How to solve an equation with big matrix?
5 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Saverio Loiacono
le 2 Fév 2024
Commenté : Saverio Loiacono
le 2 Fév 2024
Hello,
I have a problem solving this type of equation: det((-w^2)*Ma+Ka)==0
I've tried several times, using syms, fsolve, etc... Due to the sizes fo the matrix Ma and Ka that is for both 288X288, the matlab is not able to find the q solutionc of w (that is the unkown), where q is equal to tha arrays of the matrix (288). I've tried this types of code:
1)syms f(x)
f(x) = det(-(x^2)*Ma2+Ka2);
sol = vpasolve(f);
2)syms w
S = det(-(w^2)*Ma+Ka)==0;
solve = S;
3)myfun = @(x)det((-x.^2)*Ma+Ka);
[a,b]=size(Ma);
x0 = zeros(2*a,1);
C = fsolve(@(x)det((-x.^2)*Ma+Ka),x0);
None of them works, any solution? I will appreciate your help.
5 commentaires
Réponse acceptée
Torsten
le 2 Fév 2024
Modifié(e) : Torsten
le 2 Fév 2024
Looks like a generalized eigenvalue problem of the form
(A - w^2*B)*x = 0
with unknown x and w.
Use "eig" to solve it.
3 commentaires
Torsten
le 2 Fév 2024
Modifié(e) : Torsten
le 2 Fév 2024
From the documentation:
The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. The values of λ that satisfy the equation are the generalized eigenvalues. The corresponding values of v are the generalized right eigenvectors. The left eigenvectors, w, satisfy the equation w’A = λw’B.
Thus in your case:
e = eig(Ka,Ma)
The elements of e are your elements w^2. Taking the positive and negative square roots give you the 2*288 possible values for w.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Equation Solving dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!