資源簡介
傳輸矩陣法 仿真啁啾光柵的光學(xué)特性,如反射譜 透射譜 時延 相位
代碼片段和文件信息
L=0.04;
neff=1.45;
%C=0;
C=20*1e-9;
M=150001;N=50;
lamda1=1047;lamda2=1059;
lamda=linspace(lamda1lamda2M)*1e-9;
deltalamda=(lamda2-lamda1)/M*1e-9;
tic
for?k=1:M
????F=[10;01];
????for?i=1:N
????????%junyun
???????deltaneff=0.0005;
????????%fei?jun?yun?
????%deltaneff=0.0001*exp((-64*(-L/2+i*L/N)^4)/L^4);
???????lamda_D=(1053-C*L/2+C*i*L/N)*1e-9;
???????sigma=2*pi*neff*(1/lamda(k)-1/lamda_D)+2*pi*deltaneff/lamda(k)+(4*pi*neff)*C*(-L/2+i*L/N)/lamda_D^2;
???????%sigma=2*pi*neff*(1/lamda(k)-1/lamda_D)+2*pi*deltaneff/lamda(k)+(4*pi*neff)*(L/N)/lamda_D^2;
???????kac=pi*deltaneff/lamda(k);
???????RB=sqrt(kac^2-sigma^2);
???????F=F*[cosh(RB*L/N)-j*(sigma/RB)*sinh(RB*L/N)-j*(kac/RB)*sinh(RB*L/N);
???????????j*(kac/RB)*sinh(RB*L/N)cosh(RB*L/N)+j*(sigma/RB)*sinh(RB*L/N)];
?????
??????
???????
????end
????R(k)=(abs(-F(3)/F(1)))^2;
????Q(k)=phase((-F(3)/F(1)));
???
end
toc
????tao(1)=Q(1);
????tao(2)=Q(2);
????tao(3)=Q(3);
????for?i=4:M
????????if(abs(Q(i-1)-Q(i))<=1)
????????????tao(i)=((lamda1+i*0.00008)^2*1e-18/(2*pi*3e-4)*(Q(i-1)-Q
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