solns adaptive control by astrom

8/12/2019 Solns Adaptive Control by Astrom

1/46


2/46

Preface

This Solution Manual contains solutions to selected problems in the secondedition of Adaptive Control published by Addison-Wesley 1995, ISBN 0-201-55866-1.


3/46

PROBLEM SOLUTIONS

SOLUTIONS TO CHAPTER 1

1.5 Linearization of the valve shows thatv = 4v30u

The loop transfer function is then

G0( s) GPI( s)4v30

where GPIis the transfer function of a PI controller i.e.

GPI(s) = K

1 +

1sTi

The characteristic equation for the closed loop system is

sTi( s + 1)3+ K 4v

30(sTi +1) = 0

with K = 0.15 and Ti = 1 we get

( s + 1)

s(s + 1)2 +0.6v30 = 0

( s + 1)( s3 +2s2 + s +0.6v30) = 0

The root locus of this equation with respect to vo is sketched in Fig. 1.According to the Routh Hurwitz criterion the critical case is

0.6v30 = 2 v0 = 3

103 = 1.49

Since the plant G0 has unit static gain and the controller has integral

action the steady-state output is equal to v0 and the set point yr. Theclosed-loop system is stable for yr = uc = 0.3 and 1.1 but unstable foryr = uc = 5.1. Compare with Fig. 1.9.

1


4/46

2 Problem Solutions

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Real Axis

ImagAxis

Figure 1. Root locus in Problem 1.5.

1.6 Tune the controller using the Ziegler-Nichols closed-loop method. Thefrequencyu, where the process has 180 phase lag is first determined.The controller parameters are then given by Table 8.2 on page 382 where

Ku =1

G0(iu)

we have

G0( s) =e s/ q

1 + s/ q

arg G0( i) =

q arctan

q =

q G0(i) K Ti

0.5 1.0 0.45 1 5.242.0 0.45 1 2.624.1 0.45 1 1.3

A simulation of the system obtained when the controller is tuned for thesmallest flow q = 0.5 is shown Fig. 2. The Ziegler-Nichols method is notthe best tuning method in this case. In the Fig. 3 we show results for


5/46

Solutions to Chapter 1 3

0 10 20 30 40

0

1

2

Process output

0 10 20 30 40

0

1

2

Control signal

Figure 2. Simulation in Problem 1.6. Process output and control signalare shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). Thecontroller is designed for q = 0.5.

controller designed for q = 1 and in Fig. 4when the controller is designedforq = 2.

1.7 Introducing the feedbacku = k2y2

the system becomes

dx

dt =

1 0 0

0 3 0

0 0 1

x k2

0

2

1

1 0 1x +

1

0

0

u1y1 =

1 1 0xThe transfer function from u1to y1is

G( s) =

1 1 0

s +1 0 0

2k2 s +3 2k2k2 0 s +1 + k2

1

1

0

0

=s2 +( 4 k2) s + 3 + k2

( s + 1)( s+ 3)(s+ 1 + k2)


6/46

4 Problem Solutions

0 10 20 30 40

0

1

2

Process output

0 10 20 30 40

0

1

2

Control signal

Figure 3. Simulation in Problem 1.6. Process output and control signalare shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). Thecontroller is designed for q = 1.

The static gain is

G(0) =3 + k2

3(1 + k2)


7/46


0 10 20 30 40

0

1

2

Process output

0 10 20 30 40

0

1

2

Control signal

Figure 4. Simulation in Problem 1.6. Process output and control signalare shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). Thecontroller is designed for q = 2.


8/46

6 Problem Solutions


2.1 The function Vcan be written as

V(x1 xm) =

n

i,j =1 x

ixj ( aij + aji )/

2 +

n

i=1 b

ixi + c

Taking derivative with respect to xi we get

V

xi=

nj =1

( aij + aji )xj + bi

In vector notation this can be written as

gradxV(x) = (A + AT)x+ b

2.2 The model is

yt = Tt + e t =ut ut

1 b0

b1

+ etThe least squares estimate is given as the solution of the normal equation

= (T) 1TY =

u2t utut 1utut

1

u2t 1

1 utytut

1yt

(a) Input is a unit step

ut =

1 t 0

0 otherwise

Evaluating the sums we get

=

N N 1N 1 N 1

1

N1

yt

N2

yt

=

1 1 1

N

N 1

N1

yt

N2

yt

=

y1

1N 1

N2

yt y1

The estimation error is

=

e1

1N 1

N2

et e1


9/46


Hence

E( )( ) T = (T) 1 1 =

1 1

1 N

N 1

1 1

1 1

when N . Notice that the variance of the estimates do not go tozero as N . Consider, however, the estimate ofb0 + b1.

