On Thu, 10 Jul 2003 12:58:49 -0600, [email protected] wrote:
>I have a question on apertures. You may recall the formula that is frequently
given for signal attenuation through a small aperture in a large conductive
sheet. It is 20LOG(I/2L), where I is the wavelength and L is the slot length.
For example, if x is 1/2-wavelength then the attenuation is 0 dB. But I'm not
100% sure what the attenuation is referenced to. If they are referencing it to
the E-field that would be present at the aperture location if the sheet were
not there to the E-field across the length of the aperture then that makes
sense. It seems that we now have a 1/2 wavelength aperture radiating only the
signal energy that it has intercepted.
>
>Let's say it is referenced to the E-field that would be present with no
sheet. Now to say that the E-field a large distance away from the 1/2
wavelength aperture has not been attenuated by the aperture is wrong, although
this is implied by the formula. Only a fraction of the energy contained in the
total incident wave has made it through the aperture. The aperture now acts as
a dipole radiating this fraction of the total incident wave.
>
>So is the attenuation given by this formula to be referenced to the power
that would be intercepted by a dipole? 

  I'm not experienced in radiation issues, so I might be off-base.

  I remember this subject being discussed in Henry Ott's book, "Noise
Reduction Techniques in Electronic Systems".  On page 182, he starts
talking about shielding with magnetic metals, and about apertures
specifically on page 187.
  It looks like the subject concerns magnetic currents induced in the
metal radiating at the point of discontinuity (the slot).

  Is this the same situation as measuring shielding effectiveness with
two antennas?



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