自编劳斯判据判断系统稳定性的Matlab函数

问题描述：

function [routh_list,conclusion] = Routh(chara_equ)
% =======================================================
% 自编劳斯判据求解系统稳定性函数
% 输入：
% chara_equ = 特征方程向量
% 输出：
% routh_list = 劳斯表
% conclusion = 给出系统是否稳定或存在多少个不稳定的根的结论% author:xianfa110% example:
% [routh_list,con] = Routh([1 2 3 4 5]);
% return:
% routh_list =
%
% 1 3 5
% 2 4 0
% 1 5 0
% -6 0 0
% 5 0 0
% con =
%
% There is 2 unstable roots!
% =========================================================
n=length(chara_equ);
chara_equ=reshape(chara_equ,1,n);
if mod(n,2)==0
n1=n/2;
else
n1=(n+1)/2;
chara_equ=[chara_equ,0];
end
routh=reshape(chara_equ,2,n1);
routh_list=zeros(n,n1);
routh_list(1:2,:)=routh;
i=3;
while 1;
% =========特殊情况1(第一列为0，其余列不为0)=====================
if routh_list(i-1,1)==0 & sum(routh_list(i-1,2:n1))~=0
chara_equ = conv(chara_equ,[1 3]);
n=length(chara_equ);
if mod(n,2)==0
n1=n/2;
else
n1=(n+1)/2;
chara_equ=[chara_equ,0];
end
routh=reshape(chara_equ,2,n1);
routh_list=zeros(n,n1);
routh_list(1:2,:)=routh;
i=3;
end

% ==========计算劳斯表===========================================
ai=routh_list(i-2,1)/routh_list(i-1,1);
for j=1:n1-1
routh_list(i,j)=routh_list(i-2,j+1)-ai*routh_list(i-1,j+1);
end
% ==========特殊情况2(全0行)======================================
if sum(routh_list(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
routh_list(i,:)=routh_list(i-1,:).*F(1,:);
end
% =========更新==================================================
i=i+1;
if i>n
break;
end
end
% =============outhput===========
r=find(routh_list(:,1)<0);
if isempty(r)==1
conclusion='The system is stable!';
else
n2=length(r);
m=n2;
for i=1:n2-1
if r(i+1)-r(i)==1
m=m-1;
end
end
str1='There is ';
if r(n2)==n
str2=num2str(m*2-1);
else
str2=num2str(m*2);
end
str3=' unstable roots!';
conclusion = [str1,str2,str3];
end

n=length(chara_equ); 26行有错误 很急 有高手解答下吗？
加急 加急 高手请帮下我 谢谢

问题解答：

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