E( b0 + b1 b0 b1) = 1 1

1 1 1

N

N 1

1

1

= 1N 1

With a step input it is thus possible to determine the combinationb0 +b1consistently. The individual values ofb0and b1can, however,not be determined consistently.

(b) Input uis white noise with Eu2 = 1 and u is independent ofe.

Eu2t = 1 Eutut 1 = 0

cov( ) = 1 E(T) 1 =

N 00 N 1

1 =

1N

0

0 1N 1

In this case it is thus possible to determine both parameters consis-tently.

2.3 Data generating process:

y( t) = b0u( t) + b1u( t 1) + e(t) = T( t)0 + e( t)

whereT( t) = u( t), 0 = b0

ande(t) = b1u(t 1) + e(t)

Model:y(t) = bu( t)

ory( t) = bu(t) +( t) = T(t) + ( t)

where(t) = y( t) y( t)

The least squares estimate is given by

T( 0) = TEd Ed =

e(1)

...

e(N)


10/46

8 Problem Solutions

1N

T =

1N

N1

u2( t) Eu2 N

1

N

TEd =

1

N

N

1 u( t)e(t) =1

N

N

1 u(t) (b1u( t 1) + e( t)) b1E (u( t)u( t 1)) + E (u( t)e(t)) N

(a)

u(t) =

1 t 1

0 t < 1

E(u2) = 1 E (u( t)u( t 1)) = 1 Eu(t)e( t) = 0

Henceb = 0 + b1 = b0 + b1 N

i.e. bconverges to the stationary gain

(b)u(t) N(0,) Eu2 = 2 Eu(t)u( t 1) = 0 Eu(t)e( t) = 0

Henceb b0 N

2.6 The model is

yt = Tt + t =

yt

1 ut 1

Tt

ab

+

et + ce t 1

t

The least squares estimate is given by the solution to the normal equation

(2.5). The estimation error is = (T) 1T =

y2t

1

yt 1ut 1

yt

1ut

1

u2t

1

1

yt

1et c

yt 1et 1

ut 1et + c

ut

1et

1

Notice that T and are not independent. utand e tare independent, ytdepends on et, et

1, et

2, . . .and yt depends on ut

1,ut

2, . . ..

Taking mean values we get

E( ) = E(T) 1E(T)

To evaluate this expression we calculate

E

y2t 1

yt

1ut 1

yt

1ut 1

u2t 1

= N Ey2t 0

0 Eu2t


11/46


12/46


13/46


Estimator

b

1

q + a

Kuc = 0 u

y

d

y

1

Figure 5. The system redrawn.

(b) Similarly to ( a), we get

1N

Nk=1

u( k

1)2 11

N

Nk=1

u( k

1) y(k)

(u2( k 1))0

1

(u( k 1)y( k))0, as N

where ()0denote the stationary value of the argument. We haveu2( k 1)

0 = (( u( k))0)

2

(u( k 1)y( k))0 = (u( k))0b (( u( k)0+ d0)

(u( k))0 = Hud (1) d0 = K b

1 + a + K bd0

and the asymptotic LS estimate becomes

b =

(u2( k 1))0

1

(u( k 1)y( k))0 = b

1 +d0

(u( k)) 0

= b

1

1 + a + K bK b

=

1 + aK

How do we interpret this result? The system may be redrawn as inFigure 5. Since Uc = 0, we have that u = Kq+a y, and we can regard

Kq+a

as the controller for the system in Figure 5. It is then obviousthat we have estimated the negative inverse of the static controllergain.

(c) Introduction of high pass regressor filters as in Figure 6 eliminates or

at least reduces the influence from the disturbance don the estimateofb. One choice of regressor filter could be Hf(q 1) = 1 q 1, i.e. adifferentiator. Another possibility would be to introduce a constant in


14/46

12 Problem Solutions

Estimator

K

uc

u y

d

bq + a

Hf Hf

1

Figure 6. Introduction of regressor filters.

the regressor and then estimate both b and bd. The regression modelis in this case

y( t)

= u( t 1) 1b

bd = (t) T2.18 The equations for recursive least squares are

y( t) = T(t 1)0

( t) = ( t 1) + K(t)( t)

( t) = y( t) T( t 1)(t 1)

K( t) = P( t)(t 1)

=P( t 1)( t 1)

+ T(t 1)P( t 1)( t 1)

P( t) = I K( t)T( t 1)P( t 1) /

Since the quantity P( t)( t 1) appears in many places it is convenientto introduce it as an auxiliary variable w = P. The following computercode is then obtained:

Input u,y: realParameter lambda: realState phi, theta: vector

P: symmetric matrix

Local variables w: vector, den : real"Compute residuale=y-phi^T*theta

"update estimatew=P*phiden=w^T*phi+lambda


15/46


16/46



3.1 Given the process

H(z) =B(z)

A(z) =

z +1.2

z2 z +0.25Design specification: The closed system should have a pole that corre-spond to following characteristic polynomial in continuous time

s2 +2s + 1 = ( s + 1)2

This corresponds toAm(z) = z

2+ am1z + am2

with am1 = (e

1 + e 1) = 2e 1

am2 = e 2

(a) Determine an indirect STR of minimal order. The controller shouldhave an integrator and the stationary gain should be 1.Solution:Choose Bm such that

Bm(1)Am(1)

= 1

The integrator condition gives

R = R(z 1)

We get the following conditions

B T = BmAo(1)

AR + B S = AmAo(2)

As B is unstable must Bm = B B m. This makes (1) B T =B B mAo T = B

mAo. Choose B

m such that

B(1)B m(1)A(1)

= 1 B m(1) =A(1)B(1)

The simplest way is to choose

B m = bm =

A(1)B(1)

=0.252.2

Further we have

(z2 + a1z + a2)(z 1)(z+ r) +( b0z + b1)(s0z2 + s1z + s2)

= (z2 + am1z + am2)(z2+ ao1z + ao2)


17/46


with a1 = 1, a2 = 0.25 and ao1and ao2chosen so that Ao is stable.Equating coefficients give

r 1 + a1 + b0s0 = ao1 + a m1

r + a1(r 1) + a2 + b0s1 + b1s0 = ao2 + am1ao1 + am2 ar + a2( r 1) + b0s2 + b1s1 = am1ao2 + am2ao1 a2r + b1s2 = am2ao2

or1 b0 0 0

a1 1 b1 b0 0

a2 a1 0 b1 b0 a2 0 0 b1

r

s0

s1

s2

=

ao1 + am1 +1 a1aa2 + am1ao1 + a m2 + a1 a2

am1ao2 + am2ao1 + a2

am2ao2

Now choose to estimate

= b1 b1 a1 a2T

by equation 3.22 in the textbook.

(b) As H(z) is not minimum phase we cancel B = B between R andS. This is difficult, see page 118 in the textbook. An indirect STR is

given by Eq. 3.24.AoAmy = Ru + S y

with R = B R, S = B S, T = B mAo. Furthermore we have

AoAmym = AoBmuc = AoB Bmuc = B Tuc =

Tu c

y = R 1

AoAmu

=uf+S

1AoAm

y

=yfym = T

1AoAm

uc =uc f

= y ym = Ruf + S yf Tuc f

Now estimate R, S and T with a recursive method in the aboveequation. Then cancel B and calculate the control signal.

(c) Take a = 0 in Example 5.7, page 206 in the textbook. This gives

dt0

dt = uce

ds0dt

= ye

with e = y ym = y Gmuc.


18/46


3.3 The process has the transfer function

G(s) =b

s + a

q

s + p

wherepand qare known. The desired closed loop system has the transferfunction

Gm( s) =2

s2 +2 s +2

Since a discrete time controller is used the transfer functions are sampled.We get

H(z) =b0z + b1

z2 + a1z + a2

Hm(z) =bm0z + bm1

z2 + am1z + am2

The fact that p and q are known implies that one of the poles of H isknown. This information will be disregarded. In the following we willassume

G( s) =1

s( s+ 1)

Withh = 0.2 this gives

H(z) =0.0187(z+ 0.936)(z 1)(z 0.819)

Furthermore we will assume that = 2 and = 0.7.Indirect STR:The parameters of a general pulse transfer function of second order is

estimated by recursive least squares( See page 51). We have

= bo b1 a1 a2 T

( t) =u(t 1) u( t 2) y( t 1) y( t 2)

The controller is calculated by solving the Diophantine equation. We lookat two cases1.B canceled:

(z2 + a1z + a2)1 + b0( s0z + s1) = z2 + am1z + am2

z1

: a1 + b0s0 = am1 s0 =am1 a1

b0

z0 : a2 + b0s1 = am2 s1 =am2 a2

b0


19/46


The controller is thus given by

R(z) = z + b1 / b0

S(z) = s0z + s1

T(z) = t0z where t0 =1 + am1 + am2

b0

2. B not canceled:The design equation becomes

(z2 + a1z + a2)(z+ r1) +( b0z + b1)( s0z + s1) = (z2+ am1z + am2)(z + ao1)

