

W x x

zV z

k

Vk

2

0

1 0

0 0

We need to use integration by parts:

ˆ ( , )( ) , where is just a dummy variable

( , ) ( , ) | ( , )

unmeasurablet

d d f

tt

d d d d d d

measurable

z

W x x e e e d

deL W x x d W x x e W x x e d

dt

Unmeasurable part t

101


1 0

0

1

0

As seen on the previous slide:

( , ) | ( , )

Finally,

, ( )

( , ) (0), (0) (0)

The apadtive update law can now be completed and then we ca

tt

d d d d

ddt

d d d d

L W x x e W x x ed

dxdW x

dL W x x e W x x e e d

d

2 2

2 2

n say

if ( )

Our result is semi-global asymptotic. Why is it not exponential? has more

terms in it than just .

We can also write

for 2 ( )

2

n

n

T

V z k z t

V

z

V z k V t

V z

V

2 2 for 2 (0)nz k V

102


2 2

t


for 2 (0)

ˆIt can be shown that , , , , , , , . Why do we care if ?

We want z( ) , which would mean lim ( ) 0. Remember, h

n

f f

V z k V

z L e e e e L z L

t L z t z

as , ,

and in it. So, they go to zero also. This has been an example of output feedback

adaptive control. It gave us semi-global asymptotic tracking.

Why didn't we use an observer (we used a fi

fe e

lter)? We don't have exact model

knowledge (there is uncertainty in the model)!

103

Variable Structure Observer

1

Consider the system:

( , ) ( , ) , where we observe with only measurements of .

We also make the assumption that , , , , , ( , ), ( , ) , where

( , ) , ( , ) and are un

x h x x G x x u x x

x x x u u h x x G x x L

h x x G x x C

1

certain. Why do we make the assumption about

ˆ ˆboundness? We want to build a , so we want to ensure that .

For our problem, we define

ˆ

ˆ

ˆLet , where sgno

x x x

x x x

x x x

x p k x p k

2

1 2

1 2

Observer

Observation error system

( )

ˆThen, sgn( )

( , ) ( , ) sgn( )

o

o

x k x

x k x k x k x

x h x x G x x u k x k x k x

104

Variable Structure Observer (continued)

1 2

2 0

1 2

Let's create a new variable, , where

( , ) ( , ) sgn( )

Let . ( 1; ) Now, we can write

( , , ) sgn( )

So, we have

o

o ij

o

o

r

r x x

r h x x G x x u k x k x k I x

k k I (k ) i j

r N x x t k x k r

N h

0

1

2 1

( , ) ( , )

We can let our Lyapunov function be

1 , where we must prove that 02

( ) sgn( )

So, we can now write

sgn( )

tT

o o o bo o o

t

To o

T To o o

x x G x x u

V r r P t P t L d P t

L t r N k x

V r N k r k x r N k

1

( )

sgn( )

o oP L t

x

105


2 1 1

2

2

min 2


sgn( ) sgn( )

Next, we get

Using the Rayleigh-Ritz Theorem lets us write

{ }

So, 0 and ( ), where ( ) 0. If

T To o o

To

o

o o

V r N k r k x r N k x

V r k r

V k r

V V g t g t

t

min 2 min 2

1

( ) , then lim ( ) 0.

Here, ( ) { } and ( ) { }2

Therefore, , then 0 , 0!

But, we must show that ( ) 0, which requires

, where denotes the

T T

o

i oi oi

g t L g t

g t k r r g t k r r

r L r L r x x

P t

k N N i i

th component for vectors

106


0

0

0 0

0 0

1 1

1

So, our task then is to prove that

Let , so we get

sgn sgn

sgn

t

bo o

t

t

o

t

Tt tT

o o

t t

T Tt t

o

t t

L d

M L d

dxM N k x d x N k x d

d

dx dxM N d k x d x

d d

0

0 0

0

0

1

11

1

sgn

| ...

... sgn

tT

o

t

t nT T tot

o t i i tit

tT

o

t

N k x d

d NM x t N x d k x

d

x N k x d

ty

ty

ty

ty

y

ttyty

ydy

yd

d

yd

y

y

tt

t

t

t

t

2

02 y

and |y

:NotesMath Useful

0

00

107


0

1 0 0

1 11 1

1 11


sgn ...

