自控系统各种标准型整理(附例题)
1. 状态空间相关知识铺垫
在阅读本文之前,默认读者已经知晓状态空间的相关知识,包括:
- 状态空间的表达形式不止一种,可以在 x ~ = T x \tilde x = Tx x~=Tx的基础上进行变换;
- 状态空间是用来研究系统动态过程内部变量的一种手段,其所选取的状态变量 x i x_i xi的方式不止一种;
- 系统中所有状态变量 x i x_i xi并不是必须具有物理意义,有时只是为了数学表示方便而设立;同样也并不是每个状态变量都会被观测或测量到。
本文的所有状态空间表达式,都是基于如下微分方程建立的:
y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 2 y ¨ + a 1 y ˙ + a 0 y = b n u ( n ) + b n − 1 u ( n − 1 ) + ⋯ + b 1 u ˙ + b 0 u (1) y^{(n)} + a_{n-1} y^{(n-1)} + a_{n-2} y^{(n-2)} + \cdots + a_2 \ddot y + a_1 \dot y + a_0 y = \\ b_n u^{(n)} + b_{n-1} u^{(n-1)} + \cdots + b_1 \dot u + b_0 u \tag{1} y(n)+an−1y(n−1)+an−2y(n−2)+⋯+a2y¨+a1y˙+a0y=bnu(n)+bn−1u(n−1)+⋯+b1u˙+b0u(1)注意,上式中, y ( n ) y^{(n)} y(n)前的系数 a n = 1 a_n = 1 an=1。
而目的是将其写成如下状态空间形式:
X ˙ = A X + B U Y = C X + D U \dot {\bm X} = {\bm A} {\bm X} + {\bm B} {\bm U} \\ {\bm Y} = {\bm C} {\bm X} + {\bm D} {\bm U} X˙=AX+BUY=CX+DU
2. 系统输入量中【不含有】导数项
即式(1)右端仅含 u u u,式(1)变为:
y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 2 y ¨ + a 1 y ˙ + a 0 y = b 0 u (2) y^{(n)} + a_{n-1} y^{(n-1)} + a_{n-2} y^{(n-2)} + \cdots + a_2 \ddot y + a_1 \dot y + a_0 y = b_0 u \tag{2} y(n)+an−1y(n−1)+an−2y(n−2)+⋯+a2y¨+a1y˙+a0y=b0u(2)此时状态量选择为:
{ x ˙ 1 = x 2 x ˙ 2 = x 3 ⋮ x ˙ n − 1 = x n x ˙ n = − a 0 x 1 − a 1 x 2 − ⋯ − a n − 1 x n + b 0 u y = x 1 \begin{cases} \dot x_1 = x_2 \\ \dot x_2 = x_3 \\ \vdots \\ \dot x_{n-1} = x_n \\ \dot x_n = -a_0 x_1 - a_1 x_2 - \cdots - a_{n-1} x_n + b_0 u \\ y = x_1 \end{cases} ⎩
⎨
⎧x˙1=x2x˙2=x3⋮x˙n−1=xnx˙n=−a0x1−a1x2−⋯−an−1xn+b0uy=x1其状态空间矩阵为:
A = [ 0 1 0 ⋯ 0 0 0 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 1 − a 0 − a 1 − a 2 ⋯ − a n − 1 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{matrix} \right] A=
00⋮0−a010⋮0−a101⋮0−a2⋯⋯⋱⋯⋯00⋮1−an−1
B = [ 0 0 ⋮ 0 b 0 ] , C = [ 1 0 ⋯ 0 ] , D = 0 (3) {\bm B} = \left[ \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \\ b_0 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 1 & 0 & \cdots & 0 \end{matrix} \right], \quad {\bm D} = 0 \tag{3} B=
00⋮0b0
,C=[10⋯0],D=0(3)
3. 系统输入量中【含有】导数项
此时系统的微分方程即具有式(1)的形式:
y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 2 y ¨ + a 1 y ˙ + a 0 y = b n u ( n ) + b n − 1 u ( n − 1 ) + ⋯ + b 1 u ˙ + b 0 u (1) y^{(n)} + a_{n-1} y^{(n-1)} + a_{n-2} y^{(n-2)} + \cdots + a_2 \ddot y + a_1 \dot y + a_0 y = \\ b_n u^{(n)} + b_{n-1} u^{(n-1)} + \cdots + b_1 \dot u + b_0 u \tag{1} y(n)+an−1y(n−1)+an−2y(n−2)+⋯+a2y¨+a1y˙+a0y=bnu(n)+bn−1u(n−1)+⋯+b1u˙+b0u(1)此时状态量选择为:
