CONTROLLABILITY AND OBSERVABILITY

10.5 CONTROLLABILITY AND OBSERVABILITY

Consider a diagonalized state-space description of a system

and

We shall assume that all N eigenvalues ? 1 ,? 2 , ..., λ N are distinct. The state equations (10.78a) are of the form

If (the mth row in matrix ) are all zero, then and the variable z m is uncontrollable because z m is not coupled to any of the inputs. Moreover, z m is decoupled from all the remaining (N − 1) state

variables because of the diagonalized nature of the variables. Hence, there is no direct or indirect coupling of z m with any of the inputs, and the system is uncontrollable. In contrast, if at least one element in the mth row of is nonzero, z m is coupled to at least one input and is therefore controllable. Thus, a system with a diagonalized state [Eqs. (10.78)] is completely controllable if and only if the matrix has no row of zero

elements. The outputs [ Eq. (10.78b) ] are of the form

If ĉ im = 0, then the state z m will not appear in the expression for y i . Since all the states are decoupled because of the diagonalized nature of the equations, the state z m cannot be observed directly or indirectly (through other states) at the output y i . Hence the mth mode e λmt will not be observed at the output y i . If c 1m ,c 2m , ..., ĉ km (the mth column in matrix Ĉ) are all zero, the state z m will not be observable at any of the k outputs,

and the state, z m is observable. In contrast, if at least one element in the mth column of Ĉ is nonzero, z m is observable at least at one output. Thus, and the state, z m is observable. In contrast, if at least one element in the mth column of Ĉ is nonzero, z m is observable at least at one output. Thus,

If the state-space description is not in diagonalized form, it may be converted into diagonalized form using the procedure in Example 10.10 . It is also

possible to test for controllability and observability even if the state-space description is in undiagonalized form. [ 1 ] , [ 2 ]

EXAMPLE 10.11

Investigate the controllability and observability of the systems in Fig. 10.8 .

Figure 10.8

In both cases, the state variables are identified as the two integrator outputs, q 1 and q 2 . The state equations for the system in Fig. 10.8a are

and Hence

Therefore and

We shall now use the procedure in Section 10.4-1 to diagonalize this system. According to Eq. (10.74b) , we have

The solution of this equation yields

Choosing p 11 = 1 and p 21 = 1, we have

and

All the rows of are nonzero. Hence the system is controllable. Also, All the rows of are nonzero. Hence the system is controllable. Also,

The first column of Ĉ is zero. Hence the mode z 1 (corresponding to λ 1 = 1) is unobservable. The system is therefore controllable but not observable. We come to the same conclusion by realizing the system with the diagonalized state variables z 1 and z 2 , whose state equations are

According to Eqs. (10.80) and (10.81) , we have

and Figure 10.9a shows a realization of these equations. It is clear that each of the two modes is controllable, but the first mode (corresponding to λ = 1)

is not observable at the output.

Figure 10.9: Equivalents of the systems in Fig. 10.8 . The state equations for the system in Fig. 10.8b are

and Hence

so that λ 1 = −1, λ 2 = 1, and

Diagonalizing the matrix, we have

The solution of this equation yields p 11 = −p 12 and p 22 = 0. Choosing p 11 = −1 and p 21 = 1, we obtain

and

The first row of is zero. Hence the mode corresponding to λ 1 = 1 is not controllable. However, since none of the columns of Ĉ vanish, both modes are observable at the output. Hence the system is observable but not controllable.

We reach the same conclusion by realizing the system with the diagonalized state variables z 1 and z 2 . The two state equations are

From Eqs. (10.83) and (10.84) , we have

and thus

Figure 10.9b shows a realization of these equations. Clearly, each of the two modes is observable at the output, but the mode corresponding to λ 1 = 1 is not controllable.

COMPUTER EXAMPLE C10.6

Repeat Example 10.11 using MATLAB. a. System from Fig. 10.8a .

>> A = [1 0;1 -1]; B = [1; 0]; C = [1 -2]; >> [V, Lambda] = eig(A); P=inv(V); >> disp('Part (a):'), Bhat = P*B, Chat = C*inv(P) Part (a):

Bhat = -0.5000 1.1180

Chat = -2 0 Since all the rows of Bhat

are nonzero, the system is controllable. However, one column of Chat (Ĉ) is zero, so one made is unobservable.

b. System from Fig. 10.8b . >> A = [-1 0;-2 1]; B = [1; 1]; C = [0 1];

>> [V, Lambda] = eig(A); P=inv(V); >> disp('Part (b):'), Bhat = P*B, Chat = C*inv(P), Part (b):

Bhat = 0 1.4142

Chat = 1.0000 0.7071 One of the rows of Bhat

is zero, so one mode is uncontrollable. Since all of the columns of Chat (Ĉ) are nonzero, the system is observable.

10.5-1 Inadequacy of the Transfer Function Description of a System

Example 10.11 demonstrates the inadequacy of the transfer function to describe an LTI system in general. The systems in Fig. 10.8a and 10.8b both Example 10.11 demonstrates the inadequacy of the transfer function to describe an LTI system in general. The systems in Fig. 10.8a and 10.8b both

Yet the two systems are very different. Their true nature is revealed in Fig. 10.9a and 10.9b , respectively. Both the systems are unstable, but their transfer function H(s) = 1/(s + 1) does not give any hint of it. Moreover, the systems are very different from the viewpoint of controllability and observability. The system in Fig. 10.8a is controllable but not observable, whereas the system in Fig. 10.8b is observable but not controllable.

The transfer function description of a system looks at a system only from the input and output terminals. Consequently, the transfer function description can specify only the part of the system that is coupled to the input and the output terminals. From Fig. 10.9a and 10.9b we see that in both cases only a part of the system that has a transfer function H(s) = 1/(s + 1) is coupled to the input and the output terminals. This is why both systems have the same transfer function H(s) = 1/(s + 1).

The state variable description ( Eqs. 10.79 and 10.82) , on the other hand, contains all the information about these systems to describe them completely. The reason is that the state variable description is an internal description, not the external description obtained from the system behavior at external terminals.

Apparently, the transfer function fails to describe these systems completely because the transfer functions of these systems have a common factor s − 1 in the numerator and denominator; this common factor is canceled out in the systems in Fig. 10.8 , with a consequent loss of the information. Such a situation occurs when a system is uncontrollable and/or unobservable. If a system is both controllable and observable (which is the case with most of the practical systems) the transfer function describes the system completely. In such a case the internal and external descriptions are equivalent.

[ 2 ] Zadeh, L., and C. Desoer. Linear System Theory. McGraw-Hill, New York, 1963.