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EECE 574 - Adaptive ControlModel-Reference Adaptive Control - An Overview

Guy Dumont

Department of Electrical and Computer EngineeringUniversity of British Columbia

January 2013

Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 1 / 73

Model-Reference Adaptive Systems

The MRAC or MRAS is an important adaptive control methodology 1

1see Chapter 5 of the Åström and Wittenmark textbook, or H. Butler,”Model-Reference Adaptive Control-From Theory to Practice”, Prentice-Hall, 1992


Model-Reference Adaptive Systems

The MIT rule

Lyapunov stability theory

Design of MRAS based on Lyapunov stability theory

Hyperstability and passivity theory

The error model

Augmented error

A model-following MRAS


MIT Rule The Basics

The MIT Rule

Original approach to MRAC developed around 1960 at MIT foraerospace applications

With e = y− ym, adjust the parameters θ to minimize

J(θ) =12

e2

It is reasonable to adjust the parameters in the direction of the negativegradient of J:

dθ

dt=−γ

∂J∂θ

=−γe∂e∂θ

∂e/∂θ is called the sensitivity derivative of the system and is evaluatedunder the assumption that θ varies slowly


MIT Rule The Basics

The MIT Rule

The derivative of J is then described by

dJdt

= e∂e∂ t

=−γe2(

∂e∂θ

)2

Alternatively, one may consider J(e) = |e| in which case

dθ

dt=−γ

∂J∂θ

=−γ∂e∂θ

sign(e)

The sign-sign algorithm used in telecommunications where simpleimplementation and fast computations are required, is

dθ

dt=−γ

∂J∂θ

=−γsign(

∂e∂θ

)sign(e)


MIT Rule Examples

MIT Rule: Example 1

Process: y = kG(s) where G(s) is known but k is unknown

The desired response is ym = k0G(s)uc

Controller is u = θuc

Then e = y− ym = kG(p)θuc− k0G(p)uc

Sensitivity derivative

∂e∂θ

= kG(p)uc =kk0

ym

MIT ruledθ

dt= γ

′ kk0

yme =−γyme


MIT Rule Examples

MIT Rule: Example 1

Figure: MIT rule for adjustment of feedforward gain (from textbook).


MIT Rule Examples

MIT Rule: Example 1

Figure: MIT rule for adjustment of feedforward gain: Simulation results (fromtextbook).


MIT Rule Examples

MIT Rule: Example 2

Consider the first-order system

dydt

=−ay+bu

The desired closed-loop system is

dym

dt=−amym +bmuc

Applying model-following design (see lecture notes on pole placement)

degA0 ≥ 0 degS = degR = degT = 0


MIT Rule Examples

MIT Rule: Example 2

The controller is thenu(t) = t0uc(t)− s0y(t)

For perfect model-following

dydt

= −ay(t)+b[t0uc(t)− s0y(t)]

=−(a+bs0)y(t)+bt0uc

=−amym(t)+bmuc

This implies

s0 =am−a

b

t0 =bm

b


MIT Rule Examples

MIT Rule: Example 2

With the controller we can write2

y =bt0

p+a+bs0uc

With e = y− ym, the sensitivity derivatives are

∂e∂ t0

=b

p+a+bs0uc

∂e∂ s0

=b2t0

(p+a+bs0)2 uc =b

p+a+bs0y

2p is the differential operator d(.)/dtGuy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 11 / 73

MIT Rule Examples

MIT Rule: Example 2

However, a and b are unknown3. For the nominal parameters, we know that

a+bs0 = am

p+a+bs0 ≈ p+am

Then

dt0dt

= −γ′b(

1p+am

uc

)e =−γ

(am

p+amuc

)e

ds0

dt= γ

′b(

1p+am

y)

e = γ

(am

p+amy)

e

with γ = γ ′b/am, i.e. b is absorbed in γ and the filter is normalized with staticgain of one.

