[10.4 SOLUTION OF](#page-14-0) STATE EQUATIONS

← Back to LINEAR SYSTEMS AND SIGNALS Overview

10.4 SOLUTION OF STATE EQUATIONS

The state equations of a linear system are N simultaneous linear differential equations of the first order. We studied the techniques of solving linear differential equations in Chs. 2 and 4. The same techniques can be applied to state equations without any modification. However, it is more convenient to carry out the solution in the framework of matrix notation.

These equations can be solved in both the time and frequency domains (Laplace transform). The latter is relatively easier to deal with than the time-domain solution. For this reason, we shall first consider the Laplace transform solution.

10.4-1 Laplace Transform Solution of State Equations

The ith state equation [Eq. (10.14)] is of the form

$$\dot{q}_i = a_{i1}q_1 + a_{i2}q_2 + \dots + a_{iN}q_N + b_{i1}x_1 + b_{i2}x_2 + \dots + b_{ij}x_j \tag{10.27}$$

We shall take the Laplace transform of this equation. Let

$$q_i(t) \Longleftrightarrow Q_i(s)$$

so that

$$\dot{q}_i(t) \Longleftrightarrow sQ_i(s) - q_i(0)$$

Also, let

$$x_i(t) \Longleftrightarrow X_i(s)$$

The Laplace transform of Eq. (10.27) yields

$$sQ_i(s) - q_i(0) = a_{i1}Q_1(s) + a_{i2}Q_2(s) + \cdots + a_{iN}Q_N(s) + b_{i1}X_1(s) + b_{i2}X_2(s) + \cdots + b_{ij}X_j(s)$$

Taking the Laplace transforms of all N state equations, we obtain

$$s\left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] - \left[\begin{array}{c} q_1(0) \\ q_2(0) \\ \vdots \\ q_N(0) \end{array}\right] = \left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1N} \\ a_{21} & a_{22} & \cdots & a_{2N} \\ \vdots & \vdots & \cdots & \vdots \\ a_{N1} & a_{N2} & \cdots & a_{NN} \end{array}\right] \left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] + \left[\begin{array}{cccc} q_1(0) \\ \vdots \\ q_N(0) \end{array}\right] + \left[\begin{array}{cccc} b_{11} & b_{12} & \cdots & b_{1j} \\ b_{21} & b_{22} & \cdots & b_{2j} \\ \vdots & \vdots & \cdots & \vdots \\ b_{N1} & b_{N2} & \cdots & b_{Nj} \end{array}\right] \left[\begin{array}{c} X_1(s) \\ X_2(s) \\ \vdots \\ X_j(s) \end{array}\right]$$

Defining the vectors, as indicated, we have

$$s\mathbf{Q}(s) - \mathbf{q}(0) = \mathbf{A}\mathbf{Q}(s) + \mathbf{B}\mathbf{X}(s)$$

or

$$s\mathbf{Q}(s) - \mathbf{A}\mathbf{Q}(s) = \mathbf{q}(0) + \mathbf{B}\mathbf{X}(s)$$

and

$$(s\mathbf{I} - \mathbf{A})\mathbf{Q}(s) = \mathbf{x}(0) + \mathbf{B}\mathbf{X}(s)$$

where I is the N ×N identity matrix. Solving for Q(s), we have

$$Q(s) = (sI - A)^{-1}[q(0) + BX(s)]$$

= $\Phi(s)[q(0) + BX(s)]$ (10.28)

where

$$\mathbf{\Phi}(s) = (s\mathbf{I} - \mathbf{A})^{-1}$$

Thus, from Eq. (10.28),

$$\mathbf{Q}(s) = \mathbf{\Phi}(s)\mathbf{q}(0) + \mathbf{\Phi}(s)\mathbf{B}\mathbf{X}(s)$$

and

$$\mathbf{q}(t) = \underbrace{\mathcal{L}^{-1}[\Phi(s)]\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\mathcal{L}^{-1}[\Phi(s)\mathbf{B}\mathbf{X}(s)]}_{\text{zero-state response}}$$

(10.29)

Equation (10.29) gives the desired solution. Observe the two components of the solution. The first component yields q(t) when the input x(t) = 0. Hence, the first component is the zero-input response. In a similar manner, we see that the second component is the zero-state response.

