フィルターのクリア

Routh-Hurwitz criterion

28 ビュー (過去 30 日間)
Sunday
Sunday 2024 年 8 月 28 日 16:55
コメント済み: Sam Chak 2024 年 8 月 28 日 20:32
syms a b c d e f
% Define the coefficients of the polynomial
coefficients = [1 a b c d e f];
% Create the Routh-Hurwitz array
RH_array = sym(zeros(7,length(coefficients)));
RH_array(1,:) = coefficients;
RH_array(2,1) = coefficients(1);
RH_array(2,2) = coefficients(3);
RH_array(2,3) = coefficients(5);
for i=3:7
% Compute the remaining entries in the Routh-Hurwitz array
RH_array(i,1) = simplify(-det([RH_array(i-2,1) RH_array(i-2,2); RH_array(i-1,1) RH_array(i-1,2)]) / RH_array(i-2,1));
for j=2:length(coefficients)-1
RH_array(i,j) = simplify(-det([RH_array(i-2,j-1) RH_array(i-2,j); RH_array(i-1,j-1) RH_array(i-1,j)]) / RH_array(i-2,j-1));
end
end
% Check the stability criteria using the Routh-Hurwitz array
stable = true;
for i=1:size(RH_array,1)
if any(RH_array(i,:) == 0)
stable = false;
break;
end
end
if stable
disp('The polynomial is stable according to the Routh-Hurwitz criterion');
else
disp('The polynomial is unstable according to the Routh-Hurwitz criterion');
end
The polynomial is unstable according to the Routh-Hurwitz criterion
The results only shows that "The polynomial is instable according to the Routh-Hurwitz criterion". please I want to display the Roots of the polynomial and then submatrix generated from RRouth-Hurwit.
  1 件のコメント
Sam Chak
Sam Chak 2024 年 8 月 28 日 20:32
@Sunday, Your Routh-Hurwitz array is incorrectly tabulated.
num = 1;
den = [1 6 15 20 15 6 1];
G = tf(num, den)
G = 1 ------------------------------------------------ s^6 + 6 s^5 + 15 s^4 + 20 s^3 + 15 s^2 + 6 s + 1 Continuous-time transfer function.
ToF = isstable(G)
ToF = logical
1
% Define the coefficients of the polynomial
coefficients = den
coefficients = 1x7
1 6 15 20 15 6 1
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
% Create the Routh-Hurwitz array
RH_array = sym(zeros(7,length(coefficients)));
RH_array(1,:) = coefficients;
RH_array(2,1) = coefficients(1);
RH_array(2,2) = coefficients(3);
RH_array(2,3) = coefficients(5);
for i=3:7
% Compute the remaining entries in the Routh-Hurwitz array
RH_array(i,1) = simplify(-det([RH_array(i-2,1) RH_array(i-2,2); RH_array(i-1,1) RH_array(i-1,2)]) / RH_array(i-2,1));
for j=2:length(coefficients)-1
RH_array(i,j) = simplify(-det([RH_array(i-2,j-1) RH_array(i-2,j); RH_array(i-1,j-1) RH_array(i-1,j)]) / RH_array(i-2,j-1));
end
end
disp(RH_array)
% Check the stability criteria using the Routh-Hurwitz array
stable = true;
for i=1:size(RH_array,1)
if any(RH_array(i,:) == 0)
stable = false;
break;
end
end
if stable
disp('The polynomial is stable according to the Routh-Hurwitz criterion');
else
disp('The polynomial is unstable according to the Routh-Hurwitz criterion');
end
The polynomial is unstable according to the Routh-Hurwitz criterion

サインインしてコメントする。

回答 (1 件)

Torsten
Torsten 2024 年 8 月 28 日 19:57
Roots of a general polynomial of degree 6 cannot be explicitly computed. You must give numerical values to a,b,c,d,e and f and use "root" or "roots" to get numerical values for the roots.

カテゴリ

Help Center および File ExchangeStability Analysis についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by