【现代控制理论】传递函数建立状态空间表达式

对于一个控制系统的传递函数为 G ( s ) = Y ( s ) U ( s ) = b 1 s n − 1 + ⋯ + b n − 1 s + b n s n + a 1 s n − 1 + ⋯ + a n − 1 s + a n (1) G(s)=\frac{Y(s)}{U(s)}=\frac{b_{1} s^{n-1}+\cdots+b_{n-1} s+b_{n}}{s^{n}+a_{1} s^{n-1}+\cdots+a_{n-1} s+a_{n}}\tag{1} G(s)=U(s)Y(s)​=sn+a1​sn−1+⋯+an−1​s+an​b1​sn−1+⋯+bn−1​s+bn​​(1)

为了将传递函数转化为系统的状态空间表达式

由式子（1）可直接得到如下的系统状态空间表达式。

能控标准型：

x ˙ = A x + B u = [ 0 1 ⋯ 0 0 0 ⋱ 0 ⋮ ⋮ ⋱ 1 − a n − a n − 1 ⋯ − a 1 ] x + [ 0 ⋮ 0 1 ] u y = C x = [ b n b n − 1 ⋯ b 1 ] x (2) \begin{array}{c} \dot{\boldsymbol{x}}=\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B} u=\left[\begin{array}{cccc} 0 & 1 & \cdots & 0 \\ 0 & 0 & \ddots & 0\\ \vdots &\vdots & \ddots & 1 \\ -a_{n} & -a_{n-1} & \cdots & -a_{1} \end{array}\right] \boldsymbol{x}+\left[\begin{array}{l} 0 \\ \vdots \\ 0 \\ 1 \end{array}\right] u \\ \\ y=\boldsymbol{C} \boldsymbol{x}=\left[\begin{array}{llll} b_{n} & b_{n-1} & \cdots & b_{1} \end{array}\right] \boldsymbol{x} \end{array}\tag{2} x˙=Ax+Bu= ​00⋮−an​​10⋮−an−1​​⋯⋱⋱⋯​001−a1​​ ​x+ ​0⋮01​ ​uy=Cx=[bn​​bn−1​​⋯​b1​​]x​(2)

能观标准型：

x ˙ = A x + B u = [ 0 0 ⋯ 0 − a n 1 0 ⋯ 0 − a n − 1 0 1 ⋯ 0 − a n − 2 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 − a 1 ] x + [ b n b n − 1 ⋮ b 2 b 1 ] u y = C x = [ 0 0 ⋯ 1 ] x (3) \begin{array}{c} \dot{\boldsymbol{x}}=\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B} u=\left[\begin{array}{ccccc} 0 & 0 & \cdots& 0 & -a_{n} \\ 1 & 0 & \cdots & 0& -a_{n-1} \\ 0 & 1 & \cdots & 0& -a_{n-2} \\ \vdots & \vdots& \ddots & \vdots&\vdots \\ 0 & 0 & \cdots& 1 & -a_{1} \end{array}\right] \boldsymbol{x}+\left[\begin{array}{l} b_n \\ b_{n-1}\\ \vdots \\ b_2\\ b_1\\ \end{array}\right] u \\ \\ y=\boldsymbol{C} \boldsymbol{x}=\left[\begin{array}{llll} 0 & 0 & \cdots & 1 \end{array}\right] \boldsymbol{x} \end{array}\tag{3} x˙=Ax+Bu= ​010⋮0​001⋮0​⋯⋯⋯⋱⋯​000⋮1​−an​−an−1​−an−2​⋮−a1​​ ​x+ ​bn​bn−1​⋮b2​b1​​ ​uy=Cx=[0​0​⋯​1​]x​(3)