EXAMPLE 10.6 Laplace Transform Solution to State Equations

Using the Laplace transform, find the state vector q(t) for the system whose state equation is given by

q˙ = Aq+Bx

where

$$\mathbf{A} = \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \qquad \mathbf{x}(t) = u(t)$$

and the initial conditions are q1(0) = 2, q2(0) = 1.

From Eq. (10.28), we have

$$\mathbf{Q}(s) = \mathbf{\Phi}(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s)]$$

Let us first find (s). We have

$$(s\mathbf{I} - \mathbf{A}) = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} = \begin{bmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{bmatrix}$$

and

$$\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)}\\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix}$$

Now,

$$q(0)$$

is given as

$$\mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$

Also, X(s) = 1/s, and

$$\mathbf{BX}(s) = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \frac{1}{s} = \begin{bmatrix} \frac{1}{3s} \\ \frac{1}{s} \end{bmatrix}$$

Therefore,

$$\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s) = \begin{bmatrix} 2 + \frac{1}{3s} \\ 1 + \frac{1}{s} \end{bmatrix} = \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix}$$

and

$$Q(s) = \Phi(s) [q(0) + BX(s)]$$

=

$$\begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)} \\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix} \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix}$$

=

$$\begin{bmatrix} \frac{2s^2+3s+1}{s(s+4)(s+9)} \\ \frac{s-59}{(s+4)(s+9)} \end{bmatrix}$$

=

$$\begin{bmatrix} \frac{1/36}{s} - \frac{21/20}{s+4} + \frac{136/45}{s+9} \\ \frac{-63/5}{s+4} + \frac{68/5}{s+9} \end{bmatrix}$$

The inverse Laplace transform of this equation yields

$$\begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(-\frac{63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix}$$

This result is readily confirmed using MATLAB and its symbolic toolbox.

syms s >> A = [-12 2/3;-36 -1]; B = [1/3; 1]; q0 = [2;1]; X = 1/s; >> q = ilaplace(inv(s*eye(2)-A)*(q0+B*X)) q = (136*exp(-9*t))/45 - (21*exp(-4*t))/20 + 1/36 (68*exp(-9*t))/5 - (63*exp(-4*t))/5

To create a plot of the state vector, we use MATLAB’s subs command to substitute the symbolic variable t with a vector of desired values.

930 CHAPTER 10 STATE-SPACE ANALYSIS

>> t = (0:.01:2); q = subs(q); q1 = q(1,:); q2 = q(2,:);
>> plot(t,q1,'k',t,q2,'k--'); xlabel('t'); ylabel('Amplitude');
>> legend('q_1(t)','q_2(t)','Location','SE');

The resulting plot is shown in Fig. 10.7.

THE OUTPUT

The output equation is given by

$$y = Cq + Dx$$

and

$$\mathbf{Y}(s) = \mathbf{C}\mathbf{Q}(s) + \mathbf{D}\mathbf{X}(s)$$

Upon substituting Eq. (10.28) into this equation, we have

$$\mathbf{Y}(s) = \mathbf{C}[\Phi(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{x}\mathbf{X}(s)]] + \mathbf{D}\mathbf{X}(s)$$

=

$$\underbrace{\mathbf{C}\Phi(s)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{[\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s)}_{\text{zero-state response}}$$

(10.30)

The zero-state response [i.e., the response Y(s) when q(0) = 0] is given by

$$\mathbf{Y}(s) = [\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s)$$

Note that the transfer function of a system is defined under the zero-state condition [see Eq. (4.19)]. The matrix C(s)B + D is the transfer function matrix H(s) of the system, which relates the responses y1, y2, … , yk to the inputs x1, x2, … , xj:

$$\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D} \tag{10.31}$$

and the zero-state response is

$$\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s)$$

The matrix H(s) is a k × j matrix (k is the number of outputs and j is the number of inputs). The ijth element Hij(s) of H(s) is the transfer function that relates the output yi(t) to the input xj(t).

