Model Reference Adaptation Systems (MRAS)

Contents
1.1 Model Reference Adaptation Systems (MRAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.1 MIT Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.2 Determination of Adaptation Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.1.3 Normalized MIT Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.1.4 Design of MRAS Using Lyapunov Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.2 MATLAB Codes and Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.4 Contacts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1

1.1. MODEL REFERENCE ADAPTATION SYSTEMS (MRAS) Adaptive Control
1.1 Model Reference Adaptation Systems (MRAS)
MRAS is an important adaptive controller. It may be regarded as an adaptive servo system in which the desired performance is
expressed in terms of a reference model, which gives the desired response to a command signal. This is a convenient way to give
specifications for a servo problem. A block diagram of the system is shown in Figure 1.1. The system has an ordinary feedback
loop composed of the process and the controller in addition to another feedback loop that changes the controller parameters.
The parameters are changed on the basis of feedback from the error, which is the difference between the output of the system
and the output of the reference model. The ordinary feedback loop is called the inner loop, and the parameter adjustment loop
is called the outer loop. The mechanism for adjusting the parameters in a model-reference adaptive system can be obtained in
two ways: by using a gradient method or by applying stability theory.
Figure 1.1: Block diagram of a model-reference adaptive system
1.1.1 MIT Rule
The MIT rule is the original approach to model-reference adaptive control. The name is derived from the fact that it was
developed at the Instrumentation Laboratory (now the Draper Laboratory) at MIT.
To present the MIT rule, we will consider a closed-loop system in which the controller has one adjustable parameter θ.
The desired closed-loop response is specified by a model whose output is ym. Let e be the error between the output y of
the closed-loop system and the output ym of the model. One possibility is to adjust parameters in such a way that the loss
function J(θ) = 1
2e2 is minimized.
Procedure
Process : G(s) =
y
u
(1.1)
Model : Gm(s) =
ym
uc
(1.2)
Control law : u(t) = f(uc, y) (1.3)
Get closed loop from [1.1] & [1.3] :
y
uc
(1.4)
Error : e = y − ym (1.5)
∂e
∂θ
=
∂y
∂θ
(1.6)
MIT Rule :
dθ
dt
= −γe
∂e
∂θ
(1.7)
Mohamed Mohamed El-Sayed Atyya Page 2 of 40

Examples
1. Gain Adjustment
Gp(s) = θG(s) = θ
2
s2 + 2s + 4
Gm(s) = θoG(s) = θo
2
s2 + 2s + 4
; θo = 2
e = y − ym = θG(s)uc − θoG(s)uc
dθ
dt
= −γe
∂e
∂θ
= −γG(s)uce = −γ
ym
θo
e
Figure 1.2: Gain adjustment block diagram
At γ = 0.5

At γ = 0.7
At γ = 1

At γ = 1.2
At γ = 1.5

2. First Order System Adjustment
dy
dt
= −ay + bu
G(s) =
Y
U
=
b
s + a
dym
dt
= −amy + bmu
G(s) =
Ym
U
=
bm
s + am
Use the control law : u(t) = touc(t) − soy(t)
U = toUc − soY = Y
s + a
b
⇒
Y
Uc
=
to
s+a
b + so
=
boto
s + a + bso
bm = bto
am = a + bso
e = Y − Ym =
bto
s + a + bso
Uc =
bm
s + am
Uc
∂e
∂to
=
b
s + a + bso
Uc =
b
s + am
Uc
∂e
∂so
=
−b2to
(s + a + bso)2
Uc =
−b
s + a + bso
Y =
−b
s + am
Y
dθ
dt
= −γ
∂e
∂θ
e
dto
dt
= −γ1
b
s + am
Uc e
dso
dt
= γ2
b
s + am
Y e
Let : a = 1, b = 2, am = 8, bm = 8
Figure 1.3: First order system adjustment block diagram

At γ1 = 2, γ2 = 0.1
At γ1 = 2, γ2 = 1

At γ10 = 2, γ2 = 0.5
At γ1 = 1, γ2 = 0.05

3. Second Order System Adjustment
G(s) =
y
u
=
s + b
s2 + a1s + a0
Gm(s) =
ym
uc
=
s + bm
s2 + a1ms + a0m
Let the control law : u(t) = t0uc(t) − s0y(t)
y
s2 + a1s + a0
s + b
= t0uc − s0y
⇒
y
uc
=
t0s + bt0
s2 + (a1 + s0)s + a0 + bs0
e = y − ym
∂e
∂t0
=
s + b
s2 + (a1 + s0)s + a0 + bs0
uc ≈ Gm(s)uc
∂e
∂s0
=
−t0(s + b)(s + b)
[s2 + (a1 + s0)s + a0 + bs0]2
uc
=
−(s + b)
s2 + (a1 + s0)s + a0 + bs0
y =
−y2
t0uc
dθ
dt
= −γ
∂e
∂θ
e
dt0
dt
= −γ1Gmuce
ds0
dt
= γ2
y2
t0uc
e = γGmye
Figure 1.4: Second order system adjustment block diagram