Identification of coefficients of equal powers ofzgives

z2 : a1 + r1 + b0s0 = am1 + ao1

z1 : a2 + a1r1 + b1s0 + b0s1 = am1ao1 + am2

z0 : a2r1 + b1s1 = am2ao1

The solution to these linear equations is

r1 =b21n1 b0b1n2 + ao1am2b20

b20a2 a1b0b1 + b21

s0 =n1 r1

b0

s1 =b0n2 b1n1 r1( a1b0 b1)

b20

wheren1 = a1 am1 ao1

n2 = am1ao1 + am2 a2

The solution exists if the denominator of r1 is different from zero, whichmeans that there is no pole-zero cancellation. It is helpful to have access

to computer algebra for this problems e.g. Macsyma, Maple or Matem-atica! Figure 7 shows a simulation of the controller obtained when thepolynomial B is canceled. Notice the ringing of the control signal whichis typical for cancellation of a poorly damped zero. In this case we havez = 0.936. In Figure 8 we show a simulation of the controller with nocancellation. This is clearly the way to solve the problem.Direct STR:To obtain a direct self-tuning regulator we start with the design equation

AR + B S = AmAoB+

HenceB +AmAoy = ARy + B Sy = B Ru + B Sy

y = R B

AoAmu

uf

+S B

AoAmy

yf


20/46


21/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10

u

Figure 8. Simulation in Problem 3.3. Process output and control signalare shown for the indirect self-tuning regulator when the process zero isnot canceled.

by RLS. The case r0 = 0 must be taken care of separately. FurthermoreThas the form T( q) = t0qwhere

B T

AR + B S =

B toq

B + bo B Am=

toq

Am

To get unit steady state gain choose

to = Am(1)

A simulation of the system is shown in Fig. 9. We see the typical ringingphenomena obtained with a controller that cancels a poorly damped zero.To avoid this we will develop an algorithm where the process zero is notcanceled.2. No cancellation of process zero:We then haveB + = 1 and B = b0q+b1. From the analysisof the indirectSTR we know that a first order observer is required, i.e. A0 = q+ ao1. Wehave as before

y = R B

AoAmu

uf

+S B

AoAmy

yf

()


22/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10

u

Figure 9. Simulation in Problem 3.3. Process output and control signalare shown for the direct self-tuning regulator when the process zero iscanceled.

Since B is not known we can, however, not calculate uf and yf. Onepossibility is to rewrite( *)as

y = R B

R

1AoAm

u

+ S B

S

1AoAm

y

and to estimate R and S as second order polynomials and to cancelthe common factor B from the estimated polynomials. This is difficultbecause there will not be an exact cancellation. Another possibility is touse some estimate of B . A third possibility is to try to estimate B Rand B S as a bilinear problem. In Fig. 811 we show simulation whenthe model (*)is used with

B = 1

B = q

B =q +0.4

1.4

B =q 0.4

0.6


23/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10u

Figure 10. Simulation in Problem 3.3. Process output and control signalare shown for the direct self-tuning regulator when the process zero is notcanceled and when B = 1.

3.4 The process has the transfer function

G( s) =b

s( s+ 1)

with proportional feedbacku = k(uc y)

we get the closed loop transfer function

Gcl ( s) =kb

s2 + s + k b

The gain k = 1 / bgives the desired result. Idea for STR: Estimate banduse k = 1 / b. To estimate b introduce

s( s + 1)y = bu

s(s + 1)( s + a)2

yf

y = b 1( s + a)2

u


24/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10

u

Figure 11. Simulation in Problem 3.3. Process output and control signalare shown for the direct self-tuning regulator when the process zero is notcanceled and when B = q.

The equations on page 56 gives

db

dt = Pe = P(yf b)

dP

dt = P

P T

P = P

P2

2

Withb(0) = 1, P(0) = 100, = 0.1, and a = 1 we get the result shownin Fig. 14.


25/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10

u

Figure 12. Simulation in Problem 3.3. Process output and control signalare shown for the direct self-tuning regulator when the process zero is notcanceled and when B = (q + 0.4) / 1.4.


26/46


0 10 20 30 40 50

2

1

0

1

2uc y

0 10 20 30 40 50

10

0

10

u

Figure 13. Simulation in Problem 3.3. Process output and control signalare shown for the direct self-tuning regulator when the process zero is notcanceled and when B = (q 0.4) / 0.6.


27/46


0 5 10 15 202

0

2uc y

0 5 10 15 20

0

5

10u

0 5 10 15 20

0.2

0.6

1b

Figure 14. Simulation in Problem 3.4. Process output, control signaland estimated parameter b are shown for the indirect continuous-timeself-tuning regulator.