...

tT T To

o o o

t

n n

i i i ii i

noi

i i oi ii

d NM x N k x d x t N t x t N t

d

k x t k x t

d N tM k x N k

dt

0

0 0 1 11 1

n

i 1

1 0 01

...

...

The term can be written , which gives

So, if we define , then 0. Notice t

tT

o

t

n nT

o i i i ii i

T

o i oi

n

i i i oii

bo o

d x t N t

x t N t k x t k x t

x t N t x t N t

M k x t x t N t

M P

hat is not in this observer; so, we

can't exploit it for a controller!

u

108

Filtering Control, Revisited

2

Assumptions

Let's consider the following system:

( ) ( , ) , is measureable

( ), ( , )

( ), ( ) if ,

( , ), ( , ) if , ,

Let and ( ) d

M x x f x x u x

M x f x x c

M x M x L x x L

f x x f x x L x x x L

e x x M x

1 2

be such that

( ) ( ) ( ) upper and lower bounded

Let sgn( ) ( 1)

Let our error system be defined by three equations:

error system 1)

error system 2)

f f

f

f

M x M x M x

u k e e k r e

e e r

r r

2 Crafted to make the analysis work( 1)

error system 3)

Where did come from? We invented it.

f f

f f f

k e e

e e r

109

Filtering Control, Revisited (continued)

2 2

2

2

We define

( 1)( ) ( 1)

Design such that 2

Then, by multiplying through by ( ) gives

( ) ( )( 2 ) ( ) ( , )

f f f f

f

d f f

d f f

p r k e r e e r p k e

e e r

x x r e k

M x

M x M x x r e k M x f x x u

2

2

( ) ( ) ( , , ) ( )(2 )

where ( ) ( , )

Then, if we add and subtract an ( ( , , ) | is bounded apriori)

We get,

1 ( ) ( ) ( )2

Remembe

d

d

f f

d

d d x x

x x

d

M x k M x N x x t M x r e u

N M x x f x x

N N N x x t

M x k M x N N u M x

2 1 2

r that . We can now put in our control:

1 ( ) ( ) sgn( ) ( 1) ( )21where ( )(2 ) ( )2

d

d f f

d f f

N N N

M x k M x N N k e e k r e M x

N N N M x r e M x

110


2 2 2 2


1 ( )(2 ) ( )2We can show

,

Our next step is to use the Lyapunov function.

1 1 1 1 ( )2 2 2 2Where taking

d f f

f

f

f f

N N N M x r e M x

e

eN z z z

r

V M x e r e

2

the derivative yields

1 ( ) ( )2 f f f fV M x M x e e r r ee

111


)sgn(1

)sgn(

)sgn(

can write weThen, ).1(1

Let

).()( where

)sgn(

)sgn()(~

)1()(21...

...)(21)1(

)()(21

:slide previous thefrom Continuing

1

22

1

222

1

2

12

21

1

2

12222

1222

22

222

2

fdn

fdn

fdn

n

fdff

fdf

fffffffff

ffff

eekNzk

zV

eekNk

zzzV

eekNkzzzV

kM

k

xMxMM

eekNMkzzreeV

eekNxMkNerkxM

xMererkrrreeeereV

eerreexMxMV

112

22

)(

0

1

22

22

2

1

1

22

1

so, );( have We

constant. a is where,)(Let

)sgn()( where

)(1

gives Rewriting

}1),(max{}1,min{

thatmindin Keep

)sgn(1

:slide previous theFrom

zk

zV

tLVV

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V

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eekNzk

zV

nnew

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b

tP

t

bnew

fd

n

fdn


113


ykzV

zkzV

PzyyM

VyM

tNtetek

NNkP

dkNtP

deekNeeeetP

nnew

nnew

TTnew

dib

dd

t

db

t

t

fdffb

22

22

222

1

0001

1

0

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0)(

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for

for

saycan then We

where,}1,2

max{ }1,{2

min

:proof thecomplete tohave weNow,

)()()(Let

if 0 that results previous from know We

done! is Work before. thisdone ve We' )sgn()()(

)sgn()()(

have We

21

114


.0,,,,, Therefore

.0)(lim then ,)( if know weand ,2)( Here

.0)( where),( and 0 So

)0(for

)0(for

)(for

:slide previous thefrom Continuing

t

1

222

1

22

1

22

ff

T

nnew

newnnew

newnnew

reeez

tgLtgzztg

tgtgVV

ykzV

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Summary

115

Control Design Framework:

special function of everything we want to go to zero

State Error (from zero equilibrium)

