[10.6 CONTROLLABILITY AND](#page-14-0) OBSERVABILITY

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10.6 CONTROLLABILITY AND OBSERVABILITY

Consider a diagonalized state-space description of a system

$$\dot{z} = \Lambda z + \dot{B}x \quad \text{and} \quad Y = \dot{C}z + Dx \tag{10.57}$$

We shall assume that all N eigenvalues λ1, λ2, … , λ*^N* are distinct. The state equations in Eq. (10.57) are of the form

$$\dot{z}_m = \lambda_m z_m + \hat{b}_{m1} x_1 + \hat{b}_{m2} x_2 + \cdots + \hat{b}_{mj} x_j \qquad m = 1, 2, \ldots, N$$

948 CHAPTER 10 STATE-SPACE ANALYSIS

If bˆm1, bˆm2, … , bˆmj (the mth row in matrix Bˆ ) are all zero, then

$$\dot{z}_m = \lambda_m z_m$$

and the variable zm is uncontrollable because zm is not coupled to any of the inputs. Moreover, zm 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 zm with any of the inputs, and the system is uncontrollable. In contrast, if at least one element in the mth row of Bˆ is nonzero, zm is coupled to at least one input and is therefore controllable. Thus, a system with a diagonalized state [Eq. (10.57)] is completely controllable if and only if the matrix Bˆ has no row of zero elements.

The outputs [see Eq. (10.57)] are of the form

$$y_i = \hat{c}_{i1}z_1 + \hat{c}_{i2}z_2 + \cdots + \hat{c}_{iN}z_N + \sum_{m=1}^j d_{im}x_m$$

$i = 1, 2, ..., k$

If cˆim =0, then the state zm will not appear in the expression for yi. Since all the states are decoupled because of the diagonalized nature of the equations, the state zm cannot be observed directly or indirectly (through other states) at the output yi. Hence, the mth mode e^λmt will not be observed at the output yi. If cˆ1*^m, cˆ2^m, … , cˆkm* (the mth column in matrix Cˆ ) are all zero, the state zm will not be observable at any of the k outputs, and the state zm is unobservable. In contrast, if at least one element in the mth column of Cˆ is nonzero, zm is observable at least at one output. Thus, a system with diagonalized equations of the form in Eq. (10.57) is completely observable if and only if the matrix Cˆ has no column of zero elements. In this discussion, we assumed distinct eigenvalues; for repeated eigenvalues, the modified criteria can be found in the literature [1, 2].

If the state-space description is not in diagonalized form, it may be converted into diagonalized form using the procedure in Ex. 10.11. It is also possible to test for controllability and observability even if the state-space description is in undiagonalized form [1, 2].

EXAMPLE 10.12 Controllability and Observability

Investigate the controllability and observability of the systems in Fig. 10.9.

In both cases, the state variables are identified as the two integrator outputs, q¹ and q2. The state equations for the system in Fig. 10.9a are

$$\dot{q}_1 = q_1 + x \n\dot{q}_2 = q_1 - q_2$$

\n(10.58)

and

$$y = \dot{q}_2 - q_2 = q_1 - 2q_2$$

Figure 10.9 Systems for Ex. 10.12.

Hence,

$$\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & -2 \end{bmatrix} \qquad \mathbf{D} = 0$$ $$|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s-1 & 0 \\ -1 & s+1 \end{vmatrix} = (s-1)(s+1)$$

Therefore,

$$\lambda_1 = 1 \qquad \text{and} \qquad \lambda_2 = -1$$

and

$$\mathbf{\Lambda} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$

We shall now use the procedure in Sec. 10.5-1 to diagonalize this system. According to Eq. (10.55), we have

$$\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix}$$

The solution of this equation yields

$$p_{12} = 0$$

and $-2p_{21} = p_{22}$

Choosing p¹¹ = 1 and p²¹ = 1, we have

$$\mathbf{P} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix}$$

and

$$\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

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

$$\mathbf{Y} = \mathbf{C}\mathbf{q} = \mathbf{C}\mathbf{P}^{-1}\mathbf{z} = \hat{\mathbf{C}}\mathbf{z}$$

and

$$\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \end{bmatrix}$$

The first column of Cˆ is zero. Hence, the mode z¹ (corresponding to λ¹ =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 z¹ and z2, whose state equations are

$$\dot{z} = \Lambda z + Bx$$ $$y = \hat{C}z$$

Using our previous calculations, we have

$$\begin{aligned}\n\dot{z}_1 &= z_1 + x \\ \dot{z}_2 &= -z_2 + x\n\end{aligned}$$

and

$$y\,{=}\,z_2$$

Figure 10.10a 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.

