Solution Manual for Adaptive Control, 2nd Edition

46 Pages·1994·0.277 MB·English

by  Karl Johan Åström, Björn Wittenmark

Solution Manual
for
Adaptive Control
Second Edition
Karl Johan Åström
Björn Wittenmark
Preface
ThisSolutionManualcontainssolutionstoselectedproblemsinthesecond
edition of Adaptive Control published by Addison-Wesley 1995, ISBN 0-201-
55866-1.
PROBLEM SOLUTIONS
SOLUTIONS TO CHAPTER 1
1.5 Linearizationof the valve shows that
D v = 4v3D u
0
The loop transfer function is then
G (s)G (s)4v3
0 PI 0
where G is the transfer function of a PI controller i.e.
PI
(cid:18) (cid:19)
G (s) = K 1 + 1
PI sT
i
The characteristicequationfor the closed loop system is
sT(s+ 1)3 + K (cid:215) 4v3(sT + 1) = 0
i 0 i
with K = 0.15 and T = 1 we get
i
(cid:0) (cid:1)
(s+ 1) s(s+ 1)2 + 0.6v3 = 0
0
(s+ 1)(s3+ 2s2 + s + 0.6v3) = 0
0
The root locus of this equation with respect to v is sketched in Fig. 1.
o
According to the Routh Hurwitz criterion the criticalcase is
r
10
0.6v3 = 2 (cid:222) v = 3 = 1.49
0 0 3
Since the plant G has unit static gain and the controller has integral
0
action the steady-state output is equal to v and the set point y . The
0 r
closed-loop system is stable for y = u = 0.3 and 1.1 but unstable for
r c
y = u = 5.1. Compare with Fig. 1.9.
r c
1
2 Problem Solutions
2
1.5
1
0.5
s
xi
A
g 0
a
m
I
-0.5
-1
-1.5
-2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Real Axis
Figure 1. Root locus inProblem 1.5.
1.6 Tune the controller using the Ziegler-Nichols closed-loop method. The
frequencyw , where the process has 180○ phase lag is first determined.
u
ThecontrollerparametersarethengivenbyTable8.2onpage382where
1
Ku = G (iw )j
0 u
j
we have
G (s) = e−s/q
0 1 + s q
/
w w
argG (iw ) = arctan = p
0 −q − q −
q w G (iw ) K T
0 i
0.5 1.0 0.45 1 5.24
2.0 0.45 1 2.62
4.1 0.45 1 1.3
A simulationof the systemobtainedwhen the controller is tuned for the
smallestflow q = 0.5 is shown Fig. 2. The Ziegler-Nichols method is not
the best tuning method in this case. In the Fig. 3 we show results for
Solutions toChapter 1 3
Process output
2
1
0
0 10 20 30 40
Control signal
2
1
0
0 10 20 30 40
Figure 2. SimulationinProblem1.6. Process output andcontrol signal
are shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). The
controller is designed for q = 0.5.
controllerdesignedfor q = 1andinFig.4whenthecontrollerisdesigned
for q = 2.
1.7 Introducing the feedback
u = k y
2 2
−
the system becomes
8 9 8 9 8 9
>> 1 0 0>> >>0>>8 9 >>1>>
ddxt = >>>>:−0 −3 0>>>>;x−k2>>>>:2>>>>;:1 0 1;x + >>>>:0>>>>;u1
0 0 1 1 0
8 9−
y = :1 1 0;x
1
The transfer function from u to y is
1 1
8 9 8 9
8 9>>s + 1 0 0 >>−1>>1>>
G(s) = :1 1 0;>>>>: 2k2 s + 3 2k2 >>>>; >>>>:0>>>>;
k 0 s + 1+ k 0
2 2
s2 + (4 k )s+ 3+ k
= (s+ 1)(s−+ 32)(s+ 1 + k2)
2
4 Problem Solutions
Process output
2
1
0
0 10 20 30 40
Control signal
2
1
0
0 10 20 30 40
Figure 3. SimulationinProblem1.6. Process output andcontrol signal
are shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). The
controller is designed for q = 1.
The staticgainis
G(0) = 3 + k2
3(1+ k )
2
Solutions toChapter 1 5
Process output
2
1
0
0 10 20 30 40
Control signal
2
1
0
0 10 20 30 40
Figure 4. SimulationinProblem1.6. Process output andcontrol signal
are shown for q = 0.5 (full), q = 1 (dashed), and q = 2 (dotted). The
