Adaptive Control (Karl J. Astrom) 2nd Ed - Solution Manual

8/20/2019 Adaptive Control (Karl J. Astrom) 2nd Ed - Solution Manual

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LIBROS UNIVERISTARIOS

Y SOLUCIONARIOS DE

MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS

CONTIENEN TODOS LOS

EJERCICIOS DEL LIBRO

RESUELTOS Y EXPLICADOS

DE FORMA CLARA

VISITANOS PARA

DESARGALOS GRATIS.


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Solution Manualfor

Adaptive Control

Second Edition

Karl Johan ÅströmBjörn Wittenmark


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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.


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PROBLEM SOLUTIONS

SOLUTIONS TO CHAPTER 1

1.5 Linearization of the valve shows that∆v = 4v30∆u

The loop transfer function is then

G0( s) G PI ( s)4v30

where G PI is the transfer function of a PI controller i.e.

G PI (s) = K

1 +

1sT i

The characteristic equation for the closed loop system is

sT i( s + 1)3+ K ⋅ 4v

30(sT i + 1) = 0

with K = 0.15 and T i = 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 for yr = uc = 5.1. Compare with Fig. 1.9.

1


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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

I m a g A x i s

Figure 1. Root locus in Problem 1.5.

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

K u =1

G0(iω u)

we have

G0( s) = e− s/ q

1 + s/ q

arg G0( iω ) = − ω

q− arctan

ω

q = − π

q ω G0(iω ) K T i

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


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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 designedfor q = 2.

1.7 Introducing the feedbacku = − k2 y2

the system becomes

dx

dt =

− 1 0 0

0 − 3 0

0 0 − 1

x − k2

0

2

1

1 0 1 x +

1

0

0

u1 y1 =

1 1 0 xThe transfer function from u1 to y1 is

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)


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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)


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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.


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6 Problem Solutions


2.1 The function V can be written as

V ( x1 ⋅ ⋅ ⋅ xm) =

n

i, j =1 x

i x j ( aij + a ji )/

2 +

n

i=1 b

i xi + c

Taking derivative with respect to xi we get

V

xi=

n j =1

( aij + a ji ) x j + bi

In vector notation this can be written as

grad x V ( x) = ( A + AT ) x + b

2.2 The model is

yt = ϕ T t θ + e t =ut ut

− 1 b0

b1

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

θ̂ = (ΦT Φ) − 1ΦT Y =

u2t utut− 1utut

− 1

u2t− 1

− 1 ut ytut

− 1 yt

(a) Input is a unit step

ut =

1 t ≥ 0

0 otherwise

Evaluating the sums we get

θ̂ =

N N − 1 N − 1 N − 1

− 1

N 1

yt

N 2

yt

=

1 − 1− 1

N

N − 1

N 1

yt

N 2

yt

θ̂ =

y1

1 N − 1

N 2

yt − y1

The estimation error is

θ̂ − θ =

e1

1 N − 1

N 2

et − e1


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Hence

E( θ̂ − θ )( θ̂ − θ ) T = (ΦT Φ) − 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 of b0 + b1.

E( b̂0 + b̂1 − b0 − b1) = 1 1

1 − 1− 1

N

N − 1

1

1

= 1 N − 1

With a step input it is thus possible to determine the combinationb0 + b1 consistently. The individual values of b0 and b1 can, however,not be determined consistently.

(b) Input u is white noise with Eu2 = 1 and u is independent of e.

Eu2t = 1 Eutut− 1 = 0

cov( θ̂ − θ ) = 1 ⋅ E(ΦT Φ) − 1 =

N 00 N − 1

− 1 =

1 N

0

0 1 N − 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) = ϕ T ( t)θ 0 + ¯ e( t)

whereϕ T ( t) = u( t), θ 0 = b0

and¯ e(t) = b1u(t − 1) + e(t)

Model:ˆ y(t) = b̂u( t)

or y( t) = b̂u(t) + ε ( t) = ϕ T (t) θ̂ + ε ( t)

whereε (t) = y( t) − ˆ y( t)

The least squares estimate is given by

ΦT Φ( θ̂ − θ 0) = ΦT Ed Ed =

¯ e(1)

...

