Mars Adaptive Control

7/30/2019 Mars Adaptive Control

1/19

:-

s stems MRAS MRAC

Lectures, Spring 2012

-Systems

Model

my

myye Bring

Adjustment

mechanism

Controller parameters to zero!

Controller Plant

cuyu

Historical MRAS= Model Reference Adaptive SystemFormulated in continuous time, deterministic servo problem

- ...

The desi n of the ada tation mechanism differs

considerably from that of the STRs.

- the MIT rule

- use of L a unov stabilit theor

- use of input/output analysis of systems, small-gain

theorem, passivity theory, positive real functions,

Kalman-Yakubovich lemma etc.

- ...

Let ebe the error between the system and model outputs.

1

Set the cost criterion

where is the parameter vector of the

2e controller, which must be adjusted.

To make Jsmall, it is (heuristically) reasonable to changethe parameters in the direction of negative gradient ofJ.


2/19

- ...

ee

dt

eThis is the MIT rule. The term is crucial. It is called

the sensitivity derivative. It is assumed that the parameters

c ange more s ow y an e o er var a es n e sys em.

Then the sensitivity derivative can be calculated assuming

as constant.

- ...

There are other alternatives forJ e. .

eJ )(

which ives

0,1 e

ee

dtsign

0,0sign eeDef.

,

- ...

Exam le: Ada tation of a feedforward ain

)(skG

Process: where kis an unknown parameter

Underlying design problem: find a feedforward coontroller

which gives the system

)()( 0 sGksGm 0kwhere is a given parameter.

cuu With the controller the system output becomes )(skG

which gives the correct)(sGm if

/0

- ...

uGkukGe

Then calculate the sensitivity derivative

mc yk

kupkG

e)(

and apply the MIT rule

eyeykd

mm

' 0, kk are constantst

0


3/19

- ...

Model

)(0

sGk

m

s

-

e

Process

)(skG

ycu u

can be regarded as a tuning parameter. Increasingits value leads eventually to oscillations, even instability.

1( )

1

G s

s

ucsinusoidal with angular frequency 1 rad/s

01, 2k k

- ...

is a time-variable

cupGupG )())((

parameter, so that e.g.

Example: MRAS for a first-order system

buay

dt

dy

b

tytutu

m

c

0

11

21

cmmm

m ubyady

aam 022

- ...

It follows that

cubap

y2

1

myye Form the sensitivity derivatives of

u

bap

bec

21

y

bap

bu

bap

bec

2

2

2

1

2

2


4/19

- ...

But the arameters aand bare not known. Assume thatthey change slowly enough, so that the parameters can

be assumed constant in the derivation of the adaptation law.

mapbap 2

The above approximation is reasonable, if the parameter

.

the adaptation laws

- ...

eua

a

dt

dc

m

1w ere certa n constant va ues

have been included in .

ad m2

be known.

apdt m

The layout of the controlled system is on the following page.

Note that the uc and y signals must be filtered first,

.

- e)(sGm

my

++

-

yucu

)(sG

1

2

s

s

ma ma

m m

Simulation results: 1, 0.5, 2, 1m ma b a b

Correct values0 0

1 24, 2


5/19

2 1/a b

line

Simulation over 500 time units. The parameer values

The parameters do not necessarily have to adapt to their

correct values. The errorecan still become zero. Consider

again the example of the adaptation of feedforward gain. s

The s stem e uations: ukuuku ,,

The error:cc ukukke )()(

0

0 where kk /00

MIT rule gives: 022

cuk

d

Ik2

00 w c as e so u on e

wheret

2

ct

0

The estimate converges to the correct value only ifItdiverges as tapproaches infinity. This means that the

c .

tIketkute2

00

c

which will always approach zero, because either the integral

diverges or the input signal approaches zero.

rate and stability. The amplitudes of system signals, e.g. the

reference values can have a considerable effect of the possible

values of the gain.

ed

edt ,

as a gradient method for minimizing the error. To overcome

the difficulties related to signal amplitudes in the determining

, ere ex mo ca ons o e ru e, e.g.

ed

Tdt

.


6/19

ons er e au onomous eren a sys em

dx, x

dt

which is assumed to have a unique solution through a

given initial point. The function fcan be nonlinear, buts no me-vary ng. or me-var a e sys ems we

would write f= f(x,t).)