28/46



4.1 The estimate bmay be small because of a poor estimate. One possibilityis to use a projection algorithm where the estimate is restricted to be ina given range, bo b b1. This requires prior information about b:svalues. Another possibility is to replace

1

bb y

b

b2 + P

where P is the variance of the estimate. Compare with the discussion ofcautious control on pages 356358.

4.10 Using ( 4.21)the output can be written as

yt+d =R

Cut +

S

Cyt +

R1C

et+do

= rout + ft

()

Consider minimization of

J = y2t+d +u2t ( +)

Introduce the expression ( *)

J = ( rout + ft)2+ u2t

= ( r2o +)u2t +2routft + f

2t

= ( r2o +)

u2t +

2routftr2o +

+ f2t

= ( r

2

o +) ut + roftr2o +2

r2of2t

r2o + + f

2

t

Hence

J =1

r2o +

ft +

r2o +

rout

2+

r2o + f2t

=1

r2o +

ft +

ro +

ro

ut

2+

r2o + f2t

=1

r2o +

yt+d +

rout

2+

r2o + f2t

Since r2o + is a constant we find that minimizing ( +) is the same as tominimize

J1 = yt+d + ro

ut = ft +

ro + ro

ut


29/46



5.1 The plant is

G( s) =1

s(s + a) =

B

A

The desired response is

Gm( s) =2

s2 +2 s +2 =

Bm

Am

(a) Gradient method or MIT rule. Use formulas similar to (5.9). In thiscase B + = 1 and Ao is of first order. The regulator structure is

R = s + r1 S = s0s + s1 T = t0Ao

This gives the updating rules

dr1

dt = e

1

AoAmu

ds0

dt = e

p

AoAmy

ds1

dt = e

1

AoAmy

dt0

dt = e

1

AoAmuc

(b) Stability theory approach 1. First derive the error equation. If all

process zeros are cancelled we have

AoAmy = AR1y + boSy = B R1u + boSy

= bo(Ru + Sy)Further

AoAmym = AoBmuc = boTuc

HenceAoAme = AoAm(y ym) = bo(Ru + Sy Tuc)

e =b

AoAm(Ru + Sy Tuc)

Since 1 / AoAm is not SPR we introduce a polynomial D such thatD / AoAm is SPR. We then get

e =bD

AoAm Ruf + Syf Tuc fwhere

uf =1

Du yf =

1D

y uc f =1

Du c


30/46


(c) Stability theory approach 2. In this case we assume that all statesmeasurable. Process model:

x =

a 0

1 0

Ap

x +

1

0

Bp

y =

0 1 C

x

Control lawu = Lr uc Lx = 3uc 1x1 2x2

The closed loop system is

x = (Ap BpL)x + B Lr uc = Ax + Buc

y = C x

whereA() = Ap BpL =

a 1 21 0

B() = BpLr =

30

The desired response is given by

xm = Amxm + Bmuc

where

Am =

2 2

1 0

Bm =

2

0

We have(A Am)

T=

2 a 1 02 2 0

(B Bm)

T=

3 2 0Introduce the state error

e = x xm

we gete = x xm = Ax + Buc Amxm Bmuc

= Ame + (A Am)x+ (B Bm)uc()

The error goes to zero if Am is stable and

A() Am = 0 ()


31/46


B() Bm = 0 ( +)

It is thus necessary that A() and B() are such that there is a for which ( *)and ( +)hold. Introduce the Lyapunov function

V = eTPe + tr(A Am)TQa(A Am)

+ tr(B Bm)TQb(B Bm)

Notice thattr(A + B) = trA + trB

xTAx = tr(xxTA) = tr(AxxT)

tr(AB ) = tr(B A)

we get

dV

dt = tr

P eeT + Pe eT + ATQa(A Am)

+(A

Am)T

QaA +

B Qb(B

Bm) + (B

Bm)T

Qb B

( +)

But from ( *)

P eeT = P (Ame +(A Am)x +(B Bm)uc)eT

Pe eT = Pe (Ame + (A Am)x+ (B Bm)uc)T

Introducing this into ( +) and collecting terms proportional to (A Am)

T we find that they are

2tr(A Am) T

Qa A + PexT

Similarly we find that terms proportional to (B Bm) are

2tr(B

Bm)TQb B + PeuTc

HencedV

dt =eTPAme + e

TATmPe

+2tr(A Am) T

Qa A + Pe xT

+2tr(B Bm) T

Qb B + PeuTc

Hence if the symmetric matrix P is chosen so that

ATmP + PAm = Q

where Qis positive definite ( can always be done ifAm is stable!)andparameters are updated so that

