liguangrui33
Subscriber

您好,

这边比较清楚了,多谢!

还有个相关的问题,我参考的一个例子(https://optics.ansys.com/hc/en-us/articles/360042238733)里有这样的代码来变换成像(# scriptfile: calculate_standard_image.lsf):

Uz = Uz + 1e-5;
for(pz=1:length(z_image)) {
    z0 = z_image(pz);
    Ex_image(px,py,pz) = czt(Exf*exp(1i*Kz*z0)*sqrt(1/Uz/k0^2),kx,ky,x_image,y_image)*dkx*dky/(2*pi);
    Ey_image(px,py,pz) = czt(Eyf*exp(1i*Kz*z0)*sqrt(1/Uz/k0^2),kx,ky,x_image,y_image)*dkx*dky/(2*pi);
    Ez_image(px,py,pz) = czt(Ezf*exp(1i*Kz*z0)*sqrt(1/Uz/k0^2),kx,ky,x_image,y_image)*dkx*dky/(2*pi);
}
 
E2_image = abs(Ex_image)^2 + abs(Ey_image)^2 + abs(Ez_image)^2;
 
#Plot image:
image(x_image/1e-6, y_image/1e-6 ,E2_image,"x (um)","y (um)","|E|^2 image");
 
我不是很理解czt变换时这边对场分量做的修正*sqrt(1/Uz/k0^2)和需要*dkx*dky/(2*pi)的原因,这里需要您的帮助,多谢!