2.153 adaptive control lecture 2 simple adaptive …aaclab.mit.edu/material/lect/lecture2.pdf ·...

2.153 Adaptive ControlLecture 2

Simple Adaptive Systems: Identification and Control

Daniel Wiese

[email protected]

Tuesday 17-Feb-2015

([email protected]) Tuesday 17-Feb-2015 1 / 27

Introduction

Last time:

Parameter identification

Algebraic (error model 1) and dynamic (error model 3)Scalar and vector parameter systemsNon-recursive and recursive

Stability

Introduction to using Lyapunov functions

Today:

Identification of multiple parameters in first order plant

Error model 1 and 3Determining update law using Lyapunov functions

Adaptive control


Error Models

See page 273.

Relate parameter errors (cannot measure) to output/measurementerror (which can be measured)

The stability and performance of an adaptive system is dependentupon the evolution of these errors

An error model is the mathematical model which describes theevolution of these errors

By using error models, we can understand and solve adaptive controlproblems more easily, as the error models are independent of thespecific adaptive system


Identification of a Vector Parameter in an Algebraic System

Problem:

y(t) = θ>u(t)

θ: unknowny(t), u(t): measured

Construct a parameter estimate θ



Difference the two output signals as

e = y − y

Express the parameter error as

θ = θ − θReduces to Error Model 1:



Error model 1:e = θ>u

Propose the candidate Lyapunov function:

V (θ) = θ>θ

Time differentiating

V = 2θ>˙θ

Adaptive law˙θ = −eu

Gives

V = −2θ>eu

= −2(θ>u)2

Thus V ≤ 0 ⇒ stability([email protected]) Tuesday 17-Feb-2015 6 / 27

Parameter Identification: Motivating Example

Transfer function of a DC motor:

V : Voltage inputω: Angular Velocity outputK,J,B: unknown physical parameters

Express the plant transfer function as:

ω

V=

K

Js+B=

a1s+ θ1

K,J,B unknown ⇒ a1, θ1 unknown


Parameter Identification: Motivating Example

The differential equation describing the DC motor is

ω = −θ1ω + a1V

The DC motor is a first-order dynamical system

Recall: last time we assumed a1 was known

We looked at two procedures for identification

Error model 1Error model 3

Now we will consider both a1 and θ1 to be unknown

Will go through both procedures (error model 1 and 3) again


Identification: First order plant, two unknown parameters

Plant Transfer Function:xpu

=a1

s+ θ1Express the plant transfer function as

xpu

=1

s+ θ1

=1

s+ θm· s+ θms+ θ1

=1

s+ θm· 1s+θ1s+θm

=1

s+ θm· 1θm−θms+θm

+ s+θ1s+θm

=1

s+ θm· 1

1− θm−θ1s+θm

where θm > 0 is a known positive parameter selected by control designer.([email protected]) Tuesday 17-Feb-2015 9 / 27


Define θ , θm − θ1 and simplify this transfer function as

ω

u=

1

s+ θm· 1

1 + θs+θm

and realize this in the following block diagram representation, where weare using two states to represent a first order system.



Recall: when a1 known

y(t) ≡ xp(t)− a1φ(t) = θφ2(t)

⇒Error Model 1, Identify θ, a scalar

When a1 unknown:

y(t) ≡ xp(t) = a1φ(t) + θφ2(t) = θ>φ(t)

⇒Error Model 1, Identify θ, a vector([email protected]) Tuesday 17-Feb-2015 11 / 27


y(t) = θ>φ(t)

We have just seen how to determine an update law for this system onslide 5: error model 1 with vector parameter

Now we will see an alternate method of identifying the two unknownparameters


Identification: First order plant, two unknown parameters -An Alternate Method

The differential equation describing the plant is given by

xp = −θ1ω + a1u

Generate an estimate of the plant output as follows, using parameterestimates in place of the unknown parameters