3after all, we want to design an adaptive controller!Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 12 / 73

MIT Rule Examples

MIT Rule: Example 2

Figure: MIT rule for first-order (from textbook)


MIT Rule Examples

MIT Rule: Example 2

Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2 andγ = 1(from textbook)


MIT Rule Examples

MIT Rule: Example 2

Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2.Controller parameters for γ = 0.2, 1 and 5(from textbook)


MIT Rule Examples

MIT Rule: Example 2

Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2.Relation between controller parameters θ1 and θ2. The dashed line is θ2 = θ1−a/b.(from textbook)


MIT Rule Stability of the Adaptive Loop

Stability of the Adaptive Loop

Consider again the adaptation of a feedforward gain

G =1

s2 +a1s+a2

e = (θ −θ0)Guc

∂e∂θ

= Guc =ym

θ0

dθ

dt= −γ

′e∂e∂θ

=−γ′e

ym

θ0=−γeym




Thus the system can be described as

d2ym

dt2 +a1dym

dt+a2ym = θ0uc

d2ydt2 +a1

dydt

+a2y = θuc

dθ

dt= −γ(y− ym)ym

This is difficult to solve analytically




d3ydt3 +a1

d2ydt2 +a2

dydt

+ γucymy = θduc

dt+ γucy2

m

Assume uoc and yo

m constant, equilibrium is

y(t) = yom =

θ0uoc

a2

which is stable ifa1a2 > γuo

cyom =

γ

a2(uo

c)2

Thus, if γ or uc is sufficiently large, the system will be unstable




Figure: Influence on convergence and stability of signal amplitude for MIT rule(from textbook)


MIT Rule Modified MIT Rule

Modified MIT Rule

Normalization can be used to protect against dependence on the signalamplitudes

With ϕ = ∂e/∂θ , the MIT rule can be written as

dθ

dt=−γϕe

The normalized MIT rule is then

dθ

dt=−γϕe

α2 +ϕTϕ

For the previous example,

dθ

dt=−γeym/θ0

α2 + y2m/θ 2

0


MIT Rule Modified MIT Rule

Modified MIT Rule

Figure: Influence on convergence and stability of signal amplitude for modified MITrule (from textbook)


Lyapunov Design of MRAC Lyapunov Theory

Lyapunov Stability Theory

Aleksandr M Lyapunov(1857-1918, Russia)

Classmate of Markov, taught byChebyshev

Doctoral thesis The generalproblem of the stability of motionbecame a fundamentalcontribution to the study ofdynamic systems stability

He shot himself three days afterhis wife died of tuberculosis




Consider the nonlinear time-varying system

x = f (x, t) with f (0, t) = 0

If at time t = t0 ‖x(0)‖= δ , i.e. if the initial state is not the equlibriumstate x∗ = 0, what will happen?

There are four possibilities




The system is stable. For a sufficiently small δ , x stays within ε of x∗. Ifδ can be chosen independently of t0, then the system is said to beuniformly stable.

The system is unstable. It is possible to find ε which does not allow anyδ .

The system is asymptotically stable. For δ < R and an arbitrary ε , thereexists t∗ such that for all t > t∗, ‖x−x∗‖< ε . It implies that ‖x−x∗‖→ 0as t→ ∞.

If asymptotic stability is guaranteed for any δ , the system is globallyasymptotically stable.




A scalar time-varying function is locally positive definite if V(o, t) = 0and there exists a time-invariant positive definite function V0(x) such that∀t ≥ t0, V(x, t)≥ V0(x). In other words, it dominates a time-invariantpositive definite function.

V(x, t) is radially unbounded if 0 < α‖x‖ ≤ V(x, t), α > 0.

A scalar time-varying function is said to be decrescent if V(o, t) = 0 andthere exists a time-invariant positive definite function V1(x) such that∀t ≥ t0, V(x, t)≤ V1(x).

In other words, it is dominated by a time-invariant positive definitefunction.




Theorem (Lyapunov Theorem)

Stability: if in a ball BR around the equilibrium point 0, there exists a scalar functionV(x, t) with continuous partial derivatives such that

1 V is positive definite2 V is negative semi-definite

then the equilibrium point is stable.

Uniform stability and uniform asymptotic stability: If furthermore,

3 V is decrescent

then the origin is uniformly stable. If condition 2 is strengthened by requiring that V benegative definite, then the equilibrium is uniformly asymptotically stable.

Global uniform asymptotic stability: If BR is replaced by the whole state space, andcondition 1, the strengthened condition 2, condition 3 and the condition

4 V(x,t) is radially unbounded

are all satisfied, then the equilibrium point at 0 is globally uniformly asymptoticallystable.




The Lyapunov function has similarities with the energy content of the system and mustbe decreasing with time.

This result can be used in the following way to design a stable a stable adaptivecontroller

1 First, the error equation, i.e. a differential equation describing either the outputerror y− ym or the state error x− xm is derived.

