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Institutionen för
REGLERTEKNIK

FRT 050 Adaptive Control

Exam January 9, 2004, 8-13

General Instructions
This is an open book exam. You may use any book you want. However, no previous
exam sheets or solution manuals are allowed. The exam consists of 5 problems to
be solved. Your solutions and answers to the problems should be well motivated.
The credit for each problem is indicated in the problem. The total number of
credits is 25 points. Preliminary grade limits are:
Grade 3: 12 – 16 points
Grade 4: 17 – 21 points
Grade 5: 22 – 25 points

Results
The results of the exam will be posted at the latest January 16 on the note board
on the first floor of the M-building and they will also be available on the course
home page.

Do you accept publication of your grading result on our local web

page? (Godkänner du publicering av resultatet på vår lokala hemsida?)

1
1. Consider the process with the transfer function

25
G ( s) =
(s + 1)(s + 5)2
In an identification experiment, it is attempted to fit the input–output data
for the process to the model

b ( s) = b
G
s+a
Examine whether the parameters could be uniquely determined from the
experiment when the input to the process is
(
1 t≥0
a. u(t) = (2 p)
0 t<0
b. u(t) = sin ω t (2 p)

2. Consider the process

1 q
yk = uk + ek
q − au q − ae
where the noise sequence { ek} is independent white noise and uk is the
control signal.
a. Design a controller that minimizes E( y2k ). (3 p)
b. Assume that the parameter au is known and that a e is unknown. Derive a
regression model for identification of the parameter a e. (1 p)
c. Design a controller for the process when au is known and a e is slowly time-
varying with values in a large interval. Motivate your choice of controller
structure. (2 p)

3.
a. Consider the nonlinear model:

y(t) + a1 y(t − 1) = b1 u(t − 1) + b2 u(t − 1) y(t − 1)

Find a linear regression model for estimation of the parameters a1 , b1 and

b2 . (1 p)
b. It is desired to use this regression model to perform online identification of
the parameters. In addition, it is known that the parameters may be slowly
time-varying. Write down an appropriate algorithm for this estimation task,
and explain its operation. (2 p)
c. An indirect adaptive controller is to be constructed, using the estimation
method from above. Design a controll law, incorporating the reference signal
uc (t), such that the closed loop system has the transfer function:
b0
Y ( z) = Uc ( z)
z + a0
where 1 > a0 > −1. The controller may be nonlinear. (2 p)

2
4. A linear servo can be modeled by the continuous-time transfer function

b
Y ( s) = G ( s) U ( s) = U (s),
s( s + a)

where the input u(t) is the voltage to the servo and the output y(t) is the
servo angle. We would like to control the servo so that the transfer function
from reference signal uc (t) to angle becomes

ω2
Ym (s) = G m (s) Uc (s) = Uc (s). (1 )
s2 + 2ζ ω s + ω 2

As a and b may vary over time we will solve this problem with an indirect
self-tuning regulator.

a. Derive a continuous-time RST-controller

R ( s) U ( s) = − S ( s) Y ( s) + T ( s) U c ( s)

with integral action, that achieves perfect model following with respect to
(1). Introduce an observer polynomial Ao (s) = (s + ao )γ with a suitable
γ. (3 p)
b. Write down the continuous-time RLS estimator for a and b. In particular
give the regression vector ϕ (t) and a choice of the regression filter H f (s).
(2 p)
c. There may be constant load disturbances on the input of the servo. The-
refore we introduced integral action in the controller. How can you modify
the estimator in b. to reduce the effects of the disturbance on the estima-
tor? (1 p)

5.

a. Show that the system

ẋ = − x + u, x (0 ) = x 0 ,
y = x

with transfer function

1
G 1 ( s) =
( s + 1)

is strictly positive real (SPR) and that the storage function

1 T
V ( x) = x x
2
fulfills the passivity property
Z t Z t
T
V ( x(t)) = V ( x(0)) + y (τ )u(τ )dt − x T (τ ) x(τ )dτ
0 0

What is the interpretation of the three components of V ( x(T ))—i.e., the

three terms on the right-hand side? (3 p)

3
b. Show that the transfer function

1
G 2 ( s) =
( s + 1 )2

is not positive real. (1 p)