EXAMPLE 10.7 Transfer Function Matrix from State-Space Description

Let us consider a system with a state equation

$$\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

and an output equation

$$\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

Determine the transfer function matrix of the system.

In this case,

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{10.32}$$

and

$$\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 2 & s+3 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix}$$

(10.33)

Hence, the transfer function matrix H(s) is given by

$$\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}$$

\n

$$= \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix}$$

\n

$$= \begin{bmatrix} \frac{s+4}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{s+4}{s+2} & \frac{1}{s+2} \\ \frac{2(s-2)}{(s+1)(s+2)} & \frac{s^2+5s+2}{(s+1)(s+2)} \end{bmatrix}$$

\n(10.34)

and the zero-state response is

$$\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s)$$

Remember that the ijth element of the transfer function matrix in Eq. (10.34) represents the transfer function that relates the output yi(t) to the input xj(t). For instance, the transfer function that relates the output y³ to the input x² is H32(s), where

$$H_{32}(s) = \frac{s^2 + 5s + 2}{(s+1)(s+2)}$$

We can readily verify the transfer function matrix using MATLAB and its symbolic toolbox functions.

A = [0 1;-2 -3]; B = [1 0;1 1]; >> C = [1 0;1 1;0 2]; D = [0 0;1 0;0 1]; >> syms s; H = collect(simplify(C*inv(s*eye(2)-A)*B+D)) H = [ (s + 4)/(s^2 + 3*s + 2), 1/(s^2 + 3*s + 2)] [ (s + 4)/(s + 2), 1/(s + 2)] [ (2*s - 4)/(s^2 + 3*s + 2), (s^2 + 5*s + 2)/(s^2 + 3*s + 2)]

Transfer functions relating particular inputs to particular outputs, such as H32(s), can be obtained using the ss2tf and tf functions.

>> [num,den] = ss2tf(A,B,C,D,2); H_32 = tf(num(3,:),den)
    H_32 =
           s^2 + 5 s + 2
           -------------
           s^2 + 3 s + 2

CHARACTERISTIC ROOTS (EIGENVALUES) OF A MATRIX

It is interesting to observe that the denominator of every transfer function in Eq. (10.34) is (s + 1)(s + 2) except for H21(s) and H22(s), where the factor (s + 1) is canceled. This is no coincidence. We see that the denominator of every element of (s) is |sI − A| because (s) = (sI − A)−1, and the inverse of a matrix has its determinant in the denominator. Since C, B, and D are matrices with constant elements, we see from Eq. (10.31) that the denominator of (s) will also be the denominator of H(s). Hence, the denominator of every element of H(s) is |sI − A|, except for the possible cancellation of the common factors mentioned earlier. In other words, the zeros of the polynomial |sI− A| are also the poles of all transfer functions of the system. Therefore, the zeros of the polynomial |sI−A| are the characteristic roots of the system. Hence, the characteristic roots of the system are the roots of the equation

$$|s\mathbf{I} - \mathbf{A}| = 0 \tag{10.35}$$

Since |sI − A| is an Nth-order polynomial in s with N zeros λ1, λ2, … , λN, we can write Eq. (10.35) as

$$|s\mathbf{I} - \mathbf{A}| = s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N = (s - \lambda_1)(s - \lambda_2) \cdot \dots (s - \lambda_N) = 0$$

For the system in Ex. 10.7,

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

Hence,

$$\lambda_1 = -1 \quad \text{and} \quad \lambda_2 = -2$$

Equation (10.35) is known as the characteristic equation of the matrix A, and λ1, λ2, … , λ*^N* are the characteristic roots of A. The term eigenvalue, meaning “characteristic value” in German, is also commonly used in the literature. Thus, we have shown that the characteristic roots of a system are the eigenvalues (characteristic values) of the matrix A.