¯ e( N )


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8 Problem Solutions

1 N

ΦT Φ =

1 N

N 1

u2( t) → Eu2 N → ∞

1

N

ΦT Ed =

1

N

N

1 u( t) ¯ e(t) =1

N

N

1 u(t) (b1u( t − 1) + e( t))→ b1 E (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

Henceb̂ = θ̂ → θ 0 + b1 = b0 + b1 N → ∞

i.e. b̂ converges to the stationary gain

(b)u(t) ∈ N (0,σ ) ⇒ Eu2 = σ 2 Eu(t)u( t − 1) = 0 Eu(t) e( t) = 0

Henceb̂ → b0 N → ∞

2.6 The model is

yt = ϕ T t θ + ε t =

− yt

− 1 ut− 1

ϕ T t

ab

θ

+

et + ce t− 1

ε t

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

(2.5). The estimation error isθ̂ − θ = (ΦT Φ) − 1ΦT ε =

y2t

− 1 −

yt− 1ut − 1

−

yt

− 1ut

− 1

u2t

− 1

− 1 −

yt

− 1 et − c

yt− 1 et− 1

ut− 1 et + c

ut

− 1 et

− 1

Notice that ΦT and ε are not independent. ut and e t are independent, ytdepends on et, et

− 1, et

− 2, . . . and yt depends on ut

− 1,ut

− 2, . . ..

Taking mean values we get

E( θ̂ − θ ) = E(ΦT Φ) − 1 E(ΦT ε )

To evaluate this expression we calculate

E

y2t− 1 −

yt

− 1ut− 1

−

yt

− 1ut− 1

u2t− 1

= N Ey2t 0

0 Eu2t


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and

E

− yt − 1 et − c yt− 1 et− 1ut

− 1 et + c

ut

− 1 et

− 1

= − cN E yt− 1 et− 10

Eyt

− 1 et− 1 = E (ayt− 2 + but− 2 + et− 1) et − 1 = σ 2

Since

yt =b

q + a u t +

q + c

q + a e t

Ey2t = xb2

1 − a2 +

1 − 2ac + c2

1 − a2 σ 2

Eu2t = 1 Ee2t = σ

2

we get

E( â − a)2 = − σ 2c(1 − a2)

b2 + ( 1 − 2ac + c2)σ 2

E(ˆb

−

b)2

= 02.8 The model is

y(t) = a + bt + e( t) = ϕ T θ + e( t)

ϕ =

1t

θ = ab

According to Theorem 2.1 the solution is given by equation ( 2.6), i.e.

θ̂ = (ΦT Φ) − 1ΦT Y

where

ΦT =

1 1 1 11 2 3 ⋅ ⋅ ⋅ N

Y =

y(1)

y(2)...

y( N )

Hence

θ̂ =

N t=1

1 N

t=1

t

N t=1

t

N t=1

t2

− 1

N t=1

y(t)

N t=1

ty( t)

=

2

N ( N − 1) ( (2 N + 1) s0−

3s1)6

N ( N + 1)( N − 1) ( − ( N + 1) s0 + 2s1)


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10 Problem Solutions

where we have made use of

N t=1

t = N ( N + 1)

2

N t=1

t2 = N ( N + 1)( N + 2)

6

and introduced

s0 =

N t=1

y(t) s1 =

N t=1

ty( t)

The covariance of the estimate is given by

cov( θ̂ ) =σ 2(ΦT Φ) − 1 =

12 N ( N + 1)( N − 1)

( N + 1)(2 N + 1)6

−

N + 12

−

N + 12

1

Notice that the variance of b̂ decreases as N − 3 for large N but the variance of â decreases as N − 1. The reason for this is that the regressorassociated with a is 1 but the regressor associated with b is t. Notice thatthere are better numerical methods to solve for θ̂ !

2.17

(a) The following derivation gives a formula for the asymptotic LS esti-mate

b̂ = (ΦT Φ) − 1ΦY = N

k=1

φ ( k − 1)2

− 1 N k=1

φ (k − 1) ¯ y( k)

=

1 N

N k=1

u(k − 1)2

− 1 1 N

N k=1

u( k − 1) ¯ y( k)

→

E(u( k − 1)2)

− 1

E (u( k − 1) ¯ y( k)) , as N → ∞

The equations for the closed loop system are

u( k) = K ( uc( k) − y( k))

¯ y( k) = y( k) + ay( k − 1) = bu( k − 1)

The signals u( k) and y( k) are stationary signals. This follows sincethe controller gain is chosen so that the closed loop system is stable.

