Filter löschen
Filter löschen

Matrix dimensions must agree.

2 Ansichten (letzte 30 Tage)
mohammad mortezaie
mohammad mortezaie am 12 Sep. 2021
Beantwortet: per isakson am 12 Sep. 2021
I have below code for a contour plot bu I faced with "Matrix dimensions must agree.".
How can I fix it?
a=1.855;
eb=-3.276;
ep=-1.979;
t0=-1.844;
theta=0;
ee=linspace(-0.3,0.3,200);
EE=linspace(0,10,200);
[e,omega]=meshgrid(ee,EE);
sigma=0.265;
exx=e.*((cos(theta)).^2-sigma.*(sin(theta)).^2);
exy=e.*((1+sigma).*cos(theta).*sin(theta));
eyx=e.*((1+sigma).*cos(theta).*sin(theta));
eyy=e.*((sin(theta))^2-sigma.*(cos(theta)).^2);
delta1x=a/2;
delta1y=a*sqrt(3)/2;
delta2x=a/2;
delta2y=-a*sqrt(3)/2;
delta3x=-a;
delta3y=0;
delta11=a*(1+3.*eyy/4+sqrt(3).*exy/2+exx/4);
delta22=a.*(1+3.*eyy/4-sqrt(3).*exy/2+exx/4);
delta33=a.*(1+exx);
g1=-2.6275;
g2=2*g1;
del1=g1.*(exx+eyy);
del2=g2.*(exx+eyy);
on1=ep+del1 ;
on2=eb+del2;
t1=t0*exp(-3.37.*(delta11/a-1));
t2=t0*exp(-3.37.*(delta22/a-1));
t3=t0*exp(-3.37.*(delta33/a-1));
delta2=0.2;
eta=0.05;
iform=complex(0.0,1.0);
X = linspace(-1.5,1.5,200);
ky= linspace(-2,2,200);
[x,y] = meshgrid(X, ky);
u=0.0;
T=10;
phi=t1.*exp(iform.*x.*delta1x).*exp(iform.*y.*delta1y)+t2.*exp(iform.*x.*delta2x).*exp(iform.*y.*delta2y)+t3.*exp(iform.*x.*delta3x);
H2 = [on1.*ones(size(phi)) phi; conj(phi) on2.*ones(size(phi))];
[v,E]=eig(H2);
E1=E(1);
E2=E(4);
v1=v(:,1);
v2=v(:,2);
phiy=t1.*iform.*delta1y.*exp(iform.*x.*delta1x).*exp(iform.*y.*delta1y)+t2.*iform.*delta2y.*exp(iform.*x.*delta2x).*exp(iform.*y.*delta2y);
H2y=[zeros(size(phiy)) phiy;
conj(phiy) zeros(size(phiy))];
jy12=((v1'.*H2y.*v2).*conj(v1'.*H2y.*v2));
deltaE12=E2-E1;
fermi1=1/(1+exp((E1-u)/T));
fermi2=1/(1+exp((E2-u)/T));
sigma_xx=(iform.*omega.^-1).*(jy12).*(fermi2-fermi1).*( (omega+deltaE12+iform.*delta2).^-1-(omega-deltaE12+iform.*delta2).^-1 );
contour(e,EE,sigma_xx,500)

Antworten (1)

per isakson
per isakson am 12 Sep. 2021
Because (jy12) is 400x400 and omega is 200x200

Kategorien

Mehr zu Contour Plots finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by