{ x ˙ 1 = x 2 + h 1 u x ˙ 2 = x 3 + h 2 u ⋮ x ˙ n − 1 = x n + h n − 1 u x ˙ n = − a 0 x 1 − a 1 x 2 − ⋯ − a n − 1 x n + h n u y = x 1 + h 0 u \begin{cases} \dot x_1 = x_2 + h_1 u \\ \dot x_2 = x_3 + h_2 u \\ \vdots \\ \dot x_{n-1} = x_n + h_{n-1} u \\ \dot x_n = -a_0 x_1 - a_1 x_2 - \cdots - a_{n-1} x_n + h_n u \\ y = x_1 + h_0 u \end{cases} ⎩
⎨
⎧x˙1=x2+h1ux˙2=x3+h2u⋮x˙n−1=xn+hn−1ux˙n=−a0x1−a1x2−⋯−an−1xn+hnuy=x1+h0u其中的新参数 h i h_i hi为:
h 0 = b n h i = b n − i − ∑ j = 1 i a n − j h i − j (4) h_0 = b_n \\ h_i = b_{n-i} - \sum _{j=1} ^i a_{n-j} h_{i-j} \tag{4} h0=bnhi=bn−i−j=1∑ian−jhi−j(4)其状态空间矩阵为:
A = [ 0 1 0 ⋯ 0 0 0 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 1 − a 0 − a 1 − a 2 ⋯ − a n − 1 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{matrix} \right] A=
00⋮0−a010⋮0−a101⋮0−a2⋯⋯⋱⋯⋯00⋮1−an−1
B = [ h 1 h 2 ⋮ h n − 1 h n ] , C = [ 1 0 ⋯ 0 ] , D = h 0 (5) {\bm B} = \left[ \begin{matrix} h_1 \\ h_2 \\ \vdots \\ h_{n-1} \\ h_n \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 1 & 0 & \cdots & 0 \end{matrix} \right], \quad {\bm D} = h_0 \tag{5} B=
h1h2⋮hn−1hn
,C=[10⋯0],D=h0(5)可见矩阵 A {\bm A} A与式(3)中的一样。
4. 系统输入量中【含有】导数项,但 b n = 0 {\bm b}_{\bm n} {\bm =} {\bm 0} bn=0
此时系统的微分方程(1)右端从 b n − 1 u ( n − 1 ) b_{n-1} u^{(n-1)} bn−1u(n−1)开始:
y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 2 y ¨ + a 1 y ˙ + a 0 y = b n − 1 u ( n − 1 ) + ⋯ + b 1 u ˙ + b 0 u (1) y^{(n)} + a_{n-1} y^{(n-1)} + a_{n-2} y^{(n-2)} + \cdots + a_2 \ddot y + a_1 \dot y + a_0 y = \\ b_{n-1} u^{(n-1)} + \cdots + b_1 \dot u + b_0 u \tag{1} y(n)+an−1y(n−1)+an−2y(n−2)+⋯+a2y¨+a1y˙+a0y=bn−1u(n−1)+⋯+b1u˙+b0u(1)代入式(4)很容易知道, h 0 = 0 h_0 = 0 h0=0,则此时可以按照如下规则,选取新的一组状态变量:
{ x ˙ 1 = − a 0 x n + b 0 u x ˙ 2 = x 1 − a 1 x n + b 1 u ⋮ x ˙ n − 1 = x n − 2 − a n − 2 x n + b n − 2 u x ˙ n = x n − 1 − a n − 1 x n + b n − 1 u y = x n \begin{cases} \dot x_1 = -a_0 x_n + b_0 u \\ \dot x_2 = x_1 - a_1 x_n + b_1 u \\ \vdots \\ \dot x_{n-1} = x_{n-2} - a_{n-2} x_n + b_{n-2} u \\ \dot x_n = x_{n-1} - a_{n-1} x_n + b_{n-1} u \\ y = x_n \end{cases} ⎩
⎨
⎧x˙1=−a0xn+b0ux˙2=x1−a1xn+b1u⋮x˙n−1=xn−2−an−2xn+bn−2ux˙n=xn−1−an−1xn+bn−1uy=xn其状态空间矩阵为:
A = [ 0 0 ⋯ 0 − a 0 1 0 ⋯ 0 − a 1 0 1 ⋯ 0 − a 2 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 − a n − 1 ] {\bm A} = \left[ \begin{matrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{n-1} \end{matrix} \right] A=
010⋮0001⋮0⋯⋯⋯⋱⋯000⋮1−a0−a1−a2⋮−an−1
B = [ b 0 b 1 ⋮ b n − 1 ] , C = [ 0 0 ⋯ 1 ] , D = 0 (6) {\bm B} = \left[ \begin{matrix} b_0 \\ b_1 \\ \vdots \\ b_{n-1} \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 0 & 0 & \cdots & 1 \end{matrix} \right], \quad {\bm D} = 0 \tag{6} B=
b0b1⋮bn−1
