- 2074 Asoj · 2 marks
Dominant cut-off, dielectric-filled guide
Short answer
With ε_r=4, wave speed u = c/√ε_r = 1.5×10⁸ m/s. Dominant TE10 cutoff f_c = u/(2a) = 1.5×10⁸/(2×0.02) = 3.75 GHz.
A rectangular waveguide with cross-section 2 cm × 1 cm is filled with a lossless medium characterized by ε = 4ε0 and μ_r = 1. Calculate the cut-off frequency of the dominant mode.
Given: a = 2 cm, b = 1 cm, filled with ε = 4ε₀, μ_r = 1.
The wave speed in the filling is
u = 1/√(με) = c/√ε_r = 3×10⁸/√4 = 1.5×10⁸ m/s
The dominant mode is TE₁₀ (a > b):
f_c = u/(2a) = 1.5×10⁸ / (2 × 0.02) = 3.75×10⁹ Hz
Answer: f_c = 3.75 GHz (half the 7.5 GHz of the air-filled guide, since √ε_r = 2).
✓ Answer checked by Claude Opus 5.5 (high effort)
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