2.153 Adaptive Control Fall 2019 Lecture 11: Neural ...
Transcript of 2.153 Adaptive Control Fall 2019 Lecture 11: Neural ...
2.153 Adaptive ControlFall 2019
Lecture 11: Neural Networks and Adaptive Control
Anuradha Annaswamy
October 9, 2019
( [email protected]) October 9, 2019 1 / 15
Problem Statement
Problem: minxf(x)
f(x) unknown
ଵ ଶ
Identifying f(x) at all x is impossibleWe convert f(x) into basis functions
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Approximation
f(x) =
N∑i=1
ai sin(ωix)
𝑁 = 3
Fourier Series says that
|f(x)−N∑i=1
ai sin(ωix)| ≤ ε
Taylor Series says that
|f(x)−N∑i=1
aixi| ≤ ε, for x ∈ [x1, x2]
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Neural Networks∑ai sin(ωix),
∑aix
i : linearly parameterized
Departure: Neural Networks
ଵ
ଶ
ே
ଵ
ଵ
+
ଶ
ଶ
ே
ே
y =
N∑i=1
αiσ(wTi x+ θi
)x =
[x1
]Let σ
(wTi x+ θi
)= σ
(wTi x)
= σi (x); Σ = [σ1(x), σ2(x), . . . , σN (x)]T
y = αTΣ(x)
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Universal Approximation
Theorem
σ is any continuous sigmoidal function. Let
N (x) =
N∑i=1
αiσ(wTi x).
Then for any f that is continuous and ε > 0, there exists N such that
|N (x)− f(x)| < ε, ∀x ∈ X, (1)
where X: compact set
Instead of (1), one can have∫X|N (x)− f(x)|dx < ε
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Stone-Weierstrass Theorem
x
How do we adjust
wi in Σ(x)?αi in αT ?
Use the back propagation algorithm:
J = e2
α ∝ −∂J∂α
wi ∝ − ∂J
∂wi
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Back Propagation Algorithm
e = y − y =
N∑i=1
αiσ(wTi x)− y
∂e
∂αi
∣∣∣∣αi
= σ(wTi x)
; ˙αi = −γe ∂e∂αi
∣∣∣∣αi
∂e
∂wi
∣∣∣∣wi
= αiσ′∣∣∣∣wT
i x
· x; ˙wi = −γe ∂e∂wi
∣∣∣∣wi
We did not discuss stability properties
( [email protected]) October 9, 2019 7 / 15
Back Propagation Algorithm
e = y − y =
N∑i=1
αiσ(wTi x)− y
∂e
∂αi
∣∣∣∣αi
= σ(wTi x)
; ˙αi = −γe ∂e∂αi
∣∣∣∣αi
∂e
∂wi
∣∣∣∣wi
= αiσ′∣∣∣∣wT
i x
· x; ˙wi = −γe ∂e∂wi
∣∣∣∣wi
We did not discuss stability properties
( [email protected]) October 9, 2019 7 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider a very simple example:
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
( [email protected]) October 9, 2019 8 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider a very simple example:
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
( [email protected]) October 9, 2019 8 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider a very simple example:
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
( [email protected]) October 9, 2019 8 / 15
Control of a Simple Nonlinear Plant
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
= −axp + (ασ(wxp)− ασ(wxp) + ασ(wxp)− ασ(wxp)) + r
= −axp + αε− ασ(wxp) + r
Reference Model:xm = −axm + r
Error model:e = −ae− ασ(wxp) + αε
( [email protected]) October 9, 2019 9 / 15
Control of a Simple Nonlinear Plant
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
= −axp + (ασ(wxp)− ασ(wxp) + ασ(wxp)− ασ(wxp)) + r
= −axp + αε− ασ(wxp) + r
Reference Model:xm = −axm + r
Error model:e = −ae− ασ(wxp) + αε
( [email protected]) October 9, 2019 9 / 15
Control of a Simple Nonlinear Plant
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
= −axp + (ασ(wxp)− ασ(wxp) + ασ(wxp)− ασ(wxp)) + r
= −axp + αε− ασ(wxp) + r
Reference Model:xm = −axm + r
Error model:e = −ae− ασ(wxp) + αε
( [email protected]) October 9, 2019 9 / 15
Control of a Simple Nonlinear Plant
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
= −axp + (ασ(wxp)− ασ(wxp) + ασ(wxp)− ασ(wxp)) + r
= −axp + αε− ασ(wxp) + r
Reference Model:xm = −axm + r
Error model:e = −ae− ασ(wxp) + αε
( [email protected]) October 9, 2019 9 / 15
Control of a Simple Nonlinear Plant
xp = −axp + ασ(wxp) + u
u(t) = −ασ(wxp) + r
Defineε = σ(wxp)− σ(wxp) bounded
Closed-loop Equations:
xp = −axp + ασ(wxp)− ασ(wxp) + r
= −axp + (ασ(wxp)− ασ(wxp) + ασ(wxp)− ασ(wxp)) + r
= −axp + αε− ασ(wxp) + r
Reference Model:xm = −axm + r
Error model:e = −ae− ασ(wxp) + αε
( [email protected]) October 9, 2019 9 / 15
Stability
Error model:
e = −ae− ασ(wxp) + αε
˙α = −eσ(wxp))
V = 1/2(e2 + α2
)V ≤ −ae2 + αeε
Stability result:Suppose r is persistently exciting with the property
1
T
∫ t1+T
t1
|r(τ)dτ ≥ αε
a+ δ
then e(t) and α(t) are bounded for all initial conditions e(t0) and α(t0).
