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I was studying EM wave propagation in lossy dielectrics and came across two seemingly conflicting parameters of a medium. The loss tangent is defined as:

tanδ=σ/(ω*ϵ) ;

which clearly is inversely proportional to the frequency of the wave. So for a high frequency wave, the conducting medium would behave as an insulator to the propagation, and the power loss would be low.

Whereas the wave attenuation constant is defined as:

α= √(π* f* μ* σ);

which displays proportionality with the square root of the frequency. So at a high frequency, the wave would be greatly attenuated i.e. lose amplitude with propagation through the medium. But isn't this contradicting with the concept of low loss tangent thus low power loss? Am I missing something in the relation between attenuation and power loss?

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  • \$\begingroup\$ These variables are complexly interrelated. Get an advanced E&M book and examine the derivation of the general solution for wave propagation. You'll find a messy equation, with square toots inside. The index-of-refraction of metals and for dielectrics comes from this general solution, which, as you've noted, depends on the conductivity. Conductivity can vary by 20 orders of magnitude. \$\endgroup\$ – analogsystemsrf Jan 20 '18 at 4:20
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The attenuation is given by \$\alpha = \frac{\omega}{c}\sqrt[]{\frac{\epsilon'}{2}(\sqrt[]{1 + \tan^2\delta} - 1)}\$

The equation \$ tan \delta = \frac{\sigma}{\omega \epsilon}\$ assumes that the imaginary part of dielectric constant is 0, i.e. its a lossless non-ideal dielectric material, with non-zero conductivity. The more general equation for loss tangent is \$ tan \delta = \frac{\omega \epsilon" + \sigma}{\omega \epsilon'}\$

Now, if you fit the frequency into the attenuation equation, you get square-root of the frequency in the denominator, while first power of frequency in the numerator. As a consequence, with increasing frequency, the attenuation is increases. Hence no contradiction.

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