Lecture 11: Model-Reference Adaptive Systems - umu. · PDF fileLecture 11: Model-Reference...

Lecture 11: Model-Reference Adaptive Systems

Given: y(t) = Gθ(p)u(t), ym(t) = Gm(p)uc(t),

Find: u(t) = −Sθ̂(p)

Rθ̂(p)

y(t) +Tθ̂(p)

Rθ̂(p)

uc(t),d

dtθ̂ = . . .

c©Leonid Freidovich. May 14, 2010. Elements of Iterative Learning and Adaptive Control: Lecture 11 – p. 1/22

MIT rule (Example 5.1)

Consider a stable single input single output (SISO) system

y(t) = k · G(p)(

u(t))

where• y(t) is the system output,

• G(s) is a known stable transfer function,

• u(t) is a control input,• k is a constant unknown gain.



Consider a stable single input single output (SISO) system

y(t) = k · G(p)(

u(t))

where• y(t) is the system output,

• G(s) is a known stable transfer function,

• u(t) is a control input,• k is a constant unknown gain.

The problem is find a controller u(t) = T (p)R(p)

uc(t) to follow

ym(t) = Gm(p)(

uc(t))

= k0 · G(p)(

uc(t))

,

where k0 is a given constant gain.



If k were known we can solve the problem

y(t) = k · G(p)(

u(t))

−→ ym(t) = k0 · G(p)(

uc(t))

using the simple proportional controller

u(t) = θ · uc(t)




y(t) = k · G(p)(

u(t))

−→ ym(t) = k0 · G(p)(

uc(t))


u(t) = θ · uc(t)

This is, indeed, works, because if θ is chosen as

θ =k0

k




y(t) = k · G(p)(

u(t))

−→ ym(t) = k0 · G(p)(

uc(t))


u(t) = θ · uc(t)

This is, indeed, works, because if θ is chosen as

θ =k0

k

then

y(t) = k · G(p)(

θ · uc(t))

= k · G(p)

(k0

k· uc(t)

)

= ym(t)



Question: How to update (adapt) the value of θ?




Let us consider the error between the real and simulated outputs

e(t, θ) = y(t)−ym(t) = k·G(p)(

θ(t) · uc(t))

−k0·G(p)(

uc(t))





e(t, θ) = y(t)−ym(t) = k·G(p)(

θ(t) · uc(t))

−k0·G(p)(

uc(t))

Let us update θ(t) so that e(t, θ) is getting smaller.





e(t, θ) = y(t)−ym(t) = k·G(p)(

θ(t) · uc(t))

−k0·G(p)(

uc(t))


Consider the loss function that measures the size of e(t, θ)

J (t, θ) = |e(t, θ)|2

Its time derivative is given by (chain rule)

d

dtJ =

[∂

∂tJ

]

+

[∂

∂θJ

]

·d

dtθ





e(t, θ) = y(t)−ym(t) = k·G(p)(

θ(t) · uc(t))

−k0·G(p)(

uc(t))


Consider the function that measures the size of e(t, θ)

J (t, θ) = |e(t, θ)|2

Its time derivative should be made negative:

d

dtJ = · · · +

[∂

∂θJ

]

·d

dtθ ⇒

d

dtθ = −γ ·

[∂

∂θJ

]





e(t, θ) = y(t)−ym(t) = k·G(p)(

θ(t) · uc(t))

−k0·G(p)(

uc(t))


Consider the function that measures the size of e(t, θ)

J (t, θ) = |e(t, θ)|2

Its time derivative should be made negative:

d

dtJ = · · · +

[

2 e∂

∂θe

]

·d

dtθ ⇒

d

dtθ = −γ ·

[

2 e∂

∂θe

]



Computing the partial derivative of e wrt θ we have

∂

∂θe =

∂

∂θ

[

k · G(p)(

θ(t) · uc(t))]

= k · G(p)(

uc(t))

=k

k0· k0 · G(p)

(

uc(t))

=k

k0ym(t)




∂

∂θe =

∂

∂θ

[

k · G(p)(

θ(t) · uc(t))]

= k · G(p)(

uc(t))

=k

k0· k0 · G(p)

(

uc(t))

=k

k0ym(t)

Then the update law for θ becomes

d

dtθ = −γ ·

[

2 e∂

∂θe

]

= −γn · ym(t) · e(t, θ)

where γn > 0 is arbitrary since γn = γk

k0with arbitrary γ > 0.




