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
This message is from the IEEE EMC Society Product Safety
Technical Committee emc-pstc discussion list.
Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/
To cancel your subscription, send mail to:
[email protected]
with the single line:
unsubscribe emc-pstc
For help, send mail to the list administrators:
Ron Pickard: [email protected]
Dave Heald: [email protected]
For policy questions, send mail to:
Richard Nute: [email protected]
Jim Bacher: [email protected]
Archive is being moved, we will announce when it is back on-line.
All emc-pstc postings are archived and searchable on the web at:
http://www.ieeecommunities.org/emc-pstc