(A Am)T

Qa A + Pe xT =0

(B Bm)T

Qb B + Peu Tc =0

( +)


32/46


we getdV

dt = eTQe

The equations for updating the parameters derived above can now be

used. This gives2 a 1 02

2 0

Qa 1 20 0

+ Pe xT = 03 2 0Qb 30

+ Peu c = 0Hence with Qa = Iand Qb = 1

d1dt

= p11e1x1 + p12e2x1 = (p11e1 + p12e2)x1

d2dt

= p11e1x2 + p12e2x2 = (p11e1 + p12e2)x2

d3

dt = (p11e1 + p12e2)uc

where e1 = x1 xm1 and e2 = x2 xm2. It now remains to determineP such that

ATmP + PAm = Q

Choosing = 0.707, = 2 and

Q =

41.2548 11.313711.3137 16.0000

we get

P = 4 22 16The parameter update laws become

d1dt

= (4e1 +2e2)x1

d2dt

= (4e1 +2e2)x2

d3dt

= (4e1 +2e2)uc

A simulation of the system is given in Fig. 15.

5.2 The block diagram of the system is shown in Fig. 16. The PI version ofthe SPR rule isd

dt = 1

d

dt(uce) 2uce ()


33/46


0 10 20 30 40

2

1

0

1

2States x1 xm1 x2 xm2

0 10 20 30 40

0

0.5

1

Estimated parameters

Figure 15. Simulation in Problem 5.1. Top: Process and modelstates, x1 ( full), xm1 (dashed), x2 (dotted), and xm2 (dash-dotted).Bottom: Controller parameters3( full), 1 (dashed), and 2 ( dot-ted).

u c

ym

e

y

1s

1s

0

Figure 16. Block diagram in Problem 5.2.

To derive the error equation we notice that

dym

dt = 0

uc

dy

dt = uc


34/46


Hencede

dt = ( 0)uc

we getd2e

dt2 =d

dt uc +(

0

)duc

dt

Inserting the parameter update law ( *) into this we get

d2e

dt2 = 1

duc

dt e +u c

de

dt

uc 2u

2ce +(

0)duc

dt

Henced2e

dt2 +1u

2c

de

dt +

1uc

duc

dt +2u

2c

e = ( 0)

duc

dt

Assuming that u c is constant we get the following error equation

d2e

dt2 +1u

2

c

de

dt +2u

2

c

e = 0

Assuming that we want this to be a second order system with and we get

1u2c = 2 1 = 2 / u

2c

2u2c =

2 2 = 2

/ u2c

This gives an indication of how the parameters 1 and 2 should beselected. The analysis was based on the assumption that ucwas constant.To get some insight into what happens when uc changes we will givea simulation where uc is a triangular wave with varying period. Theadaptation gains are chosen for different and. Figure 17 shows whathappens when the period of the square wave is 20 and = 0.5, 1 and 2.Corresponding to the periods 12, 6 and 3. Figure 18 show what happenwhen uc is changed more rapidly.

5.6 The transfer function is

G( s) =b0s

2 + b1s + b2

s2 + a1s + a2

The transfer function has no poles and zeros in the right half plane ifa1 0, a2 0, b0 0, b1 0, and b2 0. Consider

G( i) =B( i)

A( i)

A( i)

A( i)

The condition Re G( i) 0 is equivalent to Re (B( i)A( i) ) 0. But

g() = Re

( b02+ ib1 + b2)(

2 ia1 + a2)

= b0

4+ (a1b1 b0a2 b2)

2+ a2b2


35/46


0 5 10 15 20

0

5

10

Process and model output

0 5 10 15 20

0

0.5

1

1.5Estimated parameter

0 5 10 15 20

0

5

10


0 5 10 15 20

0

0.5

1


Figure 17. Simulation in Problem 5.2 for a triangular wave of period 20.Left top: Process and model outputs, Left bottom: Estimated parameterwhen = 0.5 (full), 1 (dashed), and 2 (dotted) for = 0.7. Right top:Process and model outputs, Right bottom: Estimated parameter when = 0.4 ( full), 0.7( dashed), and 1.0( dotted) for = 1.

Completing the squares the function can be written as

g() = b0

2 +

a1b1 b0a2 b2

2b0

2+ a2b2

( a1b1 b0a2 b2)2

4b0When b0 = 0 the condition for gto be positive is that

a1b1 b0a2 b2 0 ( i)

Ifb0 > 0 the function g() is nonnegative for all if either ( i) holds orif

a1b1 b0a2 b2 < 0

and

a2b2 > ( a1b1 b0a2 b2)

2

4b0Example 1. Consider

G( s) =s2 +6s + 8s2 +4s + 3

We have a1b1 b0a2 b2 = 24 3 8 = 13 > 0. Hence the transfer

function G( s)is SPR. Example 2.

