Equivalent Impedance using Matrix, Eigen value and Eigen vector, norm
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I am solving this below problem(attachment 1). How to find the exp(theta1) and exp(theta2)...?
- Attachment pdf.pdf is the problem I am trying to solve.
- Attachment Tzeng_2006_J._Phys._A__Math._Gen._39_8579.pdf is the theory
Y11 =1-1i/sqrt(3); Y12 =1i/sqrt(3); Y13 =-1;
Y21 =1i/sqrt(3); Y22 =0; Y23 =-1i/sqrt(3);
Y31 =-1; Y32 =-1i/sqrt(3); Y33 =1+1i/sqrt(3);
Y11 = Y12 + Y13 ; Y22 = Y21 + Y23 ; Y33 = Y31 + Y32 ;
L = [Y11, Y12, Y13; Y21, Y22, Y23; Y31, Y32, Y33];
[EVector,EValue] = eig(L'*L) ; % V = Values % D = VECTORS
Sigma_2 = sqrt(EValue(2,2))
Sigma_3 = sqrt(EValue(3,3))
3 Kommentare
Walter Roberson
am 9 Nov. 2023
Q = @(v) sym(v);
Y11 = 1-1i/sqrt(Q(3)); Y12 = 1i/sqrt(Q(3)); Y13 = Q(-1);
Y21 = 1i/sqrt(Q(3)); Y22 = Q(0); Y23 = -1i/sqrt(Q(3));
Y31 = -1; Y32 = -1i/sqrt(Q(3)); Y33 = 1+1i/sqrt(Q(3));
Y = [Y11, Y12, Y13; Y21, Y22, Y23; Y31, Y32, Y33; Y41, Y42, Y43]
A = L'*L ; % e = eig(L_P*L);
[V,EVC] = eig(A) % V = Values % D = VECTOR
Antworten (0)
Siehe auch
Kategorien
Mehr zu Particle & Nuclear Physics finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!