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Calculating S-parameter of a meta-atom in a supercell of electromagnetic metasurfaces
Posted 2022年9月26日 GMT-4 23:44 Electromagnetics, RF & Microwave Engineering, LiveLink for MATLAB Version 5.6 0 Replies
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I am interested in investigating mutual coupling effect among meta-atoms in an aperiodic metasurface. Please see the figure attached below (tmp3.png) for illustration. Here I am considering some geometric perturbations of a split ring resonator (tmp1.png shows an example).
What I would like to compute is the transmission and phase delay (or S-paramter) of the center meta-atom. If there were no geometric perturbations, the computation would have been straightforward: using a single unit cell with peridoic boundary conditions, I can easily retrieve the S-parameter, using the built-in global evaluation in Comsol (e.g., emw.S21). But under my simulation setup with the perturbation, I try to calculate the S-paramater of a single meta-atom (or transmission and phase delay, equivalently), which would involve some deviation that obtained under the periodic boundary condition.
(Q1) Can I still retrieve the information? It need not be done by built-in functionalities in Comsol; it is also fine to compute the value via some snippet in Comsol with Matlab. I am aware that the equations on S-parameters are provided in User Guide of RF Module (tmp2.png, attache below). I would like to introduce a user-defined expression to employ the equations and calculate them (maybe using some surface integration?), if this works, but am not sure of how.
Lastly, I am only interested in the S-parameter of the center meta-atom. But in order to take into account the mutual coupling effect, perhaps I need to excite the whole top face (the blue region in tmp1.png) as the input port, while the S-parameter of my interest will only involve the contribution by the green region in the figure. (Q2) Can I only get the S-parameter between the two regions (green and red), without running port sweep?
Any help or comments would be greatly appreciated. I can share further details if necessary.
Thank you, Doksoo
Hello Doksoo Lee
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