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Routh-Hurwitz criterion

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Sunday
Sunday el 28 de Ag. de 2024 a las 16:55
Comentada: Sam Chak el 28 de Ag. de 2024 a las 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 comentario
Sam Chak
Sam Chak el 28 de Ag. de 2024 a las 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
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% 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

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Respuestas (1)

Torsten
Torsten el 28 de Ag. de 2024 a las 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.

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