At this point, the reader will recall that if λ1, λ2, … , λ*^N* are the poles of the transfer function, then the zero-input response is of the form

$$y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \dots + c_N e^{\lambda_N t}$$

(10.36)

This fact is also obvious from Eq. (10.30). The denominator of every element of the zero-input response matrix C(s)q(0) is |sI−A| =(s−λ1)(s−λ2)···(s−λN). Therefore, the partial fraction expansion and the subsequent inverse Laplace transform will yield a zero-input component of the form in Eq. (10.36).

10.4-2 Time-Domain Solution of State Equations

The state equation is

$$\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} \tag{10.37}$$

We now show that the solution of the vector differential Eq. (10.37) is

$$\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + \int_0^t e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{x}(\tau) d\tau$$

Before proceeding further, we must define the matrix exponential e**^A**^t . An exponential of a matrix is defined by an infinite series identical to that used in defining an exponential of a scalar. We shall define

$$e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \frac{\mathbf{A}^3 t^3}{3!} + \dots + \frac{\mathbf{A}^n t^n}{n!} + \dots = \sum_{k=0}^{\infty} \frac{\mathbf{A}^k t^k}{k!}$$

(10.38)

For example, if

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

then

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

and

$$\frac{\mathbf{A}^2 t^2}{2!} = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} t^2 & \frac{t^2}{2} \\ t^2 & \frac{3t^2}{2} \end{bmatrix}$$

and so on.

934 CHAPTER 10 STATE-SPACE ANALYSIS

We can show that the infinite series in Eq. (10.38) is absolutely and uniformly convergent for all values of t. Consequently, it can be differentiated or integrated term by term. Thus, to find (d/dt)eA*^t* , we differentiate the series on the right-hand side of Eq. (10.38) term by term:

$$\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A} + \mathbf{A}^2t + \frac{\mathbf{A}^3t^2}{2!} + \frac{\mathbf{A}^4t^3}{3!} + \cdots$$

\n

$$= \mathbf{A}\bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots\bigg] = \mathbf{A}e^{\mathbf{A}t}$$

\n

$$= \bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots + \cdots\bigg]\mathbf{A} = e^{\mathbf{A}t}\mathbf{A}$$

Hence,

$$\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A}e^{\mathbf{A}t} = e^{\mathbf{A}t}\mathbf{A}$$

Also note that from Eq. (10.38), it follows that

^{e0 = I}

where I is just the identity matrix. If we premultiply or postmultiply the infinite series for eA*^t* [Eq. (10.38)] by an infinite series for e−A*^t* , we find that

$$(e^{-At})(e^{At}) = (e^{At})(e^{-At}) = I$$

\n(10.39)

In Sec. 10.1-1, we showed that

$$\frac{d}{dt}(\mathbf{U}\mathbf{V}) = \frac{d\mathbf{U}}{dt}\mathbf{V} + \mathbf{U}\frac{d\mathbf{V}}{dt}$$

Using this relationship, we observe that

$$\frac{d}{dt}[e^{-\mathbf{A}t}\mathbf{q}] = \left(\frac{d}{dt}e^{-\mathbf{A}t}\right)\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}}$$ $$= -e^{-\mathbf{A}t}\mathbf{A}\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}}$$

(10.40)

We now premultiply both sides of Eq. (10.37) by e−A*^t* to yield

$$e^{-At}\dot{\mathbf{q}} = e^{-At}\mathbf{A}\mathbf{q} + e^{-At}\mathbf{B}\mathbf{x}$$

or

$$-e^{-At}\mathbf{A}\mathbf{q}+e^{-\mathbf{A}t}\dot{\mathbf{q}}=e^{-\mathbf{A}t}\mathbf{B}\mathbf{x}$$

