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I am referring to M. N. O. Sadiku, Elements of Electromagnetics, 7th ed:

In chapter 10, the expression for the attenuation constant α in Np/m in a lossy dielectric is derived as:

$$ \alpha = \omega\sqrt{\dfrac{\mu\epsilon}{2}\left[\sqrt{1+\left[\dfrac{\sigma}{\omega\epsilon}\right]^2}-1\right]} $$

Later in the chapter, the special case of a lossless dielectric, where \$\sigma\ll\omega\epsilon\$ is discussed. The book claims that this results in \$\alpha=0\$ for a lossless dielectric. However, I'm not convinced that the first equation goes to 0 when the condition \$\sigma\ll\omega\epsilon\$ is satisfied. Using MATLAB, I plotted the attenuation constant vs. frequency using the lossy dielectric equation:

plot of alpha vs omega

As shown by the plot, the attenuation constant increases with frequency until it levels off around 85 Np/m. Although it appears to start oscillating and drop to 0 at a very high frequency, investigation appears to reveal that this is only due to floating point precision error as the term inside the square root becomes very small. This is illustrated by the following log-log plot containing three lines: the term inside the square root, omega, and the product of omega with the square root term (equivalent to alpha):

enter image description here

At the higher frequencies, following the inflection point between \$10^2\$ and \$10^3\$, both terms have the same slope with opposite signs, resulting in a flat line for alpha. The trend ends only when floating point precision begins to degrade the accuracy of the term inside the square root.

Am I doing something wrong, or is it not possible using the lossy dielectric equation to show that the attenuation constant approaches 0 in a lossless dielectric?

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  • \$\begingroup\$ Where do you read 85 Np/m? In your diagram the attenuation constant reaches barely 3.4 × 10^-6 Np/m. \$\endgroup\$
    – Ariser
    Mar 6 at 14:04
  • \$\begingroup\$ With \$σ \ll (ω·ε)\$ we have \$\sqrt{1+σ^2/(ω·ε)^2} \approx 1+1/2·σ^2/(ω·ε)^2\$, so \$α \approx ω·\sqrt{μ·ε/2·1/2·σ^2/(ω·ε)^2} = σ·1/2·\sqrt{μ/ε}\$. Indeed, \$α\$ is finite and nonzero, notwithstanding a supposedly gross error which OP possibly made when specifying the attenuation constant. Or is it a concentration and focus ability test? \$\endgroup\$
    – V.V.T
    Mar 7 at 4:25

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Reading your first plot, I see an attenuation constant of approximately 3.5e-6 Np/m, not 85 Np/m.

In the case that you are only modeling conduction losses in your dielectric, a lossless dielectric just means "negligible conduction" (i.e. \$\sigma \approx 0\$). The inequality \$\sigma \ll \omega \varepsilon\$ tells us just how small \$\sigma\$ needs to be for \$\alpha \approx 0\$.

From an equation perspective, as \$\sigma \ll \omega \varepsilon\$, the \$\frac{\sigma}{\omega\varepsilon}\$ term in the expression for \$\alpha\$ will start to approach zero, so the term in square brackets \$[\sqrt{1 - 0} - 1] \approx 0\$.

It's fair to say that the degree to which a dielectric can be considered lossless is the degree to which \$\omega \ll \sigma \varepsilon\$. A truly lossless dielectric would have \$\sigma = 0\$.

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