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Cut-off of dominant and TE11, 2.5×1.2 cm

Short answer

Using f_c = (c/2)√[(m/a)²+(n/b)²] with a=2.5 cm, b=1.2 cm: dominant TE10 has f_c = c/(2a) = 6 GHz; TE11 has f_c = 13.87 GHz.

A rectangular waveguide has a cross-section of 2.5 cm × 1.2 cm. Find the cut-off frequencies at dominant mode and TE(1,1).

Given: a = 2.5 cm, b = 1.2 cm, air-filled (assumed), c = 3×10⁸ m/s.

f_c = (c/2) √[(m/a)² + (n/b)²]

Dominant mode (TE₁₀)

f_c(10) = c/(2a) = 3×10⁸ / (2 × 0.025) = 6×10⁹ Hz = 6 GHz

TE₁₁ mode

(1/a)² = (1/0.025)² = 1600
(1/b)² = (1/0.012)² = 6944.4
sum    = 8544.4,   √sum = 92.44 m⁻¹

f_c(11) = 1.5×10⁸ × 92.44 = 13.87×10⁹ Hz

Answer: f_c(TE₁₀) = 6 GHz; f_c(TE₁₁) = 13.87 GHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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