Tracking error

Filtered tracking error (r) is a trick to conv

V

ert a 2nd-order

system into a 1st-order system (can use with other controls)

Parameter Estimation Error

State Estimation Error

derivative of the special function

dynamics

V

V

of everything we want to go to zero control input

Feedback Linearization

Simplest case uses exact model knowledge

Adaptive Control

Observer

Filter

Summary

116

System: ( )

Let the tracking error, , be defined as

( )

1) Control Objective-make ( is a desired trajectory), assuming , .

2) Hidden Control Object

d d d

d d d d

x f x u

e

e x x e x x e x f x u

x x x x x L

ive-keep everything bounded (ie., , , ).

Design a controller based on the tracking error dynamics.

Note that if constant equilibrium point and 0 then

( ), the basicd

x x u L

x u

e x f x

Lyapunov stability analysis tools (Chapter 3) can be used.

Homework A.1

2. Design a contoller for the following system so that tracks cos( ).

where is an unknown constant. Simulate system for 1 plot the state, control, and

and the parameter estimates.

dq q t

q aq u

a a

1. Design a controller for the following system so that tracks cos( ).

where is an constant. Simulate system for 1 plot the state and control.

dq q t

q aq u

a a

Known structure but unknown parameters -> adaptive

Known structure and known parameters -> exact model knowledge control

Partially known structure (unknown component) -> robust

2

3. Design a robust controller for the following system so that tracks cos( ).

where is an unknown constant but you do know <a then < a q +1

Simulate system for 1 and a 3 plot the stat

dq q t

q aq u

a a a q

a

1 2 3e and control comparing , , controllers.

4. Design a learning control for the following system so that tracks cos( ).

where is an unknown constant. Simulate system for 1. Plot t

R R R

d

V V V

q q tq aq u

a a

he state ( ), tracking error,

and control signal ( ).

q t

u t Partially known structure (unknown component), repetitive task -> learning

Homework A.1-1 (sol)

12

1. Design a controller for the following system so that tracks cos( ).

where is an constant. Simulate system for 1 plot the state and control.

cos( )

sin( )

d

d

d

q q t

q aq u

a a

e q q t q

e t q q aq u

V e

2

2

( sin( ) )

design sin( )

V is PD, radially unbounded, is ND 0

0 , cos(t) bounded is bounded

0, is bounded, sin( ) bounded is bounded

is bounded,

d

V ee e t aq u

u t aq ke

V ke

V e

e q q q

e q t u

q

is bounded is boundedu q

Closed-loop system :

sin( ) cos( )q t k t kq

0 2 4 6 8 10-2

-1

0

1

0 2 4 6 8 10-2

-1

0

1

2

dq

q

Exact Model Knowledge, k=1

u


0 2 4 6 8 10-1

-0.5

0

0.5

1

0 2 4 6 8 10-5

0

5

10

Exact Model Knowledge, k=10

dq

q

u



exactly cancel

1. Exact model control:

0 sin( ) cos( )aq

q t k t kq

Homework A.1-2 (sol)2. Design a controller for the following system so that tracks cos( ).


and the parameter estimates..

d

d

q q t

q aq u

a a

e q

21 12 2

2 2 2

cos( )

sin( ) sin( ) sin( )

where [ ] and [ ]

ˆ where

ˆ ˆ( sin( ) )

ˆdesign sin( )

ˆ ˆ ˆ

T

q t q

e t q t aq u t W u

W q a

V e

V ee e t W u

u t W ke

V ke e W W ke eW ke

2 2

0

2

ˆ

ˆ ˆdesign ( cos( ) )

V is PD, radially unbounded, is NSD e and are bounded

T

tTd

W e

W e qe qq q q t q dt

V ke

V


2

Design a controller for the following system so that tracks cos( ).


and the parameter estimates..