∂

∂θe =

∂

∂θ

[

k · G(p)(

θ(t) · uc(t))]

= k · G(p)(

uc(t))

=k

k0· k0 · G(p)

(

uc(t))

=k

k0ym(t)

Then the update law for θ becomes

d

dtθ = −γ ·

[

2 e∂

∂θe

]

= −γn · ym(t) · e(t, θ)

where γn > 0 is arbitrary since γn = γk

k0with arbitrary γ > 0.

Remark: J (·) = |e(·)| ⇒ ddtθ = −γn · ym(t) · sign[e(t, θ)].



Suppose that

G(s) =1

s + 1

andk = 1, k0 = 2



Suppose that

G(s) =1

s + 1

andk = 1, k0 = 2

The update law for θ

d

dtθ = −γ ·

[

2 e∂

∂θe

]

= −γn · ym(t) · e(t)

is simulated for various values of

γ = 0.5, 2.5, 4.5

with uc(t) = sin t.


0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

time (sec)

θ

The behavior of θ for various values of γ.


0 5 10 15 20 25−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

time (sec)

Out

put,

y

The behavior of y(t) for various values of γ.



Suppose that the system dynamics are

d

dty = −a y + b u, y =

b

p + au




d

dty = −a y + b u, y =

b

p + au

While the desired dynamics for the closed loop system is

d

dtym = −am ym + bm uc, ym =

bm

p + am

uc




d

dty = −a y + b u, y =

b

p + au


d


bm

p + am

uc

The proportional controller that solves the problem is given by

u(t) =T (p)

R(p)uc(t) −

S(p)

R(p)y(t) = θ1 uc(t) − θ2 y(t)




d

dty = −a y + b u, y =

b

p + au


d


bm

p + am

uc

The proportional controller that solves the problem is given by

u(t) = θ1 uc(t) − θ2 y(t)

where the gains to ensure the desired system response are

θ1 = θ01 =

bm

b, θ2 = θ0

2 =am − a

b



Introduce the error signal

e(t) = y(t) − ym(t) =b θ1

p + a + b θ2uc(t) −

bm

p + am

uc




e(t) = y(t) − ym(t) =b θ1

p + a + b θ2uc(t) −

bm

p + am

uc

Computing partial derivatives of e(·) w.r.t. θ1, θ2, we have

∂e

∂θ1=

b

p + a + b θ2uc(t)

∂e

∂θ2= −

b θ1 · b

(p + a + b θ2)2uc(t) =

−b

p + a + b θ2y(t)




e(t) = y(t) − ym(t) =b θ1

p + a + b θ2uc(t) −

bm

p + am

uc


∂e

∂θ1=

b

p + a + b θ2uc(t)

∂e

∂θ2= −

b θ1 · b

(p + a + b θ2)2uc(t) =

−b

p + a + b θ2y(t)

Such formulas cannot be used for updating θ1 and θ2 values.

Indeed, constants a and b are not known!




e(t) = y(t) − ym(t) =b θ1

p + a + b θ2uc(t) −

bm

p + am

uc


∂e

∂θ1=

b

p + a + b θ2uc(t)

∂e

∂θ2= −

b θ1 · b

(p + a + b θ2)2uc(t) =

−b

p + a + b θ2y(t)

Not much can be done, we will assume that we can initialize θ2around its nominal value

θ2 ≈ θ02 =

am − a

b




e(t) = y(t) − ym(t) =b θ1

p + a + b θ2uc(t) −

bm

p + am

uc


∂e

∂θ1=

b

p + a + b θ2uc(t)

∂e

∂θ2= −

b θ1 · b

(p + a + b θ2)2uc(t) =

−b

p + a + b θ2y(t)

Not much can be done, we will assume that we can initialize θ2around its nominal value

θ2 ≈ θ02 =

am − a

b⇒ (a + b θ2) ≈ am



This results in the following relations

d

dtθ1 = −γ · e(t) ·

∂e(t)

∂θ1≈ −γ · e(t) ·

(b

s + am

uc(t)

)

= −γn · e(t) ·

(am

s + am

uc(t)