2 Second, a Lyapunov function, function of both the signal error e = x− xm and theparameter error φ = θ −θm is chosen. A typical choice is

V = eT Pe+φT

Γ−1

φ

where both P and Γ−1 are positive definite matrices.3 The time derivative of V is calculated. Typically, it will have the form

V =−eT Qe+ some terms including φ

Putting the extra terms to zero will ensure negative definiteness for V if Q ispositive definite, and will provide the adaptive law.


Lyapunov Design of MRAC Lyapunov Design of MRAC

Stability of Adaptive Loop



Lyapunov Design of MRAC

Determine controller structure

Derive the error equation

Find a Lyapunov equation

Determine adaptation law that satisfies Lyapunov theorem



Adaptation of Feedforward Gain

Process modeldydt

=−ay+ ku

Desired responsedym

dt=−aym + k0uc

Controlleru = θuc

Introduce error e = y− ym

The error equation is

dedt

=−ae+(kθ − k0)uc




System model

dydt

= −ay+ ku

dθ

dt= ??

Desired equilibrium

e = 0

θ = θ0 =k0

k




Consider the Lyapunov function

V(e,θ) =γ

2e2 +

k2(θ −θ0)

2

dVdt

= −γae2 +(kθ − k0)(dθ

dt+ γuce)

Choosing the adjustment rule

dθ

dt=−γuce

givesdVdt

=−γae2




Lyapunov rule: dθ

dt =−γuce MIT rule: dθ

dt =−γyme




Lyapunov rule: MIT rule:



First-order System

Process modeldydt

=−ay+bu

Desired responsedym

dt=−amym +bmuc

Controlleru = θ1uc−θ2y

Introduce error e = y− ym

The error equation is

dedt

=−ame− (bθ2 +a−am)y+(bθ1−bm)uc



First-order System

Candidate Lyapunov function

V(t,θ1,θ2) =12

(e2 +

1bγ

(bθ2 +a−am)2 +

1bγ

(bθ1−bm)2)

Derivative

dVdt

=dedt

+1γ(bθ2 +a−am)

dθ2

dt+

1γ(bθ1−bm)

dθ1

dt

= −ame2 +1γ(bθ2 +a−am)(

dθ2

dt− γye)+

1γ(bθ1−bm)(

dθ1

dt+ γuce)

This suggests the adaptation law

dθ1

dt= −γuce

dθ2

dt= γye



First-order System

Lyapunov Rule MIT Rule



First-order System



Lyapunov-based MRAC Design

The main advantage of Lyapunov design is that it guarantees aclosed-loop system.

For a linear, asymptotically stable governed by a matrix A, a positivesymmetric matrix Q yields a positive symmetric matrix P by the equation

ATP+PA =−Q

This equation is known as Lyapunov’s equation.



Lyapunov-based MRAC Design

The main drawback of Lyapunov design is that there is no systematicway of finding a suitable Lyapunov function V leading to a specificadaptive law.

For example, if one wants to add a proportional term to the adaptive law,it is not trivial to find the corresponding Lyapunov function.

The hyperstability approach is more flexible than the Lyapunovapproach.


Lyapunov Design of MRAC State Feedback

The Lyapunov Equation

Let the linear systemx = AX

be stable

Let Q be an arbitrary positive definite matrix

Then the Lyapunov equation

ATP+PA =−Q

always has a unique solution where P is positive definite.

The functionV(x) = xTPx

is then a Lyapunov function



State Feedback

Process: x = Ax+Bu

Desired response: xm = Amxm +Bmuc

Control law: u = Muc−Lx

Closed-loop system

x = Ax+Bu = (A−BL)x+BMuc = Ac(θ)x+Bc(θ)uc

The parametrization is Ac(θ) = Am and Bc(θ) = Bm

For compatibility we need A−Am = BL and Bm = BM



Error Equation

The error e = x− xm satisfies

e = x− xm = Ax+Bu−Axm−Bmuc

Hence

e = Ame+(A−Am−BL)x+(BM−Bm)uc

= Ame+(Ac(θ)−Am)x+(Bc(θ)−Bm)uc

= Ame+Ψ(θ −θ0)



The Lyapunov Function

Try

V(e,θ) =12(γeTPe+(θ −θ0)

T(θ −θ0))

Differentiating V

V =γ

2(eTPe+ eTPe)+(θ −θ0)

