matlab编程实现劳斯判据

Question

n1=11.96;
>> d1=conv([1 0],conv([0.1 1],[0.51 1]));
>> g=tf(n1,d1);
就上面的函数求出它稳定与否
先谢谢各位大哥大姐了

Answer 1

%程序如下:

function [RouthTable,Conclusion] = routh(polequ)
% 劳斯判据求解系统稳定性函数
% 输入：
% polequ = 特征方程向量;
% 输出：
% RouthTable = 劳斯表
% Conclusion = 系统是否稳定或存在多少个不稳定的根的结论
% 事例:
% [RouthTable,Con] = routh([1 2 3 4 5]);
% RouthTable =
% 1 3 5
% 2 4 0
% 1 5 0
% -6 0 0
% 5 0 0
% Con =
% 所判定系统有 2 个不稳定根!
%
%
n=length(polequ);
polequ=reshape(polequ,1,n);
if mod(n,2)==0
n1=n/2;
else
n1=(n+1)/2;
polequ=[polequ,0];
end
routh=reshape(polequ,2,n1);
RouthTable=zeros(n,n1);
RouthTable(1:2,:)=routh;
i=3;
while 1;
% =========特殊情况1(第一列为0，其余列不为0)=====================
if RouthTable(i-1,1)==0 && sum(RouthTable(i-1,2:n1))~=0
polequ = conv(polequ,[1 3]);
n=length(polequ);
if mod(n,2)==0
n1=n/2;
else
n1=(n+1)/2;
polequ=[polequ,0];
end
routh=reshape(polequ,2,n1);
RouthTable=zeros(n,n1);
RouthTable(1:2,:)=routh;
i=3;
end

% ==========计算劳斯表===========================================
ai=RouthTable(i-2,1)/RouthTable(i-1,1);
for j=1:n1-1
RouthTable(i,j)=RouthTable(i-2,j+1)-ai*RouthTable(i-1,j+1);
end
% ==========特殊情况2(全0行)======================================
if sum(RouthTable(i,:))==0
k=0;
l=1;
F=zeros(1,n1);
while n-i-k>=0
F(l)=n-i+1-k;
k=k+2;
l=l+1;
end
RouthTable(i,:)=RouthTable(i-1,:).*F(1,:);
end
% =========更新==================================================
i=i+1;
if i>n
break;
end
end
% =============outhput===========
r=find(RouthTable(:,1)<0);
if isempty(r)==1
Conclusion='所要判定系统稳定!';
else
n2=length(r);
m=n2;
for i=1:n2-1
if r(i+1)-r(i)==1
m=m-1;
end
end
str1='所判定系统有 ';
if r(n2)==n
str2=num2str(m*2-1);
else
str2=num2str(m*2);
end
str3=' 个不稳定根!';
Conclusion = [str1,str2,str3];
end

在MATLAB命令窗口中输入:
n=11.96;d=conv([1 0],conv([0.1 1],[0.51 1]));
equ=d;equ(4)=d(4)+n;
[RouthTable,Conclusion] = routh(equ)

则计算结果:
RouthTable =
0.0510 1.0000
0.6100 11.9600
0.0001 0
11.9600 0
Conclusion =
所要判定系统稳定!