Effacer les filtres
Effacer les filtres

Vector convergence in cartesian coordinates

5 vues (au cours des 30 derniers jours)
Yasmin Tamimi
Yasmin Tamimi le 25 Sep 2014
Modifié(e) : Yasmin Tamimi le 27 Sep 2014
Hey everyone,
How can I help my vector in cartesian coordinates converge to its resting point. The vector flips from +x-axis to the negative but instead of converging to it it keeps rotating around it!
Thanks in advance.
  2 commentaires
Geoff Hayes
Geoff Hayes le 25 Sep 2014
Yasmine - you might want to provide some context concerning what problem you are trying to solve. As well, posting some of the code may allow others to get an idea of what is happening and be in a better position to provide/offer some guidance.
Matt J
Matt J le 26 Sep 2014
Yasmine Tamimi Commented:
Ok,, so I have a magnetization vector that according to some torques flips from +x-axis to -x-axis.. the problem I'm facing is that it doesn't converge to -x-axis it keeps rotating.. So, what shall I add to help it converge?
Steps = 20000; % Arrays size
time = 100e-9; % Total time
dt = (time/(Steps)); % Time step
Tswitch = 0:dt:time; % in seconds
for t = 1:Steps %Euler's Method
dtheta(t) = dt*((C1*G0)*(hMaz(t) + alpha*hMpol(t))) ; %3.1239
dphi(t) = dt*(((C1*G0)/sin(theta(t)))*(alpha*hMaz(t)- hMpol(t))); %1.57
theta(t+1) = theta(t) + dtheta(t) ; %3.1239
phi(t+1) = phi(t) + dphi(t); %1.57
Mi(t+1) = sin(theta(t+1))*cos(phi(t+1));
Mj(t+1) = sin(theta(t+1))*sin(phi(t+1));
Mk(t+1) = cos(theta(t+1));
u(t+1) = (EAi*Mi(t+1))+(EAj*Mj(t+1))+(EAk*Mk(t+1)); % dot prod. of M and EA.
Anis(t+1) = acos(u(t+1));
epoli(t+1) = cos(theta(t+1))*cos(phi(t+1));
epolj(t+1) = cos(theta(t+1))*sin(phi(t+1));
epolk(t+1) = -sin(theta(t+1));
eazi(t+1) = -sin(phi(t+1));
eazj(t+1) = cos(phi(t+1));
eazk(t+1) = 0 ;
ddu_acos(t+1) = -1/sqrt(1-(u(t+1)^2));
du_dMpol(t+1) = (EAi*cos(theta(t+1))*cos(phi(t+1))+EAj*cos(theta(t+1))*sin(phi(t+1))- EAk*sin(theta(t+1)));
du_dMaz(t+1) = sin(theta(t+1))*(EAj*cos(phi(t+1)) - EAi*sin(phi(t+1)));
dd_Anis(t+1) = Ku*Vol*sin(2*Anis(t+1));
g(t+1) = (-4 + ((1+P)^3)*(3+u(t+1))/(4*(P^1.5))) ;
G(t+1) = 1/g(t+1) ;
Ffi(t+1) = EAj*cos(theta(t+1)) - EAk*sin(theta(t+1))*sin(phi(t+1)) ;
Ffj(t+1) = -EAi*cos(theta(t+1)) + EAk*sin(theta(t+1))*cos(phi(t+1)) ;
Ffk(t+1) = EAi*sin(theta(t+1))*sin(phi(t+1))- EAj*sin(theta(t+1))*cos(phi(t+1)) ;
hMpol(t+1) = - (dd_Anis(t+1) * ddu_acos(t+1) * du_dMpol(t+1))/C2 - ((Is*P_hbar/(2*P_Q*g(t+1)*C2))*((epoli(t+1)*Ffi(t+1))+(epolj(t+1)*Ffj(t+1))+(epolk(t+1)*Ffk(t+1))))- Ms*sin(2*theta(t+1))*((Nx-Nz)+ (Ny-Nx)*(sin(phi(t+1))^2))+((hx*cos(theta(t+1))*cos(phi(t+1))+hy*cos(theta(t+1))*sin(phi(t+1))-hz*sin(theta(t+1))));
hMaz(t+1) = - (dd_Anis(t+1) * ddu_acos(t+1) * du_dMaz(t+1))/(C2*sin(theta(t+1)))- ((Is*P_hbar/(2*P_Q*g(t+1)*C2))*((eazi(t+1)*Ffi(t+1))+(eazj(t+1)*Ffj(t+1))))- Ms*(Ny-Nx)*sin(theta(t+1))*sin(2*phi(t+1))+(hy*cos(phi(t+1)) - hx*sin(phi(t+1))) ;
end

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Réponses (1)

Youssef  Khmou
Youssef Khmou le 25 Sep 2014
There are 24 undefined variables C1;G0;hMaz;hMpol;Alpha;theta;phi;EAj;EAi;EAk;Ku;Vol;P;C2;Is;P_hbar;P_Q;Ms;Nx;Nz;Ny;hx;hy;hz;
I tired using random values but the vector hMaz gives NaN values, because we need precise values of these variables, because i think this is inside quantum box right?
  13 commentaires
Youssef  Khmou
Youssef Khmou le 26 Sep 2014
Converging M means that the magnetization is zero? what about the volume ?
Yasmin Tamimi
Yasmin Tamimi le 27 Sep 2014
Modifié(e) : Yasmin Tamimi le 27 Sep 2014
Vector M is position vector initially its M=[Mi Mj Mk]= [1 0 0] and the end result must be [-1 0 0]. The volume stays constant bcz it is the area multiplied by the thickness of the material.

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