7樓: Originally posted by cs24451 at 2016-09-01 16:13:49
額  樓主怎么實現(xiàn)的  能說下么  我這個L1 L2代表長度， 希望在我給出的值附近浮動，不能為負。。。...

function F = myfun(b)
data = load('a.txt');
x = data(:,1); yn = data(:,2);
m=20; % number of section
n1 = 2.8924;   %AlAs
n2 = 3.3702;   %GaAS
na=1;
ns=n2;
ra=(na-n2)/(na+n2);
r=(n2-n1)/(n2+n1);
ta=sqrt(1-ra^2);
t=sqrt(1-r^2);
y=zeros(1,2500);
ab1=0;
ab2=0;
for kl=1:2500

    beta1 = 2*pi*n1/x(kl)-1i*ab1/2;
    beta2 = 2*pi*n2/x(kl)-1i*ab2/2;
   
    T11_a = exp(1i*beta2*b(2))/ta;
    T12_a = ra*exp(-1i*beta2*b(2))/ta;
    T21_a = ra*exp(1i*beta2*b(2))/ta;
    T22_a = exp(-1i*beta2*b(2))/ta;
    T=[T11_a,T12_a;T21_a,T22_a];
   
        for k=1:m-1
            T11_n2 = exp(1i*beta1*b(1))/t;
            T12_n2 = r*exp(-1i*beta1*b(1))/t;
            T21_n2 = r*exp(1i*beta1*b(1))/t;
            T22_n2 = exp(-1i*beta1*b(1))/t;
            T_n2=[T11_n2,T12_n2;T21_n2,T22_n2];
            T=T*T_n2;
            
            T11_n1 = exp(1i*beta2*b(2))/t;
            T12_n1 = -r*exp(-1i*beta2*b(2))/t;
            T21_n1 = -r*exp(1i*beta2*b(2))/t;
            T22_n1 = exp(-1i*beta2*b(2))/t;  
            T_n1=[T11_n1,T12_n1;T21_n1,T22_n1];
            T=T*T_n1;

        end
        T11_n2 = exp(1i*beta1*b(1))/t;
        T12_n2 = r*exp(-1i*beta1*b(1))/t;
        T21_n2 = r*exp(1i*beta1*b(1))/t;
        T22_n2 = exp(-1i*beta1*b(1))/t;
        T_n2=[T11_n2,T12_n2;T21_n2,T22_n2];
        T=T*T_n2;
        
        T_12=[1/t,-r/t;-r/t,1/t]; % from n1 to n2, just cross the interface
        T = T*T_12;
        
        T11_c = exp(1i*beta2*2*b(2));
        T12_c = 0;
        T21_c = 0;
        T22_c = exp(-1i*beta2*2*b(2));
        T_c = [T11_c,T12_c;T21_c,T22_c]; % n2 cavity
        T = T*T_c;
        
        for k=(1:m)
            T11_n2 = exp(1i*beta1*b(1))/t;
            T12_n2 = r*exp(-1i*beta1*b(1))/t;
            T21_n2 = r*exp(1i*beta1*b(1))/t;
            T22_n2 = exp(-1i*beta1*b(1))/t;
            T_n2=[T11_n2,T12_n2;T21_n2,T22_n2];
            T=T*T_n2;
            T11_n1 = exp(1i*beta2*b(2))/t;
            T12_n1 = -r*exp(-1i*beta2*b(2))/t;
            T21_n1 = -r*exp(1i*beta2*b(2))/t;
            T22_n1 = exp(-1i*beta2*b(2))/t;  
            T_n1=[T11_n1,T12_n1;T21_n1,T22_n1];
            T=T*T_n1;
        end   
        

        T11_n2 = exp(1i*beta1*b(1))/t;
        T12_n2 = r*exp(-1i*beta1*b(1))/t;
        T21_n2 = r*exp(1i*beta1*b(1))/t;
        T22_n2 = exp(-1i*beta1*b(1))/t;
        T_n2=[T11_n2,T12_n2;T21_n2,T22_n2];
        T=T*T_n2;
        
        T_12=[1/t,-r/t;-r/t,1/t];
        T = T*T_12;
        
        T11_c = exp(1i*beta2*9*b(2));
        T12_c = 0;
        T21_c = 0;
        T22_c = exp(-1i*beta2*9*b(2));
        T_c = [T11_c,T12_c;T21_c,T22_c]; % n2 cavity
        T = T*T_c;
        
        T11_n2 = exp(1i*beta1*b(1))/t;
        T12_n2 = r*exp(-1i*beta1*b(1))/t;
        T21_n2 = r*exp(1i*beta1*b(1))/t;
        T22_n2 = exp(-1i*beta1*b(1))/t;
        T_n2=[T11_n2,T12_n2;T21_n2,T22_n2];
        T=T*T_n2;
        
        rs1=(n1-ns)/(ns+n1);
        ts1=sqrt(1-rs1^2);
        T11_s = 1/ts1;  % n1 to substrate
        T12_s = rs1/ts1;
        T21_s = rs1/ts1;
        T22_s = 1/ts1;
        T_s = [T11_s,T12_s;T21_s,T22_s];
        T=T*T_s;
        y(kl) = (abs(T(2,1)/T(1,1)))^2;
end
F = y'-yn;

b0 = [-0.5e-9,-0.5e-9];
options = optimoptions(@lsqnonlin,'TolFun',1e-20,'TolX',1e-20,...
    'Algorithm','levenberg-marquardt','Plotfcns',@optimplotfval,...
    'OptimalityTolerance',1e20,'MaxIterations',1500,...
    'Display', 'iter-detailed','MaxFunctionEvaluations',2000);
K2 = lsqnonlin(@myfun,b0,[],[],options);

2樓: Originally posted by 512095518 at 2016-08-31 19:11:07
你先給出你的擬合數(shù)據(jù)，x，y，然后給出你估計的L1 L2的值
你的這個函數(shù)function  = c(x,L1，L2)，是表示用x，L1，L2就可以計算出y值對吧

3樓: Originally posted by cs24451 at 2016-08-31 23:37:42
是的 ，我已經設置了x,y值， 想通過我的函數(shù)c（x,L1,L2）來求得合適的L1與L2值...