It then follows that E(u(k − 1)2) = E(u(k)2) and E(u(k − 1) ¯ y( k)) = E( bu( k − 1)2) = bE(u( k)2) exist and the asymptotic LS estimatebecomes b̂ = b, i.e. we have an unbiased estimate.


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Estimator

b

1

q + a

ΣK Σ uc = 0 u

y

d

y

1−

Figure 5. The system redrawn.

(b) Similarly to ( a), we get

1 N

N k=1

u( k−

1)2− 1 1

N

N k=1

u( k−

1) ¯ y(k)

→

(u2( k − 1))0

− 1

(u( k − 1) ¯ y( k))0 , as N → ∞

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

0 = (( u( k))0)

2

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

(u( k))0 = H ud (1) d0 = − K b

1 + a + K b d0

and the asymptotic LS estimate becomes

b̂ =

(u2( k − 1))0

− 1

(u( k − 1) ¯ y( k))0 = b

1 +d0

(u( k)) 0

= b

1 −

1 + a + K b K b

=

1 + a K

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

K q+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 d on the estimateof b. One choice of regressor filter could be H f (q − 1) = 1 − q − 1, i.e. adifferentiator. Another possibility would be to introduce a constant in


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Estimator

K Σ

uc

u yΣ

d

bq + a

H f H f

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 = φ (t) T θ 2.18 The equations for recursive least squares are

y( t) = ϕ T (t − 1)θ 0

θ̂ ( t) = θ̂ ( t − 1) + K (t)ε ( t)

ε ( t) = y( t) − ϕ T ( t − 1) θ̂ (t − 1)

K ( t) = P( t)ϕ (t − 1)

= P( t − 1)ϕ ( t − 1)

λ + ϕ T (t − 1) P( t − 1)ϕ ( t − 1)

P( t) = I − K ( t)ϕ T ( t − 1) P( t − 1) / λ

Since the quantity P( t)ϕ ( t − 1) appears in many places it is convenientto introduce it as an auxiliary variable w = Pϕ . 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


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theta=theta+w*e/den"Update covariance matrixP=(P-w*w^T/den)/lambda "Update regression vectorsphi=shift(phi)phi(1)=-yphi(n+1)=u


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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 to Am( z) = z

2+ am1 z + am2

with am1 = − ( e

− 1 + e− 1) = − 2 e− 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 = Bm Ao(1)

AR + B S = Am Ao(2)

As B is unstable must Bm = B B ′m. This makes (1) ⇔ B T = B B ′m Ao ⇔ T = B

′m Ao. 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 = b′m =

A(1) B(1)

=0.252.2

Further we have

( z2 + a1 z + a2)( z − 1)( z + r) + ( b0 z + b1)(s0 z2 + s1 z + s2)

= ( z2 + am1 z + am2)( z2+ ao1 z + ao2)


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with a1 = − 1, a2 = 0.25 and ao1 and ao2 chosen 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

θ = 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 and¯ S. This is difficult, see page 118 in the textbook. An indirect STR is

given by Eq. 3.24. Ao Am y = ¯ Ru + ¯ S y

with ¯ R = B − R, ¯ S = B − S, T = B ′m Ao. Furthermore we have

Ao Am ym = Ao Bmuc = Ao B B′muc = B Tuc =

T̄u c

y = ¯ R 1

Ao Amu

=u f + ¯ S

1 Ao Am

y

= y f ym = T̄

1 Ao Am

uc =uc f

ε = y − ym = ¯ Ru f + ¯ S y f − T̄uc f

Now estimate ¯ R, ¯ S and T̄ 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 = − γ uc e

ds0dt

= − γ ye

with e = y − ym = y − Gmuc.