Tθ

Then with e = Ame+Ψ(θ −θ0) we can write

V =γ

2[(

eTATm +Ψ(θ −θ0)

)Pe+ eT (Ame+Ψ(θ −θ0)

]+(θ −θ0)

Tθ



The Lyapunov Function

Now, consider the Lyapunov equation ATmP+PAm =−Q (possible

because reference model always stable)

V = −γ

2eTQe+ γ(θ −θ0)

TΨ

TPe+(θ −θ0)T

θ

= −γ

2eTQe+(θ −θ0)

T(θ + γΨTPe)

If the parameter adjustment law is chosen as

dθ

dt=−γΨ

TPe

We obtaindVdt

=−γ

2eTQe

which is negative semi-definiteNote that this assumes knowledge of the state.


Lyapunov Design of MRAC Adaptation of Feedforward Gain


Consider the error

e = (kG(p)θ − k0G(p))uc = kG(p)(θ −θ0)uc

With

x = Ax+B(θ −θ0)uc

e = Cx

A candidate Lyapunov function is

V =12(γxTPx+(θ −θ0)

2)

LetATP+PA =−Q




Differentiating V

V =γ

2(xTPx+ xTPx

)+(θ −θ0)θ

Which after using the model for x and the Lyapunov equation gives

V =−γ

2xTQx+(θ −θ0)

(θ + γucBTPx

)Using the adaptation law

dθ

dt=−γucBTPx

givesdVdt

=−γ

2xTQx



Output Feedback

The adaptation lawdθ

dt=−γucBTPx

assumes knowledge of the state x

If we can find P such thatBTP = C

then the adaptation law becomes

dθ

dt=−γuce

i.e. this is now output feedback and the state x is not required

When is it possible to do so?


Lyapunov Design of MRAC Kalman-Yakubovich Lemma

Kalman-Yakubovich Lemma

Definition: Positive real transfer functionA rational transfer function G with real coefficients is positive real (PR) if

Re G(s)≥ 0 for Re s≥ 0

A rational transfer function G with real coefficients is strictly positive real(SPR) if G(s− ε) is PR for some ε > 0

G(s) = 1/(s+1) is SPR

G(s) = 1/s is PR bur not SPR


Lyapunov Design of MRAC Kalman-Yakubovich Lemma

Kalman-Yakubovich Lemma

Theorem (Kalman-Yakubovich Lemma)Let the linear time-invariant system

x = Ax+Bu

y = Cx

be completely controllable and completely observable. The transfer function

G(s) = C(sI−A)−1B

is strictly positive real if and only if there exist positive definite matrices P andQ such that

ATP+PA = −Q

BTP = C


The Small Gain Theorem Definitions

Operator View of Dynamical Systems

L2 norm

‖u‖=(∫

∞

−∞

u2(t)dt) 1

2

L∞ norm

‖u‖= sup0≤t<∞

|u(t)|

Truncation operator:

xT(t) ={

x(t) 0≤ t ≤ T0 t > T

L2e ={

x | ‖xT‖2 = 〈xT |xT〉< ∞}



The Notion of Gain

Gain of a nonlinear system

Let the signal space be Xe. The gain γ(S) of a system S is defined as

γ(S) = supu∈Xe

‖Su‖‖u‖

where u is the input signal to the system.The gain is thus the smallest value γ such that

‖Su‖ ≤ γ(S)‖u‖ for all u ∈ Xe



The Notion of Gain

For a linear system G with signals in L2e,

γ(G) = maxω|G(iω)|

Static nonlinear system y(t) = f (u(t)), the gain of the system is

γ = maxu

|f (u)||u|



BIBO Stability

DefinitionA system is bounded-input, bounded-output (BIBO) stable if it has boundedgain

A BIBO stable system is a system for which the outputs will remainbounded for all time, for any finite initial condition and input.

A continuous-time linear time-invariant system is BIBO stable if andonly if all the poles of the system have real parts less than 0.



The Small Gain Theorem

Theorem (The Small Gain Theorem)Consider the system

Let γ1 and γ2 be the gains of the systems H1 and H2. The closed-loop system isthen BIBO stable if

γ1γ2 < 1 and its gain is less than γ =γ1

1− γ1γ2


Positivity and Passivity Background

Positivity and Passivity

Positivity and passivity are nearly equivalent concepts.

Positivity is a property of linear systems, while passivity is a moregeneral concept, applicable to both linear and nonlinear systems.

