2.153 adaptive control fall 2019 lecture 3: simple adaptive...

2.153 Adaptive ControlFall 2019

Lecture 3: Simple Adaptive Systems: Control

Anuradha Annaswamy

[email protected]

September 11, 2019

( [email protected]) September 11, 2019 1 / 14

Stability

Behavior near an Equilibrium 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 Equilibrium 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 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 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

(iii) 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



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’ signalsPose the problem as

xm = 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.

Or am = −50, km = 50, Choose r so thatxm ≈ xd


A Model-reference Approach

xp: Output of a first-order system - can only follow ’smooth’ signalsPose the problem as

xm = amxm + kmr

ï2 0 2 4 60

0.2

0.4

0.6

0.8

1

Time [s]

xdxm

Set am = −5 and km = 5. Or am = −50, km = 50, Choose r so thatxm ≈ xd


Statement of the problem

Choose u so that e(t)→ 0 as t→∞.

kp, ap are unknown.


Statement of the problem

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


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.

Step 1: u(t) = θcxp + kcr, choose θc, kc so that closed-loop transferfunction matches the reference model transfer function.Desired Parameters: θc = θ∗, kc = k∗

θ∗ =am − apkp

, k∗ =kmkp





Step 1: u(t) = θcxp + kcr, choose θc, kc so that closed-loop transferfunction matches the reference model transfer function.

Desired Parameters: θc = θ∗, kc = k∗

θ∗ =am − apkp

, k∗ =kmkp





Step 1: u(t) = θcxp + kcr, choose θc, kc so that closed-loop transferfunction matches the reference model transfer function.Desired Parameters: θc = θ∗, kc = k∗

θ∗ =am − apkp

, k∗ =kmkp


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:


Closed-loop Equations:

xp = apxp + kpu(t)

= apxp + kp[θ(t)xp + k(t)r

]=

[ap + kpθ

∗ − kpθ∗ + kpθ(t)]xp + kp

[k(t)− k∗ + k∗

]r

= amxp + kmr + kpθ(t)xp + kpk(t)r

Reference Model:xm = amxm + kmr

Error model:e = ame+ kpθ(t)xp + kpk(t)r






xp = apxp + kpu(t)


]

=[ap + kpθ


[k(t)− k∗ + k∗

]r









xp = apxp + kpu(t)


]=

[ap + kpθ


[k(t)− k∗ + k∗

]r








, k∗ =kmkp



e = ame+ kpθ(t)xp + kpk(t)r

= ame+ kpθT

(t)ω

ω =

[xpr

], θ =

[θ(t)

k(t)

]





, k∗ =kmkp




= ame+ kpθT

(t)ω

ω =

[xpr

], θ =

[θ(t)

k(t)

]( [email protected]) September 11, 2019 12 / 14

Error Model 3


= ame+ kpθT

(t)ω

ω =

[xpr

], θ =

[θ(t)

k(t)

]


Suggested Reading

Chapter 2: 2.2, 2.3, 2.4Chapter 3: 3.3


2.153 adaptive control fall 2019 lecture 3: simple adaptive...

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