Routh-Hurwitz criterion

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.

[Sam Chak]

@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

Answer 1

[Torsten]

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.

[Sunday]

I wanted to get the roots without numerical values rather use the letters in place

[Torsten]

I know - and my answer is: it's usually not possible for polynomials of degree > 4:

https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

[Sam Chak]

@Sunday may be interested in developing an application or program that analyzes the stability of arbitrary linear systems entered by the user; however, she/he may not fully understand the associated limitations.

In fact, constructing the Routh–Hurwitz array does not require finding the roots (eigenvalues) of the characteristic polynomial. Consequently, it is common for the Abel–Ruffini Impossibility Theorem and Galois theory not to be taught in typical system dynamics or undergraduate control theory courses.

Furthermore, most textbooks and professors utilize relatively low-order linear systems in their examples, and @Sunday lacks a solid reference for designing the Routh–Hurwitz program.