G( s) =3s2 + s +1s2 +3s + 4


36/46


0 5 10 15 20

0

5

10


0 5 10 15 20

0

0.5

1


0 5 10 15 20

0

5

10


0 5 10 15 20

0

0.5

1


Figure 18. Simulation in Problem 5.2 for a triangular wave of period 5.Left top: Process and model outputs, Left bottom: Estimated parameterwhen = 0.5 (full), 1 (dashed), and 2 (dotted) for = 0.7. Right top:Process and model outputs, Right bottom: Estimated parameter when = 0.4 ( full), 0.7( dashed), and 1.0( dotted) for = 1.

we have a1b1 a2b0 b2 = 3 12 1 = 10. Furthermore

a2b2 = 4

( a1b1 a2b0 b2)2

4b0=

10012

Hence the transfer function G( s)is neither PR nor SPR.

5.7 Consider the systemdx

dt = Ax + B1u

y = C1x

where

B1 =

1

0...

0

Let Qbe positive and let P be the solution of the Lyapunov equation

ATP + PA = Q ()


37/46


Define C1as

C1 = BTP =

p11 p12 . . . p1nAccording to the Kalman-Yacobuvich Lemma the transfer function

G1( s) = C1(sI A)

1B1

is then positive definite. This transfer function can, however, be writtenas

G( s) =p11s

n 1 + p12s

n 2 +. . . + p1n

sn + a1sn 1 +. . . + an

Since there are good numerical algorithms for solving the Lyapunov equa-tion we can use this result to construct transfer functions that are SPR.The method is straightforward.

1. Choose A stable.2. Solve ( *)for given Qpositive.

3. Choose B asB( s) = p11s

n 1

+ p12sn

2+. . . + p1n

5.11 Let us first solve the underlying design problem for systems with knownparameters. This can be done using pole placement. Let the plant dy-namics be

y =B

Au

and let the controller beRu = Tuc sy

The basic design equation is then

AR + B S = AmAo ()

In this case we haveA = ( s + a)( s + p)

B = bq

Am = s2+2 s +2

We need an observer of at least first order. The design equation ( *) thenbecomes

(s + a)( s+ p)( s + r1) + bq( s0s + s1) = (s2+2 s +2)( s+ a0) ( +)

where Ao = s + ao is the observer polynomial. The controller is thus ofthe formdu

dt + r1u = t0uc s0y s1

dy

dt


38/46


It has four parameters that have to be updatedr1,t0,s0, ands1. If no priorknowledge is available we thus have to update these four parameters.When parameter p is known it follows from the design equation ( +)thatthere is a relation between the parameters and it then suffices to estimatethree parameters. This is particularly easy to see when the observerpolynomial is chosen as Ao(s) = s + p. Puttings = pin ( +)gives

s0p + s1 = 0

Hences1 = ps0 ()

In this particular case we can thus update t0, s0, and r1 and computes1 from (**). Notice, however, that the knowledge of q is of no valuesince q always appear in combination with the unknown parameter b.The equations for updating the parameters are derived in the usual way.Witha0 = pequation( +) simplifies to

( s + a)( s+ r1) + bqs0 = s2

+2 s +2

IntroducingA( s) = s + a S( s) = s0 T

= t0

we get

y =B T

AR + B Suc

e

t0=

B

AR + B Suc

B

Amuc

e

s0=

B TB

(AR + B S)2uc =

B

AR + B S y

B

Amy

e r1

= A

B T

(AR + B S)2uc = A

AR + B S y A

Amy

= A

Am

B

Au =

B

Am( s + p)u

The MIT rule then gives

dr1

dt =

1

( s + p)Amu

e

ds0

dt =

1Am

y

e

dt0

dt = 1Am uce

A simulation of the controller is given in Fig. 19.


39/46


0 20 40 60 80 100

2

1

0

1

2Model and process output

0 20 40 60 80 100

0.5

1

1.5

2

Estimated parameters

Figure 19. Simulation in Problem 5.11. Top: Process (full) and model(dashed)output. Bottom: Estimated parameters r1 ( full),s0 (dashed), andt0 ( dotted).

