10.5 LINEAR [TRANSFORMATION OF A](#page-14-0) STATE VECTOR

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10.5 LINEAR TRANSFORMATION OF A STATE VECTOR

In Sec. 10.2 we saw that the state of a system can be specified in several ways. The sets of all possible state variables are related—in other words, if we are given one set of state variables, we should be able to relate it to any other set. We are particularly interested in a linear type of relationship. Let q1,q2,…,qN and w1,w2,…,wN be two different sets of state variables specifying the same system. Let these sets be related by linear equations as

$$w_1 = p_{11}q_1 + p_{12}q_2 + \dots + p_{1N}q_N$$

\n

$$w_2 = p_{21}q_1 + p_{22}q_2 + \dots + p_{2N}q_N$$

\n

$$\vdots$$

\n

$$w_N = p_{N1}q_1 + p_{N2}q_2 + \dots + p_{NN}q_N$$

or

⎡ w1 w2 ⎢	⎤ ⎥	⎡ p11 p21 ⎢	p12 p22	··· ···	⎤ p1N p2N ⎥	⎡ ⎤ q1 q2 ⎢ ⎥
⎢ ⎢ ⎣	⎥ = ⎥ ⎦	⎢ ⎢ ⎣			⎥ ⎥ ⎦	⎢ ⎥ ⎢ ⎥ ⎣ ⎦
wN		pN1	pN2	···	pNN	qN
w				P		q

Defining the vector w and matrix P as just shown, we obtain the compact matrix representation

$$\mathbf{w} = \mathbf{P}\mathbf{q} \tag{10.46}$$

and

$$\mathbf{q} = \mathbf{P}^{-1}\mathbf{w} \tag{10.47}$$

Thus, the state vector q is transformed into another state vector w through the linear transformation in Eq. (10.46).

If we know ^w, we can determine ^q from ^{q = P}−¹ w, provided P−¹ exists. This is equivalent to saying that P is a nonsingular matrix† (|P| = 0). Thus, if P is a nonsingular matrix, the vector w defined by Eq. (10.46) is also a state vector. Consider the state equation of a system

q˙ = Aq+Bx

$$\mathbf{w} = \mathbf{P}\mathbf{q}$$ $$\mathbf{q} = \mathbf{P}^{-1}\mathbf{w}$$

and

then

If

Hence, the state equation now becomes

$$\mathbf{P}^{-1}\dot{\mathbf{w}} = \mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{B}\mathbf{x}$$

^q˙ ^{= P}−¹

w˙

or

$$\dot{\mathbf{w}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{P}\mathbf{B}\mathbf{x}$$

= $\hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x}$ (10.48)

where

$$\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} \qquad \text{and} \qquad \hat{\mathbf{B}} = \mathbf{P}\mathbf{B} \tag{10.49}$$

Equation (10.48) is a state equation for the same system, but now it is expressed in terms of the state vector w.

^† This condition is equivalent to saying that all N equations in Eq. (10.46) are linearly independent; that is, none of the N equations can be expressed as a linear combination of the remaining equations.

The output equation is also modified. Let the original output equation be

$$y = Cq + Dx$$

In terms of the new state variable w, this equation becomes

$$\mathbf{y} = \mathbf{C}(\mathbf{P}^{-1}\mathbf{w}) + \mathbf{D}\mathbf{x}$$ $$= \hat{\mathbf{C}}\mathbf{w} + \mathbf{D}\mathbf{x}$$ $$\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1}$$

(10.50)

EXAMPLE 10.10 Linear Transformation of the State Vector

The state equations of a certain system are given by

$$\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \end{bmatrix} x(t)$$

Find the state equations for this system when the new state variables w¹ and w² are given as

$$\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix}$$

(10.51)

According to Eq. (10.48), the state equation for the state variable w is given by

$$\dot{\mathbf{w}} = \hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x}$$

where [see Eqs. (10.49) and (10.50)]

$$\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}$$ $$= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$$ $$= \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix}$$

and

where

$$\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$$

Therefore,

$$\begin{bmatrix} \dot{w}_1 \\ \dot{w}_2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} + \begin{bmatrix} 3 \\ -1 \end{bmatrix} x(t)$$

This is the desired state equation for the state vector w. The solution of this equation requires a knowledge of the initial state w(0). This can be obtained from the given initial state q(0) by using Eq. (10.51).

We can obtain the same result with less effort using MATLAB.

