Hi,
Another way of thinking about this is to consider the effective aperture of
the slot antenna on the source side of the shield this will more easily give
the amount of energy coupled into the antenna. However, remember that the
shielding effectiveness equation is specific to planes wave sources and uses
the plane wave or far field from the shield as the measurement the transmitted
field.
Your comment about wanting to increase energy transmission implies that you
need to influence the field around the aperture. There are ways of doing this
by adding features to a shield that will degrade the shielding by 6 or more dB
depending upon what frequency and bandwidth you need, however the basic
shielding equations are not applicable as they make far field (plane wave)
assumptions which do not apply. If you would like to send me an email off the
list with more details I'll give you some means of achieving this.
Colin..
From: drcuthb...@micron.com [mailto:drcuthb...@micron.com]
Sent: Thursday, July 10, 2003 7:03 PM
To: ed.pr...@cubic.com; emc-p...@majordomo.ieee.org
Subject: RE: apertures
Ed,
thanks you did give me the answer I was looking for. I think you are right on
the radiation from the aperture. It should have a dipole pattern and
illuminate only half the hemisphere. I did not include this in the
calculations and will add the extra 3 dB. I am investigating this because I
actually want to increase the signal that passes through an aperture. I will
be interested to see what others tell us.
Dave
From: Price, Ed [mailto:ed.pr...@cubic.com]
Sent: Thursday, July 10, 2003 3:08 PM
To: emc-p...@majordomo.ieee.org
Subject: RE: apertures
>-----Original Message-----
>From: drcuthb...@micron.com [ mailto:drcuthb...@micron.com]
>Sent: Thursday, July 10, 2003 11:59 AM
>To: emc-p...@majordomo.ieee.org
>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
>
>
>
Dave:
Allow me to follow the power model. If the aperture has a long dimension of
1/2 wavelength, then the RF power illuminating the source side of the aperture
will propagate through the aperture with very little loss.
The total power propagating through the aperture is dependent on the area of
the aperture, as the aperture allows through all of the power that the
illuminating plane wave presents to the aperture area. For instance, if the
plane wave had a power density of 1 mW/sq cm, and the aperture had a 1 sq cm
area, then 1 mW would be propagating through the aperture, and that 1 mW would
then radiate out the far side of the aperture.
Now here the model gets a little foggy to me. Should I consider this 1 mW to
now be an isotropic radiator? I don't think so, because the barrier (that
contains the aperture) would block half the radiation. Indeed, the reflection
off the barrier would look like gain over isotropic. Should I now model the 1
mW as applied to a dipole (the end of the 1/2 wave aperture)?
Despite my floundering at the relaunching of the power that got through the
aperture, at least I can now imagine this power propagating out in a
hemispherical wavefront, spreading its 1 mW over greater and greater areas.
Hmmm, did I answer anything along the way?
Regards,
Ed
Ed Price
ed.pr...@cubic.com WB6WSN
NARTE Certified EMC Engineer & Technician
Electromagnetic Compatibility Lab
Cubic Defense Systems
San Diego, CA USA
858-505-2780 (Voice)
858-505-1583 (Fax)
Military & Avionics EMC Is Our Specialty