)




d

dtθ1 ≈ −γn · e(t) ·

(am

s + am

uc(t)

)

d

dtθ2 = −γ · e(t) ·

∂e(t)

∂θ2≈ −γ · e(t) ·

(−b

s + am

y(t)

)

= γn · e(t) ·

(am

s + am

y(t)

)

where

γn = γb

am

and should be positive!




d

dtθ1 ≈ −γn · e(t) ·

(am

s + am

uc(t)

)

d

dtθ2 = −γ · e(t) ·

∂e(t)

∂θ2≈ −γ · e(t) ·

(−b

s + am

y(t)

)

= γn · e(t) ·

(am

s + am

y(t)

)

where

γn = γb

am

and should be positive!

To implement this algorithm we need to know the sign of b!



Consider the static system with unknown gain k

y(t) = k · u(t), G(s) ≡ 1

and the problem of amplifying uc(t) so that we match

ym(t) = k0 · uc(t)




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)

With u(t) = θ uc(t) introduce the error

e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)


e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.

d

dtθ(t) = −γ · e(t) ·

∂e(t)

∂θ= −γ ·k

(θ(t) − θ0

)uc(t) ·k uc(t)




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)


e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.

d

dtθ(t) = −γ · k2 · (uc(t))

2 ·(θ(t) − θ0

)




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)


e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.

d

dt

(θ(t) − θ0

)= −γn · k · (uc(t))

2 ·(θ(t) − θ0

)




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)


e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.

(θ(t) − θ0

)= exp

{

−γn · k ·

∫ t

0(uc(t))

2 dτ

}

·(θ(0) − θ0

)




y(t) = k · u(t), G(s) ≡ 1


ym(t) = k0 · uc(t)


e(t) = y(t)−ym(t) = k·(

θ uc(t))

−k0·uc(t) = k(θ − θ0

)uc(t)

with θ0 = k0/k.

e(t) = k ·exp

{

−γn · k ·

∫ t

0(uc(t))

2 dτ

}

·(θ(0) − θ0

)

︸︷︷︸

θ(t)−θ0

·uc(t)


MIT rule (Example 5.3), cont’d

For the system and model given by

y(t) = k · u(t), ym(t) = k0 · uc(t)

we define e(t) = y(t) − ym(t) and take

u(t) = θ(t)uc(t),d

dtθ(t) = −γn ·k ·(uc(t))

2 ·(θ(t) − θ0

)




y(t) = k · u(t), ym(t) = k0 · uc(t)


u(t) = θ(t)uc(t),d

dtθ(t) = −γn · uc(t) · e(t)




y(t) = k · u(t), ym(t) = k0 · uc(t)


u(t) = θ(t)uc(t),d


As the result we obtain

θ(t) = θ0 + σ(t), e(t) = k · σ(t) · uc(t)

σ(t) = exp{

−γn · k · It

}(

θ(0) − θ0)

, It =

∫ t

0(uc(t))

2 dτ




y(t) = k · u(t), ym(t) = k0 · uc(t)


u(t) = θ(t)uc(t),d



θ(t) = θ0 + σ(t), e(t) = k · σ(t) · uc(t)

σ(t) = exp{

−γn · k · It

}(

θ(0) − θ0)

, It =

∫ t

0(uc(t))

2 dτ

If θ(0) 6= θ0, for e(t) → 0 as t → ∞ we need:

exp{

−γn · k · It

}

→ 0 or uc(t) → 0




y(t) = k · u(t), ym(t) = k0 · uc(t)


u(t) = θ(t)uc(t),d



θ(t) = θ0 + σ(t), e(t) = k · σ(t) · uc(t)

σ(t) = exp{

−γn · k · It

}(

θ(0) − θ0)

, It =

∫ t

0(uc(t))

2 dτ

If θ(0) 6= θ0, for e(t) → 0 as t → ∞ we need:

It =

∫ t

0(uc(t))

2 dτ → ∞ or uc(t) → 0

.c©Leonid Freidovich. May 14, 2010. Elements of Iterative Learning and Adaptive Control: Lecture 11 – p. 15/22

Tuning the Gain for MIT rule

Consider again the problem with scaling the reference

y = k · G(p)u, ym = k0 · G(p)uc, u = θ uc

where θ(t) is determined by MIT rule:

d

dtθ = −γ · ym · e, e = y − ym






d

dtθ = −γ · ym · e, e = y − ym

The equation for θ can be re-written as follows

d

dtθ = −γ · ym ·

(

y − ym

)

= −γ · ym ·(

k · G(p) θ uc − ym

)






d

dtθ = −γ · ym · e, e = y − ym


d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)






d

dtθ = −γ · ym · e, e = y − ym


d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)

Here• the functions ym(t) and uc(t) are known!• the range of the constant gain γ, for which the nominal value

θ0 (its stationary point) is stable, should be determined.