Stability definition: The solution x(t) = 0 is called stable,s a e n e sense o yapunov

if for any given > 0 there exists a number > 0 such that

)()0( txx

The solution is unstable, if it is not stable. The solution is,

stable and

tx as

If the system is asymptotically stable and convergent

0)( tx( ) for all initial values, it is then globally

asymp o ca y s a e orasymp o ca y s a e n e arge.The definition of stability is defined for the null solution,

not the system itsel . Equivalently, it can be de ined in

terms of an equilibrium point xe, where f(xe) = 0.

non near system can ave many equ r um po nts. e

equations can always be scaled such that a particular equi-

.

asymptotically stable, it can only have one equilibrium point.

Forlinears stems the above definitions can be sim lifiedconsiderably:

e equ r um s s a e, ere ex s a cons an suc a

0,)( 0 txtx where )0(xx

It is asymptotically stable, if it is stable and

0)( tx as tapproaches infinity.

Note that for linear systems stability is always global.


7/19

Another in ut stabilit conce t is the in ut-out ut stabilit(BIBO stability, bounded input-bounded output).

,

gives a bounded output signal.

n asymp o ca y s a e sys em s a ways s a e.

A BIBO system is asymptotically stable, if the system is

both reachable and observable. For example, i there are

unobservable modes, these might be unstable, but this does

.

The second method of L a unov direct method:

System: 0)0()),(()( ftxftx

If a (scalar-valued) Lyapunov function V(x) is found such that

. , xx

2. 0)0( V

3. Vis continuously differentiable with respect to all xi

VV. 0))(()()(

txx

txx

tV

then the s stem is stable at the ori in or the null solution

is stable).

.4: The time derivative ofV(calculated along the systemtra ector must be ne ative semidefinite.

If (in 4) 0)( tV (negative definite)

then the system is asymptotically stable at the origin.

If, additionall , )(xV when x then the system

is globally asymptotically stable.

The solution is unstable if there exists a ositive definiteV(x), such that its time derivative is positive definite.

the energy of a system, which must decrease in order the

solution to conver e to a stable e uilibrium oint.

To find a Lyapunov function is generally very difficult. If a

,

regarding stability. The equilibrium point can then be stable

or unstable.


8/19

211 4 xxx

212 22 xxx

The origin xe= 0 is the equilibrium point. Consider theossible L a unov function 22 21

1. V(x) > 0, for all xnot equal to zero. Ok!. .

3. Ok!

4. next a e

22

22)()(21

21

21

xx

xxtx

x

xV

4428

222

2

22121

2

1 xxxxxx

2121 xxxx

Ok!Moreover, xV as x

The system is globally asymptotically stable. That could

,which have negative real parts.

Exam le: )()( 12

2

2

1121 xfxxxxx

)()( 22

2

2

1212 xfxxxxx

The system is nonlinear. Investigate the stability of theorigin, which is clearly an equilibrium point. Try

2

2

2

1)( xxxV as a candidate for a Lyapunov function.

ear y , an are sa s e . or , ca cu a e

)(2221

xfV

)(

212211

2

21

xfx

.

The system is also globally asymptotically stable, which

implies that the origin is the only equilibrium point.

Note that if the time derivative weree.g.

0)()( 2212

2 xxxxV (negative semidefinite)

wou on y mp y s a y e er va ve s zero a ways

when

21 xx


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T ical e uilibrium curves and s stem tra ector in the

phase plane2x

321VVV

s stem tra ector

1x

1V

2V

3V

Exam le: Consider a linear s stem

0)0(),()( xxtAxtx

where A is a constant matrix. Investigate the quadratic form

PxxxV )( , where Pis a symmetric positive definite

matrix , as a candidate for a Lyapunov function.

Recall that a square matrix Pis positive definite, if the

xx s pos ve or a nonzero vec ors x.

T

AxPPxxV

tV TT

AxPxPAxxx

TTT

AxPxxPAx TTTTT

xPAPAx

xxxx

TT if QPAPAT

0 QxxT and Qis positive definite.

In the calculation the fact that Pis symmetric, was used.Also, note that a scalar can be transposed without changing

.

Note that Qis symmetric.


10/19

- .. .

Lyapunov theory for nonlinear time-varying systems:

or e sys ems , txtx

the equilibrium pointsare defined by

*

0,,x

-

txtAtx has a unique equilibrium point at theorigin,0, provided that is non-singular.

