# Calculate attenuation constant of an aperture

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1. Apr 5, 2016

### wu_weidong

1. The problem statement, all variables and given/known data
The shielding efficiency of an aperture depends on e−αd, where α is the frequency-dependent attenuation constant of the aperture and d is the thickness of the material (or the cutoff waveguide) at frequencies below cutoff.

where ωc=2πfc.

Calculate α for an air-vent (unshielded) of dimension 0.5×0.3×0.8 m3.

The solution given is

Cutoff frequency = 300MHz (cutoff wavelength = 1 m), α = 5.92

3. The attempt at a solution
I've tried to apply the formula above, with

ωc = 2π(300×106), μ (free space) = 4π×10−7, ϵ (free space) = 8.85×10−12, f = 100×106, fc = 300×106

But I get α = 37.25, not the value given. What am I missing?

2. Apr 10, 2016