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it is known that in an ideal waveguide (with a perfect electric conductor and with non-dispersive and non-conductive dielectric) there is a cut-off frequency, for each mode, such that for higher frequencies the electromagnetic wave propagates without attenuation (the wave vector is purely imaginary), for minor frequencies it propagates with attenuation (the wave vector is purely real).

Briefly, the wave guide is a high pass filter, for each mode.

Now my question is: let's consider the situation in which the frequency is lower than the cut-off frequency. There is total attenuation of the EM wave. Which is the reason for that? Where does the EM energy go, since the dielectric is non - conductive?

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  • \$\begingroup\$ There won't be total attenuation. \$\endgroup\$ – Andy aka Oct 2 '19 at 9:58
  • \$\begingroup\$ You should say what kind of waveguide you're working with. For example, coaxial cable is a waveguide with no low frequency cut-off. \$\endgroup\$ – The Photon Oct 2 '19 at 16:15
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Below cutoff frequency, there is no energy flowing in to the waveguide. Everything you try to feed in is reflected back. And as there is no energy flowing, no losses are needed. You only see so called evanescent waves that decay exponentially when going inside the waveguide.

https://en.wikipedia.org/wiki/Evanescent_field

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