Tuning the Gain for MIT rule (cont’d)

d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)

In general the analysis of stability is difficult!



d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)


Consider the case when ym(t) ≡ yom, uc(t) = uo

c , then ODE

d

dtθ(t) + γ · k · yo

m · u0c · G(p)

[

θ(t)]

= γ(yom)2

is linear and time-invariant!



d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)



c , then ODE

d


m · u0c · G(p)

[

θ(t)]

= γ(yom)2


Stability is determined by the roots of the algebraic equation

s + µ · G(s) = 0, µ = γ · k · yom · u0

c



d

dtθ(t) + γ · k · ym(t) · G(p)

[

θ(t)uc(t)]

= γy2m(t)



c , then ODE

d


m · u0c · G(p)

[

θ(t)]

= γ(yom)2


Stability is determined by the roots of the algebraic equation

s + µ · G(s) = 0, µ = γ · k · yom · u0

c

Root locus analysis (variation of zeros with µ) can be used.A reasonable value for γ can be obtained from this analysis andmight work for slowly varying signals.


Example 5.4

Let, as in Example 5.1

G(s) =1

s + 1, k = 1, k0 = 2


Example 5.4


G(s) =1

s + 1, k = 1, k0 = 2

The characteristic equation

s + µ1

s + 1= 0 ⇔ s2 + s + µ = 0

has stable zeros if and only if

µ = γ · k · yom · u0

c = γ ·(

k0 G(0)u0c

)

· u0c = 2 γ (u0

c)2 > 0


Example 5.4


G(s) =1

s + 1, k = 1, k0 = 2


s + µ1

s + 1= 0 ⇔ s2 + s + µ = 0


µ = γ · k · yom · u0

c = γ ·(

k0 G(0)u0c

)

· u0c = 2 γ (u0

c)2 > 0

So, γ > 0 will work.

Note, however, that the transient depends on u0c !

The relative damping is ζ = 12√

µ= 1

2√

2 γ |u0c |

.

µ ≈ 1 is reasonable ⇐ take γ ≈ 0.5 for u0c ≈ 1 in average.


Example 5.5

Consider the stable system with relative degree 2:

G(s) =1

s2 + a1 s + a2, a1 > 0, a2 > 0


Example 5.5


G(s) =1

s2 + a1 s + a2, a1 > 0, a2 > 0


s + µ1

s2 + a1 s + a2= 0 ⇔ s3 + a1 s

2 + a2 s + µ = 0


µ > 0 and a1 a2 > µ = γ · k · yom · u0

c


Example 5.5


G(s) =1

s2 + a1 s + a2, a1 > 0, a2 > 0


s + µ1

s2 + a1 s + a2= 0 ⇔ s3 + a1 s

2 + a2 s + µ = 0


µ > 0 and a1 a2 > µ = γ · k · yom · u0

c

Conclusion: with any choice of γ > 0, stability is lost forsufficiently large magnitudes of the reference signal u0

c !


Normalized MIT rule

d

dtθ = −γ · e(t, θ) ·

φ

α + φT φ, φ =

∂

∂θe(t, θ), α > 0


Next Lecture / Assignments:

Next meeting (May 24, 13:00-15:00, in A208Tekn):Lyapunov-based design.

Homework problem: The process and model are described by

G(s) =1

s, Gm(s) =

2

s + 2

For the control law

u(t) = θ1 uc(t) − θ2 y(t)

design an MIT-like adaptation law such that

θi ≈ −

(

γ1 + γ21

p

) [

e∂

∂θie

]

, i ∈ {1, 2} .

Simulate the MRAS with various gains.

Consider γ1,2 ∈ {0, 1, 5} and a unit square wave for uc(t).Compare performance for different combinations.


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