In a rectangular waveguide with cutoff frequency f_c for a given mode, how does attenuation behave relative to the operating frequency f?

Introduction / Context:
Waveguides support propagation only above a cutoff frequency f_c that depends on the mode and cross-section. Below cutoff, fields are evanescent and decay rapidly with distance.

Given Data / Assumptions:

• Uniform rectangular waveguide.
• Single operating mode considered with cutoff f_c.
• Low-loss metallic walls (typical microwave practice).

Concept / Approach:

For f < f_c, the propagation constant becomes imaginary and fields attenuate exponentially. For f > f_c, the mode propagates with finite attenuation primarily due to conductor and dielectric losses, which are relatively small.

Step-by-Step Solution:

Verification / Alternative check:

Measurement with a network analyzer shows negligible transmission below cutoff, while above cutoff S21 rises sharply and only modestly degrades with wall losses.

Why Other Options Are Wrong:

• Lowest at f = f_c: At cutoff, the guide is at the edge of propagation; attenuation tends to be large.
• Lowest when f < f_c: Below cutoff there is no propagation.
• Independent of f: Contradicts the defining property of cutoff.
• Always the same: Not true in dispersive, lossy structures.

Common Pitfalls:

Assuming modest leakage exists below cutoff; confusing conductor loss trends with cutoff behavior.

Final Answer:

Attenuation is low when f > f_c and very high when f < f_c