RE: apertures
Yes, the potential across the closer of the the two sides of a slot is saying the same thing as the potential across the longest sides. i.e. the closer of the two sides of a slot are the longest sides of that slot. And technically speaking, babinet's principle is concerned with optics. Taking the same method used in optics by babinet and applied to elctromagnetics gives the duality between slots and wires. In other words, a wire in space has the dual of a long thin slot cut into an infinite metal sheet. And the explanation as to how the ploarity of the e and h fields getting flopped was excellent. Regards, Doug McKean _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
RE: apertures
One approach to take for small apertures in solid sheets is to reverse the model. That is model the equivalent dipole, you will have far fewer elements and no meshing issues. Due to the duality between E and H known as Babinet's priciple, this is vaild method. Check out slide 8 on this link: http://www-ece2.engr.ucf.edu/~tomwu/course/eel6492/notes00/emc9.pdf and pages 50 - 54 on this one: Example 5 is just what I mean. http://faculty.uml.edu/aakyurtlu/16.571/Aperture%20Antennas.pdf (these were the first to show up on Google that showed the points I wanted to make, another search might reveal more information) cheers, Colin.. From: drcuthb...@micron.com [mailto:drcuthb...@micron.com] Sent: Wednesday, July 16, 2003 11:23 AM To: abx...@hotmail.com; emc-p...@majordomo.ieee.org Subject: RE: apertures I thought a slot is driven with a potential across the closest sides. