- 2076 Asoj · 3 marks
Single-mode range, air-filled 2×1 cm
Short answer
TE_10 cut-off: f_c = c/(2a) = 3×10⁸/(2×0.02) = 7.5 GHz. Next higher modes (TE_20, TE_01) start at 15 GHz. Single-mode operation range: 7.5 GHz < f < 15 GHz.
An air-filled rectangular waveguide has dimensions a = 2 cm and b = 1 cm. Determine the range of frequencies over which the guide will operate in single mode (TE10).
Given: a = 2 cm, b = 1 cm, air-filled.
f_c(TE₁₀) = c/(2a) = 3×10⁸ / (2 × 0.02) = 7.5 GHz
f_c(TE₂₀) = c/a = 15 GHz
f_c(TE₀₁) = c/(2b) = 3×10⁸ / (2 × 0.01) = 15 GHz
f_c(TE₁₁, TM₁₁) = (c/2)√[(1/0.02)² + (1/0.01)²] = 1.5×10⁸ × √12500 = 16.77 GHz
The next higher modes after TE₁₀ start at 15 GHz (TE₂₀ and TE₀₁, equal because a = 2b).
Only the dominant TE₁₀ mode propagates for
7.5 GHz < f < 15 GHz
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Cut-off and single-mode range, 4×2 cm · Electromagnetism, chapter 6
- Dominant cut-off, dielectric-filled guide · Electromagnetism, chapter 6
- Define TM mode; TM11 cut-off 5×3 cm · Electromagnetism, chapter 6
- First four modes, 2.5×1.2 cm guide · Electromagnetism, chapter 6
- Why no TEM mode; first four modes · Electromagnetism, chapter 6
- Air-filled 3 cm waveguide at 6 GHz; coupler parameters · RF and Microwave Engineering, chapter 4
- Dominant mode and 2.254 cm waveguide at 6 GHz · RF and Microwave Engineering, chapter 4
- Rectangular waveguide TM mode field equations · RF and Microwave Engineering, chapter 4 · asked 11×
- Rectangular waveguide TE mode field equations · RF and Microwave Engineering, chapter 4 · asked 11×