Substituting this result into Eq. (10.40) yields

$$\frac{d}{dt}[e^{-At}\mathbf{q}] = e^{-At}\mathbf{B}\mathbf{x}$$

The integration of both sides of this equation from 0 to t yields

$$e^{-\mathbf{A}t}\mathbf{q}\big|_{0}^{t} = \int_{0}^{t} e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau$$

or

$$e^{-\mathbf{A}t}\mathbf{q}(t) - \mathbf{q}(0) = \int_0^t e^{-\mathbf{A}\tau} \mathbf{B}\mathbf{x}(\tau) d\tau$$

Hence,

$$e^{-\mathbf{A}t}\mathbf{q} = \mathbf{q}(0) + \int_0^t e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau$$

Premultiplying this result by eA*^t* and using Eq. (10.39), we have

$$\mathbf{q}(t) = \underbrace{e^{\mathbf{A}t}\mathbf{q}(0)}_{\text{ZIR}} + \underbrace{\int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{B}\mathbf{x}(\tau) d\tau}_{\text{ZSR}}$$

(10.41)

This is the desired solution. The first term on the right-hand side represents q(t) when the input x(t) = 0. Hence, it is the zero-input component. The second term, by a similar argument, is seen to be the zero-state component.

The results of Eq. (10.41) can be expressed more conveniently in terms of the matrix convolution. We can define the convolution of two matrices in a manner similar to the multiplication of two matrices, except that the multiplication of two elements is replaced by their convolution. For example,

$$\begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} * \begin{bmatrix} g_1 & g_2 \ g_3 & g_4 \end{bmatrix} = \begin{bmatrix} (x_1 * g_1 + x_2 * g_3) & (x_1 * g_2 + x_2 * g_4) \\ (x_3 * g_1 + x_4 * g_3) & (x_3 * g_2 + x_4 * g_4) \end{bmatrix}$$

By using this definition of matrix convolution, we can express Eq. (10.41) as

$$\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t)$$

\n(10.42)

Note that the limits of the convolution integral [Eq. (10.41)] are from 0 to t. Hence, all the elements of eA*^t* in the convolution term of Eq. (10.42) are implicitly assumed to be multiplied by u(t).

The result of Eqs. (10.41) and (10.42) can be easily generalized for any initial value of t. It is left as an exercise for the reader to show that the solution of the state equation can be expressed as

$$\mathbf{q}(t) = e^{\mathbf{A}(t-t_0)}\mathbf{q}(t_0) + \int_{t_0}^t e^{\mathbf{A}(t-\tau)} \mathbf{B}\mathbf{x}(\tau) d\tau$$

DETERMINING eAt

The exponential e**^At required in Eqs. (10.41) and (10.42) can be computed from the definition in Eq. (10.38). Unfortunately, this is an infinite series, and its computation can be quite laborious. Moreover, we may not be able to recognize the closed-form expression for the answer. There are several efficient methods of determining e^A**^t in closed form. It was shown in Sec. 10.1-3 that for an N ×N matrix A,

$$e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{N-1} \mathbf{A}^{N-1}$$

(10.43)

where

$$\begin{bmatrix}\n\beta_0 \\ \beta_1 \\ \vdots \\ \beta_{N-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{N-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{N-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \lambda_N & \lambda_N^2 & \cdots & \lambda_N^{N-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\ne^{\lambda_1 t} \\ ne^{\lambda_2 t} \\ \vdots \\ e^{\lambda_N t}\n\end{bmatrix}$$

and λ1,λ2,…,λ*^N* are the N characteristic values (eigenvalues) of A.