,C=[00⋯1],D=0(6)
5. 可控标准型
此时系统的微分方程即具有式(1)的形式:
y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 2 y ¨ + a 1 y ˙ + a 0 y = b n u ( n ) + b n − 1 u ( n − 1 ) + ⋯ + b 1 u ˙ + b 0 u (1) y^{(n)} + a_{n-1} y^{(n-1)} + a_{n-2} y^{(n-2)} + \cdots + a_2 \ddot y + a_1 \dot y + a_0 y = \\ b_n u^{(n)} + b_{n-1} u^{(n-1)} + \cdots + b_1 \dot u + b_0 u \tag{1} y(n)+an−1y(n−1)+an−2y(n−2)+⋯+a2y¨+a1y˙+a0y=bnu(n)+bn−1u(n−1)+⋯+b1u˙+b0u(1)此时状态量选择为:
{ x ˙ 1 = x 2 x ˙ 2 = x 3 ⋮ x ˙ n = − a 0 x 1 − a 1 x 2 − ⋯ − a n − 1 x n + u y = − β 0 x 1 − β 1 x 2 − ⋯ − β n − 1 x n \begin{cases} \dot x_1 = x_2 \\ \dot x_2 = x_3 \\ \vdots \\ \dot x_n = -a_0 x_1 - a_1 x_2 - \cdots - a_{n-1} x_n + u \\ y = -\beta_0 x_1 - \beta_1 x_2 - \cdots - \beta_{n-1} x_n \end{cases} ⎩
⎨
⎧x˙1=x2x˙2=x3⋮x˙n=−a0x1−a1x2−⋯−an−1xn+uy=−β0x1−β1x2−⋯−βn−1xn其中的新参数 β i \beta_i βi为:
{ β 0 = b 0 − a 0 b n β 1 = b 1 − a 1 b n ⋮ β i = b i − a i b n ⋮ β n − 2 = b n − 2 − a n − 2 b n β n − 1 = b n − 1 − a n − 1 b n (7) \begin{cases} \beta_0 = b_0 - a_0 b_n \\ \beta_1 = b_1 - a_1 b_n \\ \vdots \\ \beta_i = b_i - a_i b_n \\ \vdots \\ \beta_{n-2} = b_{n-2} - a_{n-2} b_n \\ \beta_{n-1} = b_{n-1} - a_{n-1} b_n \end{cases} \tag{7} ⎩
⎨
⎧β0=b0−a0bnβ1=b1−a1bn⋮βi=bi−aibn⋮βn−2=bn−2−an−2bnβn−1=bn−1−an−1bn(7)其状态空间矩阵为:
A = [ 0 1 0 ⋯ 0 0 0 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 1 − a 0 − a 1 − a 2 ⋯ − a n − 1 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{matrix} \right] A=
00⋮0−a010⋮0−a101⋮0−a2⋯⋯⋱⋯⋯00⋮1−an−1
B = [ 0 0 ⋮ 0 1 ] , C = [ β 0 β 1 ⋯ β n − 1 ] , D = 0 (8) {\bm B} = \left[ \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} \beta_0 & \beta_1 & \cdots & \beta_{n-1} \end{matrix} \right], \quad {\bm D} = 0 \tag{8} B=
00⋮01
,C=[β0β1⋯βn−1],D=0(8)
6. 可观标准型
可观标准型即为可控标准型的转置:
A c = A o T , B c = C o T , C c = B o T {\bm A_c} = {\bm A_o}^T, {\bm B_c} = {\bm C_o}^T, {\bm C_c} = {\bm B_o}^T Ac=AoT,Bc=CoT,Cc=BoT其中下标 c , o c,o c,o分别表示可控和可观矩阵。由此得到:
A = [ 0 0 ⋯ 0 − a 0 1 0 ⋯ 0 − a 1 0 1 ⋯ 0 − a 2 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 − a n − 1 ] {\bm A} = \left[ \begin{matrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{n-1} \end{matrix} \right] A=
010⋮0001⋮0⋯⋯⋯⋱⋯000⋮1−a0−a1−a2⋮−an−1
B = [ β 0 β 1 β 2 ⋮ β n − 1 ] , C = [ 0 0 ⋯ 1 ] , D = 0 (9) {\bm B} = \left[ \begin{matrix} \beta_0 \\ \beta_1 \\ \beta_2 \\ \vdots \\ \beta_{n-1} \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 0 & 0 & \cdots & 1 \end{matrix} \right], \quad {\bm D} = 0 \tag{9} B=
β0β1β2⋮βn−1
,C=[00⋯1],D=0(9)
7. 传递函数特征方程【无重根时】的约当型
注意:此类型只能在传递函数的特征方程没有重根时使用!
