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## Homework Statement

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 ω

_{c}=2πf

_{c}.

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

^{3}.

The solution given is

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

## The Attempt at a Solution

I've tried to apply the formula above, with

ω

_{c}= 2π(300×10

^{6}), μ (free space) = 4π×10

^{−7}, ϵ (free space) = 8.85×10

^{−12}, f = 100×10

^{6}, f

_{c}= 300×10

^{6}

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