˙xp = −θ1xp + a1u

Define the following output error and parameter errors

e = xp − xpθ1 = θ1 − θ1a1 = a1 − a1



The output error dynamics are given by

e = −θ1xp + a1u+ θ1xp − a1u

Add and subtract θ1xp

e = −θ1xp + θ1xp − θ1xp + a1u+ θ1xp − a1u= (θ1 − θ1)xp − θ1(xp − xp) + a1u

= −θ1e− θ1xp + a1u

= −θ1e+ θ>φ

where

θ =

[θ1a1

]and φ =

[−xpu

]([email protected]) Tuesday 17-Feb-2015 14 / 27


Error Model 3:e = −θ1e+ θ>φ

We saw error model 3 last lecture for a system with one unknown

We will again determine an update law using a Lyapunov function

Choose a quadratic function V of the dominant errors in the system

V (e, θ) =1

2

(e2 + θ>θ

)Goal: Choose

˙θ so that V ≤ 0



Take the time derivative of V

V = ee+ θ˙θ

= −θ1e2 + θ>eφ+ θ>˙θ

Choose˙θ = −eφ

⇒ V = −θ1e2

Thus V ≤ 0⇒ stability.


Adaptive Control

Everything we have seen thus far has been identification of unknownparameters

Now we introduce adaptive control: how to control systems withunknown parameters


Adaptive Control of a First-Order Plant

Problem:

Plant:xp = apxp + kpu

Find u such that xp follows a desired command.


A Model-reference Approach

xp: Output of a first-order system - can only follow ’smooth’ signals

Ensure xd is a ‘smooth’ signal essentially by filtering the desiredcommand

Pose the problem asxm = amxm + kmr

−2 0 2 4 60

0.2

0.4

0.6

0.8

1

Time [s]

xdxm

Set am = −5 and km = 5, say. Choose r so that xm ≈ xd([email protected]) Tuesday 17-Feb-2015 19 / 27

Statement of the problem

Choose u so that e(t)→ 0 as t→∞.ap unknown

kp unknown, but with known sign


Certainty Equivalence Principle

Step 1: Algebraic Part: Find a solution to the problem when parametersare known.

Step 2: Analytic Part: Replace the unknown parameters by their estimates.Ensure stable and convergent behavior.

The use of the parameter estimates in place of the true parameters isknown as the certainty equivalence principle.


Certainty Equivalence Principle- Step 1

Step 1: Algebraic Part: Propose the control law

u(t) = θcxp + kcr

and choose θc, kc so that closed-loop transfer function matches thereference model transfer function.

xp = apxp + kp(θcxp + kcr)

= (ap + kpθc)xp + kpkcr

Now compare this to the reference model equation

xm = amxm + kmr

Desired Parameters: θc = θ∗ and kc = k∗ must satisfy

ap + kpθ∗ = am and kpk

∗ = km


Certainty Equivalence Principle- Step 1

Solving for the nominal or ideal parameters

θ∗ =am − apkp

and k∗ =kmkp

This is represented with the following block diagram


Certainty Equivalence Principle - Step 2

Step 2: Analytic Part: Replace the unknown parameters by their estimates.Ensure stable and convergent behavior. From Step 1, we have

u(t) = θ∗xp + k∗r, θ∗ =am − apkp

, k∗ =kmkp

Replace θ∗ and k∗ by their estimates θ(t) and k(t).

u(t) = θ(t)xp + k(t)r

θ(t) =?? k(t) =??



Adaptive control input:

u(t) = θ(t)xp + k(t)r

Define the parameter errors as

θ(t) = θ(t)− θ∗

k(t) = k(t)− k∗

Plug the control law into the plant equation:

xp = apxp + kpu(t)

= apxp + kp[θ(t)xp + k(t)r

]= apxp + kp

[θ(t)xp + θ∗xp + k(t)r + k∗r

]=

[ap + kpθ

∗]xp + kpθ(t)xp + kpk∗r + kpk(t)r

= amxp + kpθ(t)xp + kmr + kpk(t)r



Reference Model:xm = amxm + kmr

Define the tracking error as

e = xp − xm

Error model:

e = ame+ kpθ(t)xp + kpk(t)r

= ame+ kpθT(t)ω


ω =

xp

r

θ =

θ(t)

k(t)


e = ame+ kpθ>(t)ω

This is again error model 3!


2.153 adaptive control lecture 2 simple adaptive …aaclab.mit.edu/material/lect/lecture2.pdf ·...

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