What is the attenuation below cut-off frequency for TE10?

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tworitdash
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It is not useful to talk about the attenuation below cut-off frequency, but I have this doubt about what happens to the wave below cut off for an electric conductor. As we know if we derive the propagation constant, it becomes imaginary saying that there should not be any wave propagating in the medium. The discussion ends here.

For frequencies above the cut-off, we consider the finite conductivity of the conductor to determine the attenuation based on the formulae below.

$$ P_{loss} = (1/2) \iint Rs(\vec{r}) \left| \vec{J_s}(\vec{r}) \right| ^ 2ds$$

$$ \vec{J_s}(\vec{r}) = \hat{n} \times \vec{H}_{PEC}(\vec{r}) $$

Where ##R_s## is a function of conductance and frequency. From the above equations with the poynting vector, we find the attenuation constant.

However, we do not consider the attenuation due to the lossy but conducting medium when the frequency is below the cut-off. Does it affect below the cut-off? If so, why and how can we compute it? What should be the ##\vec{H}_{PEC}## below the cut-off?
 
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I do not know how to calculate the loss due to the walls of the tube or the dielectric but the attenuation of the evanescent wave is in any case very high. Terman, in Radio Engineering, quotes 20.9dB per distance/radius for capacitive plates in a round copper tube. For magnetic coupling he quotes 33.3dB per radius/distance. He also gives a correction for skin depth.
 
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