根据状态变量的选取方式不同,一般约当型有2种表达方式。
由式(1)可以写出复平面内的传递函数关系式:
W ( s ) = b n s n + b n − 1 s n − 1 + ⋯ + b 1 s + b 0 s n + a n − 1 s n − 1 + ⋯ + a 1 s + a 0 = N ( s ) D ( s ) W(s) = \frac{ b_n s^n + b_{n-1} s^{n-1} + \cdots + b_1 s + b_0 }{ s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0 } = \frac{ N(s) }{ D(s) } W(s)=sn+an−1sn−1+⋯+a1s+a0bnsn+bn−1sn−1+⋯+b1s+b0=D(s)N(s)其中 D ( s ) = 0 D(s)=0 D(s)=0是传递函数的特征方程,其特征根为 λ i \lambda_i λi。那么分母可以写为:
D ( s ) = ( s − λ 1 ) ( s − λ 2 ) ⋯ ( s − λ n ) D(s) = \left(s - \lambda_1 \right) \left(s - \lambda_2 \right) \cdots \left(s - \lambda_n \right) D(s)=(s−λ1)(s−λ2)⋯(s−λn)此时传递函数即:
W ( s ) = N ( s ) D ( s ) = b n s n + b n − 1 s n − 1 + ⋯ + b 1 s + b 0 ( s − λ 1 ) ( s − λ 2 ) ⋯ ( s − λ n ) W(s) = \frac{ N(s) }{ D(s) } = \frac{ b_n s^n + b_{n-1} s^{n-1} + \cdots + b_1 s + b_0 }{ \left(s - \lambda_1 \right) \left(s - \lambda_2 \right) \cdots \left(s - \lambda_n \right) } W(s)=D(s)N(s)=(s−λ1)(s−λ2)⋯(s−λn)bnsn+bn−1sn−1+⋯+b1s+b0展开即得:
W ( s ) = N ( s ) D ( s ) = Y ( s ) U ( s ) = ∑ i = 1 n c i s − λ i (10) W(s) = \frac{ N(s) }{ D(s) } = \frac{ Y(s) }{ U(s) } = \sum _{i=1} ^n \frac{ c_i }{ s - \lambda_i } \tag{10} W(s)=D(s)N(s)=U(s)Y(s)=i=1∑ns−λici(10)其中 c i c_i ci为传递函数 W ( s ) W(s) W(s)在极点 λ i \lambda_i λi处的留数:
c i = ( s − λ i ) W ( s ) ∣ s = λ i (11) c_i = \left(s - \lambda_i \right) W(s) \Big\rvert _{s = \lambda_i} \tag{11} ci=(s−λi)W(s)
s=λi(11)从而
Y ( s ) = ∑ i = 1 n c i s − λ i U ( s ) Y(s) = \sum _{i=1} ^n \frac{ c_i }{ s - \lambda_i } U(s) Y(s)=i=1∑ns−λiciU(s)
7.1 第一型
设状态变量为
X i ( s ) = 1 s − λ i U ( s ) X_i(s) = \frac{1}{s - \lambda_i} U(s) Xi(s)=s−λi1U(s)则有
{ x ˙ 1 = λ 1 x 1 + u x ˙ 2 = λ 2 x 2 + u ⋮ x ˙ n = λ n x n + u y = c 1 x 1 + c 2 x 2 + ⋯ + c n x n \begin{cases} \dot x_1 = \lambda_1 x_1 + u \\ \dot x_2 = \lambda_2 x_2 + u \\ \vdots \\ \dot x_n = \lambda_n x_n + u \\ y = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n \end{cases} ⎩
⎨
⎧x˙1=λ1x1+ux˙2=λ2x2+u⋮x˙n=λnxn+uy=c1x1+c2x2+⋯+cnxn其状态空间矩阵为:
A = [ λ 1 λ 2 ⋱ λ n − 1 λ n ] {\bm A} = \left[ \begin{matrix} \lambda_1 & & & & \\ & \lambda_2 & & & \\ & & \ddots & & \\ & & & \lambda_{n-1} & \\ & & & & \lambda_n \end{matrix} \right] A=