Definition: The equilibrium point 0 is stableat t0, if for anyR> 0 there exists a ositive scalarr R t such that

, ttRtxrtx

.

can be chosen independent oft0, the equilibrium point isuniformly stable.

Definition: The equilibrium point is uniformly

uniformly to the equilibrium point.

,

but the converse is not generally true.

Example: The system xx

has the solution )(1

)( 00 tx

ttx

which converges asymptotically to the origin. The

, 0

longer time to get close to the origin.

- ,

definite ifV(0,t) = 0 and there exists a time-invariant positivedefinite function V0(x) such that

)(),( 0 xVtxV

Naturally, V(x,t) is negative definite, if - V(x,t) is positivedefinite; V(x,t) is positive semi-definite, if it dominates apositive semi-definite function etc.


11/19

e n t on: sca ar unc on x, s ecrescen V(0,t) = 0, and if there exists a time-invariant positive

1

, 1

xamp e: e unc on21)(sin1),( xxttxV

is ositive definite, because it dominates the function

22

)( xxxV It is also decrescent, because it is dominatedby 22211 2)( xxxV

ven a me-vary ng sca ar unc on x, , s er va vealong a system trajectory is

),( txfx

V

t

V

dt

dV

Stability: If in a region around the equilibrium point thereexists a scalar function V(x,t) with continuous partialderivatives such that 1. Vis positive definite, 2. dV/dtis

v - , u u .

, , . ,

stable. If, (in 2), dV/dtis negative definite, then thee uilibrium is uniforml as m toticall stable.

If,additionally, the stability region is the whole state-space

and Vis radially unbounded,then the equilibrium isglobally uniformly asymptotically stable.

Example: Consider the system

)()()( 22

11 txetxtxt

212

and examine the stability of the equilibrium point (0,0).

Choose the Lyapunov function candidate

22221 1),( xextxV t

This function is positive definite, because it dominates thetime-invariant positive definite function

2

2

2

1 xx It is also decrescent, because it is dominated bythe time-invariant positive definite function

2

2

2

1 2xx


12/19

Furthermore texxxxtxV22

221

2

1 212),(

and212122112),( xxxxxxxxtxV

w c s nega ve e n e. ere ore e or g n s

globally asymptotically stable.

Usually it is easier to prove a derivative to be negative

semi-definite not ne ative definite which would be neededfor asymptotic stability.

A special theorem called Barbalats lemma, helps. First,note the following:

1. 0f does not imply that fconverges as tapproaches.

))(log(sin)( ttf tt

tf as,0

))(log(cos

)(

t

2. converges does not imply that 0f

2tttends to zero, but its derivative is

s n ee unbounded.

3. Iffis lower bounded and decreasin 0

converges to a limit. (But nothing can be said about

.

Question: Given that a function tends to a finite limit, what

actually converges to zero? Barbalats lemma indicates that

the derivative itself should have some smoothness.

finite limit, as tapproaches infinity, and if df/dtisuniformly continuous, then

ttf as0)(

The term uniforml is crucial here. A sufficient condition

for a function to be uniformly continuous is that itsderivative is bounded. This can be seen from the finite

difference theorem: Let gbe a differentiable function. Forall tand t1 there exists a t2 (between the two other points)

))(()()( 121 tttgtgtg

If dg/dtis bounded, the uniformity condition can beestablished.


13/19

s mp e coro ary o e emma: e eren a e

function f(t) has a finite limit as tapproaches infinity, and,

ttf as0)(

A Lyapunov-Like lemma: If 1. V(x,t) is lower bounded,. x, s nega ve sem - e n e, . x, s

uniformly continuous is time, then ttxV as0),(

The theoretical results can now be used to establish adaptive

w u y y.

Design of MRAS using Lyapunov theory:

x :

Process and the desired res onse are

buady

0m aubady

dt

dt

c 21

Introduce the error myye

orm t e er vat ve o t e error

decmmm uyaaeadt

12

The error goes to zero, if the parameters are in the correct

values. Tr to find an ad ustment mechanism that will

drive the parameters to these values.

Assume that b > 0 and try the candidate for a Lyapunovfunction

212

2

2

21

111),,( mm bbaabeeV

The function is zero when eis zero and the controllerparameters are in their correct values. Calculate the


14/19

dt

bbdt

aabdt

ee

dtmm

11

22

ye

dt

daabea mm

22

2 1

eudt

dbb cm

11

1

Update the parameters as

221 , eadV

yed

eud

mc

u e er va ve o s nega ve sem - e n e, no

negative definite. Thus )0()( VtV and thus the variables

21,, e are bounded. Then y= e+ ymis also bounded.

deVd2

cmmmmm uyaaeaeadt

eadt

122

which is bounded, because yeuc and, are bounded.

