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?
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A15 mm
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B10 mm
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C9 mm
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D7.5 mm
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E6 mm
Answer
Correct Answer: 7.5 mm
Explanation
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:
1) Compute fc from f0 = 1.5 * fc ⇒ fc = f0 / 1.5 = 30 GHz / 1.5 = 20 GHz.2) Use fc = c / (2a) ⇒ a = c / (2 * fc).3) Substitute: a = (3 * 10^8) / (2 * 20 * 10^9) = 3e8 / 4e10 = 7.5e−3 m.4) Convert to millimeters: a = 7.5 mm.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