Let : G(s) = s+5
s2+4s+3 (stable), Gm(s) = s+10
s2+8s+10
At γ1 = γ2 = 10
At γ1 = 10, γ2 = 5

At γ1 = 1.2, γ2 = 0.5
At γ1 = 1.2, γ2 = 0.5

Let : G(s) = s+5
s2+2s−3 (unstable), Gm(s) = s+10
s2+8s+10
At γ1 = γ2 = 20
At γ1 = 20, γ2 = 10

At γ1 = 20, γ2 = 40
4. Second Order System Adjustment with First Order Controller
G(s) =
y
u
=
s + b
s2 + a1s + a0
=
N
D
Gm(s) =
ym
uc
=
s + bm
s3 + a2ms2 + a1ms + a0m
Let the control law : u(t) =
t0
1 + r1s
uc(t) −
s0
1 + r1s
y(t)
y =
t0G(s)
1 + r1s
uc −
s0G(s)
1 + r1s
y
y
uc
=
t0N
D(1 + r1s) + s0N
e = y − ym
∂e
∂t0
=
N
D(1 + r1s) + s0N
uc ≈ Gmuc
∂e
∂s0
=
−N2
[D(1 + r1s) + s0N]2
uc ≈ −Gm
y
uc
uc = −Gmy
∂e
∂r1
=
−NDs
[D(1 + r1s) + s0N]2
uc ≈ −Gm
Ds
D(1 + r1s) + s0N
uc
≈ −Gmuc
∂θ
∂t
= −γ
∂e
∂θ
e
∂t0
∂t
= −γ1[Gmuc]e ⇒ t0 = −
γ1
s
[Gmuc]e
∂s0
∂t
= γ2[Gmy]e ⇒ s0 =
γ2
s
[Gmy]e

∂r1
∂t
= γ3[Gmuc]e ⇒ r1 =
γ3
s
[Gmuc]e
Let:
G(s) =
s + 2
s2 + s + 6
Gm(s) =
s + 24
s3 + 9s2 + 26s + 24
Figure 1.5: Second order system adjustment with ﬁrst order controller block diagram
For Step Input:
γ1 = 50, γ2 = 25, γ3 = −100

For square Input:
γ1 = 7.5, γ2 = 3.75, γ3 = −15

For Sinusoidal Input:
γ1 = 7.5, γ2 = 3.75, γ3 = −15

1.1.2 Determination of Adaptation Gain
• Consider the plant transfer function G(s).
• Multiply the denominator of G(s) by s and add the term µ to get the characteristic equation
sG(s) + µ = 0
where, µ = γymuck.
• Find µ that places all the roots in left half of S − plane.
• If ymuck = constant ⇒ γ =
µ
ymuck
Examples
1. First order system
Let :
B = 1, A = s + 1, µ = 1

2. Second order system
Let :
B = 1, A = s2
+ s + 1, µ = 0.4

1.1.3 Normalized MIT Rule
Procedure
Process : G(s) =
y
u
(1.8)
Model : Gm(s) =
ym
uc
(1.9)
Control law : u(t) = f(uc, y) (1.10)
Get closed loop from [1.8] & [1.10] :
y
uc
(1.11)
Error : e = y − ym (1.12)
∂e
∂θ
=
∂y
∂θ
= −ϕ (1.13)
Normalized MIT Rule :
dθ
dt
= γ
ϕe
α + ϕT ϕ
(1.14)
α > 0 (1.15)
Examples
1. Gain Adjustment
Gp(s) = θG(s) = θ
2
s2 + 2s + 4
Gm(s) = θoG(s) = θo
2
s2 + 2s + 4
; θo = 2
e = y − ym = θG(s)uc − θoG(s)uc
ϕ = −
∂e
∂θ
= −G(s)uc = −
ym
θ0
dθ
dt
= γ
ϕe
α + ϕT ϕ
= −γ
yme/θ0
α + (ym/θ0)2
= −γ
yme
α + y2
m