注意点

（1）系统传递函数的极点都不同 那么式（1）可展现成部分分式的形式：

G ( s ) = c 1 s − p 1 + c 2 s − p 2 + ⋯ + c n s − p n (4) G(s)=\frac{c_{1}}{s-p_{1}}+\frac{c_{2}}{s-p_{2}}+\cdots+\frac{c_{n}}{s-p_{n}}\tag{4} G(s)=s−p1​c1​​+s−p2​c2​​+⋯+s−pn​cn​​(4) 其中 c i = lim ⁡ t → p i ( s − p i ) G ( s ) , i = 1 , 2 , ⋯   , n (5) c_{i}=\lim _{t \rightarrow p_{i}}\left(s-p_{i}\right) G(s), \quad i=1,2, \cdots, n\tag{5} ci​=t→pi​lim​(s−pi​)G(s),i=1,2,⋯,n(5) 可得系统状态空间表达式 x ˙ = [ p 1 0 ⋯ 0 0 p 2 ⋮ ⋮ ⋱ 0 0 ⋯ 0 p n ] x + [ 1 1 ⋮ 1 ] u y = [ c 1 c 2 ⋯ c n ] x (6) \begin{array}{c} \dot{\boldsymbol{x}}=\left[\begin{array}{cccc} p_{1} & 0 & \cdots & 0 \\ 0 & p_{2} & & \vdots \\ \vdots & & \ddots & 0 \\ 0 & \cdots & 0 & p_{n} \end{array}\right] \boldsymbol{x}+\left[\begin{array}{c} 1 \\ 1 \\ \vdots \\ 1 \end{array}\right] u \\ \\ y=\left[\begin{array}{llll} c_{1} & c_{2} & \cdots & c_{n} \end{array}\right] x \end{array}\tag{6} x˙= ​p1​0⋮0​0p2​⋯​⋯⋱0​0⋮0pn​​ ​x+ ​11⋮1​ ​uy=[c1​​c2​​⋯​cn​​]x​(6) （2）系统传递函数的极点有相同

此时式（1）可以展开成下面的形式（7） g ( s ) = c 11 ( s − p 1 ) r + c 12 ( s − p 1 ) r − 1 + ⋯ + c 1 r ( s − p 1 ) + c r + 1 ( s − p r + 1 ) + ⋯ + c n ( s − p n ) (7) \begin{aligned} g(s)=& \frac{c_{11}}{\left(s-p_{1}\right)^{r}}+\frac{c_{12}}{\left(s-p_{1}\right)^{r-1}}+\cdots+\frac{c_{1 r}}{\left(s-p_{1}\right)} &+\frac{c_{r+1}}{\left(s-p_{r+1}\right)}+\cdots+\frac{c_{n}}{\left(s-p_{n}\right)}\tag{7} \end{aligned} g(s)=​(s−p1​)rc11​​+(s−p1​)r−1c12​​+⋯+(s−p1​)c1r​​​+(s−pr+1​)cr+1​​+⋯+(s−pn​)cn​​​(7)

其中单极点 p i p_i pi​ 对应的系数 c i ( i = r + 1 , ⋯   , n ) c_i(i=r+1, \cdots, n) ci​(i=r+1,⋯,n) 仍按照式 (5) 计算, 而 r r r 重极点 p 1 p_1 p1​ 对应的系数 c i j ( j = 1 , 2 , ⋯   , r ) c_{i j}(j=1,2, \cdots, r) cij​(j=1,2,⋯,r) 则按式 (8) 计算: c 1 j = 1 ( j − 1 ) ! lim ⁡ t → p 1 d j − 1 d s j − 1 { ( s − p 1 ) r g ( s ) } , j = 1 , 2 , ⋯   , r (8) c_{1 j}=\frac{1}{(j-1) !} \lim _{t \rightarrow p_1} \frac{\mathrm{d}^{j-1}}{\mathrm{d} s^{j-1}}\left\{\left(s-p_1\right)^r g(s)\right\}, \quad j=1,2, \cdots, r \tag{8} c1j​=(j−1)!1​t→p1​lim​dsj−1dj−1​{(s−p1​)rg(s)},j=1,2,⋯,r(8) 可得系统状态空间表达式为: [ x ˙ 1 x ˙ 2 ⋮ x ˙ r x ˙ r + 1 ⋮ x ˙ n ] = [ p 1 1 0 p 1 ⋱ ⋮ 0 ⋱ 1 0 0 ⋯ p 1 p r + 1 0 ⋱ p n ] [ x 1 x 2 ⋮ x r x r + 1 ⋮ x n ] + [ 0 0 ⋮ 1 1 ⋮ 1 ] u (9) \left[\begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \\ \vdots \\ \dot{x}_r \\ \dot{x}_{r+1} \\ \vdots \\ \dot{x}_n \end{array}\right]=\left[\begin{array}{ccccccc} p_1 & 1 & & 0 & & & \\ & p_1 & \ddots & \vdots & & 0 & \\ & & \ddots & 1 & & & \\ 0 & 0 & \cdots & p_1 & & & \\ & & & & p_{r+1} & & \\ & 0 & & & & \ddots & \\ & & & & & & p_n \end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_r \\ x_{r+1} \\ \vdots \\ x_n \end{array}\right]+\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \\ 1 \\ \vdots \\ 1 \end{array}\right] u\tag{9} ​x˙1​x˙2​⋮x˙r​x˙r+1​⋮x˙n​​ ​= ​p1​0​1p1​00​⋱⋱⋯​0⋮1p1​​pr+1​​0⋱​pn​​ ​ ​x1​x2​⋮xr​xr+1​⋮xn​​ ​+ ​00⋮11⋮1​ ​u(9) y = [ c 11 c 12 ⋯ c 1 r c r + 1 ⋯ c n ] [ x 1 x 2 ⋮ x r x r + 1 ⋮ x n ] (10) y=\left[\begin{array}{lllllll} c_{11} & c_{12} & \cdots & c_{1 r} & c_{r+1} & \cdots & c_n \end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_r \\ x_{r+1} \\ \vdots \\ x_n \end{array}\right]\tag{10} y=[c11​​c12​​⋯​c1r​​cr+1​​⋯​cn​​] ​x1​x2​⋮xr​xr+1​⋮xn​​ ​(10) 可以看出, 状态空间表达式中, A \boldsymbol{A} A 矩阵具有约当型, 因此又称为约当标准型。

