Rectangular waveguide (dielectric filled): εr = 4, inside dimensions a = 3.0 cm, b = 1.2 cm. Find the cut-off frequency for the dominant TE10 mode.

Introduction / Context:The dominant mode in a rectangular waveguide is TE10. Its cut-off frequency depends on the broader dimension a and the medium inside the guide. Knowing the cut-off lets you determine operating bands and dispersion characteristics for guided waves.

Given Data / Assumptions:

• Rectangular waveguide dimensions: a = 3.0 cm = 0.03 m, b = 1.2 cm (b does not affect TE10 cut-off).
• Filled with dielectric of relative permittivity εr = 4 (assume μr ≈ 1).
• Speed of light c ≈ 3 × 10^8 m/s.

Concept / Approach:

For TE10 mode, cut-off frequency fc = c / (2 a √εr). Dielectric filling reduces phase velocity and lowers the cut-off compared with an air-filled guide by factor √εr.

Step-by-Step Solution:

Verification / Alternative check:

If the guide were air-filled (εr = 1), fc would be 3 × 10^8 / (2 × 0.03) ≈ 5 GHz. Filling with εr = 4 halves fc to 2.5 GHz, matching the computed result.

Why Other Options Are Wrong:

• 5 GHz: correct for air-filled, not εr = 4.
• 10 GHz and 12.5 GHz: far above the correct fc; would require much smaller 'a' or different medium.
• 1.25 GHz: would imply εr ≈ 16, not 4.

Common Pitfalls:

• Using b instead of a for the TE10 cut-off.
• Forgetting the √εr factor when the guide is dielectric filled.

Final Answer:

2.5 GHz