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3.3 The process has the transfer function

G(s) =b

s + a ⋅

q

s + p

where p and q are known. The desired closed loop system has the transferfunction

Gm( s) =ω 2

s2 + 2ξ ω s + ω 2

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

H ( z) =b0 z + b1

z2 + a1 z + a2

H m( z) =bm0 z + bm1

z2 + am1 z + 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)

With h = 0.2 this gives

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

Furthermore we will assume that ω = 2 and ζ = 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

θ = bo b1 a1 a2 T

ϕ ( 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 + a1 z + a2)1 + b0( s0 z + s1) = z2 + am1 z + am2

z1

: a1 + b0s0 = am1 s0 =am1 − a1

b0

z0 : a2 + b0s1 = am2 s1 =am2 − a2

b0


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The controller is thus given by

R( z) = z + b1 / b0

S( z) = s0 z + s1

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

b0

2. B not canceled:The design equation becomes

( z2 + a1 z + a2)( z + r1) + ( b0 z + b1)( s0 z + s1) = ( z2+ am1 z + am2)( z + ao1)

Identification of coefficients of equal powers of z gives

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 have z = − 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 = Am Ao B+

Hence B + Am Ao y = ARy + B Sy = B Ru + B Sy

y = R B −

Ao Amu

u f

+ S B −

Ao Am y

y f


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0 10 20 30 40 50

−2

−1

0

1

2 uc y

0 10 20 30 40 50

−10

0

10

u

Figure 7. Simulation in Problem 3.3. Process output and control signalare shown for the indirect self-tuning regulator when the process zero iscanceled.

From this model R and S can be estimated. The polynomial T is thengiven by

T =to Ao Bm

B −

where to is chosen to give the correct steady state gain. Again we separate

two cases.1. Cancel B:If the polynomial B is canceled we have B + = z + b1 / b0, B − = b0. Fromthe analysis of the indirect STR we know that no observer is needed inthis case and that the controller has the structure deg R = deg S = 1.Hence

y(t) = R

bo

Amu( t)

+ S

bo

Am y(t)

Since bo is not known we include it in the polynomial R and S andestimate it. The polynomial R then is not monic. We have

y( t) = ( r0q + r1) 1 Am

u( t)

+ ( s0q + s1)

1 Am

y( t)

To obtain a direct STR we thus estimate

θ = r0 r1 s0 s1T


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0 10 20 30 40 50

−2

−1

0

1

2 uc 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. FurthermoreT has the form T ( q) = t0q where

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 have B + = 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 −

Ao Amu

u f

+ S B −

Ao Am y

y f

(∗)


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0 10 20 30 40 50

−2

−1

0

1

2 uc 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 u f and y f . Onepossibility is to rewrite ( *) as

y = R B −

R′

1 Ao Am

u

+ S B −

S′

1 Ao Am

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. 8–11 we show simulation whenthe model (*) is used with

B − = 1

B − = q

B − =q + 0.4

1.4

B − =q − 0.4

0.6


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0 10 20 30 40 50

−2

−1

0

1

2 uc y

0 10 20 30 40 50

−10

0

10 u

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 / b gives the desired result. Idea for STR: Estimate b anduse k = 1 / b̂. To estimate b introduce

s( s + 1) y = bu

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

y f

y = b 1( s + a)2

ϕ

u


25/47


0 10 20 30 40 50

−2

−1

0

1

2 uc 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

db̂

dt = Pϕ e = Pϕ ( y f − b̂ϕ )

dP

dt = α P−

Pϕ ϕ T

P = α P−

P2

ϕ 2

With b̂(0) = 1, P(0) = 100, α = − 0.1, and a = 1 we get the result shownin Fig. 14.


26/47


0 10 20 30 40 50

−2

−1

0

1

2 uc 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.


27/47


0 10 20 30 40 50

−2

−1

0

1

2 uc 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.


28/47


0 5 10 15 20−2

0

2 uc y

0 5 10 15 20

0

5

10 u

0 5 10 15 20

0.2

0.6

1 b

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


29/47



4.1 The estimate b̂ may 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 ≤ b̂ ≤ b1. This requires prior information about b:s values. Another possibility is to replace

1

b̂b y

b̂

b̂2 + P

where P is the variance of the estimate. Compare with the discussion of cautious control on pages 356–358.