λ1λ2⋱λn−1λn
B = [ 1 1 ⋮ 1 1 ] , C = [ c 1 c 2 ⋯ c n ] , D = 0 (12) {\bm B} = \left[ \begin{matrix} 1 \\ 1 \\ \vdots \\ 1 \\ 1 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} c_1 & c_2 & \cdots & c_n \end{matrix} \right], \quad {\bm D} = 0 \tag{12} B=
11⋮11
,C=[c1c2⋯cn],D=0(12)其中 A {\bm A} A为对角矩阵,主对角线上元素为所有特征根 λ i \lambda_i λi,而其余元素均为0。
7.2 第二型
设状态变量为
X i ( s ) = c i s − λ i U ( s ) X_i(s) = \frac{c_i}{s - \lambda_i} U(s) Xi(s)=s−λiciU(s)与第一型相比, X i ( s ) X_i(s) Xi(s)的分子不再是1而是 c i c_i ci。则有
{ x ˙ 1 = λ 1 x 1 + c 1 u x ˙ 2 = λ 2 x 2 + c 2 u ⋮ x ˙ n = λ n x n + c n u y = x 1 + x 2 + ⋯ + x n \begin{cases} \dot x_1 = \lambda_1 x_1 + c_1 u \\ \dot x_2 = \lambda_2 x_2 + c_2 u \\ \vdots \\ \dot x_n = \lambda_n x_n + c_n u \\ y = x_1 + x_2 + \cdots + x_n \end{cases} ⎩
⎨
⎧x˙1=λ1x1+c1ux˙2=λ2x2+c2u⋮x˙n=λnxn+cnuy=x1+x2+⋯+xn其状态空间矩阵为:
A = [ λ 1 λ 2 ⋱ λ n − 1 λ n ] {\bm A} = \left[ \begin{matrix} \lambda_1 & & & & \\ & \lambda_2 & & & \\ & & \ddots & & \\ & & & \lambda_{n-1} & \\ & & & & \lambda_n \end{matrix} \right] A=
λ1λ2⋱λn−1λn
B = [ c 1 c 2 ⋮ c n − 1 c n ] , C = [ 1 1 ⋯ 1 ] , D = 0 (13) {\bm B} = \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_{n-1} \\ c_n \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 1 & 1 & \cdots & 1 \end{matrix} \right], \quad {\bm D} = 0 \tag{13} B=
c1c2⋮cn−1cn
,C=[11⋯1],D=0(13)其中 A {\bm A} A为对角矩阵,与第一型相同。
8. 传递函数特征方程【有重根时】的约当型
这里以一个含有三重根的特征方程为例。设此时的特征方程为
D ( s ) = ( s − λ 1 ) 3 ( s − λ 4 ) ⋯ ( s − λ n ) D(s) = \left(s - \lambda_1 \right) ^3 \left(s - \lambda_4 \right) \cdots \left(s - \lambda_n \right) D(s)=(s−λ1)3(s−λ4)⋯(s−λn)其中 λ 1 \lambda_1 λ1为三重实极点,其余 λ i ( i ≠ 1 ) \lambda_i ( i \neq 1) λi(i=1)为单实极点。则传递函数此时可以分解为:
W ( s ) = N ( s ) D ( s ) = Y ( s ) U ( s ) = c 11 ( s − λ 1 ) 3 + c 12 ( s − λ 1 ) 2 + c 13 ( s − λ 1 ) + ∑ i = 4 n c i s − λ i (14) W(s) = \frac{ N(s) }{ D(s) } = \frac{ Y(s) }{ U(s) } = \frac{ c_{11} }{ \left(s - \lambda_1 \right)^3 } + \frac{ c_{12} }{ \left(s - \lambda_1 \right)^2 } + \frac{ c_{13} }{ \left(s - \lambda_1 \right) } + \sum_{i=4} ^n \frac{ c_i} {s - \lambda_i } \tag{14} W(s)=D(s)N(s)=U(s)Y(s)=(s−λ1)3c11+(s−λ1)2c12+(s−λ1)c13+i=4∑ns−λici(14)其状态变量选取方法和第一型、第二型相同。此时的状态空间矩阵为:
A = [ λ 1 1 λ 1 1 0 λ 1 λ 4 0 ⋱ λ n ] , B = [ 0 0 1 1 ⋮ 1 ] {\bm A} = \left[ \begin{array}{ c c c | c c c } \lambda_1 & 1 & {} & {} & {} & {} \\ {} & \lambda_1 & 1 & {} & {\bm 0} & {} \\ {} & {} & \lambda_1 & {} & {} & {} \\ \hline {} & {} & {} & \lambda_4 & {} & {} \\ {} & {\bm 0} & {} & {} & \ddots & {} \\ {} & {} & {} & {} & {} & \lambda_n \\ \end{array} \right], \quad {\bm B} = \left[ \begin{array}{ c } 0 \\ 0 \\ 1 \\ \hline 1 \\ \vdots \\ 1 \end{array} \right] A=
λ11λ101λ1λ40⋱λn
,B=
0011⋮1
C = [ c 11 c 12 c 13 c 4 ⋯ c n ] , D = 0 (15) {\bm C} = \left[ \begin{array}{ c c c | c c c } c_{11} & c_{12} & c_{13} & c_4 & \cdots & c_n \end{array} \right], \quad {\bm D} = 0 \tag{15} C=[c11c12c13c4⋯cn],D=0(15)或者如下形式亦可:
A = [ λ 1 1 λ 1 0 1 λ 1 λ 4 0 ⋱ λ n ] , B = [ c 11 c 12 c 13 c 4 ⋮ c n ] {\bm A} = \left[ \begin{array}{ c c c | c c c } \lambda_1 & {} & {} & {} & {} & {} \\ 1 & \lambda_1 & {} & {} & {\bm 0} & {} \\ {} & 1 & \lambda_1 & {} & {} & {} \\ \hline {} & {} & {} & \lambda_4 & {} & {} \\ {} & {\bm 0} & {} & {} & \ddots & {} \\ {} & {} & {} & {} & {} & \lambda_n \\ \end{array} \right], \quad {\bm B} = \left[ \begin{array}{ c } c_{11} \\ c_{12} \\ c_{13} \\ \hline c_4 \\ \vdots \\ c_n \end{array} \right] A=
λ11λ110λ1λ40⋱λn
,B=
c11c12c13c4⋮cn
C = [ 0 0 1 1 ⋯ 1 ] , D = 0 (16) {\bm C} = \left[ \begin{array}{ c c c | c c c } 0 & 0 & 1 & 1 & \cdots & 1 \end{array} \right], \quad {\bm D} = 0 \tag{16} C=[0011⋯1],D=0(16)应当解释的是,在式(15)和(16)中, A {\bm A} A不再是完全的对角阵,由于重根 λ 1 \lambda_1 λ1的存在, λ 1 \lambda_1 λ1在 A {\bm A} A中对应有一个子块,其内部为上三角或下三角的带有1的矩阵(如 A {\bm A} A左上角的子块)。除此之外,在B或C中,与之对应的也有其留数的子块。 A {\bm A} A、 B {\bm B} B、 C {\bm C} C中的子块阶数均与 λ 1 \lambda_1 λ1的重数相同,如本例中 λ 1 \lambda_1 λ1为三重根,则 A {\bm A} A左上角的子块为三阶, B {\bm B} B、 C {\bm C} C中的子块也为三阶。本文均以细线将子块分隔出以便标明。
9. 例题
9.1 系统输入量中【不含有】导数项
设系统的微分方程为
y ( 3 ) + 2 y ¨ + 5 y ˙ + 2 y = 2 u y^{(3)} + 2 \ddot y + 5 \dot y + 2y= 2u y(3)+2y¨+5y˙+2y=2u可见 n = 3 n=3 n=3,则其系数可以立即写出: a 0 = 2 , a 1 = 5 , a 2 = 2 , a 3 = 1 ; b 0 = 2 a_0 = 2, a_1 = 5, a_2 = 2, a_3 = 1;b_0 = 2 a0=2,a1=5,a2=2,a3=1;b0=2。代入式(3)有:
