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:[email protected]] Sent: Thursday, July 10, 2003 3:08 PM To: [email protected] Subject: RE: apertures >-----Original Message----- >From: [email protected] [ mailto:[email protected]] >Sent: Thursday, July 10, 2003 11:59 AM >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 > > > 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 [email protected] 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
