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Calculate attenuation constant of an aperture

  1. Apr 5, 2016 #1
    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.

    2mxj4fq.png
    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. jcsd
  3. Apr 10, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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