On Sat, 24 Feb 2007, Chad Huskow wrote:
My calculation is the transmission of the PC slabs (waveguide-cavity system). The cavity is located beside the waveguide. According to the MEEP material and your suggestion, I define the Gaussian source in exactly the same way (2D PC case) except that my source lies in the z=0 plane. The region centered at z=0 with a thickness similar to the slab thickness. The source have the center frequency (fcen) and the width (df). The frequency of this source is a single mode waveguide, and the resonant mode of the cavity is inside this frequency range. The flux region is a plane (big enough) to catch all of the energy propagating in the waveguide mode. I split my job into 2 steps: First, I calculate the transmission in straight waveguide, and second with resonator. I have checked my data of the second step file (with resonator), I am suprised that the resonator do not couple the modes. I don't know why it is? Could you help me?
Chad, there is nothing wrong in principle with the sequence of steps you describe to compute the transmission spectrum in a slab structure. However, there are a few things to be careful of.
First, how do you know that you have a resonant mode in the cavity? The best way is to directly excite the cavity mode with a source in the cavity, and then analyze the response with harminv.
Second, how do you know that your cavity mode is the correct symmetry to couple to the waveguide mode? (The best thing is to look at the field pattern after you excited the mode with a source inside the cavity.)
Third, are you sure the losses are not so great that you just can't see the peak because the transmission is too low? The key thing here is the ratio of the radiative loss rate (radiative 1/Q_r) to the decay rate into the waveguides (1/Q_w). The losses go like ~ 2 Q_w / Q_r, so if you make Q_w too big (by putting too many periods of the crystal between the cavity and the waveguide) then the transmission will be small.
Fourth, you should also be careful if the Q is very large because then the peak may be too narrow to see with the frequency resolution you are using. (In general, it is difficult to see very narrow peaks in a transmission spectrum, due to the Fourier uncertainty principle.)
Steven _______________________________________________ meep-discuss mailing list [email protected] http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/meep-discuss