.lemma, the error will go to zero.

But note that the parameters do not necessarily converge

to their correct values; we know onl that the are

bounded. But in spite of that the adaptation works.

The fact that an adaptive control law can work perfectly,although the parameters do not converge to their true

va ues, s a p enomenon w c we ave seen e ore.

ot ce t e s m ar ty etween t e resu ts y us ng t e

rule and the Lyapunov-based method. Both have a similar

ed

T ac yu c

m

m yuap

(Lyapunov) (MIT)


15/19

ea za on:

+

-

+

e

yucu

)(sGm

m

-

)(sG

1

s

s

2

Lyapunov

0.5 / ( 1)G s s

2 / ( 2)mG s s

a=1, b= 0.5, am= bm= 2

1

LTI case (discussed earlier also): Consider the homogenous

linear time-invariant system

0)0(, xxAxdt

dx

which is known to be asymptotically stable. Consider the

possibility for using PxxxV T)( as a Lyapunov function,

where Pis a symmetric positive definite matrix.

a cu a e e me er va ve a ong e so u on a o er

conditions are trivially fulfilled).

VdV

xxxxxxx

xdtTTTTT

xPAPAxPAxxPxAx TTTTT

0 QxxxQx TT

in which Qis a positive definite matrix. Hence, if thereexist positive definite matrices Qand P, which satisty theyapunov equa on

T


16/19

then Vis a L a unov function and the ori inal s stem isasymptotically stable (even globally).

Note that Qis necessarily symmetric, because Pwas

assumed to be symmetric.

In fact,we can go a bit further. The following can be

roved: IfA is as m toticall stable, then foreachsymmetric positive definite matrix Qthere exists aunique symmetric positive definite matrix Pthat satisfies

e yapunov equa on.

It follows that if we choose any symmetric pos. def. matrixQand from Lyapunov equation we get a solution P, which

. ., .

We could try e.g. Q= I.

State-space models: Adaptation of the feedforward gain

Consider again the same example discussed in the beginning

of MRAC theory. The plant has the transfer function kG(s),where G(s) is known but kunknown. The desired response

0

.

0 cc

0 If a realization ofkG is given by the triplet0 (A,B,C) then

uBAxdt

xc

0

Cxe

is the realization,which generates the error function. Assume

that the hono enous s stem Axx is as m toticall stable.

Then there exist positive definite matrices Pand Qsuch that

QPAPAT


17/19

Choose the candidate for a Lyapunov function

21

2 PxxV no e a s a sca ar n s case

The time derivative ofValong the solution becomes

xPPxdt

TT 0)(2

uBAxPPx cTT 00)(

2

uBPxAxPxuPBxPAxx cTTTTcTT 000

2

PxBuxAPPAx TcTTT 002)(2

PxBudt

dQxx

T

c

T

0

2

dPxBu

dtc

so that the derivative ofVis negative definite. The statevector and the errorewill go to zero as tapproachesinfinity. However, the parameter error

0 will not

necessarily converge.

Drawback: in order to im lement the ada tation law, the

states have to be measurable.

But ifPcan be chosen such that

TT exx

d

dtc


18/19

,

if

s s s r c y pos ve rea ,

such that sG is PR.

Im s G(s)Im

e e

Example:1

1)(

ssG is SPR,

ssG

1)(

is PR, but notSPR

Im s G(s)Im

Re Re

1

1s

Nyquist curve in the RHP!

Im G(s) G(s)Im

Re Re

1

1.0

1

ss

1

By using any small the curve will move into the LHP.

The question on the realizability of the adaptation law

is given by the Kalman-Yakubovich lemma:

Lemma: Let the time-invariant system

BuAxdx

Cxy

.

Then the transfer function BAsICsG 1)()(

,


19/19

Pand Qsuch that

QPAPA

and CPBT

In the adaptive feedforward gain problem, ifGis SPR,then the parameter adjustment rule

eu

d

makes the output errorego to zero.t

k0G(s)

uc

ym

- e

Process

kG(s)y +

s

k0G(s)

ym

- e

Model

u

kG(s)y +

sProcess

c

Mars Adaptive Control

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