( [email protected]) October 9, 2019 10 / 15
Stability
Error model:
e = −ae− ασ(wxp) + αε
˙α = −eσ(wxp))
V = 1/2(e2 + α2
)V ≤ −ae2 + αeε
Stability result:Suppose r is persistently exciting with the property
1
T
∫ t1+T
t1
|r(τ)dτ ≥ αε
a+ δ
then e(t) and α(t) are bounded for all initial conditions e(t0) and α(t0).( [email protected]) October 9, 2019 10 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider the plant:
xp = −axp +
N∑j=1
αjσ(wjxp)
+ u
u(t) = −
N∑j=1
αjσ(wjxp)
+ r
ε =
N∑j=1
σ(wjxp)− σ(wjxp)
bounded
( [email protected]) October 9, 2019 11 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider the plant:
xp = −axp +
N∑j=1
αjσ(wjxp)
+ u
u(t) = −
N∑j=1
αjσ(wjxp)
+ r
ε =
N∑j=1
σ(wjxp)− σ(wjxp)
bounded
( [email protected]) October 9, 2019 11 / 15
Control of a Nonlinear Dynamic System
xp = −axp + f(xp) + u
f(xp) ≈
N∑j=1
αjσ(wjxp)
Consider the plant:
xp = −axp +
N∑j=1
αjσ(wjxp)
+ u
u(t) = −
N∑j=1
αjσ(wjxp)
+ r
ε =
N∑j=1
σ(wjxp)− σ(wjxp)
bounded
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contd.
xp = −axp +
N∑j=1
αjσ(wjxp)
+ u
u(t) = −
N∑j=1
αjσ(wjxp)
+ r
ε =
N∑j=1
σ(wjxp)− σ(wjxp)
bounded
xp = −axp +
N∑j=1
[αjσ(wjxp)− αjσ(wjxp)] + r
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contd.
xp = −axp +
N∑j=1
[αjσ(wjxp)− αjσ(wjxp)] + r
= −axp +
N∑j=1
αjε− αjσ(wjxp) + r
Error model:
e = −ae−N∑j=1
αjε− αjσ(wjxp)
˙αj = eσ(wjxp))
( [email protected]) October 9, 2019 13 / 15
Stability Analysis
Error model:
e = −ae−N∑j=1
αjε− αjσ(wjxp)
˙αj = eσ(wjxp))
V = 1/2
e2 +
N∑j=1
α2j
V ≤ −ae2 + e
N∑j=1
αjε
= −ae2 + ed0; d0 : bounded
( [email protected]) October 9, 2019 14 / 15
Stability Analysis
Error model:
e = −ae−N∑j=1
αjε− αjσ(wjxp)
˙αj = eσ(wjxp))
V = 1/2
e2 +
N∑j=1
α2j
V ≤ −ae2 + e
N∑j=1
αjε
= −ae2 + ed0; d0 : bounded
( [email protected]) October 9, 2019 14 / 15
Stability Result
Error model:
e = −ae−N∑j=1
αjε− αjσ(wjxp)
˙αj = eσ(wjxp))
Suppose r is persistently exciting with the property
1
T
∫ t1+T
t1
|r(τ)dτ ≥ d0a
+ δ
then e(t) and α(t) are bounded for all initial conditions e(t0) and α(t0).
( [email protected]) October 9, 2019 15 / 15