Example Problems and Solutions 717
The same system may be represented by the following state-space equation, which is in the
observable canonical form:
(9–122)
(9–123)
Show that the state-space representation given by Equations (9–120) and (9–121) gives a sys-
tem that is state controllable, but not observable. Show, on the other hand, that the state-space rep-
resentation defined by Equations (9–122) and (9–123) gives a system that is not completely state
controllable, but is observable. Explain what causes the apparent difference in the controllability
and observability of the same system.
Solution.Consider the system defined by Equations (9–120) and (9–121). The rank of the
controllability matrix
is 2. Hence, the system is completely state controllable. The rank of the observability matrix
is 1. Hence the system is not observable.
Next consider the system defined by Equations (9–122) and (9–123). The rank of the
controllability matrix
is 1. Hence, the system is not completely state controllable. The rank of the observability matrix
is 2. Hence, the system is observable.
The apparent difference in the controllability and observability of the same system is caused
by the fact that the original system has a pole-zero cancellation in the transfer function. Referring
to Equation (2–29), for D=0we have
If we use Equations (9–120) and (9–121), then
[Note that the same transfer function can be obtained by using Equations (9–122) and (9–123).]
Clearly, cancellation occurs in this transfer function.
=
s+0.8
(s+0.8)(s+0.5)=
1
s^2 +1.3s+0.4[0.8 1]B
s+1.3- 0.4
1
sRB
0
1
R
G(s)=[0.8 1]B
s
0.4- 1
s+1.3R
- 1
B
0
1
R
G(s)=C(s I-A)-^1 B[CA C*]=B
0
1
1
- 1.3
R
[BAB]= B
0.8
1
- 0.4
- 0.5
R
[CA C*]= B
0.8
1
- 0.4
- 0.5
R
[BAB]=B
0
1
1
- 1.3
R
y=[ 0 1 ]B
x 1
x 2R
B
x# 1
x# 2R =B
0
1
- 0.4
- 1.3
RB
x 1
x 2R + B
0.8
1
Ru