- 2073 Shrawan · 4 marks
First four modes, 2.5×1.2 cm guide
Short answer
Apply f_c = (c/2)√[(m/a)²+(n/b)²] with a=2.5 cm, b=1.2 cm. The first four cutoff frequencies are TE10: 6.00 GHz, TE20: 12.00 GHz, TE01: 12.50 GHz, and TE11/TM11: 13.87 GHz.
Calculate the cut-off frequencies of the first four propagating modes for an air-filled copper waveguide with dimension a = 2.5 cm, b = 1.2 cm.
Given: a = 2.5 cm, b = 1.2 cm, air-filled, c = 3×10⁸ m/s.
f_c(mn) = (c/2) √[(m/a)² + (n/b)²] = 1.5×10⁸ √[(m/0.025)² + (n/0.012)²]
TE₁₀: 1.5×10⁸ × 40 = 6.00 GHz
TE₂₀: 1.5×10⁸ × 80 = 12.00 GHz
TE₀₁: 1.5×10⁸ × 83.33 = 12.50 GHz
TE₁₁ / TM₁₁: 1.5×10⁸ × 92.44 = 13.87 GHz
(Next: TE₂₁/TM₂₁ = 17.33 GHz.)
| Order | Mode | f_c (GHz) |
|---|---|---|
| 1 | TE₁₀ (dominant) | 6.00 |
| 2 | TE₂₀ | 12.00 |
| 3 | TE₀₁ | 12.50 |
| 4 | TE₁₁ and TM₁₁ (degenerate) | 13.87 |
Answer: 6 GHz, 12 GHz, 12.5 GHz and 13.87 GHz. Because the walls are copper (a good conductor), they can be treated as perfect conductors when finding the cut-off frequencies.
✓ Answer checked by Claude Opus 5.5 (high effort)
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