Air-filled circular waveguides: which attenuation component is typically negligible, and which dominates?

Difficulty: Easy

Correct Answer: Attenuation from imperfect dielectric (air) is negligible

Explanation:


Introduction:
Loss mechanisms in hollow waveguides include conductor (wall) loss and dielectric loss. For air-filled guides, dielectric loss is extremely small, so conductor loss dominates practical attenuation.


Given Data / Assumptions:

  • Air as dielectric (very low loss tangent).
  • Standard copper/aluminum walls with finite conductivity.
  • Operation above cutoff with smooth interior surfaces.


Concept / Approach:

Conductor loss arises from surface currents and finite skin-depth resistive effects, producing attenuation αc that decreases with increasing frequency (away from cutoff). Air contributes negligible dielectric loss αd compared to αc, so it is routinely ignored in calculations unless special dielectrics are inserted.


Step-by-Step Solution:

1) Identify loss channels: α = αc + αd.2) For air: tanδ ≈ 0 → αd ≈ 0.3) Finite wall conductivity ⇒ αc > 0, typically the dominant term.4) Therefore, dielectric attenuation is negligible relative to conductor attenuation.


Verification / Alternative check:

Manufacturer data for standard waveguides quotes attenuation per meter dominated by conductor loss; changes in gas filling (air vs dry nitrogen) barely affect α.


Why Other Options Are Wrong:

  • Wall loss negligible or total attenuation nearly zero: unrealistic.
  • Both significant or dielectric dominant: inconsistent with air’s very low loss tangent.


Common Pitfalls:

Assuming “air-filled” means lossless. Conductor loss sets the floor in real systems.


Final Answer:

Attenuation from imperfect dielectric (air) is negligible

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