串联分解适用于传递函数已被分解为因式相乘的形式： g ( s ) = b 1 ( s − z 1 ) ( s − z 2 ) ⋯ ( s − z n − 1 ) ( s − p 1 ) ( s − p 2 ) ⋯ ( s − p n ) (11) g(s)=\frac{b_1\left(s-z_1\right)\left(s-z_2\right) \cdots\left(s-z_n-1\right)}{\left(s-p_1\right)\left(s-p_2\right) \cdots\left(s-p_n\right)}\tag{11} g(s)=(s−p1​)(s−p2​)⋯(s−pn​)b1​(s−z1​)(s−z2​)⋯(s−zn​−1)​(11) 现以一个三阶系统传递函数为例予以说明。设 g ( s ) = Y ( s ) U ( s ) = b 1 ( s − z 2 ) ( s − z 3 ) ( s − p 1 ) ( s − p 2 ) ( s − p 3 ) = b 1 ( s − p 1 ) × ( s − z 2 ) ( s − p 2 ) × ( s − z 3 ) ( s − p 3 ) (12) \begin{aligned} g(s) &=\frac{Y(s)}{U(s)} \\ &=\frac{b_1\left(s-z_2\right)\left(s-z_3\right)}{\left(s-p_1\right)\left(s-p_2\right)\left(s-p_3\right)} \\ &=\frac{b_1}{\left(s-p_1\right)} \times \frac{\left(s-z_2\right)}{\left(s-p_2\right)} \times \frac{\left(s-z_3\right)}{\left(s-p_3\right)} \end{aligned}\tag{12} g(s)​=U(s)Y(s)​=(s−p1​)(s−p2​)(s−p3​)b1​(s−z2​)(s−z3​)​=(s−p1​)b1​​×(s−p2​)(s−z2​)​×(s−p3​)(s−z3​)​​(12) 显然,这个系统可以看作为三个一阶系统串联而成,其模拟结构如图所示 若指定图中每个积分器的输出为状态变量, 可得系统状态空间表达式为 [ x ˙ 1 x ˙ 2 x ˙ 3 ] = [ p 1 0 0 1 p 2 0 1 p 2 − z 2 p 3 ] [ x 1 x 2 x 3 ] + [ b 1 0 0 ] u y = [ 1 p 2 − z 2 p 3 − z 3 ] [ x 1 x 2 x 3 ] (13) \begin{aligned} {\left[\begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{array}\right] } &=\left[\begin{array}{ccc} p_1 & 0 & 0 \\ 1 & p_2 & 0 \\ 1 & p_2-z_2 & p_3 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]+\left[\begin{array}{c} b_1 \\ 0 \\ 0 \end{array}\right] u \\ y &=\left[\begin{array}{lll} 1 & p_2-z_2 & p_3-z_3 \end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right] \end{aligned}\tag{13} ​x˙1​x˙2​x˙3​​ ​y​= ​p1​11​0p2​p2​−z2​​00p3​​ ​ ​x1​x2​x3​​ ​+ ​b1​00​ ​u=[1​p2​−z2​​p3​−z3​​] ​x1​x2​x3​​ ​​(13)