910 CHAPTER 10 STATE-SPACE ANALYSIS

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910 CHAPTER 10 STATE-SPACE ANALYSIS

Since differentiation is a linear operation, it is easy to show that

$$\frac{d}{dt}(\mathbf{A} + \mathbf{B}) = \frac{d\mathbf{A}}{dt} + \frac{d\mathbf{B}}{dt} \quad \text{and} \quad \frac{d}{dt}(c\mathbf{A}) = c\frac{d\mathbf{A}}{dt}$$

The derivative of a matrix product is given as

$$\frac{d}{dt}(\mathbf{AB}) = \frac{d\mathbf{A}}{dt}\mathbf{B} + \mathbf{A}\frac{d\mathbf{B}}{dt} = \dot{\mathbf{A}}\mathbf{B} + \mathbf{A}\dot{\mathbf{B}}\tag{10.2}$$

We can prove Eq. (10.2) as follows. Let A be an m×n matrix and B an n×p matrix. Then, if

C = AB

from Eq. (B.33), we have

$$c_{ik} = \sum_{j=1}^{n} a_{ij} b_{jk}$$

and

$$\dot{c}_{ik} = \underbrace{\sum_{j=1}^{n} \dot{a}_{ij} b_{jk}}_{d_{ik}} + \underbrace{\sum_{j=1}^{n} a_{ij} \dot{b}_{jk}}_{e_{ik}}$$

or

$$\dot{c}_{ik} = d_{ik} + e_{ik}$$

(10.3)

Equation (10.3) along with the multiplication rule clearly indicates that dik is the ikth element of matrix AB˙ and eik is the ikth element of matrix AB˙ . Equation (10.2) then follows.

If we let ^{B = A}−¹ in Eq. (10.2), we obtain

$$\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d\mathbf{A}}{dt}\mathbf{A}^{-1} + \mathbf{A}\frac{d}{dt}\mathbf{A}^{-1}$$

But since

$$\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d}{dt}\mathbf{I} = 0$$

we have

$$\frac{d}{dt}(\mathbf{A}^{-1}) = -\mathbf{A}^{-1}\frac{d\mathbf{A}}{dt}\mathbf{A}^{-1}$$

10.1-2 The Characteristic Equation of a Matrix: The Cayley–Hamilton Theorem

For an (n×n) square matrix A, any vector x (x = 0) that satisfies the equation

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{x} \tag{10.4}$$

is an eigenvector (or characteristic vector), and λ is the corresponding eigenvalue (or characteristic value) of A. Equation (10.4) can be expressed as

$$(\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = 0 \qquad \text{or} \qquad (\lambda \mathbf{I} - \mathbf{A})\mathbf{x} = 0$$

The solution for this set of homogeneous equations exists if and only if

$$|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix} = 0$$

(10.5)

Equation (10.5) is known as the characteristic equation of matrix A and can be expressed as

$$Q(\lambda) = |\lambda \mathbf{I} - \mathbf{A}| = \lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_1\lambda + a_0\lambda^0 = 0$$

(10.6)

Q(λ) is called the characteristic polynomial of matrix A. The n zeros of the characteristic polynomial are the eigenvalues of A and, corresponding to each eigenvalue, there is an eigenvector that satisfies Eq. (10.4).

The Cayley–Hamilton theorem states that every n×n matrix A satisfies its own characteristic equation. In other words, Eq. (10.6) is valid if λ is replaced by A:

$$\mathbf{Q}(\mathbf{A}) = \mathbf{A}^n + a_{n-1}\mathbf{A}^{n-1} + \dots + a_1\mathbf{A} + a_0\mathbf{A}^0 = 0$$

(10.7)

FUNCTIONS OF A MATRIX

We now demonstrate the use of the Cayley–Hamilton theorem [Eq. (10.7)] to evaluate functions of an n×n square matrix A.