dq q t

q aq u

a a

V ke


bounded cos( ) is bounded

Closed loop error system:

sin( ) ( ) ( ) and , , are bounded is bounded

2

d

V

e q q e t q q

e t W q u ke W q e q e

V

and e, are bounded is bounded 0 0

ˆ0, is bounded, is bounded is bounded

is bounded, is bounded is bounded

kee e V V e

e q u

q u q

0 1 2 3 4 5 6 7 8 9 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

time

a,

ahat-

:

Paremter Estimate

0 1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

time

q,

qd-:

States x1, x2(:) with k=1

a

a

dq

q

Homework A.1-2 (sol)Adaptive, k=1

0 1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5

time

q,

qd-:

States x1, x2(:) with k=10

0 1 2 3 4 5 6 7 8 9 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

time

a,

ahat-

:

Paremter Estimate

a

a

Adaptive, k=10

dq

q



exactly cancel

2

0


0 sin( ) cos( )

2. Adaptive closed-loop system:

( cos( ) ) sin( ) cos( )

aq

t

aq

q t k t kq

q aq q q t q dt t k t kq


2

3. Design a robust controller for the following system so that tracks cos( ).


Simulate system for 1 and a 3 plot the stat

dq q tq aq u

a a a q

a

1

212

2

2 2

22 2 2 2

e, control. controller.

cos( )

sin( ) sin( )

( sin( ) )

design sin( ) where a q +1

a q +1

(a a q +1 ) (a a q +1

R

d

V

e q q t q

e t q t aq u

V e

V ee e t aq u

eu t ke

e

eV ke e aq

e

eV ke q e ke q

e

2

0 by definition of the bounding function

)

V is PD, radially unbounded, is ND e 0

e ke

V

k=1

0 2 4 6 8 10-10

-5

0

5

100 2 4 6 8 10

-2

-1

0

1

2R1Robust V , 1k

dq

q

u



exactly cancel

2

0

2


0 sin( ) cos( )


( cos( ) ) sin( ) cos( )

3.a Robust - Sliding Mode

cos( )a q +1

cos(

aq

t

aq

q t k t kq


t qq aq

t

compensation for unknown

sin( ) cos( ))

aq

t k t kqq


2

Design a robust controller for the following system so that tracks cos( ).


Simulate system for 1 and a 3 plot the state,

dq q tq aq u

a a a q

a

2

212

2 2

22 2

control, controller.cos( )

sin( ) sin( )

( sin( ) )

1design sin( ) where a q +1

1a q +1

R

d

Ve q q t qe t q t aq uV eV ee e t aq u

u t e ke

V ke e aq e


22 2

22 2 2 2 2 2 2

2 2

2 2

( )

1a q +1

1 1( a q +1 a q +1 ) a q +1 (1 a q +1 )

if a q +1 , then

if a q +1 , then

2

follow deriv

positive

cont

V ke e aq e

V ke e e ke e e

e V ke

e V ke

V kV

ation in notes to show the system is

Globally Uniformly Ultimately Bounded (GUUB),


0 1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5

time

q, q

d

q, qd(:) with k=1, eps=2

0 1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2

3

4

5

6

time

u

u

R2Robust V , 2, 1k

dq

q

u

Note that the analysis only guaranteed Ultimate Bounded tracking error.


0 1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5

time

q,

qd

q, qd(:) with k=1, eps=.1

0 1 2 3 4 5 6 7 8 9 10-20

0

20

40

60

80

100

time

u

u

R2Robust V , .1, 1k

dq

q

u


0 1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

time

u

u


exactly cancel

2

0

2

1. Exact model control:0 sin( ) cos( )


( cos( ) ) sin( ) cos( )

3.a Robust - Sliding Modecos( )

a q +1cos(

aq

t

aq

q t k t kq


t qq aq

t


22


sin( ) cos( ))

3.b Robust - High Gain1

q +1 cos( ) sin( ) cos( )

aq

aq

t k t kqq

q aq t q t k t kq


2

Design a robust controller for the following system so that tracks cos( ).