>> A = [0 1;-2 -3]; B = [1; 2];
>> P = [1 1;1 -1];
>> Ahat = P*A*inv(P), Bhat = P*B
   Ahat = -2 0
             3 -1
   Bhat = 3
            -1

INVARIANCE OF EIGENVALUES

We have seen that the poles of all possible transfer functions of a system are the eigenvalues of the matrix A. If we transform a state vector from q to w, the variables w1, w2, … , wN are linear combinations of q1, q2, … , qN and therefore may be considered to be outputs. Hence, the poles of the transfer functions relating w1, w2, … , wN to the various inputs must also be the eigenvalues of matrix A. On the other hand, the system is also specified by Eq. (10.48). This means that the poles of the transfer functions must be the eigenvalues of Aˆ . Therefore, the eigenvalues of matrix A remain unchanged for the linear transformation of variables represented by Eq. (10.46), and the eigenvalues of matrix ^A and matrix ^Aˆ (A^{ˆ =} PAP−¹ ) are identical, implying that the characteristic equations of A and Aˆ are also identical. This result also can be proved alternately as follows.

Consider the matrix ^P(sI−A)P−¹ . We have

$$\mathbf{P}(s\mathbf{I} - \mathbf{A})\mathbf{P}^{-1} = \mathbf{P}s\mathbf{IP}^{-1} - \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = s\mathbf{P}\mathbf{IP}^{-1} - \hat{\mathbf{A}} = s\mathbf{I} - \hat{\mathbf{A}}$$

Taking the determinants of both sides, we obtain

$$|\mathbf{P}||s\mathbf{I} - \mathbf{A}||\mathbf{P}^{-1}| = |s\mathbf{I} - \hat{\mathbf{A}}|$$

The determinants ^|P^| and ^|P−¹ | are reciprocals of each other. Hence,

$$|s\mathbf{I} - \mathbf{A}| = |s\mathbf{I} - \hat{\mathbf{A}}|$$

This is the desired result. We have shown that the characteristic equations of A and Aˆ are identical. Hence, the eigenvalues of A and Aˆ are identical.

In Ex. 10.10, matrix A is given as

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$$

The characteristic equation is

$$|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s & -1 \\ 2 & s+3 \end{vmatrix} = s^2 + 3s + 2 = 0$$

Also,

$$\hat{\mathbf{A}} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix}$$

and

$$|s\mathbf{I} - \hat{\mathbf{A}}| = \begin{bmatrix} s+2 & 0\\ -3 & s+1 \end{bmatrix} = s^2 + 3s + 2 = 0$$

This result verifies that the characteristic equations of A and Aˆ are identical.

10.5-1 Diagonalization of Matrix A

For several reasons, it is desirable to make matrix A diagonal. If A is not diagonal, we can transform the state variables such that the resulting matrix Aˆ is diagonal.† One can show that for any diagonal matrix A, the diagonal elements of this matrix must necessarily be λ1, λ2, … , λ*^N* (the eigenvalues) of the matrix. Consider the diagonal matrix A:

$$\mathbf{A} = \begin{bmatrix} a_1 & 0 & 0 & \cdots & 0 \\ 0 & a_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & a_N \end{bmatrix}$$

The characteristic equation is given by

$$|\mathbf{sI} - \mathbf{A}| = \begin{bmatrix} (s - a_1) & 0 & 0 & \cdots & 0 \\ 0 & (s - a_2) & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & (s - a_N) \end{bmatrix} = 0$$

or

$$(s - a_1)(s - a_2) \cdots (s - a_N) = 0$$

The nonzero (diagonal) elements of a diagonal matrix are therefore its eigenvalues λ1, λ2, … , λN. We shall denote the diagonal matrix by the symbol, A:

$$\mathbf{\Lambda} = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_N \end{bmatrix}$$

(10.52)

Let us now consider the transformation of the state vector A such that the resulting matrix Aˆ is a diagonal matrix .

^† In this discussion we assume distinct eigenvalues. If the eigenvalues are not distinct, we can reduce the matrix to a modified diagonalized (Jordan) form.

944 CHAPTER 10 STATE-SPACE ANALYSIS

Consider the system

$$\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x}$$

We shall assume that λ1, λ2, … , λN, the eigenvalues of A, are distinct (no repeated roots). Let us transform the state vector q into the new state vector z, using the transformation

$$z = Pq \tag{10.53}$$

Then, after the development of Eq. (10.48), we have

$$\dot{z} = PAP^{-1}z + PBx$$

We desire the transformation to be such that PAP−¹ is a diagonal matrix given by Eq. (10.52), or

⁼ PAP−¹

$$\dot{\mathbf{z}} = \mathbf{\Lambda}\mathbf{z} + \mathbf{B}\mathbf{x} \tag{10.54}$$

Hence,

or

$$\Lambda P = PA \tag{10.55}$$

We know and A. Equation (10.55) therefore can be solved to determine P.

EXAMPLE 10.11 Diagonal Form of the State Equations

Find the diagonalized form of the state equations for the system in Ex. 10.10.

