Vent / Honeycomb Waveguide-Below-Cutoff Shielding
Compute cutoff frequency and waveguide-below-cutoff shielding for vent or honeycomb openings from their size and depth.
Worked example
Input: Frequency 1000 MHz, Opening shape Rectangular (longest side), Opening size w (longest side or diameter) 5 mm, Opening depth (tube length) t 10 mm
Result: Cutoff frequency f_c 29979.2 MHz, Shielding of one opening (absorption term) 54.5447 dB, Depth / size (t/w) 2
Result: Cutoff frequency f_c 29979.2 MHz, Shielding of one opening (absorption term) 54.5447 dB, Depth / size (t/w) 2
Formula
Rectangular: SE ≈ 27.3 · (t/w) · √(1 − (f/f_c)²), f_c = c/(2w)
Circular: SE ≈ 32.0 · (t/d) · √(1 − (f/f_c)²), f_c = 1.841c/(πd)
For f ≥ f_c this formula does not apply (shown as SE ≈ 0 dB)
How it works
A deep opening attenuates signals like a waveguide below its cutoff frequency. For a rectangle f_c = c/(2w) and SE ≈ 27.3·(t/w)·√(1 − (f/f_c)²); for a circle f_c = 1.841c/(πd) and SE ≈ 32.0·(t/d)·√(1 − (f/f_c)²). With depth twice the size and f well below f_c, a rectangle gives about 55 dB. For the example (5 mm rectangle, 10 mm deep, 1 GHz), f_c is about 29.98 GHz and SE about 54.5 dB.
Practical tipThe value covers only the absorption term of one opening and excludes reflection loss at the entrance. Many openings increase total leakage, and a thin plate (small t/w) gives little benefit, so use honeycomb filters or tubes to gain depth. The formula is most accurate when f is well below f_c; for seams and slots use the aperture shielding calculator.
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Results and summaries are for reference. For certification and test reports use the latest official standard text and calibrated instrument data. Last updated: 2026-10-10