if I take d= lambda/2 and m =1. The angle cos theta =1 and similarly theta = 0 degree. then how wave propagate in parallel plate waveguide.
A signal cannot propagate in parallel plate waveguide 1/2 wavelength wide with the E-field parallel to the plates. It is at cut-off.
With the E-field normal to the plates there is no cut-off for the fundamantal mode which is a TEM mode. With half-wavelength separation, no other mode can propagate.
I agree with Dr. M. White. Signal propagates in transverse manner (mode concept), longitudinal direction has no electric field.
Thanks,
One way to understand this is to think about what would happen to a plane wave if it arrived at the edge of a stack of metal plates.
If there was a plane wave, vertically polarised (electric field vertical), travelling in a horizontal direction, then any number of horizontal thin metal plates could be put in place and it would make no difference to the wave, whatever the separation. This would be a vertical polariser, for instance, and each gap is a parallel-plate waveguide in the fundamental mode m=0.
If the wave was horizontally polarized, then there would be a reflection from the metal edges, but if the plates are more than a half-wavelength apart some power will travel between the plates. The modes that would be excited would be the m=1 (more than 1/2 wavelength) m=3 (more than 3/2 wavelengths), m=5 (more than 5/2 wavelengths) etc. The modes for even m won't be excited unless the wave travels at an angle to the horizontal, and they will appear when the separation is more than 1 wavelength, 2 wavelengths, etc.
For the vertically polarised wave, if the wave is at an angle to the horizontal, other modes also appear, in the same order.
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