Simulate system for 1 and a 3 plot the state,

dq q tq aq u

a a a q

a

3

212

22

2 22 2 2

2 2 2

2 2

control, controller.cos( ) ; sin( ) sin( )

( sin( ) )

design sin( ) where a q +1

a q +1 a q +1a q +1

a q +1 a q +1

R

d

Ve q q t q e t q t aq uV eV ee e t aq u

eu t ke

e

e eV ke e aq ke e

e

2 222 2 2 2 2

2 2

2 2

1

2

a q +1 a q +1 a q +1 a q +1

a q +1 a q +1

follow derivation from notes.

e

e e e eV ke ke

e e

V ke

0 2 4 6 8 10-2

0

2

0 2 4 6 8 10-2

0

2

R3Robust V , 0.5

dq

q

u

R3Robust V , 0.5


0 2 4 6 8 10-2

-1

0

1

2

0 2 4 6 8 10-2

-1

0

1

2

R3Robust V , 0.05

dq

q

R3Robust V , 0.05

u



exactly cancel

2

0

2

1. Exact model control:0 sin( ) cos( )


( cos( ) ) sin( ) cos( )

3.a Robust - Sliding Modecos( )

a q +1cos(

aq

t

aq

q t k t kq


t qq aq

t


22 2


2 2

sin( ) cos( ))

3.b Robust - High Gain1

a q +1 cos( ) sin( ) cos( )

3.c Robust - High Frequency

a q +

aq

aq

t k t kqq

q aq t q t k t kq

q aq

2

2


1 cos( )sin( ) cos( )

a q +1 cos( )

aq

t qt k t kq

t q

4. Design a learning control for the following system so that tracks cos( ).

where is an unknown constant. Simulate system for 1. Plot the state, tracking error, control.

dq q tq aq u

a a


One of the advantagesof the repetitive learning scheme is that the requirement thatthe robot return to the exact same initial condition after each learningtrial is replaced by the less restrictive requirement that the desired trajectoryof the robot be periodic.

Homework A.1-2 (sol)a=1, k=kd=5

R3Robust V , 0.5

dqq

Homework A.2

2. Design an adaptive tracking contoller for the following system:

where is an unknown constant.

q q au

a

21 1 2

2

3. Use backsteppping to design an adaptive contoller for the following system:


q aq q

q u

a

2 3 2 2

1. In preparation for designing an adaptive contoller,

write a linear parameterization for the following system:

1sin( ) cos( )

where , , , , are unknown constants.

q aq aq b q q d q e q abq uc

a b c d e


2 3 2 2

1. In preparation for designing an adaptive contoller,

write a linear parameterization for the following system:

1sin( ) cos( )

where , , , , are unknown constants.

q aq aq b q q d q l q abq uc

a b c d l

2 3 2

2

Linear parameterization for the system:

1

[( ) sin( ) cos( ) ] ( )

a

b

q q q q q q q q u W q ucd

l

ab

21Note, will adapt for " " not for " ", for " " not for , for " " in addition to " " and " "

individually

c l l ab a bc

Homework A.2-2 (sol)2. Design an adaptive tracking contoller for the following system:


Looks harmless but note that anything we put in will get multiplied by " ".

1Can't include i

q q au

a

u a

a

n since is unknown.u a

1 1

2 12

1 1 1 12 2 2 2

Rewrite as:1 1

1 1 1 1where [ ] and

1 ˆ where 21 1 1ˆ ˆ ˆ

substitue -dynamics:

d d d

T

T T T T

q q u W ua a

e q q q W u W u W q qa a a a

V ea

V ee ee eea a a

e

1

1

2 2 21 1 1

12

ˆ( )ˆdesign

ˆ ˆ ˆ

ˆdesign [ ]


T T

Td

V e W u

u W ke

V ke eW ke W e ke W e

W e q q eV ke

V

Homework A.2-3 (sol)2

1 1 2

2

3. Use backsteppping to design an adaptive tracking controller for the following system:

where is unknown constants.

q aq q

q u

a

1 1 1

21 1 1 1 1 2

21 1 1 2 2 2 2 2

2

21 11 12 2

Tracking in upper subsytem:

Introduce the embedded control:

where

Design adaptive "control input" :

ˆ where and

d

d d

d d d

d

T

e q q

e q q q aq q

e q aq q q q

q

V e

21 1 1 1 1 1 2 2

22 1 1 1 1

2 2 2 21 1 1 1 1 1 2 1 1 1 1 1 2 1 1

21 1

21 1 1 1 2

ˆ ˆlet and

ˆ ˆ( )