At
γ = 1.2, α = 0.1

dy
dt
= −ay + bu
G(s) =
Y
U
=
b
s + a
dym
dt
= −amy + bmu
G(s) =
Ym
U
=
bm
s + am
Use the control law : u(t) = touc(t) − soy(t)
U = toUc − soY = Y
s + a
b
⇒
Y
Uc
=
to
s+a
b + so
=
boto
s + a + bso
bm = bto
am = a + bso
e = Y − Ym =
bto
s + a + bso
Uc =
bm
s + am
Uc
∂e
∂to
=
b
s + a + bso
Uc =
b
s + am
Uc
≈ Gm(s)Uc = ym
∂e
∂so
=
−b2to
(s + a + bso)2
Uc =
−b
s + a + bso
Y =
−b
s + am
Y
≈ −Gm(s)Y
dθ
dt
= γ
ϕe
α + ϕT ϕ
dto
dt
= −γ1
yme
α1 + y2
m
dso
dt
= γ2
Gm(s)ye
α2 + (Gm(s)y)2
Let : a = 1, b = 2, am = 8, bm = 8

At
γ1 = 5, γ2 = 10, α1 = 0.1, α2 = 5

G(s) =
y
u
=
s + b
s2 + a1s + a0
Gm(s) =
ym
uc
=
s + bm
s2 + a1ms + a0m
Let the control law : u(t) = t0uc(t) − s0y(t)
y
s2 + a1s + a0
s + b
= t0uc − s0y
⇒
y
uc
=
t0s + bt0
s2 + (a1 + s0)s + a0 + bs0
e = y − ym
∂e
∂t0
=
s + b
s2 + (a1 + s0)s + a0 + bs0
uc ≈ Gm(s)uc = ym
∂e
∂s0
=
−t0(s + b)(s + b)
[s2 + (a1 + s0)s + a0 + bs0]2
uc
=
−(s + b)
s2 + (a1 + s0)s + a0 + bs0
y =
−y2
t0uc
dθ
dt
= γ
ϕe
α + ϕT ϕ
dt0
dt
= −γ1
yme
α1 + y2
m
ds0
dt
= γ2
y2uc
α2u2
c + y4
Figure 1.9: Second order system adjustment block diagram

Let :
G(s) =
s + 5
s2 + 4s + 3
(stable), Gm(s) =
s + 10
s2 + 8s + 10
, γ1 = 15, γ2 = 0.1, α1 = 15,
α2 = 20

Let :
G(s) =
s + 5
s2 + 2s − 3
(unstable), Gm(s) =
s + 10
s2 + 8s + 10
, γ1 = 400, γ2 = 1,
α1 = α2 = 0.001

4. Second Order System Adjustment with First Order Controller
G(s) =
y
u
=
s + b
s2 + a1s + a0
=
N
D
Gm(s) =
ym
uc
=
s + bm
s3 + a2ms2 + a1ms + a0m
Let the control law : u(t) =
t0
1 + r1s
uc(t) −
s0
1 + r1s
y(t)
y =
t0G(s)
1 + r1s
uc −
s0G(s)
1 + r1s
y
y
uc
=
t0N
D(1 + r1s) + s0N
e = y − ym
∂e
∂t0
=
N
D(1 + r1s) + s0N
uc ≈ Gmuc = ym
∂e
∂s0
=
−N2
[D(1 + r1s) + s0N]2
uc ≈ −Gm
y
uc
uc = −Gmy
∂e
∂r1
=
−NDs
[D(1 + r1s) + s0N]2
uc ≈ −Gm
Ds
D(1 + r1s) + s0N
uc
≈ −Gmuc = −ym
dθ
dt
= γ
ϕe
α + ϕT ϕ
∂t0
∂t
= −γ1
yme
α1 + y2
m
∂s0
∂t
= γ2
Gmye
α2 + (Gmy)2
∂r1
∂t
= γ3
yme
α3 + y2
m
Let:
G(s) =
s + 2
s2 + s + 6
, Gm(s) =
s + 24
s3 + 9s2 + 26s + 24
,
γ1 = 400, γ2 = 33.3333, γ3 = −50, α1 = 100, α2 = 100, α3 = 400

