- 2068 Chaitra · 2+2 marks
Cut-off of TM11 and dominant mode
Short answer
With ε_r=2, speed u = c/√2. Dominant TE10 cutoff f_c = u/(2a) = 9.91 GHz. TM11 cutoff using (u/2)√[(1/a)²+(1/b)²] yields 26.58 GHz.
Consider a rectangular waveguide with ε_r = 2, μ = μ0 with dimensions a = 1.07 cm, b = 0.43 cm. Find the cut-off frequency for TM11 mode and the dominant mode.
Given: ε_r = 2, μ = μ₀, a = 1.07 cm, b = 0.43 cm.
Speed in the filling: u = c/√ε_r = 3×10⁸/√2 = 2.121×10⁸ m/s.
f_c = (u/2) √[(m/a)² + (n/b)²], with u/2 = 1.0607×10⁸.
TM₁₁ mode
(1/a)² = (1/0.0107)² = 8734.4
(1/b)² = (1/0.0043)² = 54083.8
sum = 62818.2, √ = 250.64 m⁻¹
f_c(TM₁₁) = 1.0607×10⁸ × 250.64 = 26.58×10⁹ Hz
Dominant mode (TE₁₀)
f_c(TE₁₀) = u/(2a) = 2.121×10⁸ / (2 × 0.0107) = 9.91×10⁹ Hz
Answer: f_c(TM₁₁) = 26.58 GHz; dominant mode TE₁₀ with f_c = 9.91 GHz.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Define TM mode; TM11 cut-off 5×3 cm · Electromagnetism, chapter 6
- Cut-off of dominant and TE11, 2.5×1.2 cm · Electromagnetism, chapter 6
- Cut-off and single-mode range, 4×2 cm · Electromagnetism, chapter 6
- Dominant cut-off, dielectric-filled guide · Electromagnetism, chapter 6
- Define cutoff frequency, dominant mode · Electromagnetism, chapter 6
- Single-mode range, air-filled 2×1 cm · Electromagnetism, chapter 6
- Rectangular waveguide cut-off and phase velocity · RF and Microwave Engineering, chapter 4
- Dominant mode and 2.254 cm waveguide at 6 GHz · RF and Microwave Engineering, chapter 4
- Circular waveguide TM field strength and dominant mode · RF and Microwave Engineering, chapter 4
- Short note: Dominant mode in waveguide · RF and Microwave Engineering, chapter 4