A = [ 0 1 0 0 0 1 − 2 − 5 − 2 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -2 \end{matrix} \right] A=
00−210−501−2
B = [ 0 0 b 0 ] , C = [ 1 0 0 ] , D = 0 {\bm B} = \left[ \begin{matrix} 0 \\ 0 \\ b_0 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 1 & 0 & 0 \end{matrix} \right], \quad {\bm D} = 0 B=
00b0
,C=[100],D=0
9.2 系统输入量中【含有】导数项
设系统的微分方程为
y ( 3 ) + 2 y ¨ + 5 y ˙ + 2 y = 2 u ¨ + 3 u ˙ + u y^{(3)} + 2 \ddot y + 5 \dot y + 2y= 2 \ddot u +3 \dot u + u y(3)+2y¨+5y˙+2y=2u¨+3u˙+u可见 n = 3 n=3 n=3,则其系数可以立即写出: a 0 = 2 , a 1 = 5 , a 2 = 2 , a 3 = 1 ; b 0 = 1 , b 1 = 3 , b 2 = 2 , b 3 = 0 a_0 = 2, a_1 = 5, a_2 = 2, a_3 = 1;b_0 = 1, b_1 = 3, b_2 = 2, b_3 = 0 a0=2,a1=5,a2=2,a3=1;b0=1,b1=3,b2=2,b3=0。代入式(4)先计算 h i h_i hi:
h i = b n − i − ∑ j = 1 i a n − j h i − j ⟹ h 0 = b n = b 3 = 0 , h 1 = b 3 − 1 − ∑ j = 1 1 a 3 − j h 1 − j = b 2 − a 2 h 0 = 2 , h 2 = b 3 − 2 − ∑ j = 1 2 a 3 − j h 2 − j = b 1 − a 2 h 1 − a 1 h 0 = 3 − 2 × 2 = − 1 h 3 = b 3 − 3 − ∑ j = 1 3 a 3 − j h 3 − j = b 0 − a 2 h 2 − a 1 h 1 − a 0 h 0 = 1 + 2 × 1 − 5 × 2 = − 7 h_i = b_{n-i} - \sum _{j=1} ^i a_{n-j} h_{i-j} \Longrightarrow \\ h_0 = b_n = b_3 = 0, \\ h_1 = b_{3-1} - \sum _{j=1} ^1 a_{3-j} h_{1-j} = b_2 - a_2 h_0 = 2, \\ h_2 = b_{3-2} - \sum _{j=1} ^2 a_{3-j} h_{2-j} = b_1 - a_2 h_1 - a_1 h_0 = 3 - 2 \times 2 = -1 \\ h_3 = b_{3-3} - \sum _{j=1} ^3 a_{3-j} h_{3-j} = b_0 - a_2 h_2 - a_1 h_1 - a_0 h_0 = 1 + 2 \times 1 - 5 \times 2 = -7 hi=bn−i−j=1∑ian−jhi−j⟹h0=bn=b3=0,h1=b3−1−j=1∑1a3−jh1−j=b2−a2h0=2,h2=b3−2−j=1∑2a3−jh2−j=b1−a2h1−a1h0=3−2×2=−1h3=b3−3−j=1∑3a3−jh3−j=b0−a2h2−a1h1−a0h0=1+2×1−5×2=−7代入式(5)有:
其状态空间矩阵为:
A = [ 0 1 0 0 0 1 − 2 − 5 − 2 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -2 \end{matrix} \right] A=
00−210−501−2
B = [ h 1 h 2 h 3 ] = [ 2 − 1 − 7 ] , C = [ 1 0 ⋯ 0 ] , D = h 0 = 0 {\bm B} = \left[ \begin{matrix} h_1 \\ h_2 \\ h_3 \end{matrix} \right] = \left[ \begin{matrix} 2 \\ -1 \\ -7 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 1 & 0 & \cdots & 0 \end{matrix} \right], \quad {\bm D} = h_0 = 0 B=
h1h2h3
=
2−1−7
,C=[10⋯0],D=h0=0
9.3 系统输入量中【含有】导数项,但 b n = 0 {\bm b}_{\bm n} {\bm =} {\bm 0} bn=0