The scalar, controllable system

x = Ax+bu

y = cTx

with a transfer function

H(s) = cT(sI−A)−1b

is said to be positive real (PR) if Re[H(s)]≥ 0 for all Re(s)≥ 0.




If s = jω , it means that the Nyquist diagram of H(jω) must lies in theright half plane, including the imaginary axis.

Thus, the phase shift must be between −90deg and +90deg.

The Kalman-Yakubovich lemma states that the system H(s) is strictlypositive real if there exist positive definite matrices P and Q such that:

ATP+PA = −Q

Pb = c




Passivity is a general form of positivity

A passive linear system must be positive real and vice-versa.

Consider a system with input u and output y. The energy of the system issuch that

ddt

[Stored energy]=[external power input]+[internal power generation]

If the internal power generation is negative, then the system isdissipative or strictly passive. If the internal power generation is lessthan or equal to zero the system is passive.




If the input u and output y are such that the external power input isexpressed as uTy, then

The system is passive if 〈y|u〉 ≥ 0The system is input strictly passive (ISP) if there exists ε > 0 such that〈y|u〉 ≥ ‖u‖2

The system is output strictly passive (OSP) if there exists ε > 0 such that〈y|u〉 ≥ ‖y‖2

Examples include electrical networks (power proportional to product ofcurrent and voltage), mass-spring systems (mechanical power is force ×velocity).



The Passivity Theorem

Theorem (Passivity Theorem)Consider the system

Let H1 be strictly output passive and H2 be passive. The closed-loop system isthen BIBO stable



Positive Real Transfer Functions

A rational transfer function G(s) with real coefficients is PR if and only if thefollowing conditions hold

1 The function has no poles in the right half-plane2 If the function has poles on the imaginary axis or at infinity, they are

simple poles with positive residues3 The real part of G is nonnegative along the iω axis

Re(G(iω))≥ 0

G is SPR if conditions 1 and 3 hold and if G(s) has no poles or zeros on theimaginary axis




Positive real (PR): Re G(jω)≥ 0

Input strictly passive (ISP): Re G(jω)≥ ε > 0

Input strictly passive (ISP): Re G(jω)≥ ε|G(jω)|2

ExamplesG(s) = s+1 is SPR and ISP but not OSPG(s) = 1/(s+1) is SPR and OSP but not ISPG(s) = (s2 +1)/(s+1)2 is OSP and ISP but not SPRG(s) = 1/s is PR, but not SPR, OSP or ISP



Positive Real Transfer Functions

Theorem (Lemma)

Let r be a bounded square integrable function, and let G(s) be a positive realtransfer function. The system whose input-output relation is given by

y = r(G(p)ru)

is then passive.


Positivity and Passivity Application to Adaptive Control


MIT rule:

Lyapunov rule:



Analysis

Redraw (b) as:

Because γ/s is PR, H is passiveThus, if G is SPR, the passivity theorem implies that the system is L2 (BIBO) stableIf non-zero initial conditions, e(t) is still in L2 and thus goes to zero as t→ ∞

This is stable for all γ > 0, so the adaptation can be made arbitrarily fast as long as G isSPRIf G is not SPR, then may be possible to introduce a compensator Gc such that GGc isSPR



Analysis

The Kalman-Yakubovich lemma can be used to find Gc such that GGc isPR

However, when relative degree of G is greater than 1, Gc will containderivatives

Augmented error



Augmented Error

Factor G asG = G1G2

where G1 is SPR

e = G(θ −θ0)uc = (G1G2)(θ −θ0)uc

= G1 (G2(θ −θ0)uc(θ −θ0)G2uc− (θ −θ0)G2uc)

Introduce the augmented error

ε = e+η

where η is the error augmentation defined by

η = G1(θ −θ0)G2uc−G(θ −θ0)uc = G1(θG2uc)−Gθuc

The correction term η vanishes when θ is constant



Augmented Error

The adaptation lawθ =−γεG2uc

gives a closed-loop system where e(t) goes to zero as t goes to infinity



Augmented Error

Error augmentation is a fundamental idea in MRAC

Details can sometimes be messy...See example in Section 5.8 for an output feedback design

Error signal

e =b0

A0Am(Ru+Sy−Tuc)

Key is introduction of a filter Q such that b0Q/A0Am is SPR andef = Qe/PThen the closed-loop system is BIBO stable


Positivity and Passivity Example

Active Suspension



Active Suspension


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