model reference adaptive control - adaptive control

EE 264 SIST, ShanghaiTech

Model Reference Adaptive Control

YW 9-1

Contents

Introduction

Motivating Examples

MRC for SISO plant

Model Reference Adaptive Control 9-2

General Structure

Figure: General Structure of MRAC

Categories:

i) Direct and Indirect MRAC

ii) Adaptive law with and without normalizationModel Reference Adaptive Control 9-3

General Structure

Figure: General Structure of MRAC

Categories:

i) Direct and Indirect MRAC

ii) Adaptive law with and without normalizationModel Reference Adaptive Control 9-4

Contents

Introduction

Motivating Examples

MRC for SISO plant


Adaptive regulation

Example: one parameter case adaptive regulation

Consider

x = ax+ u, x(0) = x0

where a is a constant but unknown. The control objective is find

proper input signal u such that x→ 0 as t→∞.

Reference model

x = −amx, am > 0

Control law

u = −k(t)x

with k∗ = am + a.Model Reference Adaptive Control 9-6

Adaptive regulation


Consider

x = ax+ u, x(0) = x0



Reference model

x = −amx, am > 0

Control law

u = −k(t)x


Adaptive Regulation

Indirect approach:

k(t) = am + θ, θ = Φ(x, u)

Direct approach: rewrite the plant as

x = −amx+ k∗x+ u = −amx− kx˙k = k := ϕ(x)

with k = k − k∗. Consider a Lyapunov candidate function

V (x, k) = x2

2 + k2

2γ


Adaptive Regulation

Indirect approach:

k(t) = am + θ, θ = Φ(x, u)




V (x, k) = x2

2 + k2

2γ


Adaptive Regulation

To generate the adaptive law for k. Taking the time derivative of

V as

V = −amx2 − kx2 + kϕ

γ

Take ϕ(x) = γx2, we have

V = −amx2

indicates

x, k, k ∈ L∞

x ∈ L2, x ∈ L∞, therefore x→ 0 as t→∞.

Note, we cannot establish that k → k∗.Model Reference Adaptive Control 9-12

Adaptive Regulation


V as


γ


V = −amx2

indicates

x, k, k ∈ L∞



Adaptive Tracking

Example 2: Two parameter case adaptive tracking. Consider

x = −ax+ bu

a, b unknown, but sgn(b) is assume to be known. Reference model

xm = bm

s+ amr

It is assumed that am, bm, and r are chosen so that xm represents

the desired state response of the plant. The control goal is

x(t)→ xm(t)


Adaptive Tracking

The optimal control law

u = −k∗x+ l∗r

where k∗ and l∗ verifyx(s)r(s) = bl∗

s+ a+ bk∗= bm

s+ am= xm(s)

r(s)results in

l∗ = bm

b, k∗ = am − a

b

Since a, b unknown, we propose the control law

u = −kx+ lr

where k, l are estimates for k∗, l∗ respectively.Model Reference Adaptive Control 9-17

Adaptive Tracking


u = −k∗x+ l∗r


s+ a+ bk∗= bm

s+ am= xm(s)

r(s)results in

l∗ = bm

b, k∗ = am − a

b


u = −kx+ lr


Adaptive Tracking

Express the plant equation in terms of k∗ and l∗

x = −amx+ bmr + b(k∗x− l∗r + u)

Define the error e := x− xm, we have a B-SPM

e = b

s+ am(k∗x− l∗r + u)

Alternatively, we can express the error equation as

e = −ame+ b(k∗x− l∗r + u)

= −ame+ b(−kx+ lr)

with k := k − k∗, l := l − l∗.Model Reference Adaptive Control 9-20

Adaptive Tracking




e = b






Adaptive Tracking

This motivates the following Lyapunov candidate function for

designing the adaptive laws

V (e, k, l) = e2

2 + k2

2γ1|b|+ l2

2γ2|b|

where γ1 > 0, γ2 > 0. The time derivative is given by

V = −ame2 − bkex+ bler + |b|k

γ1

˙k + |b|lγ2

˙l = −ame2

Because |b| = b sgn(b), the indefinite terms disappear if we choose

k = γ1ex sgn(b), l = −γ2er sgn(b)


Adaptive Tracking

Properties:

e, k, l ∈ L∞, x, u ∈ L∞

e ∈ L2 and e, k, l ∈ L∞ if r ∈ L∞

if r is s.r. of order 2, e→ 0, k → k∗ , l→ l∗ exponentially fast.