ˆDesign

ˆ ˆ ˆ

ˆDesign

d d

d d

a a

V e e e q aq q aa

q q aq k e

V e e k e e e aq aa k e e e q a a

a e q

V k e e


22 2 2 1 1 1 1 1 1

2 2 2 22 2 2 1 1 1 1 1 1 2 1 1 1 2

22 1 2

2 2 2 2 22 1 2 2 1 1 1 2 2 1 1 1 1 1 1 2 1 1 1 2

1 1

ˆ ˆ2

ˆ2

ˆ2

Design

d d

d d d

d d

d

q q u q aq aq q k e

q q u q e q q aq aq q k q aq q

V V

V V k e e u q e q q aq aq q k q aq q

u q e q

41 1 2 1 1 2 1 2 2

2 2 3 22 1 1 2 2 1 1 1

ˆ2

ˆ2

d aux

aux

aq q k q q e u

V k e aq a ak q u

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

2 2 3 22 1 1 2 2 1 1 1

ˆWhat if we just use our that we already designed?

ˆ ˆ ˆ2 2

ˆ2

This is a problem because we can't deal with

a

V k e aq k q a aq k q a

V k e aq k q a

a


23 2 2 2 2

3 2 2 2

2 2 3 23 1 1 2 2 1 1 1 2 2

3 21 1 1 2

2 2 3 23 1 1 2 1 1 1 2 2

What if we repeat our previous adaptation approach?

1ˆ where

2

ˆ

ˆ ˆ2

ˆ ˆDesign 2

ˆ ˆ2

aux

aux

V V a a a a

V V a a

V k e aq a ak q u a a

u aq k q a

V k e aq k q a

2

3 22 1 1 1 2

2 22 1 1 2

ˆ ˆDesign 2

a

a aq k q

V k e

Homework A.3

Homework A.3

2.

Homework A.33.

4.

Homework A.3-2 (sol)2.

Homework A.4

2

1. Design an observer to estimate in the open-loop system:

2cos( )


x

x x u

x x

2

2. Design an observer to estimate and a tracking controller for x in the system:

2cos( )


x

x x u

x x

2

3. Design a filter and a tracking controller for x in the system:

2cos( )


x x u

x x

0

2

1. Design an observer to estimate in the open-loop system:

2cos( )


Define:

ˆ


u

x

x x u

x x

x x x

s x x r s x

2 21 12 2

2

ˆ

propose

ˆ


ˆ

ˆ

substitute the open-loop system ( with u=0

x x x x

V x s

V xx ss xx s x x x

x s x

V x s x s x x x

x sx s x x x

x

2 2

2 2

2

2 2

):

ˆ2cos( )


ˆ 2cos( )stabilize cancel cross termcancel

V x sx s x x x

x s V x

x x x s x

V x s


2

1. (cont) Design an observer to estimate in the open-loop system:

2cos( )


x

x x u

x x


2 21 12 2

2

2 2

Designed:

ˆ 2cos( )

is PD, is ND

ˆ, 0

ˆ0

observer is bounded if , are bounded

stabilize cancel cross termcancel

V x s

x x x s x

V x s

V V

x s x x

x s x x x

x x

2ˆ 2cos( )


ˆ ˆ(terms to get differentiated to make )

ˆterms that don't get differentiated to make

Rewrite the observer by replacing s=

x x x s x

x p x

p x

x

2

2

ˆ

2

2

and regrouping

ˆ 2cos( )

2cos( ) 1 1


ˆ 1

2cos( ) 1

Prove that it works:

ˆ 1 2cos( ) 1

put in p put in x

x

x x x x x x

x x x

x p x

p x x

x p x x

1x x

But that estimate has velocity measurement in it?

