- 2076 Chaitra · 6 marks
Cut-off and single-mode range, 4×2 cm
Short answer
Cut-off of TE_10: f_c = c/(2a) = 3×10⁸/(2×0.04) = 3.75 GHz. Next modes TE_20 and TE_01 have f_c = 7.5 GHz. Single-mode operation range: 3.75 GHz < f < 7.5 GHz.
A rectangular waveguide has dimension a = 4 cm and b = 2 cm. Determine the cut-off frequency and range of frequencies over which the guide will operate in single mode.
Given: a = 4 cm, b = 2 cm (air-filled, c = 3×10⁸ m/s).
Cut-off frequency of mode (m, n):
f_c(mn) = (c/2) √[(m/a)² + (n/b)²]
Cut-off of the dominant mode (TE₁₀)
f_c(10) = c/(2a) = 3×10⁸ / (2 × 0.04) = 3.75×10⁹ Hz
Cut-off frequency = 3.75 GHz (dominant mode TE₁₀).
Other low modes
f_c(20) = c/a = 3×10⁸/0.04 = 7.5 GHz
f_c(01) = c/2b = 3×10⁸/(2×0.02) = 7.5 GHz
f_c(11) = (c/2)√[(1/0.04)² + (1/0.02)²] = 1.5×10⁸ × √(625 + 2500) = 8.385 GHz
The next higher mode after TE₁₀ is TE₂₀ (and TE₀₁, equal at 7.5 GHz because a = 2b).
Single-mode range
Only TE₁₀ propagates when its cut-off is below f and the next cut-off (7.5 GHz) is above f:
3.75 GHz < f < 7.5 GHz
(Bandwidth 3.75 GHz.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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