The state equations for the system in Fig. 10.9b are

$$\dot{q}_1 = -q_1 + x \n\dot{q}_2 = \dot{q}_1 - q_1 + q_2 = -2q_1 + q_2 + x$$

\n(10.59)

and

y = q²

Hence,

$$\mathbf{A} = \begin{bmatrix} -1 & 0 \\ -2 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = 0$$ $$|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s+1 & 0 \\ -1 & s-1 \end{vmatrix} = (s+1)(s-1)$$

$x_2 = 1$ , and

so that λ¹ = −1, λ² = 1, and

$$\Lambda = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$

Diagonalizing the matrix, we have

$$\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} -1 & 0 \ -2 & 1 \end{bmatrix}$$

The solution of this equation yields p¹¹ = −p¹² and p²² = 0. Choosing p¹¹ = −1 and p²¹ = 1, we obtain

$$\mathbf{P} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix}$$

and

$$\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ $$\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \end{bmatrix}$$

The first row of Bˆ is zero. Hence, the mode corresponding to λ¹ = 1 is not controllable. However, since none of the columns of Cˆ 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 z¹ and z2. The two state equations are

$$\dot{z} = \Lambda z + \dot{B}x$$ $$y = \hat{C}z$$

Using our previous calculations, we have

$$\begin{aligned}\n\dot{z}_1 &= z_1\\ \n\dot{z}_2 &= -z_2 + x\n\end{aligned}$$

and thus,

y = z¹ +z²

Figure 10.10b shows a realization of these equations. Clearly, each of the two modes is observable at the output, but the mode corresponding to λ¹ = 1 is not controllable.

USING MATLAB TO DETERMINE CONTROLLABILITY AND OBSERVABILITY

As demonstrated in Ex. 10.11, we can use MATLAB’s eig function to determine the matrix P that will diagonalize A. We can then use P to determine Bˆ and Cˆ , from which we can determine the controllability and observability of a system. Let us demonstrate the process for the two present systems.

First, let us use MATLAB to compute Bˆ and Cˆ for the system in Fig. 10.9a.

A = [1 0;1 -1]; B = [1; 0]; C = [1 -2]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P*B, Chat = C*inv(P) Bhat = -0.5000 1.1180 Chat = -2 0

Since all the rows of Bˆ are nonzero, the system is controllable. However, one column of Cˆ is zero, so one mode is unobservable.

Next, let us use MATLAB to compute Bˆ and Cˆ for the system in Fig. 10.9b.

A = [-1 0;-2 1]; B = [1; 1]; C = [0 1]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P*B, Chat = C*inv(P) Bhat = 0 1.4142 Chat = 1.0000 0.7071

One of the rows of Bˆ is zero, so one mode is uncontrollable. Since all of the columns of Cˆ are nonzero, the system is observable.

As expected, the MATLAB results confirm our earlier conclusions regarding the controllability and observability of the systems of Fig. 10.9.

10.6-1 Inadequacy of the Transfer Function Description of a System

Example 10.12 demonstrates the inadequacy of the transfer function to describe an LTI system in general. The systems in Figs. 10.9a and 10.9b both have the same transfer function

$$H(s) = \frac{1}{s+1}$$

Yet the two systems are very different. Their true nature is revealed in Figs. 10.10a and 10.10b, 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. 10.9a is controllable but not observable, whereas the system in Fig. 10.9b 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 Figs. 10.10a and 10.10b, 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. (10.58) and (10.59)], 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. 10.9, 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.