We can also determine eA*^t* by comparing Eqs. (10.41) and (10.29). It is clear that

$$e^{\mathbf{A}t} = \mathcal{L}^{-1}[\Phi(s)] = \mathcal{L}^{-1}[(s\mathbf{I} - \mathbf{A})^{-1}]$$

\n(10.44)

Thus, eA*^t* and (s) are a Laplace transform pair. To be consistent with Laplace transform notation, eA*^t* is often denoted by φ(t), the state transition matrix (STM):

^eA^{t = φ}(t)

EXAMPLE 10.8 Time-Domain Method to Solve State Equations

Use the time-domain method to solve Ex. 10.6.

For this case, the characteristic roots are given by

$$|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{vmatrix} = s^2 + 13s + 36 = (s+4)(s+9) = 0$$

The roots are λ¹ = −4 and λ² = −9, so

$$\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 1 & -9 \end{bmatrix}^{-1} \begin{bmatrix} e^{-4t} \\ e^{-9t} \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 9e^{-4t} - 4e^{-9t} \\ e^{-4t} - e^{-9t} \end{bmatrix}$$

and

$$e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A}$$

= $\left(\frac{9}{5}e^{-4t} - \frac{4}{5}e^{-9t}\right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \left(\frac{1}{5}e^{-4t} - \frac{1}{5}e^{-9t}\right) \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix}$
= $\begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t}) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix}$

The zero-input response is given by [see Eq. (10.41)]

$$e^{\mathbf{A}t}\mathbf{q}(0) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t})\\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix} \begin{bmatrix} 2\\ 1 \end{bmatrix}$$ $$= \begin{bmatrix} \left(\frac{-16}{15}e^{-4t} + \frac{46}{15}e^{-9t}\right)u(t) \\ \left(\frac{-64}{5}e^{-4t} + \frac{69}{5}e^{-9t}\right)u(t) \end{bmatrix}$$

Note here the presence of u(t), indicating that the response begins at t = 0.

The zero-state component is eA*^t* ∗Bx [see Eq. (10.42)], where

$$\mathbf{B}\mathbf{x} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} u(t) = \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix}$$

and

$$e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t) = \begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}u(t)) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) \end{bmatrix} * \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix}$$

Note again the presence of the term u(t) in every element of eA*^t* . This is the case because the limits of the convolution integral run from 0 to t [Eq. (10.41)]. Thus,

$$e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) * \frac{1}{3}u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) * u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t})u(t) * \frac{1}{3}u(t) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) * u(t) \end{bmatrix}$$ $$= \begin{bmatrix} -\frac{1}{15}e^{-4t}u(t) * u(t) + \frac{2}{5}e^{-9t}u(t) * u(t) \\ -\frac{4}{5}e^{-4t}u(t) * u(t) + \frac{9}{5}e^{-9t}u(t) * u(t) \end{bmatrix}$$

Substitution for the preceding convolution integrals from the convolution table (Table 2.1) yields

$$e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} -\frac{1}{60}(1 - e^{-4t})u(t) + \frac{2}{45}(1 - e^{-9t})u(t) \\ -\frac{1}{5}(1 - e^{-4t})u(t) + \frac{1}{5}(1 - e^{-9t})u(t) \end{bmatrix}$$ $$= \begin{bmatrix} (\frac{1}{36} + \frac{1}{60}e^{-4t} - \frac{2}{45}e^{-9t})u(t) \\ \frac{1}{5}(e^{-4t} - e^{-9t})u(t) \end{bmatrix}$$

The sum of the two components now gives the desired solution for q(t):

$$\mathbf{q}(t) = \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(\frac{-63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix}$$

This result confirms the solution obtained by using the frequency-domain method [see Ex. 10.6]. Once the state variables q¹ and q² have been found for t≥0, all the remaining variables can be determined from the output equation.

