drevier created the topic: WG Excitation
Expanding on my waveguide knowledge here. If I were to drive a rectangular waveguide with a TE10 mode source (think HFSS/CST waveport setup to give a TE10 mode field distribution) at a frequency higher than the TE20 mode cutoff would that necessarily induce a TE20 mode or would all the energy at that frequency stay in a TE10 mode because that's how I was driving it there.
In essence, will energy that *can* be coupled into a higher order mode *always* go to that mode or can it remain in the lower order mode?
Picture attached to help
Desert Sage replied to the topic: WG Excitation
In a theoretically perfect world you could excite one mode without exciting the other. However, in the real world this is near if not exactly impossible. Any small deviations from perfection will excite the other mode(s). You could try and suppress the mode but it will still result in energy lost in that mode.
An old VP of engineering once told me if it can (excite the) mode, it will (excite the) mode. I have found that to be sage advice.
drevier replied to the topic: WG Excitation
Excellent and thanks for the advice. I did a brief experiment in HFSS and ramped a WR340 from 2-18 GHz to see the effect. I was shocked to see something that has a TE20 mode at 14 GHz (and only 14 GHz, not any other frequency) while all the other frequencies exhibited something much closer to TE10. This, per your explanation, would be due to meshing/calculation imperfections then. I'll try meshing it more and seeing what it yields.
parkerMV replied to the topic: WG Excitation
This is an interesting question from a theoretical perspective, having read up a little on the field of nonlinear optics.
In general, it is MUCH MUCH easier for radio-frequency electromagnetic waves to jump to a second harmonic than visible wavelengths.
Any time you have a resonance cavity and energy of a different wavelength bounces around in that cavity, it's going to want to convert into the type of wave that is most fitting to the cavity. That is why [in other applications] it is often said that the cavity "amplifies" a certain wavelength.
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