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2.153 Adaptive ControlFall 2019
Lecture 3: Simple Adaptive Systems: Control
Anuradha Annaswamy
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.
( [email protected]) September 11, 2019 2 / 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.
( [email protected]) September 11, 2019 2 / 14
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!
( [email protected]) September 11, 2019 3 / 14
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!
( [email protected]) September 11, 2019 3 / 14
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!
( [email protected]) September 11, 2019 3 / 14
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.
( [email protected]) September 11, 2019 4 / 14
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.
( [email protected]) September 11, 2019 4 / 14
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.
( [email protected]) September 11, 2019 5 / 14
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 = 0Choose 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.
( [email protected]) September 11, 2019 5 / 14
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 = 0Choose 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.
( [email protected]) September 11, 2019 5 / 14
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 = 0Choose 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.
( [email protected]) September 11, 2019 5 / 14
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 = 0Choose 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.
( [email protected]) September 11, 2019 5 / 14
Adaptive Control of a First-Order Plant
Problem:
Plant:xp = apxp + kpu
Find u such that xp follows a desired command.
( [email protected]) September 11, 2019 6 / 14
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
( [email protected]) September 11, 2019 7 / 14
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
( [email protected]) September 11, 2019 7 / 14
Statement of the problem
Choose u so that e(t)→ 0 as t→∞.
kp, ap are unknown.
( [email protected]) September 11, 2019 8 / 14
Statement of the problem
Choose u so that e(t)→ 0 as t→∞. kp, ap are unknown.
( [email protected]) September 11, 2019 8 / 14
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
( [email protected]) September 11, 2019 9 / 14
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
( [email protected]) September 11, 2019 9 / 14
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
( [email protected]) September 11, 2019 9 / 14
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
( [email protected]) September 11, 2019 9 / 14
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) =??
( [email protected]) September 11, 2019 10 / 14
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) =??
( [email protected]) September 11, 2019 10 / 14
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) =??
( [email protected]) September 11, 2019 10 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
Certainty Equivalence Principle - Step 2
Adaptive control input:
u(t) = θ(t)xp + k(t)r
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
( [email protected]) September 11, 2019 11 / 14
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
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 12 / 14
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
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 12 / 14
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
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 12 / 14
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
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]( [email protected]) September 11, 2019 12 / 14
Error Model 3
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 13 / 14
Error Model 3
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 13 / 14
Error Model 3
e = ame+ kpθ(t)xp + kpk(t)r
= ame+ kpθT
(t)ω
ω =
[xpr
], θ =
[θ(t)
k(t)
]
( [email protected]) September 11, 2019 13 / 14
Suggested Reading
Chapter 2: 2.2, 2.3, 2.4Chapter 3: 3.3
( [email protected]) September 11, 2019 14 / 14