In this case,

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$$

We found λ¹ = −1 and λ² = −2. Hence,

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

and Eq. (10.55) becomes

$$\begin{bmatrix} -1 & 0 \ 0 & -2 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 0 & 1 \ -2 & -3 \end{bmatrix}$$

Equating the four elements on two sides, we obtain

$$-p_{11} = -2p_{12}$$

\n

$$-p_{12} = p_{11} - 3p_{12}$$

\n

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

\n

$$-2p_{22} = p_{21} - 3p_{22}$$

The reader will immediately recognize that the first two equations are identical and that the last two equations are identical. Hence, two equations may be discarded, leaving us with only two equations [p¹¹ = 2p¹² and p²¹ = p22] and four unknowns. This observation means that there is no unique solution. There is, in fact, an infinite number of solutions. We can assign any value to p¹¹ and p²¹ to yield one possible solution.† If p¹¹ = k¹ and p²¹ = k2, then we have p¹² = k1/2 and p²² = k2:

$$\mathbf{P} = \begin{bmatrix} k_1 & \frac{k_1}{2} \\ k_2 & k_2 \end{bmatrix}$$

We may assign any values to k¹ and k2. For convenience, let k¹ = 2 and k² = 1. This substitution yields

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

The transformed variables [Eq. (10.53)] are

$$\begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \begin{bmatrix} 2q_1 + q_2 \\ q_1 + q_2 \end{bmatrix}$$

This expression relates the new state variables z¹ and z² to the original state variables q¹ and q2. The system equation with z as the state vector is given by [see Eq. (10.54)]

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

where

$$\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$$ $$\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 4 \\ 4 \end{bmatrix}.$$

z2

4 3

x (10.56)

−1 0 0 −2

z˙1 z˙2

=

Hence,

^† If, however, we want the state equations in diagonalized form, as in Eq. (10.26), where all the elements of

$$\hat{B}$$

matrix are unity, there is a unique solution. The reason is that the equation $\hat{B} = PB$ , where all the elements of $\hat{B}$ are unity, imposes additional constraints. In the present example, this condition will yield $p_{11} = 1/2$ , $p_{12} = 1/4$ , $p_{21} = 1/3$ , and $p_{22} = 1/3$ . The relationship between z and q is then

$$z_1 = \frac{1}{2}q_1 + \frac{1}{4}q_2$$

and $z_2 = \frac{1}{3}q_1 + \frac{1}{3}q_2$

or

$$\begin{aligned}\n\dot{z}_1 &= -z_1 + 4x \\ \dot{z}_2 &= -2z_2 + 3x\n\end{aligned}$$

Note the distinctive nature of these state equations. Each state equation involves only one variable and therefore can be solved by itself. A general state equation has the derivative of one state variable equal to a linear combination of all state variables. Such is not the case with the diagonalized matrix . Each state variable zi is chosen so that it is uncoupled from the rest of the variables; hence, a system with N eigenvalues is split into N decoupled systems, each with an equation of the form

$$\dot{z}_i = \lambda_i z_i + (\text{input terms})$$

This fact also can be readily seen from Fig. 10.8a, which is a realization of the system represented by Eq. (10.56). In contrast, consider the original state equations [see Ex. 10.10]

$$\dot{q}_1 = q_2 + x(t) \n\dot{q}_2 = -2q_1 - 3q_2 + 2x(t)$$

A realization for these equations is shown in Fig. 10.8b. It can be seen from Fig. 10.8a that the states z¹ and z² are decoupled, whereas the states q¹ and q² (Fig. 10.8b) are coupled. It should be remembered that Figs. 10.8a and 10.8b are realizations of the same system.†

Figure 10.8 Two realizations of the second-order system.

^† Here we have only a simulated state equation; the outputs are not shown. The outputs are linear combinations of state variables (and inputs). Hence, the output equation can be easily incorporated into these diagrams.

MATRIX DIAGONALIZATION VIA MATLAB

The key to diagonalizing matrix A is to determine a matrix P that satisfies P = PA [Eq. (10.55)], where is a diagonal matrix of the eigenvalues of A. This problem is directly related to the classic eigenvalue problem, stated as

AV = V

where V is a matrix of eigenvectors for A. If we can find V, we can take its inverse to determine ^P. That is, ^{P = V}−¹ . This relationship is more fully developed in Sec. 10.8.

MATLAB’s built-in function eig can determine the eigenvectors of a matrix and, therefore, can help us determine a suitable matrix P. Let us demonstrate this approach for the current case.

>> A = [0 1;-2 -3]; B = [1; 2];
>> [V, Lambda] = eig(A);
>> P = inv(V), Lambda, Bhat = P*B
   P = 2.8284 1.4142
          2.2361 2.2361
   Lambda = -1 0
               0 -2
   Bhat = 5.6569
            6.7082

Therefore,

$$\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2.8284 & 1.4142 \\ 2.2361 & 2.2361 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \mathbf{P}\mathbf{q}$$

and

$$\dot{\mathbf{z}} = \begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 5.6569 \\ 6.7082 \end{bmatrix} x(t) = \Lambda \mathbf{z} + \hat{\mathbf{B}} \mathbf{x}$$

Recall that neither P nor Bˆ are unique, which explains why the MATLAB output does not need to match our previous solution. Still, the MATLAB results do their job and successfully diagonalize matrix A.