A TE10 rectangular waveguide must operate from 25 to 35 GHz. If the band center is 1.5 times the cutoff frequency, what is the required broad dimension a?

Introduction:Rectangular waveguides are band-pass structures whose lower edge is set by the TE10 cutoff frequency fc = c / (2a). Given an operating band and a design rule tying the band center to cutoff, we can determine the broad wall dimension a directly.

Given Data / Assumptions:

• Operating band: 25–35 GHz → center f0 = (25 + 35) / 2 = 30 GHz.
• Design rule: f0 = 1.5 * fc.
• TE10 cutoff: fc = c / (2a), where c ≈ 3 * 10^8 m/s.

Concept / Approach:

Use the center-to-cutoff relationship to find fc, then solve for a from the TE10 cutoff formula. Ensuring f0 exceeds fc by a safe factor guarantees single-mode operation over most of the band while avoiding excessive proximity to cutoff.

Step-by-Step Solution:

Verification / Alternative check:

A = 7.5 mm corresponds to an fc of 20 GHz; operating between 25–35 GHz keeps the guide well above cutoff, consistent with low dispersion near the middle of the band.

Why Other Options Are Wrong:

• 10–15 mm: would lower fc below 20 GHz, not meeting the 1.5× rule.
• 9 mm: gives fc ≈ 16.7 GHz, again too low given the specified relation.
• 6 mm: fc ≈ 25 GHz, making the low edge equal to cutoff, which violates the given design constraint.

Common Pitfalls:

Using free-space wavelength at band edges instead of the TE10 cutoff formula; cutoff depends only on a for the dominant mode.

Final Answer:

7.5 mm