4.10 Using ( 4.21) the output can be written as

yt+d = R∗

C∗ut +

S∗

C∗ yt +

R∗1C∗

et+do

= rout + f t

(∗)

Consider minimization of

J = y2t+d + ρ u2t ( +)

Introduce the expression ( *)

J = ( rout + f t)2+ ρ u2t

= ( r2o + ρ )u2t + 2rout f t + f

2t

= ( r2o + ρ )

u2t +

2rout f tr2o + ρ

+ f 2t

= ( r

2

o + ρ ) ut + ro f tr2o + ρ 2

−

r2o f 2t

r2o + ρ + f

2

t

Hence

J =1

r2o + ρ

f t +

r2o + ρ

rout

2+

ρ

r2o + ρ f 2t

=1

r2o + ρ

f t +

ro +

ρ

ro

ut

2+

ρ

r2o + ρ f 2t

=1

r2o + ρ

yt+d +

ρ

rout

2+

ρ

r2o + ρ f 2t

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

J 1 = yt+d + ρ ro

ut = f t +

ro + ρ ro

ut


30/47



5.1 The plant is

G( s) =1

s(s + a) =

B

A

The desired response is

Gm( s) =ω 2

s2 + 2ζ ω s + ω 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 = t0 Ao

This gives the updating rules

dr1

dt = γ e

1

Ao Amu

ds0

dt = γ e

p

Ao Am y

ds1

dt = γ e

1

Ao Am y

dt0

dt = − γ e

1

Ao Amuc

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

process zeros are cancelled we have

Ao Am y = AR1 y + bo Sy = B R1u + bo Sy

= bo( Ru + Sy)Further

Ao Am ym = Ao Bmuc = boTuc

Hence Ao Am e = Ao Am( y − ym) = bo( Ru + Sy − Tuc)

e =b

Ao Am( Ru + Sy − Tuc)

Since 1 / Ao Am is not SPR we introduce a polynomial D such that D / Ao Am is SPR. We then get

e =bD

Ao Am Ru f + Sy f − Tuc f where

u f =1

D u y f =

1 D

y uc f =1

D u c


31/47


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

˙ x =

− a 0

1 0

A p

x +

1

0

B p

y =

0 1 C

x

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

The closed loop system is

˙ x = ( A p − B p L) x + B Lr uc = Ax + Buc

y = C x

where A(θ ) = A p − B p L =

− a − θ 1 − θ 21 0

B(θ ) = B p Lr =

θ 30

The desired response is given by

˙ xm = Am xm + Bmuc

where

Am =

− 2ζ ω − ω 2

1 0

Bm =

ω 2

0

We have( A − Am)

T =

2ζ ω − a − θ 1 0ω 2 − θ 2 0

( B − Bm)

T =

θ 3 − ω 2 0Introduce the state error

e = x − xm

we get˙ e = x − xm = Ax + Buc − Am xm − Bmuc

= Am e + ( A − Am) x + ( B − Bm)uc(∗)

The error goes to zero if Am is stable and

A(θ ) − Am = 0 (∗)


32/47


B(θ ) − Bm = 0 ( +)

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

V = eT Pe + tr( A − Am)T Qa( A − Am)

+ tr( B − Bm)T Qb( B − Bm)

Notice thattr( A + B) = trA + trB

xT Ax = tr( xxT A) = tr( AxxT )

tr( AB ) = tr( B A)

we get

dV

dt = tr

P ˙ eeT + Pe ˙ eT + ˙ AT Qa( A − Am)

+ ( A−

Am)T

Qa ˙ A +

˙ B Qb( B

−

Bm) + ( B−

Bm)T

Qb ˙ B

( +)

But from ( *)

P ˙ eeT = P ( Am e + ( A − Am) x + ( B − Bm)uc) eT

Pe ˙ eT = Pe ( Am e + ( 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)T Qb ˙ B + PeuT c

HencedV

dt = eT PAm e + e

T AT m Pe

+ 2tr( A − Am) T

Qa ˙ A + Pe xT

+ 2tr( B − Bm) T

Qb ˙ B + PeuT c

Hence if the symmetric matrix P is chosen so that

AT m P + PAm = − Q

where Q is positive definite ( can always be done if Am is stable!) andparameters are updated so that