THE OUTPUT

The output equation is given by

$$\mathbf{y}(t) = \mathbf{C}\mathbf{q}(t) + \mathbf{D}\mathbf{x}(t)$$

The substitution of the solution for q [Eq. (10.42)] in this equation yields

$$\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t)$$

Since the elements of B are constants,

$$e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t) = e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t)$$

938 CHAPTER 10 STATE-SPACE ANALYSIS

With this result, the output equation becomes

$$\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} \mathbf{B} * \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t)$$

Now recall that the convolution of x(t) with the unit impulse δ(t) yields x(t). Let us define a j × j diagonal matrix δ(t) such that all its diagonal terms are unit impulse functions. It is then obvious that

$$\delta(t) * \mathbf{x}(t) = \mathbf{x}(t)$$

and the output equation can be expressed as

$$\mathbf{y}(t) = \mathbf{C}[e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t)] + \mathbf{D}\delta(t) * \mathbf{x}(t)$$ $$= \mathbf{C}e^{\mathbf{A}t}\mathbf{q}(0) + [\mathbf{C}e^{\mathbf{A}t}\mathbf{B} + \mathbf{D}\delta(t)] * \mathbf{x}(t)$$

With the notation φ(t) for eA*^t* , the output equation may be expressed as

$$\mathbf{y}(t) = \underbrace{\mathbf{C}\boldsymbol{\phi}(t)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\left[\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)\right] * \mathbf{x}(t)}_{\text{zero-state response}}$$

The zero-state response, that is, the response when q(0) = 0, is

$$\mathbf{y}(t) = [\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)] * \mathbf{x}(t) = \mathbf{h}(t) * \mathbf{x}(t)$$

where

$$\mathbf{h}(t) = \mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t) \tag{10.45}$$

The matrix h(t) is a k × j matrix known as the impulse response matrix. The reason for this designation is obvious. The ijth element of h(t) is hij(t), which represents the zero-state response yi when the input xj(t) = δ(t) and when all other inputs (and all the initial conditions) are zero. Not surprisingly, vectors h(t) and H(s) form a Laplace transform pair,

$$\mathcal{L}[\mathbf{h}(t)] = \mathbf{H}(s)$$

EXAMPLE 10.9 State Transition Matrix by Inverse Laplace Transform

For the system described in Ex. 10.7, use Eq. (10.44) to determine e**^A**^t :

$$\boldsymbol{\phi}(t) = e^{\mathbf{A}t} = \mathcal{L}^{-1} \boldsymbol{\Phi}(s)$$

This problem was solved earlier with frequency-domain techniques. From Eq. (10.33), we have

$$\begin{aligned} \n\phi(t) &= \mathcal{L}^{-1} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)}\\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \\ \n&= \mathcal{L}^{-1} \begin{bmatrix} \frac{2}{s+1} - \frac{1}{s+2} & \frac{1}{s+1} - \frac{1}{s+2} \\ \frac{-2}{s+1} + \frac{2}{s+2} & \frac{-1}{s+1} + \frac{2}{s+2} \end{bmatrix} \\ \n&= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \n\end{aligned}$$

The same result is obtained in Ex. 10.1 (Sec. 10.1-3) by using Eq. (10.43) [see Eq. (10.13)]. Also, δ(t) is a diagonal j×j or 2×2 matrix:

$$\delta(t) = \begin{bmatrix} \delta(t) & 0\\ 0 & \delta(t) \end{bmatrix}$$

Substituting the matrices φ(t), δ(t), C, D, and B [Eq. (10.32)] into Eq. (10.45), we have

$$\mathbf{h}(t) = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \delta(t) & 0 \\ 0 & \delta(t) \end{bmatrix}$$ $$= \begin{bmatrix} 3e^{-t} - 2e^{-2t} & e^{-t} - e^{-2t} \\ \delta(t) + 2e^{-2t} & e^{-2t} \\ -6e^{-t} + 8e^{-2t} & \delta(t) - 2e^{-2t} + 4e^{-2t} \end{bmatrix}$$

As the reader can verify, the Laplace transform of this equation yields the transfer function matrix H(s) in Eq. (10.34).