Difficulty: Easy
Correct Answer: High-pass filter
Explanation:
Introduction / Context:Metallic waveguides support propagation only above their cutoff frequency. Below cutoff, fields are evanescent and power does not propagate. This frequency-selective behavior mirrors a classic filter response in RF design.
Given Data / Assumptions:
Concept / Approach:
For a given mode, the propagation constant β is real for f > fc and imaginary for f < fc. Thus, the section 'passes' energy above fc and 'blocks' it below fc, which is the hallmark of a high-pass filter behavior.
Step-by-Step Solution:
Identify dominant mode with cutoff fc.For f < fc: β becomes imaginary; fields decay exponentially → no transmission.For f > fc: β is real; power propagates → transmission occurs.Therefore, the waveguide section functions as a high-pass filter.Verification / Alternative check:
Plotting S21 versus frequency for an ideal waveguide shows near-zero transmission below fc and increasing transmission above fc, matching a high-pass characteristic.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing cavity resonators (band-pass) with uniform guides; overlooking that mode conversion or discontinuities can create additional filtering not present in a straight section.
Final Answer:
High-pass filter
Discussion & Comments