Here is another way to view an aperture, referenced to a resonant 1/2
wavelength dipole. On one side of a very small aperture, in a large conductive
sheet, there is a dipole. It is connected to an identical dipole on the other
side of the aperture. The energy intercepted by the first dipole is
re-radiated by the second dipole. This represents the case of a 1/2 wavelength
aperture. A smaller aperture is down by 20LOG(D/d) where d is the length of
the small aperture and D is 1/2 wavelength.
Dave Cuthbert
Micron Technology
> -----Original Message-----
> From: drcuthbert
> Sent: Thursday, July 10, 2003 12:59 PM
> To: [email protected]
> Subject: apertures
>
> 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?
>
> Dave Cuthbert
> Micron Technology
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