( A − Am)T

Qa ˙ A + Pe xT =0

( B − Bm)T

Qb ˙ B + Peu T c =0

( +)


33/47


we getdV

dt = − eT Qe

The equations for updating the parameters derived above can now be

used. This gives2ζ ω − a − θ 1 0ω 2

− θ 2 0

Qa − θ̇ 1 − θ̇ 20 0

+ Pe xT = 0θ 3 − ω 2 0Qb θ̇ 30

+ Peu c = 0Hence with Qa = I and Qb = 1

dθ 1dt

= p11 e1 x1 + p12 e2 x1 = ( p11 e1 + p12 e2) x1

dθ 2dt

= p11 e1 x2 + p12 e2 x2 = ( p11 e1 + p12 e2) x2

dθ 3

dt = − ( p11 e1 + p12 e2)uc

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

AT m P + PAm = − Q

Choosing ζ = 0.707,ω = 2 and

Q =

41.2548 11.313711.3137 16.0000

we get

P = 4 22 16The parameter update laws become

dθ 1dt

= (4 e1 + 2 e2) x1

dθ 2dt

= (4 e1 + 2 e2) x2

dθ 3dt

= − (4 e1 + 2 e2)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 of the SPR rule isdθ

dt = − γ 1

d

dt(uc e) − γ 2uc e (∗)


34/47


0 10 20 30 40

−2

−1

0

1

2 States 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 parameters θ 3 ( full), θ 1 (dashed), and θ 2 ( dot-ted).

u c

ym

e

y

1s

1s

Σ

−

θ

Π

θ 0

Figure 16. Block diagram in Problem 5.2.

To derive the error equation we notice that

dym

dt = θ 0

uc

dy

dt = θ uc


35/47


Hencede

dt = (θ − θ 0)uc

we getd2 e

dt2 =dθ

dt uc + (θ −

θ 0

)duc

dt

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

d2 e

dt2 = − γ 1

duc

dt e + u c

de

dt

uc − γ 2u

2c e + (θ − θ

0)duc

dt

Henced2 e

dt2 + γ 1u

2c

de

dt +

γ 1uc

duc

dt + γ 2u

2c

e = (θ − θ 0)

duc

dt

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

d2 e

dt2 + γ 1u

2

c

de

dt + γ 2u

2

c

e = 0

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

γ 1u2c = 2ζ ω γ 1 = 2ζ ω / u

2c

γ 2u2c = ω

2 γ 2 = ω 2

/ u2c

This gives an indication of how the parameters γ 1 and γ 2 should beselected. The analysis was based on the assumption that uc was 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 ω and ζ . Figure 17 shows whathappens when the period of the square wave is 20 and ω = 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 if a1 ≥ 0, a2 ≥ 0, b0 ≥ 0, b1 ≥ 0, and b2 ≥ 0. Consider

G( iω ) = B( iω )

A( iω ) ⋅

A( − iω )

A( − iω )

The condition Re G( iω ) ≥ 0 is equivalent to Re ( B( iω ) A( − iω ) ) ≥ 0. But

g(ω ) = Re

( − b0ω 2+ iω b1 + b2)( − ω

2− iω a1 + a2)

= b0ω

4+ (a1b1 − b0a2 − b2)ω

2+ a2b2


36/47


0 5 10 15 20

0

5

10

Process and model output

0 5 10 15 20

0

0.5

1

1.5 Estimated 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 parameter θ when ω = 0.5 (full), 1 (dashed), and 2 (dotted) for ζ = 0.7. Right top:Process and model outputs, Right bottom: Estimated parameter θ whenζ = 0.4 ( full), 0.7 ( dashed), and 1.0 ( dotted) for ω = 1.

Completing the squares the function can be written as

g(ω ) = b0

ω 2 +

a1b1 − b0a2 − b2

2b0

2+ a2b2 −

( a1b1 − b0a2 − b2)2

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

a1b1 − b0a2 − b2 ≥ 0 ( i)

If b0 > 0 the function g(ω ) is nonnegative for all ω 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


37/47


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 parameter θ when ω = 0.5 (full), 1 (dashed), and 2 (dotted) for ζ = 0.7. Right top:Process and model outputs, Right bottom: Estimated parameter θ whenζ = 0.4 ( full), 0.7 ( dashed), and 1.0 ( dotted) for ω = 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 = C1 x

where

B1 =

1

0...