1.1.4 Design of MRAS Using Lyapunov Theory
We will now show how Lyapunovs stability theory can be used to construct algorithms for adjusting parameters in adaptive
systems. To do this, we first derive a differential equation for the error, e = y − ym. This differential equation contains the
adjustable parameters. We then attempt to find a Lyapunov function and an adaptation mechanism such that the error will
go to zero.When using the Lyaponov theory for adaptive systems, we find that dV/dt is usually only negative semi-definite.
The procedure is to determine the error equation and a Lyapunov function with a bounded second derivative.
General Case
Given a continuous time system and the target dynamics
˙x = Ax + Bu, ˙xm = Amxm + Bmuc
Consider the controller and the error signals
u(t) = Muc(t) − Lx(t), e(t) = x(t) − xm(t)
If the model-matching problem is solvable, then the error dynamics is
de
dt
= Ax + Bu − Amxm − Bmuc
= Ame + (A − Am − BL) x + (BM − Bm) uc
= Ame + Ψ(x, uc) • θ − θ0
Consider the following Lyapunov function candidate
V =
1
2
eT
Pe +
1
γ
θ − θ0 T
θ − θ0
The time-derivative of V is
˙V =
1
2
eT
PAm + AT
mP e + θ − θ0 T
ΨT
Pe +
1
γ
θ − θ0 T ˙θ
If we solve the Lyapunov equation for P = PT > 0
PAm + AT
mP = −Q, Q > 0
and choose the update law as
˙θ = −γΨT
Pe = −γΨT
(x, uc) • P • (x − xm)
then
˙V = −
1
2
eT
(t)Qe(t)
and we conclude that e(t) → 0

Examples
1. Gain Adjustment
Process : ˙x = −ax + bu
Model : ˙xm = −amxm + bmuc
a = am
Control law : u(t) = m1uc(t)
de
dt
= −ame + (−a + am)x + (bm1 − bm)uc = −ame + (bm1 − bm)uc
= −ame + Ψ(x, uc) • θ − θ0
Ψ =
uc
b
θ = m1
θ0
=
bm
b
˙θ = −γ
uc
b
Pe = −γuce
Let
a = am = 4, b = 2, bm = 4, γ = 3

Process : ˙x = −ax + bu
Model : ˙xm = −amxm + bmuc
Control law : u(t) = m1uc(t) − l1x(t)
de
dt
= −ame + (−a + am − bl1)x + (bm1 − bm)uc
= −ame + Ψ(x, uc) • θ − θ0
Ψ =
−x
b
uc
b
θ = [l1 m1]T
θ0
=
am − a
b
bm
b
˙l1
˙m1
=
γ1xe
−γ2uce
Let
a = 2, am = 8, b = 1, bm = 8, γ1 = 0.1, γ2 = 3

Process : G(s) =
s + 5
s2 + 4s + 3
Model : Gm(s) =
s + 10
s2 + 8s + 10
A =
0 1
−3 −4
B =
0
1
Am =
0 1
−10 −8
Bm =
0
1
Control law : u(t) = m1uc − l1 0 x(t)
de
dt
=
0 1
−10 −8
e +
0 0
7 − l1 0
x(t) +
0
m1 − 1
uc
Ψ = [−x uc]
θ = [l1 m1]T
˙l1
˙m1
=
γ1xe
−γ2uce
Figure 1.13: second order system adjustment block diagram
At
γ1 = 0.01, γ2 = 0.4

4. Second Order System Adjustment (another solution)
Process : G(s) =
s + 5
s2 + 4s + 3
Model : Gm(s) =
s + 10
s2 + 8s + 10
A =
0 1
−3 −4
B =
0
1
Am =
0 1
−10 −8
Bm =
0
1
Control law : u(t) = m1uc − l1 l2 x(t)
de
dt
=
0 1
−10 −8
e +
0 0
7 − l1 4 − l2
x(t) +
0
m1 − 1
uc
Ψ = [−x − ˙x uc]
θ = [l1 l2 m1]T


˙l1
˙l2
˙m1

 =


γ1xe
γ2 ˙xe
−γ3uce


Figure 1.14: second order system adjustment block diagram
At
γ1 = 0.01, γ2 = 0.8, γ2 = 0.5

1.2. MATLAB CODES AND SIMULATION Adaptive Control
1.2 MATLAB Codes and Simulation
1 http://goo.gl/2nhYkk
2 http://goo.gl/RqIX4D
3 http://goo.gl/rOCcS6
4 http://goo.gl/Bx5Tnn
1 http://goo.gl/trRRzu
2 http://goo.gl/QrSHSW
1 http://goo.gl/hdrI90
2 http://goo.gl/DDXsR9
3 http://goo.gl/G67l2g
4 http://goo.gl/QJWWvw
1 http://goo.gl/YzgkEL
2 http://goo.gl/86ZvGi
3 http://goo.gl/cSxmF3
4 http://goo.gl/MW7zTw
1.3 References
1. Karl Johan Astrom, Adaptive Control, 2nd
Edition.
2. Leonid B. Freidovich, lecture 12.
1.4 Contacts
mohamed.atyya94@eng-st.cu.edu.eg

Model Reference Adaptation Systems (MRAS)

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