A question about cut-off frequency in waveguide, thanks!
For example a rectangular waveguide has its first two cutoff freq to be TE10 ~ 6.56GHz and TE20 ~ 13.12GHz, and I feed in a wave of freq 14GHz (does this wave have to be a TE or TM wave so that it can actually propagate?) , then there should have at least two modes(TE10&TE20) co-exist inside the waveguide right? and, although the wave bears two different modes for propagation, it still maintains its own frequency, i.e. 14GHz, right? if I'm right so far, then the thing bothers me a lot is that, how can this monochromatic wave having two different wavelengths inside the waveguide? Thank you so much for your patient explanation!
Try [Comms] Waveguide Handbook (Microwave 1986)
check your PM.
best regard
It depends how you feed yours waveguide. You can exite any propagating modes purely or as a mixture of them.
It's not necessary. If you will launch only TE10 (or TE20) mode , for example, it will propagate
alone inside perfect waveguide even the frequency is higher than the second (third, forth...) cutoffs.
OK.
Where is a contradiction ? The monochromaticity doesn't depend on modal pattern.
You will have just more complicate field distribution inside waveguide comparing to single mode excitation.
Moreover single mode signal could be easily converted to multimoded on a waveguide imperfections (flanges, bends, ...)
Where is a contradiction ? The monochromaticity doesn't depend on modal pattern.
You will have just more complicate field distribution inside waveguide comparing to single mode excitation.
Moreover single mode signal could be easily converted to multimoded on a waveguide imperfections (flanges, bends, ...)
To the last point, it just makes me feel pretty stranged that inside the waveguide, the wave bears different wavelength while having the same frequency.
Thanks again!