0

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

AT P + PA = − Q (∗)


38/47


Define C1 as

C1 = BT P =

p11 p12 . . . p1n According to the Kalman-Yacobuvich Lemma the transfer function

G1( s) = C1(sI − A)−

1 B1

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 Q positive.

3. Choose B as B( 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

A u

and let the controller be Ru = Tuc − sy

The basic design equation is then

AR + B S = Am Ao (∗)

In this case we have A = ( s + a)( s + p)

B = bq

Am = s2+ 2ζ ω s + ω 2

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

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

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

dt + r1u = t0uc − s0 y − s1

dy

dt


39/47


It has four parameters that have to be updated r1, t0, s0, and s1. 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. Putting s = − p in ( +) gives

− s0 p + 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.With a0 = p equation ( +) simplifies to

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

+ 2ζ ω s + ω 2

Introducing A′( s) = s + a S′( s) = s0 T

′= t0

we get

y = B T ′

A′ R + B S′ uc

e

t0=

B

A′ R + B S′ uc

B

Amuc

e

s0= −

B T ′ B

( A′ R + B S′)2 uc = −

B

A′ R + B S′ y −

B

Am y

e r1

= − A′ B T

′

( A′ R + B S′)2 uc = − A

′

A′ R + B S′ y − A

′

Am y

= − A′

Am

B

A u = −

B

Am( s + p) u

The MIT rule then gives

dr1

dt = γ

1

( s + p) Amu

e

ds0

dt = γ

1 Am

y

e

dt0

dt =− γ 1 Am uc e

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


40/47


0 20 40 60 80 100

−2

−1

0

1

2 Model 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 + kb uc − ym

Hence e

k =

b

p2 + p + kb uc −

b2k

( p2 + p + k b)2 uc

=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 = − γ ′

e

k e = − γ ′b

γ

1

p2 + p + 1 ( uc − y)

e


41/47


0 20 40 60−2

0

2 ym y, gamma = 0.05

0 20 40 600

1

2 k, gamma = 0.05

0 20 40 60

−2

0

2 ym y, gamma = 0.3

0 20 40 60

0

1

2 k, gamma = 0.3

0 20 40 60

−2

0

2 ym y, gamma = 1

0 20 40 60

0

1

2 k, gamma = 1

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

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


42/47



6.1 The process is given by

G( s) =b

s( s + a)

and the regressor filter should be

G f (s) =1

A f (s) =

1 Am( s)

=1

s2 + 2ζ ω s + ω 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 ξ . The process parame-ters are contained in θ , and the controller parameters in ϑ . The relationbetween these are given by

ϑ =

r1

s0

s1

t0

=

2ζ ω + ao − â

( ao2ζ ω + ω 2 − âr1) / b̂

ω 2 ao / b̂

ω 2 / b̂

= χ (θ )

To find A(ϑ ), B(ϑ ), C(ϑ ) and D(ϑ ) we start by finding realizations for y, y f , u f and the controller. We get

d

dt ˙ y

y = − a 0

1 0 ˙ y y +

b

0u

d

dt

˙ y f y f

= − 2ζ ω − ω 21 0

˙ y f y f

+ 10

yand

d

dt

u̇ f u f

= − 2ζ ω − ω 21 0

u̇ f u f

+ 10

uand the control law can be rewritten as

u = − s0 y + t0uc +− s1 + r1s0

p + r1 y +

aot0 − t0r1

p + r1uc

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

˙ x = − r1 x + ( − s1 + r1s0) y + ( aot0 − r1t0)uc

u = x − s0 y + t0uc


43/47


Combining the states results in

d

dt

˙ y

y

˙ y f

y f

u̇ f

u f

x

=

− a − bs0 0 0 0 0 b

1 0 0 0 0 0 0

0 1 − 2ζ ω − ω 2 0 0 0

0 0 1 0 0 0 0

0 − s0 0 0 − 2ζ ω − ω 2 1

0 0 0 0 1 0 0

0 − s1 + r1s0 0 0 0 0 − r1

×

˙ y

y

˙ y f y f

u̇ f

u f

x

+

bt0

0

0

0

t0

0

aot0 − r1t0

uc

which defines the relation

dξ

dt = A(ϑ )ξ + B(ϑ )uc

Now we need to express e and ϕ in the states so that we find the C and D matrices. The estimator tries to find the parameters in

