Single-Mode Upper Frequency for a 1 cm × 2 cm Air-Filled Rectangular Waveguide For an air-filled rectangular waveguide with internal dimensions 1 cm (height b) × 2 cm (width a), estimate the practical maximum operating frequency for single-mode (TE10 only) operation.

Introduction / Context:
To maintain single-mode operation in a rectangular waveguide, we operate above the TE10 cutoff but below the cutoff of the next higher mode (usually TE20 or TE01). Estimating the upper frequency limit ensures stable patterns and predictable impedance.

Given Data / Assumptions:

• Dimensions: a = 2 cm = 0.02 m (broad wall), b = 1 cm = 0.01 m (narrow wall).
• Air fill: wave speed c ≈ 3 * 10^8 m/s.
• Single-mode band bounded above by the lowest of fc(TE20) and fc(TE01).

Concept / Approach:

Cutoff frequency formula: fc(mn) = (c / 2) * sqrt((m / a)^2 + (n / b)^2). For TE10: fc10 = c / (2a). For TE20: fc20 = c / a. For TE01: fc01 = c / (2b). With a = 2 cm and b = 1 cm, fc20 = c / 0.02 = 15 GHz and fc01 = c / 0.02 = 15 GHz, both equal. A practical upper edge is set slightly below these, often ~95% of the next-mode cutoff, giving about 14–14.25 GHz.

Step-by-Step Solution:

Verification / Alternative check:

Standard waveguide single-mode recommendations place the upper limit a few percent below next-mode cutoff to avoid excitation due to tolerances or bends, consistent with ~14 GHz.

Why Other Options Are Wrong:

1000 MHz and 5000 MHz are well below TE10 cutoff; 140,000 MHz is far above multi-mode onset; 7500 MHz equals TE10 cutoff and is not an upper limit.

Common Pitfalls:

Confusing cutoff frequencies with operating band edges; ignoring that both TE20 and TE01 can define the upper bound depending on a and b.

Final Answer:

14000 MHz