仍以9.3中微分方程为例,注意到其 b n = b 3 = 0 b_n = b_3 = 0 bn=b3=0。那么可以代入式(6):
A = [ 0 0 − 2 1 0 − 5 0 1 − 2 ] {\bm A} = \left[ \begin{matrix} 0 & 0 & -2 \\ 1 & 0 & -5 \\ 0 & 1 & -2 \\ \end{matrix} \right] A=
010001−2−5−2
B = [ b 0 b 1 ⋮ b n − 1 ] = [ 1 3 2 ] , C = [ 0 0 ⋯ 1 ] , D = 0 (6) {\bm B} = \left[ \begin{matrix} b_0 \\ b_1 \\ \vdots \\ b_{n-1} \end{matrix} \right] = \left[ \begin{matrix} 1 \\ 3 \\ 2 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 0 & 0 & \cdots & 1 \end{matrix} \right], \quad {\bm D} = 0 \tag{6} B=
b0b1⋮bn−1
=
132
,C=[00⋯1],D=0(6)
9.4 可控标准型
设系统微分方程为:
y ( 3 ) + 2 y ¨ + 3 y ˙ + 3 y = 2 u ( 3 ) + u ¨ + 3 u ˙ + 4 u y^{(3)} + 2 \ddot y + 3 \dot y + 3y = 2 u^{(3)} + \ddot u + 3 \dot u + 4u y(3)+2y¨+3y˙+3y=2u(3)+u¨+3u˙+4u可见 n = 3 , a 0 = 3 , a 1 = 3 , a 2 = 2 , a 3 = 1 ; b 0 = 4 , b 1 = 3 , b 2 = 1 , b 3 = 2 n=3, a_0 = 3, a_1 = 3, a_2 = 2, a_3 = 1; b_0 = 4, b_1 = 3, b_2 = 1, b_3 = 2 n=3,a0=3,a1=3,a2=2,a3=1;b0=4,b1=3,b2=1,b3=2。代入式(7)计算参数 β i \beta_i βi:
{ β 0 = b 0 − a 0 b n = 4 − 3 × 2 = − 2 , β 1 = b 1 − a 1 b n = 3 − 3 × 2 = − 3 , β 2 = b 2 − a 2 b n = 1 − 2 × 2 = − 3 (7) \begin{cases} \beta_0 = b_0 - a_0 b_n = 4 - 3 \times 2 = -2, \\ \beta_1 = b_1 - a_1 b_n = 3 - 3 \times 2 = -3, \\ \beta_2 = b_2 - a_2 b_n = 1 - 2 \times 2 = -3 \end{cases} \tag{7} ⎩
⎨
⎧β0=b0−a0bn=4−3×2=−2,β1=b1−a1bn=3−3×2=−3,β2=b2−a2bn=1−2×2=−3(7)再代入式(8)即得:
A = [ 0 1 0 0 0 1 − 3 − 3 − 2 ] {\bm A} = \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -3 & -3 & -2 \end{matrix} \right] A=
00−310−301−2
B = [ 0 0 1 ] , C = [ − 2 − 3 − 3 ] , D = 0 (8) {\bm B} = \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} -2 & -3 & -3 \end{matrix} \right], \quad {\bm D} = 0 \tag{8} B=
001
,C=[−2−3−3],D=0(8)
9.5 可观标准型
由于可观标准型即为可控标准型的转置,则可以立即写出:
A = [ 0 0 − 3 1 0 − 3 0 1 − 2 ] {\bm A} = \left[ \begin{matrix} 0 & 0 & -3 \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{matrix} \right] A=
010001−3−3−2
B = [ − 2 − 3 − 3 ] , C = [ 0 0 1 ] , D = 0 (8) {\bm B} = \left[ \begin{matrix} -2 \\ -3 \\ -3 \end{matrix} \right], \quad {\bm C} = \left[ \begin{matrix} 0 & 0 & 1 \end{matrix} \right], \quad {\bm D} = 0 \tag{8} B=
−2−3−3
,C=[001],D=0(8)