Alternatively, we can generate a, b first, then combined with

u = kx+ lr where

k(t) = am + a(t)b(t)

, l(t) = bm

b(t)

This is a typical indirect MRAC method.

Note, for now, methods presented do not have normalization.Model Reference Adaptive Control 9-26

Adaptive Tracking

Properties:

e, k, l ∈ L∞, x, u ∈ L∞

e ∈ L2 and e, k, l ∈ L∞ if r ∈ L∞



u = kx+ lr where


, l(t) = bm

b(t)



Contents

Introduction

Motivating Examples

MRC for SISO plant


Problem Statement

Consider the SISO LTI plant

xp = Apxp +Bpup, xp(0) = x0 ∈ Rnp

yp = C>p xp

where yp, up ∈ R and Ap, Bp, Cp have the appropriate dimensions.

The transfer function of the plant :

yp = Gp(s)up = kpZp(s)Rp(s)up

where Zp(s), Rp(s) are monic known polynomials, constant kp is

a.k.a. high-frequency gain.Model Reference Adaptive Control 9-32

Problem Statement

Reference model

xm = Amxm +Bmr, xm(0) = xm0 ∈ Rnm

ym = C>mxm

where ym ∈ R and r ∈ R is the reference input.

The transfer function

ym = Wm(s)r

is expressed as

Wm(s) = kmZm(s)Rm(s)

where Zm(s), Rm(s) are monic polynomials and km is a constant.Model Reference Adaptive Control 9-33

Problem Statement

Reference model

xm = Amxm +Bmr, xm(0) = xm0 ∈ Rnm

ym = C>mxm

where ym ∈ R and r ∈ R is the reference input.

The transfer function

ym = Wm(s)r

is expressed as

Wm(s) = kmZm(s)Rm(s)

where Zm(s), Rm(s) are monic polynomials and km is a constant.Model Reference Adaptive Control 9-34

Problem Statement

MRC problem: The MRC objective is to determine the plant

input up so that all signals(xp, yp, up) are bounded and the plant

output yp tracks the reference model output ym as close as

possible for any given reference input r(t) which is a uniformly

bounded piecewise continuous function of time.


Problem assumptions

Plant :

P1. Zp(s) is a monic Hurwitz polynomial with degree of mp.

P2. An upper bound n of the degree np of Rp(s) is known.

P3. The relative degree n∗ = np −mp of Gp(s) is known.

P4. The sign of the high-frequency gain kp is known.

Reference model:

M1. Zm(s), Rm(s) are monic Hurwitz polynomials of degree

qm, pm, respectively, where pm ≤ n

M2. The relative degree n∗m = pm − qm of Wm(s) is the same as

that of Gp(s), i.e., n∗m = n∗


MRC scheme

In addition, for MRC problem, we assume Gp(s) is known exactly.

A trivial choice for up is

up = km

kp

Zm(s)Rm(s)

Rp(s)Zp(s) r

which leads to the closed-loop transfer function

yp

r= km

kp

Zm

Rm

Rp

Zp

kpZp

Rp= Wm(s)

Drawbacks: This control law may involve zero-pole cancellations

outside C− when Rp(s) is not Hurwitz.

Core design logic : transfer function matchingModel Reference Adaptive Control 9-38

MRC scheme



up = km

kp

Zm(s)Rm(s)

Rp(s)Zp(s) r


yp

r= km

kp

Zm

Rm

Rp

Zp

kpZp

Rp= Wm(s)




MRC scheme

Let us consider the feedback control law

up = θ∗>1α(s)Λ(s)up + θ∗>2

α(s)Λ(s)yp + θ∗3yp + c∗0r

where Λ(s) = Λ0(s)Zm(s) and Λ0(s) is monic, Hurwitz, and of

degree n0 = n− 1− qm.

