状态向量的线性变换

状态空间表达式进行线性变换变为标准型的基础。

Description

{\begin{cases} \dot{x} = A x + B u \\ y = C x + D u \end{cases}

系统的特征值：就是系统矩阵 $A$ 的特征值，也就是特征方程 $| λ I - A | = 0$ 的根。
系统的不变量：特征多项式 $| λ I - A | = 0$ 的系数（唯一地确定系统的特征值）
经过非奇异的线性变换后，系统的特征值和系统的不变量均不改变。

求变换矩阵 $T$ ， $x = T z$ ，将系统矩阵 $A$ 变换为 Jordan 标准型 $J = T^{- 1} A T$ ，得到变换后的状态空间表达式：

\begin{array}{r} {\begin{cases} T \dot{z} = A T z + B u \\ y = C T z + D u \end{cases} \Rightarrow {\begin{cases} \dot{z} = T^{- 1} A T z + T^{- 1} B u \\ y = C T z + D u \end{cases} \end{array}

\dot{z} = T^{- 1} A T z + T^{- 1} B u = {\begin{cases} \dot{z} = Λ z + T^{- 1} B u 特征值无重根，化为对角阵 \\ \dot{z} = J z + T^{- 1} B u 特征值有重根，化为约旦标准型 \end{cases}

由 $n$ 个互异特征值求得的 $n$ 个特征向量构成变换矩阵：
$| λ I - A | = 0$ 计算出特征值， $(λ_{i} I - A) P_{i} = 0$ 计算第 $i$ 个特征值对应的特征向量 $P_{i}$

\begin{array}{r} (λ_{i} I - A) P_{i} = 0 \Rightarrow T = (\begin{array}{c} P_{1}, P_{2}, \dots, P_{n} \end{array}) \end{array}

\begin{array}{r} Λ = T^{- 1} A T = (\begin{array}{c} λ_{1} \\ λ_{2} \\ ⋱ \\ λ_{n} \end{array}) \end{array}

假设有 $q$ 个 $λ_{1}$ 的重根，其余 $(n - q)$ 个互异的根
$P_{1}$ 为特征值 $λ_{1}$ 对应的特征向量， $P_{2}, \dots, P_{q}$ 为广义特征向量

\begin{array}{r} T = (P_{1}, P_{2}, \dots, P_{q}, P_{q + 1}, \dots, P_{n}) \end{array}

{\begin{cases} λ_{1} P_{1} - A P_{1} = 0 \\ λ_{1} P_{2} - A P_{2} = - P_{1} \\ ⋮ \\ λ_{1} P_{q} - A P_{q} = - P_{q - 1} \end{cases} {\begin{cases} (λ_{1} I - A) P_{1} = 0 \\ (λ_{1} I - A) P_{2} = - P_{1} \\ (λ_{1} I - A) P_{3} = - P_{2} \\ ⋮ \\ (λ_{1} I - A) P_{q} = - P_{q - 1} \end{cases}

{\begin{array}{r} J = T^{- 1} A T = {(\begin{array}{c} J_{1} \\ λ_{q + 1} \\ ⋱ \\ λ_{n} \end{array})}_{n \times n} J_{1} = (\begin{array}{c} λ_{1} & 1 \\ λ_{1} & 1 \\ ⋱ & 1 \\ λ_{1} \end{array}) \end{array}}_{q \times q}

如果系统矩阵为友矩阵

\begin{array}{r} A = (\begin{array}{c} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & 0 \\ - a_{0} & - a_{1} & - a_{2} & \dots & - a_{n - 1} \end{array}) \end{array}

\begin{array}{r} T = (\begin{array}{c} 1 & 1 & \dots & 1 \\ λ_{1} & λ_{2} & \dots & λ_{n} \\ λ_{1}^{2} & λ_{2}^{2} & \dots & λ_{n}^{2} \\ ⋮ & ⋮ & ⋮ \\ λ_{1}^{n - 1} & λ_{2}^{n - 1} & \dots & λ_{n}^{n - 1} \end{array}) \end{array}

\begin{array}{r} T_{3} = (\begin{array}{c} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{array}) \end{array}

假设三重根：

\begin{array}{r} T = (\begin{array}{c} 1 & 0 & 0 & \dots & 1 \\ λ_{1} & 1 & 0 & \dots & λ_{n} \\ λ_{1}^{2} & 2 λ_{1} & 1 & \dots & λ_{n}^{2} \\ ⋮ & ⋮ & ⋮ & \dots & ⋮ \\ λ_{1}^{n - 1} & \frac{d}{d λ_{1}} (λ_{1}^{n - 1}) & \frac{1}{2} \frac{d^{2}}{d λ_{1}^{2}} (λ_{1}^{n - 1}) & \dots & λ_{n}^{n - 1} \end{array}) \end{array}

$\frac{1}{n!} \frac{d^{n}}{d λ_{1}^{n}} (λ_{1}^{n})$

\begin{array}{r} T_{3} = (\begin{array}{c} 1 & 0 & 0 \\ λ_{1} & 1 & 0 \\ λ_{1}^{2} & 2 λ_{1} & 1 \end{array}) \end{array}

将传递函数分解为部分分式之和。有理分式的分解

\begin{array}{r} W (s) = \frac{c_{1}}{s - λ_{1}} + \frac{c_{2}}{s - λ_{2}} + \dots + \frac{c_{n}}{s - λ_{n}} = \sum_{i = 1}^{n} \frac{c_{i}}{s - λ_{i}} \end{array}

\begin{array}{r} A = (\begin{array}{c} λ_{1} \\ λ_{2} \\ ⋱ \\ λ_{n} \end{array}) B = (\begin{array}{c} 1 \\ 1 \\ ⋮ \\ 1 \end{array}) C = (\begin{array}{c} c_{1} c_{2} \dots c_{n} \end{array}) \end{array}

\begin{array}{r} A = (\begin{array}{c} λ_{1} \\ λ_{2} \\ ⋱ \\ λ_{n} \end{array}) B = (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) C = (\begin{array}{c} 1 1 \dots 1 \end{array}) \end{array}

\begin{array}{r} W (s) = \frac{c_{1 q}}{(s - λ_{1})^{q}} + \frac{c_{1 (q - 1)}}{(s - λ_{1})^{q - 1}} + \dots + \frac{c_{11}}{s - λ_{1}} + \sum_{i = q + 1}^{n} \frac{c_{i}}{s - λ_{i}} \end{array}

\begin{array}{r} A = (\begin{array}{c} λ_{1} & 1 \\ λ_{1} & 1 \\ ⋱ & 1 \\ λ_{1} \\ λ_{q + 1} \\ ⋱ \\ λ_{n} \end{array}) B = (\begin{array}{c} 0 \\ 0 \\ ⋮ \\ 1 \\ 1 \\ ⋮ \\ 1 \end{array}) C = (\begin{array}{c} c_{1 q} c_{1 (q - 1)} \dots c_{11} c_{q + 1} \dots c_{n} \end{array}) \end{array}