状态空间表达式的能控标准型和能观标准型

寻找线性变换，得到单输入单输出系统的能控能控标准型。

Description

判断能控能观性：线性定常系统的能控性线性定常系统的能观性
化为标准型：状态向量的线性变换
写为传递函数：状态空间表达式的建立#二、从外部描述建立

\begin{matrix} \dot{z} = T^{- 1} A T z + T^{- 1} B u \\ y = C T z + D u \end{matrix}

能控性与能观性的对偶关系

互为对偶系统的两个系统：系统能控性分析 $\Leftrightarrow$ 能观性分析

系统的能控性等价于对偶系统的能观测性
对偶系统的传递函数矩阵相互转置
对偶系统的特征方程相同。

1. 定常系统

\begin{array}{r} Σ_{1} (A, B, C) Σ_{2} (A^{T}, C^{T}, B^{T}) \end{array}

\begin{array}{r} S y m 1 = {\begin{cases} \dot{x} = A x + B u \\ y = C x \end{cases} \Leftrightarrow S y m 2 = {\begin{cases} \dot{x} = A^{T} + C^{T} u \\ y = B^{T} x \end{cases} \end{array}

2. 时变系统

\begin{array}{r} Σ_{1} (A, B, C) Σ_{2} (- A^{T}, C^{T}, B^{T}) \end{array}

\begin{array}{r} S y m 1 = {\begin{cases} \dot{x} = A (t) x + B (t) u \\ y = C (t) x \end{cases} \Leftrightarrow S y m 2 = {\begin{cases} \dot{x} = - A^{T} (t) + C^{T} (t) u \\ y = B^{T} (t) x \end{cases} \end{array}

一、能控标准型

首先判断系统是否能控，构造能控性矩阵 $r a n k M = r a n k (\begin{matrix} b & A b & \dots & A^{n - 1} b \end{matrix}) = n$
其次利用系统矩阵计算特征多项式：

\begin{array}{r} | λ I - A | = λ^{n} + a_{n - 1} λ^{n - 1} + \dots + a_{1} λ_{1} + a_{0} \end{array}

1.1 能控标准 I 型

找线性变换使得系统矩阵为友矩阵，可以直接写出对应的传递函数。

\begin{array}{r} T_{c_{1}} = (\begin{array}{c} A^{n - 1} b & A^{n - 2} b & \dots & A b & b \end{array}) (\begin{array}{c} 1 \\ a_{n - 1} & 1 \\ ⋮ & ⋮ & ⋱ \\ a_{2} & a_{3} & \dots & 1 \\ a_{1} & a_{2} & \dots & a_{n - 1} & 1 \end{array}) \end{array}

\begin{aligned} \bar{A} & = T_{c_{1}}^{- 1} A T_{c_{1}} = (\begin{array}{c} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & 1 \\ - a_{0} & - a_{1} & - a_{2} & \dots & - a_{n - 1} \end{array}) \\ \bar{b} & = T_{c_{1}}^{- 1} b = (\begin{array}{c} 0 \\ 0 \\ ⋮ \\ 1 \end{array}) \bar{c} = c T_{c_{1}} = (\begin{array}{c} β_{0}, β_{1}, \dots, β_{n - 1} \end{array}) \end{aligned}

可以直接得到系统的传递函数：

\begin{array}{r} W (s) = \bar{c} (s I - \bar{A})^{- 1} \bar{b} = \frac{β_{n - 1} s^{n - 1} + β_{n - 2} s^{n - 2} + \dots + β_{1} s + β_{0}}{s^{n} + a_{n - 1} s^{n - 1} + \dots + a_{1} s + a_{0}} \end{array}

分母的系数多项式直接来自系统矩阵计算的特征多项式
分子的系数多项式来自计算的输出矩阵

1.2 能控标准 II 型

\begin{array}{r} T_{c_{2}} = (\begin{array}{c} b & A b & \dots & A^{n - 1} b \end{array}) = M \end{array}

\begin{aligned} \bar{A} & = T_{c_{2}}^{- 1} A T_{c_{2}} = {(\begin{array}{c} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & 1 \\ - a_{0} & - a_{1} & - a_{2} & \dots & - a_{n - 1} \end{array})}^{T} \\ \bar{b} & = T_{c_{2}}^{- 1} b = (\begin{array}{c} 1 \\ 0 \\ ⋮ \\ 0 \end{array}) \bar{c} = c T_{c_{2}} = (\begin{array}{c} β_{0}, β_{1}, \dots, β_{n - 1} \end{array}) \end{aligned}

二、能观标准型

首先判断系统是否能观，构造能观性矩阵

\begin{array}{r} r a n k N = r a n k (\begin{array}{c} C \\ C A \\ ⋮ \\ C A^{n - 1} \end{array}) = n \end{array}

或利用能控性和能观性的对偶关系，直接先求得对偶系统

\begin{array}{r} S y m 1 = {\begin{cases} \dot{x} = A x + B u \\ y = C x \end{cases} \Leftrightarrow S y m 2 = {\begin{cases} \dot{x} = A^{T} + C^{T} u \\ y = B^{T} x \end{cases} \end{array}