Consider a function f(λ) in the form of an infinite power series:

$$f(\lambda) = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda_2^2 + \dots = \sum_{i=0}^{\infty} \alpha_i \lambda^i$$

(10.8)

Since λ, being an eigenvalue (characteristic root) of A, satisfies the characteristic equation [Eq. (10.6)], we can write

$$\lambda^{n} = -a_{n-1}\lambda^{n-1} - a_{n-2}\lambda^{n-2} - \dots - a_1\lambda - a_0$$

(10.9)

If we multiply both sides by λ, the left-hand side is λ*ⁿ+1, and the right-hand side contains the terms λⁿ, λⁿ−1, … , λ. Using Eq. (10.9), we substitute λⁿ* in terms of λ*ⁿ−1, λⁿ−2,…, λ so that the highest power on the right-hand side is reduced to n − 1. Continuing in this way, we see that λⁿ+^k* can be expressed in terms of λ*ⁿ−1, λⁿ−2,…,λ for any k. Hence, the infinite series on the right-hand side of Eq. (10.8) can always be expressed in terms of λⁿ−1, λⁿ*−2,…,λ and a constant as

$$f(\lambda) = \beta_0 + \beta_1 \lambda + \beta_2 \lambda^2 + \dots + \beta_{n-1} \lambda^{n-1}$$

(10.10)

If we assume that there are n distinct eigenvalues λ1, λ2, … , λn, then Eq. (10.10) holds for these n values of λ. The substitution of these values in Eq. (10.10) yields n simultaneous equations

$$\begin{bmatrix} f(\lambda_1) \\ f(\lambda_2) \\ \vdots \\ f(\lambda_n) \end{bmatrix} = \begin{bmatrix} 1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1} \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_{n-1} \end{bmatrix}$$

912 CHAPTER 10 STATE-SPACE ANALYSIS

Solving for the β coefficients yields

$$\begin{bmatrix}\n\beta_0 \\ \beta_1 \\ \vdots \\ \beta_{n-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\nf(\lambda_1) \\ f(\lambda_2) \\ \vdots \\ f(\lambda_n)\n\end{bmatrix}$$

\n(10.11)

Since A also satisfies Eq. (10.9), we may advance a similar argument to show that if f(A) is a function of a square matrix A expressed as an infinite power series in A, then

$$f(\mathbf{A}) = \alpha_0 \mathbf{I} + \alpha_1 \mathbf{A} + \alpha_2 \mathbf{A}^2 + \cdots = \sum_{i=0}^{\infty} \alpha_i \mathbf{A}^i$$

and, as argued earlier, the right-hand side can be expressed by using terms of power less than or equal to n−1,

$$f(\mathbf{A}) = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{n-1} \mathbf{A}^{n-1} = \sum_{i=0}^{n-1} \beta_i \mathbf{A}^i$$

(10.12)

in which the coefficients βis are found from Eq. (10.11). If some of the eigenvalues are repeated (multiple roots), the results are somewhat modified.

We shall demonstrate the utility of this result with the following two examples.

10.1-3 Computation of an Exponential and a Power of a Matrix

Let us compute eA*^t* defined by

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

From Eq. (10.12), we can express

$$e^{\mathbf{A}t} = \sum_{i=1}^{n-1} \beta_i(\mathbf{A})^i$$

in which the βis are given by Eq. (10.11), with f(λi) = e^λit .

EXAMPLE 10.1 Computing the Exponential of a Matrix

Compute e**^A**^t for the case

$$\mathbf{A} = \left[ \begin{array}{cc} 0 & 1 \\ -2 & -3 \end{array} \right]$$

The characteristic equation is

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

Hence, the eigenvalues are λ¹ = −1, λ² = −2, and

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

in which

$$\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -2 \end{bmatrix}^{-1} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2e^{-t} - e^{-2t} \\ e^{-t} - e^{-2t} \end{bmatrix}$$

and

$$e^{\mathbf{A}t} = (2e^{-t} - e^{-2t}) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + (e^{-t} - e^{-2t}) \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$$ $$= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix}$$

(10.13)

COMPUTATION OF A^k As Eq. (10.12) indicates, we can express A*^k* as

$$\mathbf{A}^{k} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \cdots + \beta_{n-1} \mathbf{A}^{n-1}$$

in which the βis are given by Eq. (10.11) with f(λi) = λ*^{k i}* . For a completed example of the computation of A*^k* by this method, see Ex. 10.13.