function [Eg]= test_cal(L,U,P) for t=-0.2:0.01:0; xx=1:length(t);x3 = 0.20*t^2 + 5.65e-5/cos(cos(8.92*t)) ... + 0.0003*t*cos(cos(8.922*t)) + ... 0.037*t^3*cos(cos(8.92*t)) - ... 0.031 - 0.31*t - 0.36*t^4;x2 = 0.31*t + 0.002*sin(14.83*sin(t))^3 ... + 0.0012*sin(14.83*sin(t))^2 + ... 0.0008*sin(14.83*sin(t))/t - ... 0.0072 - 0.098*t^2 - ... 0.0015*cos(t)*sin(14.83*sin(t));x1 = 0.36+ 2.46*t + 28.82*t^3 + ... 3.46*t^4 + 2.46*t^5*sin(sin(sin(99.88*t^2))) - ... 0.011*sin(sin(99.87*t^2));x0 = 2.66 + 0.43*t*cos(30.07*t^2) + ... 0.015*t*cos(30.08*t^2)^2 - 0.48*t ... - 0.73*t^2 - 0.097*cos(30.08*t^2) ... - 1.19*t^3*cos(30.08*t^2);a0 = 1.42e-10; for tt=0.05:0.05:1alpha0 = x0; alpha1 = x1; alpha3 = x3; c = 1; U=0;kx = -2*pi/(3*a0):2*pi/(3*499.5*a0):2*pi/(3*a0);for c = 1:1000hk(c) = (exp(0)+exp(i*3*a0/2*kx(c))+exp(-i*(3*a0/2*kx(c)))); %h(k) for ky = 0.
z=sin(c);H = [U,-alpha0*hk(c),-alpha1,-alpha3*hk(c); %H matrix
-alpha0*conj(hk(c)),U,-alpha3*conj(hk(c)),-alpha1;-alpha1,-alpha3*hk(c),0,-alpha0*hk(c);-alpha3*conj(hk(c)),-alpha1,-alpha0*conj(hk(c)),0];E(:,c) = eig(H); %Solve the eigenvalue problem.
endE3 =E(3,:); E4 = E(4,:);E2 = E(2,:); Min = abs(E2-E3);Eg = min(Min) endend end
The output is like the following
Eg = 0.0069Eg = 0.0069Eg = 0.0069Eg = 0.0069Eg = 0.0069Eg = 0.0069Eg = 0.0069Eg = 0.0069
So I need it as
Eg=[0.0069 0.0069 ...]
Best Answer