y f =b

p( p + a) u f

which is rewritten as

p2 y f =

− py f u f

ab

= ϕ T θ 0Clearly

ϕ =

0 0 − 1 0 0 0 00 0 0 0 0 1 0

˙ y

y

˙ y f y f

u̇ f

u f

x

and e = p2 y f − ϕ

T θ̂ = − 2ζ ω ˙ y f − ω 2 y f + y + â ˆ y f − b̂u f


44/47


If we use the relations â = − r1 + 2ζ ω + ao and b̂ = ω 2 / t0 then e can bewritten as

e = − 2ζ ω ˙ y f − ω 2 y f + y + ( − r1 + 2ζ ω + ao) ˙ y f − ω 2

t0u f

=

0 1 − r1 + ao 0 0 − ω 2t0 0

˙ y

y

˙ y f y f

u̇ f

u f

x

Combining the expressions for ϕ and e gives

eϕ

=

0 1 − r1 + ao 0 0 − ω 2

t00

0 0 − 1 0 0 0 0

0 0 0 0 0 1 0

˙ y

y

˙ y f y f

u̇ f

u f

x

= C(ϑ )ξ

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

dθ

dt = Pϕ e

dP

dt = α P − Pϕ ϕ T P

where P is a 2 × 2 matrix and e and ϕ 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


45/47


To use the averaged equations we need

avg

( Gmuc) ( Guc) = avg

v

b

( p + b)2 v

=12v2 ⋅

b2

b2 + ω 2 cos2ϕ =a2u2

0b2

2( a2 + ω 2)( b2 + ω 2)2cos2 ϕ − 1

=

a2b2u202(a2 + ω 2)( b2 + ω 2)

2b2

ω 2 + b2− 1

=

a2b2u20( b2

− ω 2)

2( a2 + ω 2)( b2 + ω 2)2

where we have introduced

uc = u0 sin ω t

v =a

p + auc

ϕ = atanω

b

Similarly we have

avg ( uc Guc) =u202

G cos (2ϕ + ϕ 1)

=u202 ⋅

ab2√

a2 + ω 2 ⋅ ( b2 + ω 2)(cos2ϕ cos ϕ 1 − sin2ϕ sin ϕ 1)

=u20ab

2( ab2 − ω 2(a + 2b))2( a2 + ω 2)( b2 + ω 2)2

whereϕ = atan

ω

b ϕ 1 = atan

ω

a

It follows from the analysis on page 302–304 that the MIT rule gives astable system as long as ω < b while the stability condition for the SPRrule is

ω <

a

a + 2b b

with b = 10a we getω MI T = 10a

ω SP R = 2.18a

6.10 The adaptive system was designed for a process with the transfer function

Ĝ(s) =b

s + a (1)

The controller has the structure

u = θ 1uc − θ 2 y (2)


46/47


The desired response is

Gm( s) =bm

s + am(3)

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

Gcl =bθ 1

s + a + bθ 2

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

bθ 1 = bm

a + bθ 2 = am

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

a = am−

bmθ 2/

θ 1b = bm / θ 1

The parameters a and b can thus be interpreted as the parameters of themodel the controller believes in. Inserting the expressions for θ 1 and θ 2from page 318 we get

a(ω ) =

229 − 31ω 2

259 − ω 2

b(ω ) =458

259 − ω 2

( +)

When

ω = 22931 = 2.7179we get a(ω ) = 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 of the transfer function

Ĝ =b

s + a (∗∗)

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

Ĝ(2.7179i) = − 0.6697i.When ω =√

259 = 16.09 we get infinite values of a and b. Notice that G( i

√

259) = − 0.0587 that is real and negative. Theonly way to make Ĝ( iω ) negative and real is to have infinite large values


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of a and b. 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( +).

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Adaptive Control (Karl J. Astrom) 2nd Ed - Solution Manual

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