线性定常状态方程的解

Solution of linear time-invariant state equations

当初始时刻为 $t_{0}$ ，初始状态 $x (t_{0})$ 时，矩阵微分方程的解为：

\begin{array}{r} \dot{x} (t) = A x (t) \Rightarrow x (t) = e^{A (t - t_{0})} x (t_{0}) = Φ (t - t_{0}) x (t_{0}) \end{array}

求解核心：利用状态向量的线性变换将系统矩阵化为标准型，再结合特殊矩阵的指数函数

计算出特征值： $| λ I - A | = 0$

求解的核心：计算逆矩阵，分解因式进行拉普拉斯逆变换，求得时间域上的解。

由拉普拉斯变换得到：

\begin{matrix} s X (s) = A X (s) + x (0) \\ (s I - A) X (s) = x (0) \\ X (s) = (s I - A)^{- 1} x (0) \\ x (t) = L^{- 1} [(s I - A)^{- 1}] x (0) \end{matrix}

对比矩阵指数得到：

\begin{array}{r} e^{A t} = L^{- 1} [{(s I - A)}^{- 1}] \end{array}

求解核心：将矩阵指数表达为 $n - 1$ 及 $n - 1$ 以下幂次的线性组合，记住系数的计算公式

计算出特征值： $| λ I - A | = 0$
由凯莱-哈密顿定理得到：

\begin{aligned} e^{A t} & = \sum_{m = 0}^{n - 1} α_{m} (t) A^{m} = α_{n - 1} A^{n - 1} + α_{n - 2} A^{n - 2} + \dots + α_{1} A + α_{0} I \end{aligned}

\begin{array}{r} (\begin{array}{c} α_{0} \\ α_{1} \\ ⋮ \\ α_{n - 1} \end{array}) = {(\begin{array}{c} 1 & λ_{1} & λ_{1}^{2} & \dots & λ_{1}^{n - 1} \\ 1 & λ_{2} & λ_{2}^{2} & \dots & λ_{2}^{n - 1} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & λ_{n} & λ_{n}^{2} & \dots & λ_{n}^{n - 1} \end{array})}^{- 1} (\begin{array}{c} e^{λ 1 t} \\ e^{λ_{2} t} \\ ⋮ \\ e^{λ_{n} t} \end{array}) \end{array}

\begin{array}{r} (\begin{array}{c} α_{0} \\ α_{1} \\ α_{2} \end{array}) = {(\begin{array}{c} 1 & λ_{1} & λ_{1}^{2} \\ 1 & λ_{2} & λ_{2}^{2} \\ 1 & λ_{3} & λ_{3}^{2} \end{array})}^{- 1} (\begin{array}{c} e^{λ 1 t} \\ e^{λ_{2} t} \\ e^{λ_{3} t} \end{array}) \end{array}

\begin{array}{r} (\begin{array}{c} α_{0} \\ α_{1} \\ ⋮ \\ α_{n - 2} \\ α_{n - 1} \end{array}) = {(\begin{array}{c} 0 & 0 & 0 & \dots & 0 & 1 \\ 0 & 0 & 0 & \dots & 1 & (n - 1) λ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 1 & \dots & \dots & \frac{(n - 1) (n - 2)}{2!} λ^{n - 3} \\ 0 & 1 & 2 λ & \dots & (n - 2) λ^{(n - 3)} & (n - 1) λ^{n - 2} \\ 1 & λ & λ^{2} & \dots & λ^{n - 2} & λ^{n - 1} \end{array})}^{- 1} (\begin{array}{c} \frac{1}{(n - 1)!} t^{n - 1} e^{λ t} \\ \frac{1}{(n - 2)!} t^{n - 2} e^{λ t} \\ ⋮ \\ \frac{1}{2!} t^{2} e^{λ t} \\ t e^{λ t} \\ e^{λ t} \end{array}) \end{array}

$α_{n - i} = \frac{d α_{n - i + 1}}{d λ} \cdot \frac{1}{i - 1}$

\begin{array}{r} (\begin{array}{c} α_{0} \\ α_{1} \\ α_{2} \end{array}) = {(\begin{array}{c} 0 & 0 & 1 \\ 0 & 1 & 2 λ \\ 1 & λ & λ^{2} \end{array})}^{- 1} (\begin{array}{c} \frac{1}{2!} t^{2} e^{λ t} \\ t e^{λ t} \\ e^{λ t} \end{array}) \end{array}

当初始时刻为 $t_{0}$ ，初始状态 $x (t_{0})$ 时，线性非齐次状态方程的解由两部分组成：

\begin{array}{r} \dot{x} (t) = A x (t) + B u (t) \Rightarrow x (t) = Φ (t - t_{0}) x (t_{0}) + \int_{t_{0}}^{t} Φ (t - τ) B u (τ) d τ \end{array}

\begin{array}{r} \dot{x} (t) = A x (t) + B u (t) \Rightarrow x (t) = Φ (t) x (0) + \int_{0}^{t} Φ (t - τ) B u (τ) d τ \end{array}

脉冲响应： $u (t) = K δ (t)$ $x (t) = e^{A t} x_{0} + e^{A t} B K$
阶跃响应： $u (t) = K \times 1 (t)$ $x (t) = e^{A t} x_{0} + A^{- 1} (e^{A t} - I) B K$
~~斜坡响应~~： $u (t) = K t \times 1 (t)$ $x (t) = e^{A t} x_{0} + [A^{- 2} (e^{^{A t}} - I) - A^{- 1 t}] B K$