先判断对偶系统的能控性，再按照能控标准型的计算方法进行计算

能观标准 I 型也是对偶系统的能控标准 II 型
能观标准 II 型也是对偶系统的能控标准 I 型

2.2 能观标准 I 型

2.2 能观标准 II 型

\begin{aligned} \tilde{A} & = T_{o_{2}}^{- 1} A T_{o_{2}} = (\begin{array}{c} 0 & 0 & 0 & \dots & 0 & - a_{0} \\ 1 & 0 & 0 & \dots & 0 & - a_{1} \\ 0 & 1 & 0 & \dots & 0 & - a_{2} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & 1 & - a_{n - 1} \end{array}) \\ \tilde{b} & = T_{o_{2}}^{- 1} b = (\begin{array}{c} β_{0} \\ β_{1} \\ ⋮ \\ β_{n - 1} \end{array}) \tilde{c} = c T_{o_{2}} = (\begin{array}{c} 0, 0, \dots, 1 \end{array}) \end{aligned}

\begin{array}{r} W (s) = \bar{c} (s I - \bar{A})^{- 1} \bar{b} = \frac{β_{m} s^{m} + β_{m - 1} s^{m - 1} + \dots + β_{1} s + β_{0}}{s^{n} + a_{n - 1} s^{n - 1} + \dots + a_{1} s + a_{0}} \end{array}

三、实际例题

将下列标准型写为四种标准型

\begin{array}{r} \dot{x} = (\begin{array}{c} 1 & 2 & 0 \\ 3 & - 1 & 1 \\ 0 & 2 & 0 \end{array}) x + (\begin{array}{c} 2 \\ 1 \\ 1 \end{array}) u y = (\begin{array}{c} 0 & 0 & 1 \end{array}) x \end{array}

计算系统的特征多项式：

\begin{array}{r} | λ I - A | = λ^{n} + a_{n - 1} λ^{n - 1} + \dots + a_{1} λ_{1} + a_{0} = λ^{3} - 9 λ + 2 \end{array}

能控标准型

判断系统能控性：

\begin{array}{r} M = (\begin{array}{c} b & A b & A^{2} b \end{array}) = (\begin{array}{c} 2 & 4 & 16 \\ 1 & 6 & 8 \\ 1 & 2 & 12 \end{array}) r a n k M = 3 \end{array}

能控标准 I 型

\begin{array}{r} \bar{A} = (\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - 2 & 9 & 0 \end{array}) \bar{b} = (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) \end{array}

\begin{array}{r} \bar{C} = C (\begin{array}{c} A^{2} b & A b & b \end{array}) (\begin{array}{c} 1 \\ a_{2} & 1 \\ a_{1} & a_{2} & 1 \end{array}) = (\begin{array}{c} 0 & 0 & 1 \end{array}) (\begin{array}{c} 16 & 4 & 2 \\ 8 & 6 & 1 \\ 12 & 2 & 1 \end{array}) (\begin{array}{c} 1 \\ 0 & 1 \\ - 9 & 0 & 1 \end{array}) = (\begin{array}{c} 3 & 2 & 1 \end{array}) \end{array}

能控标准 II 型

\begin{array}{r} \bar{A} = (\begin{array}{c} 0 & 0 & - 2 \\ 1 & 0 & 9 \\ 0 & 1 & 0 \end{array}) \bar{b} = (\begin{array}{c} 1 \\ 0 \\ 0 \end{array}) \end{array}

\begin{array}{r} \bar{C} = C (\begin{array}{c} b & A b & A^{2} b \end{array}) = (\begin{array}{c} 0 & 0 & 1 \end{array}) (\begin{array}{c} 2 & 4 & 16 \\ 1 & 6 & 8 \\ 1 & 2 & 12 \end{array}) = (\begin{array}{c} 1 & 2 & 12 \end{array}) \end{array}

能观标准型

判断系统能观性：

\begin{array}{r} N = (\begin{array}{c} C \\ C A \\ C A^{2} \end{array}) = (\begin{array}{c} 0 & 0 & 1 \\ 0 & 2 & 0 \\ 6 & - 2 & 2 \end{array}) r a n k N = 3 \end{array}

能观标准 I 型

\begin{array}{r} \tilde{A} = (\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - 2 & 9 & 0 \end{array}) \tilde{b} = (\begin{array}{c} 1 \\ 2 \\ 12 \end{array}) \tilde{C} = (\begin{array}{c} 1 & 0 & 0 \end{array}) \end{array}

能观标准 II 型

\begin{array}{r} \tilde{A} = (\begin{array}{c} 0 & 0 & - 2 \\ 1 & 0 & 9 \\ 0 & 1 & 0 \end{array}) \tilde{b} = (\begin{array}{c} 3 \\ 2 \\ 1 \end{array}) \tilde{C} = (\begin{array}{c} 0 & 0 & 1 \end{array}) \end{array}

注意

首先判断 能控性/能观性，再进行特征多项式计算，得到系统不变量，马上可以直接写出系统矩阵（以及能控标准型的输入矩阵、能观标准型的输出矩阵）。