第8章 MATLAB系统空间分析法

1
Z A Z B u y C Z Du
EX3
A=[0 -2;1 -3];B=[2 0]';C=[0 3]; A1 = 0 P=[6 2;2 0];%变换矩阵x=Pz -2 P1=inv(P); A1=P1*A*P %z坐标系的模型 B1=P1*B C1=C*P B1 = 0 1 -3
X A X B u Equals to Model A , B , C Model y CX X AX Bu A, B, C y CX
u
u
B
x
x
C
y
A
But x in model is different from x in the origin system A, B, C, because x0 is/are unmeasured directly.
A=[0 1 0;0 0 1;2 -5 4]; [V,D]=eig(A)
P=[1 0 1;1 1 2;1 2 4]
P1=inv(P)； J=P1*A*P
符号计算
Jo=jordan(A) Jo = 2 0 0 0 1 0 0 1 1
8. 4 系统的可控性和可观性
MATLAB提供函数分别计算能控性矩阵和能观测性矩阵
s=solve(det(s*I-A)) %求解
EX2 求控制系统的特征值及特征向量
[V,D]=eig(A) V= -0.4082 0.7071 0.5774 -0.4082 -0.7071 0.5774 -0.8165 0 -0.5774 Veig=inv(V)*A*V Veig = 1.0000 -0.0000 -0.0000 0 1.0000 0 -0.0000 0.0000 4.0000
elseif rOB<n
end
rOB = disp('System is unobservable') 1
System is unobservable
可控标准形
S b Ab A2b An 1b
若S为非奇异，逆矩阵存在，设为 则，变换矩阵为P


sT n T sn A P T n 1 s n A
n – the number of state variable
The eigenvalues of the closed-loop system will be given by
sI ( A BG) 0
(G [ g1, g2 , , gn ])
n
If the desired location of the closed-loop poles are desired characteristic equation will be
若状态方程是齐次的,即有:
x Ax x(t ) e x(0)
At
x Ax EX1
a=[0 1 0;0 0 1;-6 -11 -6];
x0=[1;1;1]; t=0:0.1:10;
1 0 0 A 0 0 1 6 11 6
1 x 0 1 1
X AX Bu y CX Du
1
z Tx x T 1 z
A TAT 1 B TB C CT
1
T 1Z AT 1Z Bu y CT 1Z Du
Байду номын сангаас
z [TAT ]z [TB]u y [CT ]z Du
8.5 系统状态反馈与状态观测器
•利用反馈结构，研究在什么条件下能 实现闭环系统极点的任意配置，以达 到预期要求。 •状态反馈与状态观测器原理
参见《线性控制系统工程》Module24，25
24.1 The Structure of State – Space Feedback Control （状态反馈控制的结构）
(s 1 )(s 2 )(s n ) 0
We let
i
sI ( A BG) (s 1 )(s 2 )(s n ) 0
)n , ,2 ,1 i( g
, to make the closed-loop poles to be
We can obtained located in desired position.
8.2 传递矩阵G
C[sI-A]-1B+D
A=[0 1;0 -2];B=[1 0;0 1];C=[1 0;0 1];D=0; syms s I=[1 0;0 1]; G= [ [ 1/s, 1/s/(s+2)] 0, 1/(s+2)]
G=C*inv(s*I-A)*B
8.3 线性变换

状态方程的线性变换 ss2ss(sys,T)
T s1 T 1 s 2 S c T s n
A=[-2 2 -1;0 -2 0;1 -4 0];B=[0 1 1]';n=length(A); CAM=ctrb(A,B); if det(CAM)~=0 CAM1=inv(CAM); end P=[CAM1(3,:);CAM1(3,:)*A;CAM1(3,:)*A*A]; P1=inv(P); A1 = 0 1 0 A1=P*A*P1 0 0 1 B1=P*B -2 -5 -4
u
Where, x are assumed to be unmeasured directly.
The approach taken to solve the problem is as following : (1) To construct a model of the system under study; (2) Assume (subject to certain restrictions ) that the computed state variables are good approximations to the true state variables; (3) From these computed state variables, a suitable controller for the actual system may be constructed using the techniques described in Module 24.
The principle of designing a state – space controller
, , 2 ,1
, the
2. The sufficient and necessary condition of state feedback for closed-loop placement: (状态反馈实现极点配置的充要条件）
for i=1:length(t)
x(:,i)=expm(a*t(i))*x0; end plot3(x(1,:),x(2,:),x(3,:)); grid on
系统的特征方程、特征值及特征向量

特征方程：|sI-A|=0 特征值及特征向量：
特征值矩阵
特征向量矩阵
[V，D]=eig（A） A*V = V*D
Now, we construct a model A, B , C to simulate the origin system A, B, C, and assume the parameter matrix A, B , C are good approximations to A, B, C
EX4利用特征值及范德蒙特矩阵求约当阵
V= -0.5774 0.5774 -0.2182 -0.5774 0.5774 -0.4364 -0.5774 0.5774 -0.8729 D= 1.0000 0 0 0 1.0000 0 0 0 2.0000 P= 1 0 1 1 1 2 1 2 4 J= 1 1 0 0 1 0 0 0 2
第8章 系统状态空间分析法
8.4节和8.5节
内容

系统特征方程及解 关于系统相似变换 关于系统可观性、可控性判别的 状态反馈极点配置 状态观测器
8.1 系统状态方程的解
X(t ) e X0 e
At 0 t A ( t )
Bu()d

状态转移矩阵
(t ) e At L1{[sI A]1}
EX2 已知控制系统 求控制系统的特征方程
A=[2 1 -1;1 2 -1;-1 -1 2]; I=[1 0 0;0 1 0;0 0 1];
2 1 1 A 1 2 1 1 1 2
syms s %符号计算
det(s*I-A)
ans = s^3-6*s^2+9*s-4 s= [ 4] [ 1] [ 1]
The state variables of system are all controllable.
状态观测器设计
25.1 Observer – A model of the system under study （P550 Section 2）
X AX Bu y CX
1
C1 =
6
0
The eigenvalues of system are unchanged by the linear transformation: (线性变 换不改变系统的特征值）
det(sI A) det(sI A)
约当标准形

canon(sys,’model’) canon(sys,’companion’)
B1 = 0 0 1
可观测标准形
V CT

ATCT ( AT ) n 1 CT