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Reflection and Loss Tangents

by jmckennon
Tags: loss, reflection, tangents
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May1-12, 03:21 PM
P: 42
I've been reading over Balanis' Advanced Engineering Electromagnetics in an effort to teach myself a bit about Electromagnetism.

I've stumbled across a question (not one in the book) that I can't seem to wrap my head around.

If there's a lossy dielectric or medium, how does the reflection coefficient change/get effected as opposed to a lossless medium or dielectric?

It's easy to determine the reflection coefficient for a lossless medium, I'm fine with that portion, but I'm having trouble understanding how to do it for a lossy medium.

I've googled around, but can't seem to find an answer. As a side, but possibly unrelated note, is reflection coefficient the same as the complex propagation constant for the case of a lossy dielectric/medium? That's about all I was able to find on the topic. Hope someone can clear this up!
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Bob S
May1-12, 05:59 PM
P: 4,663
If a dielectric is lossy, the refraction coefficient must be dispersive.

At a single frequency, the index of refraction (and reflection coefficient) and the absorption are independent. So I don't believe at a single frequency, a change in the absorption (attenuation) will affect the reflection coefficient.

However, the absorption (loss tangent) in a dielectric is very closely tied to the dispersion (frequency dependence of index of refraction). The refractive index is complex when absorption is present, and the absorption is determined by the dispersion in the refractive index. Mathematically, this relation is shown in the Kramers-Kronig relations. See

These relations apply to lossy microwave dielectrics, and even to comples impedances in passive electric circuits. See book by Bode Network Analysis and Feedback Amplifier Design, and Morse and Feshbach Methods of Theoretical Physics Vol 2.
May2-12, 01:30 AM
P: 1,781
Get a copy of Ruck volumes one and two. It's very hard to find but itself presents the solution you seek in its most general and perfect form. Sorry I no longer have access to it so I can't give you the correct formula.

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