Chapfer 7
An Illustrative Example 7.1 Sfatemenf of the Problem A thirdorder, timeinvariant process is considered for illustration. The reference input Y to the control system is assumed to be a step function of magnitude ranging from  n to n. At the control input, there occurs an independent random disturbance n with mean value equal to E . The disturbance n may be expressed as n = v + E , where v denotes an independent random variable with zero mean. This control process is characterized by
+
x(t) = Ax(t)
+ dm,(t) + n(t)
(7.11)
in which the state vector x is
x = [c
XI
xz
Y]’
the disturbance vector n is n = [0 0
z,
01’
and the other symbols are defined by m, 1
0 ‘0l
=m
+
E,
0 0
2
and
0
d = [0 0 1 01’
0 O 0 I
The component c of the state vector x is the output variable. A state diagram describing this digital control system is shown in Fig. 7.11, 95
96
7. A N ILLUSTRATIVE EXAMPLE
where the state variable x2 is inaccessible for measurement and observation. It is required to determine the optimum control law so that the following error criterion (7.12)
. r
c
Process
m
U
U
(b)
FIG.7.11.
(a) Block diagram for the illustrative system. ( b ) State variable diagram for the system of the illustrative example.
is minimized. The output c shown in Fig. 7.11 can be directly measured, but the measurement of x1 introduces some noise which can be considered as an independent random variable with zero mean and standard deviation equal to 4.
7.2 Design of Optimum Digital Control The statetransition equation describing this control process is 
~ ( +k 1T) = $(T)x(kT)+ g(T)m,(kT)+ u(kT)
(7.21)
7.2 DESIGN O F OPTIMUM DIGITAL CONTROL
97
in which
g ( T )= [g,(T) g z m g,(T) 01’ u ( k T ) = [u,(kT) U , ( k T ) U 3 ( k T ) 01’
The elements of these matrices are y12(T ) = 1  e
*
vZ2= eP)~,(T ) =
+ &e
4  e
qZ3(T) = e
*  e**
P),, T() = e
2T
z7’
+ T + 2e  &eZT) g,(T) = &(I  2e + e2*)
g,(T) = &(#
g3(T)= t(l  eZT)
u,(kT) =
(k+lJT
e
‘1
dt,
l k 1’
u,(kT) =  ( k U 3 ( k T )=
+ l)T
[l,
e
dt,
[*
(k+l)T
e
dt,
lk1’
@+I)* e%klTTs)
!I*
e2ra v(t3) dz,
e2Tnv ( t Z )d t ,
~ ( t )
IkT
Assuming T = 1, the transition matrix is 1 0.632 0 . 2
0
(7.22)
98
7. A N ILLUSTRATIVE EXAMPLE
and g ( T ) = [1.582 0.2 0.433 01'
(7.23)
The expected value of u(kT)is zero, since v ( t ) has zero mean. The performance index of Eq. (7.12) may be expressed as (7.24)
[.
where the matrix Q is
1 0 0  1
0 O 0 O 0 O 1
0 0
;]
(7.25)
If all the state variables are measurable, the optimum control law is obtained from Eqs. (3.318) and (3.44), m,"(k) = B x ( k )
(7.26)
where =
[g'(T)Qg(T)Ilg'(T)Q~(T)
(7.27)
and m"(k)= m,"(k) &
(7.28)
Using Eqs. (7.22), (7.23) and (7.25), the matrix B is determined: B
=
[ 0.597
 0.398
 1.26
0.5971
(7.29)
Thus, if all the components of the state vector can be directly measured, the optimum control is given by mlo(k)= 0.597 [ ~ ( k) r(k)]  0.398~,(k) 1.26~2(k) (7.210)
Since the state variable x2 is not accessible for measurement, Eq. (7.210) does not provide a complete solution.
7 . 2 DESIGN OF OPTIMUM DIGITAL CONTROL
99
Making use of the results of Chapters 4 and 5, one needs to determine the best estimate of the state variables from the measurable quantities. It has been shown in Chapter 4 that the optimum control law is h l o ( k / k )= B f i ( k / k )
(7.21 1)
where i ( k / k ) denotes the best estimate of the state variables, Let the measurable state vector be
where
Y W = [Y,(k)
YzW
Y,(k)l'
w(k) = [O w,(k) 01'
[a E E ,I
(7.213) (7.214)
and the measurement matrix is
M = O l 0 0
(7.215)
The recurrence relationships for the optimum estimator which determines the best estimate i ( k / k ) from the measured quantities of y ( k ) are C(k
+ 1) = +[I  A " ( k ) M ] C ( k ) [I A"(k)M]'+' + +A"(k)W(k)A"'(K)+'+ R ( k ) A " @ )= [C(k)M'][MC(k)M' + W ( k ) ]  l
(7.216) (7.21 7)
where W ( k )= Ew(k)w'(k)
R(k) = Eu(k)u'(k)
Assume that the measurement starts at t = 0. The quantities c(0) and r(0) are both independent random variables, uniformly dis
100
7 . AN ILLUSTRATIVE EXAMPLE
tributed between  n and +n. At this starting point, x,(O) and x2(0) are zero. E{c(O)}= E{r(O)}= 0 E{c2(0)}=
1:
c2 P(c) dc = n2/3= 3.29
n
E{r2(0)}= 3.29
Since, initially, i ( O /  1) = 0, 3.29 0 0 0 (7.21 8)
0 0 3.29
For the measurement noise, E{ml(k)}= 0
and
E{m12(k)}= 4
Hence
The external random disturbance v(d) has zero mean and its autocorrelation function is given by
R(k) = Eu(k)u’(k)
:I
7 . 2 DESIGN O F OPTIMUM DIGITAL CONTROL 711
‘12
‘13
721
‘22
‘23
‘31
732
733
0
0
0
=I,
101
0
0
.1
0.0073
0.016 0.026 0
0.010
0.040 0.075 0
0.056
0.116 0.199 0 0
0
Starting with the known values of C(O), one obtains 3.29 0 0
0 3.29 0.304 0 0

4 0 0 0.304
Hence
(7.219)
Then from Eq. (7.216),
.I
C(1) = d [ l  A ” ( O ) M ] C ( O )[ I  A o ( 0 ) M ] ’ ~ ’ +~ A o ( 0 ) W ( O ) A o ’ ( O )R( ~ ’0) +
=[.
0.0073 0.016 0.026 0 0.010
0.040 0.075 0
0.056
0.116 0.199 0 0
0
(7.220)
102
7. AN ILLUSTRATIVE EXAMPLE
Substituting into Eq. (7.217) yields the value of A"(1). Thus, by repeated application of Eqs. (7.216) and (7.217), all values of A " ( k ) can be found. The best estimate ? ( k / k ) is then determined from Eqs. (5.312), (5.323), and (5.324).