This is a simple example because there is no term in the system dynamicsx


2. Design an observer to estimate and a tracking controller for x in the system:

2cos( )


x

x x u

x x

2 21 12 2

Define:

ˆ

ˆ (similar to the filtered tracking error ) then

Follow the same approach as previous problem with u 0

ˆ

d

O

O

x x x

s x x r s x x x x x

e x x

V x s

V xx ss xx s x x

2 2

2

2 2

ˆ2cos( )


ˆ 1

2cos( ) 1

O

O

x

V x sx s x u x x

x p x

p x x u

V x s


2. (cont) Design an observer to estimate and a tracking controller for x in the system:

2cos( ) ( is measureable but is not).

x

x x u x x

2 21 12 2

2 2

1

Define:

ˆ

ˆ (similar to the filtered tracking error ) then

Follow the same approach as previous problem with u 0

Control design:

ˆ

O

O

d

x x x

s x x r s x x x x x

V x s

V x s

e x x

1

1

1

ˆ (note that this is a measureable signal)

Substitute from the filter equation (implementable form of the observer) and inject

1

Define

1

d

d

d d d

d d

d

e x x

p

e x p x p p

p p p p p p

e x x p

2 2 2 2 2 21 1 1 1 1 11 12 2 2 2 2 2

2 21 1 1 1

2 21

Propose:

1

d

O

O

d d

p

V V e p x s e p

V V e e pp x s e e pp

x s e x x p p pp

2

Reminder of the implementable observer from previous problem:

ˆ 1

2cos( ) 1

x p x

p x x u


2 21

1

2 2 2 2 2 21 1 1 1

1

1

Design 1

Differentiate : 1 and sub. and sub. from observe

d d

d d

stabilizecancel

d

d d d

V x s e x x p p pp

p x x e

V x s e e p pp x s e e p p p p

p p x x e

2 2 2 21 1 1

2 2 2 21 1 1

r:

1 2cos( ) 1

2cos( ) 1 1

Replace with since we have in Lyapunov function but not (b

d

d

V x s e e p p x x e x x u

x s e e p p x e x x x u

x x s x s x

2 2 2 21 1 1

2 2 2 21 1 1

21

oth unmeasureable):

2cos( ) 1 1

2cos( ) 1 1 1

Design part of :

2cos( ) 1 1

d

d

d

cancel

V x s e e p p x e x x s x u

x s e e p p x e x x x s u

u

u x e x x x

1 1

2 2 2 21 11

stabilizecancel crossterm

e p u

V x s e p p s pu


2 2 2 21 1

2 2 2 21 1

2

1 1

222 2 2 21 1

22

11

From previous slide: 1

1

Design 1

1 1

worst case: 1 11

Drop the negative te

n

n

nn

V x s e p p s pu

V x s e p p s pu

u k p

V x s e p p s k p

sp s k p p

k

2 2 2 21

2 2 2 21

1

2

2 2 2 21

1

2 2 2 21

1

1

rm to find a new upper bound

1

11

11

choose 1 tracking

n

n

n

n

V x s e p p s

sV x s e p s

k

sV x s e p

k

V x s e pk

k GES



2

2 2 21 1 12 2 2

2 2 21 1 12 2 2

Define:

2cos( )

Propose:

d d d

f f

f f

f d f

d f

f

f

f f

f f f n

e x x e x x e x x

e e e e e e

e e k e

e e e x x e e

x x u e e

V e e

V e e

V ee e e

e e e e e k

2 2

2 2 2

2 2 2

2 2 2 2

2cos( )

2cos( )

1 2cos( ) 2

f f

f f d f

f f d

f f

f d f

e e e e ke

e e e ke x x u e e

e e e ke x x u

e e e k e

e e k x x u k e e

2

3. Design a filter and a tracking controller for x in the system:

2cos( ) ( is measureable but is not).x x u x x


2 2 2 2

2

2 2 2

Assume for now is measureable:

1 2cos( ) 2

Design 2cos( ) 2

1

Choose 1 GES tracking

f

f d f

cancel cancel

d f

f

e

V e e k x x u k e e

u x x k e e

V e e k

k


Is measureable? (have to ask since we defined )

1 1


(terms to get differentiated to make )

terms that don't

f f f

f f f f

f f

e e e k e

e e k e e e e k e k e ke

e p e

p

get differentiated to make

1 1

f

f

f

e

e p ke

p k e k e

contents 1.motivation for nonlinear controlmotivation for nonlinear control 2.the tracking...

Documents

control design slide

control objective

single control

motion nonlinear control

adaptive control design

tracking problem slide

nonlinear feedback

filter slide