controller is designed for q = 2.
6 Problem Solutions
SOLUTIONS TO CHAPTER 2
2.1 The function V can be written as
Xn Xn
V(x (cid:215)(cid:215)(cid:215) x ) = x x (a + a ) 2+ b x + c
1 m i j ij ji i i
/
i,j=1 i=1
Takingderivative with respect to x we get
i
Xn
(cid:127)V = (a + a )x + b
x ij ji j i
(cid:127) i j=1
In vector notationthis can be written as
grad V(x) = (A+ AT)x+ b
x
2.2 The model is
8 9
8 9
y = j Tq + e = :u u ;>>:b0>>; + e
t t t t t−1 b t
1
Theleastsquaresestimateisgivenasthesolutionofthenormalequation
8 P P 9 8 P 9
qˆ = (F TF )−1F TY = >>:P u2t Putut−1>>;−1>>:P utyt >>;
uu u2 u y
t t−1 t−1 t−1 t
( )
a Input is a unit step (cid:26)
1 t ‡ 0
u =
t
0 otherwise
Evaluatingthe sums we get
8 9 8 9
>>XN >> >>XN >>
8 9 >> y >> 8 9>> y >>
qˆ = >>>: N N −1>>>;−1>>>>>> 1 t>>>>>> = >>>>: 1 −N1 >>>>;>>>>>> 1 t>>>>>>
N 1 N 1 >>>XN >>> 1 >>>XN >>>
− − >: y >; − N 1 >: y >;
t − t
8 9 2 2
>>> y1 >>>
qˆ = >>>>>: 1 XN y y >>>>>;
t 1
N 1 −
− 2
The estimationerror is
8 9
>>> e1 >>>
qˆ−q = >>>>>: 1 XN e e >>>>>;
t 1
N 1 −
− 2
Solutions toChapter 2 7
Hence
8 9 8 9
E(qˆ q )(qˆ q )T = (F TF )−1(cid:215) 1 = >>>>: 1 −N1 >>>>; >>: 1 −1>>;
− − 1 → 1 1
− N 1 −
−
when N . Notice thatthe varianceof the estimatesdo not go to
zero as N→∞ . Consider, however, the estimate of b + b .
0 1
→∞ 8 98 98 9
E(bˆ0 + bˆ1 b0 b1) = :1 1;>>>>: 1 −N1 >>>>;>>:1>>; = 1
− − 1 1 N 1
− N 1 −
−
With a step input it is thus possible to determine the combination
b +b consistently.The individualvaluesof b and b can,however,
0 1 0 1
not be determined consistently.
(b) Input u is white noise with Eu2 = 1 and u is independent of e.
Eu2 = 1 Euu = 0
t t t−1
8 9
cov(qˆ q ) = 1(cid:215) E(F TF )−1 = 8>>:N 0 9>>;−1 = >>>>>>> N1 0 >>>>>>>
− 0 N 1 : 1 ;
− 0
N 1
−
In this case it is thus possible to determine both parameters consis-
tently.
2.3 Data generatingprocess:
y(t) = b u(t) + b u(t 1) + e(t) = j T(t)q 0+ e¯(t)
0 1
−
where
j T(t) = u(t), q 0 = b
0
and
e¯(t) = b u(t 1) + e(t)
1
−
Model:
yˆ(t) = bˆu(t)
or
y(t) = bˆu(t) + e(t) = j T(t)qˆ + e(t)
where
e(t) = y(t) yˆ(t)
−
The leastsquares estimateis given by
8 9
( )
>> e¯ 1 >>
F TF (qˆ−q 0) = F TEd Ed = >>>>>>: ... >>>>>>;
( )
e¯ N
8 Problem Solutions
XN
1 F TF = 1 u2(t) Eu2 N
N N → →∞
1
XN XN
1 F TE = 1 u(t)e¯(t) = 1 u(t)(b u(t 1) + e(t))
N d N N 1 −
1 1
b E(u(t)u(t 1)) + E(u(t)e(t)) N
1
→ − →∞
( )
a (cid:26)
u(t) = 1 t ‡ 1
0 t < 1
E(u2) = 1 E(u(t)u(t 1)) = 1 Eu(t)e(t) = 0
−
Hence
bˆ = qˆ q 0 + b = b + b N
1 0 1
→ →∞
ˆ
i.e. b converges to the stationarygain
( )
b
u(t) ˛ N(0,s )(cid:222) Eu2 = s 2 Eu(t)u(t 1) = 0 Eu(t)e(t) = 0
−
Hence
ˆ
b b N
0
→ →∞
2.6 The model is
8 98 9 8 9
y = j Tq + e = : y u ;>>:a>>;+:e + ce ;
t t t | − t−1{z t−1 } b | t {z t−1 }
| {z }
j tT q et
Theleastsquaresestimateisgivenbythesolutiontothenormalequation
( )
2.5 .The estimationerror is
qˆ q = (F TF )−1F Te =
− 8 P P 9 8 P P 9
>>>: P y2t−1 − Pyt−1ut−1>>>;−1 >>>:−P yt−1et−cP yt−1et−1>>>;
y u u2 u e + c u e
− t−1 t−1 t−1 t−1 t t−1 t−1
Notice that F T and e are not independent. u and e are independent, y
t t t
depends on e, e , e ,... and y depends on u ,u ,....
t t−1 t−2 t t−1 t−2
Takingmean values we get
E(qˆ q ) = E(F TF )−1E(F Te)
−
To evaluatethis expression we calculate
8 P P 9 8 9
E>>>: P y2t−1 − Pyt−1ut−1>>>; = N>>>:Ey2t 0 >>>;
y u u2 0 Eu2
− t−1 t−1 t−1 t