2.153 Adaptive Control Lecture 1 Simple Adaptive …aaclab.mit.edu/material/lect/lecture1.pdf2.153...

2.153 Adaptive ControlLecture 1

Simple Adaptive Systems: Identification

Anuradha Annaswamy

[email protected]

( [email protected] ) 1 / 17

Parameter Adaptation - Recursive Schemes

Adaptive Control: The control of Uncertain Systems

Adaptive Control (in this Course):The control of Linear Time-invariant Plants with Unknown Parameters


Adaptive Control: A Parametric Framework

Nonlinear, time-varying, with unknown parameter θ

x = f(x, u, θ, t) y = h(x, u, θ, t)

Linear Time-Varying (LTV) with unknown parameter θ

x = A(θ, t)x+B(θ, t)u y = C(θ, t)x+D(θ, t)u

Linear Time-Invariant (LTI) with unknown parameter θ

x = A(θ)x+B(θ)u y = C(θ)x+D(θ)u

System to be controlled (open-loop): PlantControlled System (closed-loop): System




x = f(x, u, θ, t) y = h(x, u, θ, t)





System to be controlled (open-loop): Plant

Controlled System (closed-loop): System




x = f(x, u, θ, t) y = h(x, u, θ, t)





System to be controlled (open-loop): PlantControlled System (closed-loop): System


Direct and Indirect Adaptive Control

θp: Plant parameter - unknown; θc: Control parameter

Indirect Adaptive Control: Estimate θp as θp. Compute θc using θp.

θp → θp → θc

Direct Adaptive Control: Directly estimate θc as θc. Compute the plantestimate θp using θc

θp → θc → θc


Identification of a Single Parameter

θ: Unknown, scalar

y(t) = θu(t)

Identify θ using measurements {u(t), y(t)}.


Identification of a Vector Parameter

y(t) = θTu(t)

y ∈ R, θ ∈ Rn, u : R+ → Rn




y(t) = θTu(t)

y ∈ R,

θ ∈ Rn, u : R+ → Rn




y(t) = θTu(t)

y ∈ R, θ ∈ Rn,

u : R+ → Rn




y(t) = θTu(t)

y ∈ R, θ ∈ Rn, u : R+ → Rn



Identification of a Single Parameter - Recursive Scheme

y(t) = θu(t)

θ: Unknown, scalar

Identify θ as θ(t) at every instant


Identification of a Single Parameter - Recursive Scheme

y(t) = θu(t)

θ: Unknown, scalar Identify θ as θ(t) at every instant


Identification of a Vector Parameter - Recursive Scheme

y(t) = θTu(t)

y ∈ R, θ ∈ Rn, u : R+ → Rn

Identify θ as θ(t) at every instant


Error Model 1

θ: Unknown, u(t) and e(t) can be measured at each instant t.


Identification of a Parameter in a Dynamic System

Simplest Transfer Function of a Motor:

V : Voltage input ω: Angular Velocity output

K,J,B: Physical parameters

Plant:K

Js+B=

a1s+ θ1

K,J,B unknown ⇒ a1, θ1 unknown


One way of identifying parameters a1 and θ1

Assume that a1 is known.

Identify θ1 as θ. θ = θ − θ1

Plant: ω = −θ1ω + u u = a1V



Assume that a1 is known. Identify θ1 as θ.

θ = θ − θ1




Assume that a1 is known. Identify θ1 as θ. θ = θ − θ1



Error Model 3

e = −θ1e+ θu


An alternate procedure for identifying θ1:

a1s+ θ1

=

a1s+ θm

1 +θm − θ1s+ θm

θ ≡ θ1 − θm


Stability

Behavior near an Equilirbium Point.

Consider the following dynamical system

x(t) = f(x(t), t)

x(t0) = x0(1)

Definition: equilibrium point (pg 45) The state xeq is an equilibriumpoint of (1) if it satisfies:

f(xeq, t) = 0 (2)

for all t ≥ t0.


Stability

Behavior near an Equilirbium Point.Consider the following dynamical system

x(t) = f(x(t), t)

x(t0) = x0(1)

Definition: equilibrium point (pg 45) The state xeq is an equilibriumpoint of (1) if it satisfies:

f(xeq, t) = 0 (2)

for all t ≥ t0.


Stability of LTI Plants

A motivating example: determine the stability of the origin for thefollowing scalar system

x(t) = Ax(t)

Equilibrium point: x = 0

Can determine the stability of the origin by evaluating eigenvalues of A

x(t) = eA(t−t0)x(t0)

A = V ΛV −1; V : from eigenvector; Λ : diag(λi) : from eigenvalues

Stability follows if Re(λi) ≤ 0Asymptotic stability follows if Re(λi) < 0.

Lyapunov’s methods allow us to determine the stability of an equilibriumfor such a system without solving the differential equation!


Stability of LTI Plants

A motivating example: determine the stability of the origin for thefollowing scalar system

x(t) = Ax(t)

Equilibrium point: x = 0Can determine the stability of the origin by evaluating eigenvalues of A

x(t) = eA(t−t0)x(t0)

A = V ΛV −1; V : from eigenvector; Λ : diag(λi) : from eigenvalues

Stability follows if Re(λi) ≤ 0Asymptotic stability follows if Re(λi) < 0.

Lyapunov’s methods allow us to determine the stability of an equilibriumfor such a system without solving the differential equation!


Lyapunov Stability

For the systemx = f(x)

Let

(i) V (x) > 0, ∀x 6= 0, and V (0) = 0

(ii) V (x) =(∂V∂x

)Tf(x) < 0

(ii) V (x)→∞ as ‖x‖ → ∞

Then x = 0 is asymptotically stable.

If instead of (ii), we have

(ii’) V ≤ 0

Then x = 0 is stable.


Error Model 1

Error Model 1 leads to the following

x(t) = A(t)x(t) A(t) = −u(t)uT (t)

Equilibrium point: x = 0

Choose a quadratic function

V =1

2xTx

V = xTA(t)x = −xTu(t)uT (t)x = −(xTu(t)

)2 ≤ 0

⇒ stability.A later lecture will show that if u(t) is ” persistently exciting”, x(t)→ 0.We therefore conclude that error model 1 leads to a stable parameterestimation. Asymptotic stability will be shown later.


Error Model 1


x(t) = A(t)x(t) A(t) = −u(t)uT (t)

Equilibrium point: x = 0Choose a quadratic function

V =1

2xTx


)2 ≤ 0



Error Model 1


x(t) = A(t)x(t) A(t) = −u(t)uT (t)


V =1

2xTx


)2 ≤ 0

⇒ stability.

A later lecture will show that if u(t) is ” persistently exciting”, x(t)→ 0.We therefore conclude that error model 1 leads to a stable parameterestimation. Asymptotic stability will be shown later.


Error Model 1


x(t) = A(t)x(t) A(t) = −u(t)uT (t)


V =1

2xTx


)2 ≤ 0

⇒ stability.A later lecture will show that if u(t) is ” persistently exciting”, x(t)→ 0.

We therefore conclude that error model 1 leads to a stable parameterestimation. Asymptotic stability will be shown later.


Error Model 1


x(t) = A(t)x(t) A(t) = −u(t)uT (t)


V =1

2xTx


)2 ≤ 0



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