5.12 The closed loop transfer function is

Gcl ( s) =kb

s2 + s + k b

This is compatible with the desired dynamics. The error is

e = y ym =kb

p2 + p + kbuc ym

Hence e

k =

b

p2 + p + kbuc

b2k

(p2 + p + k b)2uc

=b

p2 + p + kb( uc y)

b 1

p2 + p +1( uc y) p =

d

dt

The following adjustment rule is obtained from the MIT rule

dk

dt =

e

ke = b

1

p2 + p +1( uc y)

e


40/46


0 20 40 602

0

2ym y, gamma = 0.05

0 20 40 600

1

2k, gamma = 0.05

0 20 40 60

2

0

2ym y, gamma = 0.3

0 20 40 60

0

1

2k, gamma = 0.3

0 20 40 60

2

0

2ym y, gamma = 1

0 20 40 60

0

1

2k, gamma = 1

Figure 20. Simulation in Problem 5.12. Left: Process (full) and model(dashed) output. Right: Estimated parameter kfor different values of.

A simulation of the system is given Fig. 20. This shows the behavior ofthe system when u c is a square wave.


41/46



6.1 The process is given by

G( s) =b

s( s + a)

and the regressor filter should be

Gf(s) =1

Af(s) =

1Am( s)

=1

s2 +2 s +2

The controller is given by

U( s) =s0s + s1

s + r1Y(s) +

t0(s + ao)

s + r1uc(s)

For the estimation of process parameters we use a continuous-time RLSalgorithm. The process is of second order and the controller is of firstorder. The regressor filter is of second order and both inputs and outputsshould be filtered. Hence we need seven states in. The process parame-ters are contained in , and the controller parameters in. The relationbetween these are given by

=

r1

s0

s1

t0

=

2 + ao a

( ao2 +2 ar1) / b

2 ao / b

2 / b

= ()

To find A(), B(), C() and D()we start by finding realizations fory, yf, ufand the controller. We get

d

dt y

y = a 0

1 0 yy +

b

0u

d

dt

yfyf

= 2 21 0

yfyf

+ 10

yand

d

dt

ufuf

= 2 21 0

ufuf

+ 10

uand the control law can be rewritten as

u = s0y + t0uc + s1 + r1s0

p + r1y +

aot0 t0r1

p + r1uc

We need one state for the controller and it can be realized as

x = r1x + ( s1 + r1s0)y + ( aot0 r1t0)uc

u = x s0y + t0uc


42/46


43/46


If we use the relations a = r1 +2 + aoandb = 2 / t0 then ecan bewritten as

e = 2 yf 2yf + y +( r1 +2 + ao)yf 2

t0uf

=

0 1 r1 + ao 0 0 2t0 0

y

y

yfyf

uf

uf

x

Combining the expressions for and egives

e

=

0 1 r1 + ao 0 0 2

t00

0 0 1 0 0 0 0

0 0 0 0 0 1 0

y

y

yfyf

uf

uf

x

= C()

i.e. D() = 0. As given in the problem description, the estimator isdefined by

d

dt = Pe

dP

dt = P P TP

where P is a 2 2 matrix and eand are given above.

6.3 The averaged equations for the parameter estimates are given by (6.54)on page 303. In this particular case we have

G( s) =

ab2

( s + a)(s + b)2

Gm( s) =a

s + a


44/46


45/46


The desired response is

Gm( s) =bm

s + am(3)

Combining( 1) and( 2)gives the closed loop transfer function

Gcl =b1

s + a + b2

Equating this with Gm( s)given by ( 3) gives

b1 = bm

a + b2 = am

If these equations are solved for 1 and 2 we obtain the controllerparameters that give the desired closed loop system. Conversely if theequations are solved for aand b we obtain the parameters of the processmodel that corresponds to given controller parameters. This gives

a = am

bm2/

1b = bm / 1

The parametersa and bcan thus be interpreted as the parameters of themodel the controller believes in. Inserting the expressions for1 and2from page 318 we get

a() =

229 312

259 2

b() =458

259 2

( +)

When

= 22931 = 2.7179we get a() = 0. The reson for this is that the value of the plant transferfunction

G(s) =458

(s + 1)( s2 +30s + 229) ()

is G(2.7179i) = 0.6697i. The transfer function of the plant is thuspurely imaginary. The only way to obtain a purely imaginary value ofthe transfer function

G =b

s + a ()

is to make a = 0. Also notice that b(2.7179i) = 1.8203 which gives

G(2.7179i) = 0.6697i.When =

259 = 16.09 we get infinite values ofaand b. Notice that G( i

259) = 0.0587 that is real and negative. Theonly way to make G( i)negative and real is to have infinite large values


46/46


ofaandb. It is thus easy to explain the behavior of the algorithm from thesystem identification point of view. The controller can be interpreted asif it is fitting a model (**) to the process dynamics ( *). With a sinusoidalinput it is possible to get a perfect fit and the parameters are given by( +).

solns adaptive control by astrom

Documents