α(s) , αn−2(s) =[sn−2, sn−3, . . . , s, 1

]> for n ≥ 2

α(s) , 0 for n = 1

c∗0, θ∗3 ∈ R; θ∗1, θ∗2 ∈ Rn−1 are constant parameters.

Hence, the controller parameter vector to be designed is

θ∗ =[θ∗>1 , θ∗>2 , θ∗3, c

∗0

]>∈ R2n


MRC scheme

We can now meet the control objective if we select θ∗ so that the

closed-loop poles are stable and the closed-loop transfer functiony(s)r(s) = Gc(s) = Wm(s), i.e., the matching equation

c∗0kpZpΛ(Λ− θ∗>1 α

)Rp − kpZp

(θ∗>2 α+ θ∗3Λ

) = kmZm

Rm

is satisfied for all s ∈ C. Choosing

c∗0 = km

kp

and using Λ(s) = Λ0(s)Zm(s), the matching equation becomes(Λ− θ∗>1 α

)Rp − kpZp

(θ∗>2 α+ θ∗3Λ

)= ZpΛ0Rm


MRC scheme

Equating the coefficients of the powers of s on both sides, we can

express the matching equation in terms of the algebraic equation

Sθ∗ = p

where θ∗ =[θ∗>1 , θ∗>2 , θ∗3

]>; S is an (n+ np − 1)× (2n− 1)

matrix that depends on the coefficients of Rp, kp, Zp, and Λ; and

p is an n+ np − 1 vector with the coefficients of ΛRp − ZpΛ0Rm.

Lemma Let the degrees of Rp, Zp,Λ,Λ0, and Rm be as specified

in assumptions. Then the solution θ∗ of Sθ∗ = p always exists.In

addition, if Rp, Zp are coprime and n = np, then the solution θ∗ is

unique.Model Reference Adaptive Control 9-45

MRC scheme



Sθ∗ = p

where θ∗ =[θ∗>1 , θ∗>2 , θ∗3

]>; S is an (n+ np − 1)× (2n− 1)







MRC scheme

A state-space realization of the control law

ω1 = Fω1 + gup, ω1(0) = 0 ∈ Rn−1

ω2 = Fω2 + gyp, ω2(0) = 0 ∈ Rn−1

up = θ∗>ω

where θ∗ =[θ∗>1 , θ∗>2 , θ∗3, c

∗0

]T, ω = [ω>1 , ω>2 , yp, r]> and

F =

−λn−2 −λn−3 −λn−4 · · · −λ0

1 0 0 · · · 0

0 1 0 · · · 0...

.... . .

. . ....

0 0 · · · 1 0

, g =

1

0...

0

λi are the coefficients of Λ(s)

Λ(s) = sn−1 + λn−2sn−2 + · · ·+ λ1s+ λ0 = det(sI − F )


MRC scheme

Example: Let us consider the second-order plant

yp = −3(s+ 4)s2 − 3s+ 2up

and the reference model

ym = 1s+ 1r

1) check assumptions of plant and reference: np = 2., n∗ = 1 is

equal to that of the reference model.

2) choose the polynomial Λ(s) = s+ 2 = Λ0(s) and the control

input

up = θ∗11

s+ 2up + θ∗21

s+ 2yp + θ∗3yp + c∗0r


MRC scheme


yp = −3(s+ 4)s2 − 3s+ 2up


ym = 1s+ 1r




input

up = θ∗11

s+ 2up + θ∗21

s+ 2yp + θ∗3yp + c∗0r


MRC scheme

3) solve θ∗ by matching equationyp(s)r(s) = −3c∗0(s+ 4)(s+ 2)

(s+ 2− θ∗1) (s− 1)(s− 2) + 3(s+ 4) (θ∗2 + θ∗3(s+ 2)) = Gc(s)

Forcing Gc(s) = 1s+1 , we have c∗0 = −1

3 , and the matching

equation becomes

(θ∗1 − 3θ∗3) s2+(−3θ∗1 − 3θ∗2 − 18θ∗3) s+2θ∗1−12θ∗2−24θ∗3 = −8s2−18s−4

4) Implement control law as

ω1 = −2ω1 + up

ω2 = −2ω2 + yp

up = −2ω1 − 4ω2 + 2yp −(1

3

)r.


model reference adaptive control - adaptive control

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