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• 2.153 Adaptive Control Lecture 9

Closed-loop Reference Models and Transients

( [email protected] ) 1 / 11

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

u(t) = θ(t)xp + k(t)r

θ̇(t) = −sign(kp)exp k̇(t) = −sign(kp)er

( [email protected] ) 2 / 11

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

u(t) = θ(t)xp + k(t)r

θ̇(t) = −sign(kp)exp k̇(t) = −sign(kp)er

( [email protected] ) 2 / 11

• Stability and Convergence Leads to Error Model 3: ė = ame+ θ̃

T ω

V = 1

2

( e2 + |kp|θ̃

T θ̃

) V̇ = eė+ θ̃

T ˙̃ θ

= ame 2 + kpeθ̃

Tω + |kp|θ̃ T ˙̃ θ

= ame 2 + θ̃T (kpeω + |kp|

˙̃ θ) = ame

2 ≤ 0

⇒ e(t) and θ̃(t) are bounded for all t ≥ t0; e(t)→ 0

( [email protected] ) 3 / 11

• Stability and Convergence Leads to Error Model 3: ė = ame+ θ̃

T ω

V = 1

2

( e2 + |kp|θ̃

T θ̃

) V̇ = eė+ θ̃

T ˙̃ θ

= ame 2 + kpeθ̃

Tω + |kp|θ̃ T ˙̃ θ

= ame 2 + θ̃T (kpeω + |kp|

˙̃ θ) = ame

2 ≤ 0

⇒ e(t) and θ̃(t) are bounded for all t ≥ t0; e(t)→ 0

( [email protected] ) 3 / 11

• Adaptive Gain Example Simulation Parameters: am = −1, km = 1, ap = 1, kp = 2

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =1

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =10

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =100

( [email protected] ) 4 / 11

• Adaptive Gain Example Simulation Parameters: am = −1, km = 1, ap = 1, kp = 2

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =1

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s] S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =10

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =100

( [email protected] ) 4 / 11

• Adaptive Gain Example Simulation Parameters: am = −1, km = 1, ap = 1, kp = 2

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =1

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s] S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s]

P a ra m et er

θ k

γ =10

0 10 20 30

0

0.5

1

1.5

2

2.5

3

time [s]

S ta te

xm xp

0 10 20 30 −3

−2

−1

0

1

time [s] P a ra m et er

θ k

γ =100

( [email protected] ) 4 / 11

• Closed-Loop Reference Model

Plant: ẋp = apxp + kpu

Closed-loop Reference Model: ẋcm = amx c m + kmr − `ec

Controller: u = θ(t)xp + k(t)r

Adaptive law: ˙̄̃ θ = −γsgn(bp)ecφ ˜̄θ> =

[ θ̃ k̃

] and φ> =

[ xp r

] 1 Stability is guaranteed

2 limt→∞ e c(t) = 0

( [email protected] ) 5 / 11

• Closed-Loop Reference Model

Plant: ẋp = apxp + kpu

Closed-loop Reference Model: ẋcm = amx c m + kmr − `ec

Controller: u = θ(t)xp + k(t)r

Adaptive law: ˙̄̃ θ = −γsgn(bp)ecφ ˜̄θ> =

[ θ̃ k̃

] and φ> =

[ xp r

] 1 Stability is guaranteed 2 limt→∞ e

c(t) = 0

( [email protected] ) 5 / 11

• Closed-Loop Reference Model

Plant: ẋp = apxp + kpu

Closed-loop Reference Model: ẋcm = amx c m + kmr − `ec

Controller: u = θ(t)xp + k(t)r

Adaptive law: ˙̄̃ θ = −γsgn(bp)ecφ ˜̄θ> =

[ θ̃ k̃

] and φ> =

[ xp r

] 1 Stability is guaranteed 2 limt→∞ e

c(t) = 0

( [email protected] ) 5 / 11

• Closed-Loop Reference Model

Plant: ẋp = apxp + kpu

Closed-loop Reference Model: ẋcm = amx c m + kmr − `ec

Controller: u = θ(t)xp + k(t)r

Adaptive law: ˙̄̃ θ = −γsgn(bp)ecφ ˜̄θ> =

[ θ̃ k̃

] and φ> =

[ xp r

]

1 Stability is guaranteed 2 limt→∞ e

c(t) = 0

( [email protected] ) 5 / 11

• Closed-Loop Reference Model

Plant: ẋp = apxp + kpu

Closed-loop Reference Model: ẋcm = amx c m + kmr − `ec

Controller: u = θ(t)xp + k(t)r

Adaptive law: ˙̄̃ θ = −γsgn(bp)ecφ ˜̄θ> =

[ θ̃ k̃

] and φ> =

[ xp r

] 1 Stability is guaranteed 2 limt→∞ e

c(t) = 0 ( [email protected] ) 5 / 11

• Transient Performance With CRM

CRM gain ` affects:

L2 norm of ec(t) L∞ norm of xcm(t)

L2 norm of θ̇(t), k̇(t) L2 norm of u(t) (under investigation)

( [email protected] ) 6 / 11

• Transient Performance With CRM: L2 norm of ec(t)

Lyapunov Function: V (ec, ˜̄θ) = 12e c2 + 12γ

−1|kp| ˜̄θ> ˜̄θ

Derivative of V : V̇ (ec, ˜̄θ) = (am + `)e c2 ≤ 0

Integrate V̇ : ∫∞ 0 V̇ (e

c(τ), ˜̄θ(τ))dτ = V (∞)− V (0)

⇒ −(am + `) ∫∞ 0 (e

c(τ))2dτ = V (0)− V (∞)

⇒ ∫∞ 0 e

c(t)2dτ ≤ V (0)|am+`|

⇒ ‖ec(t)‖L2 =

√ V (0)

|am + `|

where: V (0) = 12e(0) 2 +

|kp| 2γ

˜̄θ>(0)˜̄θ(0)

( [email protected] ) 7 / 11

• Transient Performance With CRM: L2 norm of ec(t)

Lyapunov Function: V (ec, ˜̄θ) = 12e c2 + 12γ

−1|kp| ˜̄θ> ˜̄θ

Derivative of V : V̇ (ec, ˜̄θ) = (am + `)e c2 ≤ 0

Integrate V̇ : ∫∞ 0 V̇ (e

c(τ), ˜̄θ(τ))dτ = V (∞)− V (0)

⇒ −(am + `) ∫∞ 0 (e

c(τ))2dτ = V (0)− V (∞)

⇒ ∫∞ 0 e

c(t)2dτ ≤ V (0)|am+`|

⇒ ‖ec(t)‖L2 =

√ V (0)

|am + `|

where: V (0) = 12e(0) 2 +

|kp| 2γ

˜̄θ>(0)˜̄θ(0)

( [email protected] ) 7 / 11

• Transient Performance With CRM: L2 norm of ec(t)

Lyapunov Function: V (ec, ˜̄θ) = 12e c2 + 12γ

−1|kp| ˜̄θ> ˜̄θ

Derivative of V : V̇ (ec, ˜̄θ) = (am + `)e c2 ≤ 0

Integrate V̇ : ∫∞ 0 V̇ (e

c(τ), ˜̄θ(τ))dτ = V (∞)− V (0)

⇒ −(am + `) ∫∞ 0 (e

c(τ))2dτ = V (0)− V (∞)

⇒ ∫∞ 0 e

c(t)2dτ ≤ V (0)|am+`|

⇒ ‖ec(t)‖L2 =

√ V (0)

|am + `|

where: V (0) = 12e(0) 2 +

|kp| 2γ

˜̄θ>(0)˜̄θ(0)

( [email protected] ) 7 / 11

• Transient Performance With CRM: L2 norm of ec(t)

Lyapunov Function: V (ec, ˜̄θ) = 12e c2 + 12γ

−1|kp| ˜̄θ> ˜̄θ

Derivative of V : V̇ (ec, ˜̄θ) = (am + `)e c2 ≤ 0

Integrate V̇ : ∫∞ 0 V̇ (e

c(τ), ˜̄θ(τ))dτ = V (∞)− V (0)

⇒ −(am + `) ∫∞ 0 (e

c(τ))2dτ = V (0)− V (∞)

⇒ ∫∞ 0 e

c(t)2dτ ≤ V (0)|am+`|

⇒ ‖ec(t)‖L2 =

√ V (0)

|am + `|

where: V (0) = 12e(0) 2 +

|kp| 2γ

˜̄θ>(0)˜̄θ(0)

( [email protected] ) 7 / 11

• Transient Performance With CRM: L2 norm of ec(t)

Lyapunov Function: V (ec, ˜̄θ) = 12e c2 + 12γ

−1|kp| ˜̄θ> ˜̄θ

Derivative of V : V̇ (ec, ˜̄θ) = (am + `)e c2 ≤ 0

Integrate V̇ : ∫∞ 0 V̇ (e

c(τ), ˜̄θ(τ))dτ = V (∞)− V (0)

⇒ −(am + `) ∫∞ 0 (e

c(τ))2dτ = V (0)− V (∞)

⇒ ∫∞ 0 e

c(t)2dτ ≤ V (0)|am+`|

⇒ ‖ec(t)‖L2 =

√ V (0)

|am + `|

where: V (0) = 12e(0) 2 +

|kp| 2γ

˜̄θ>(0)˜̄θ(0)

( [email protected] ) 7 / 11

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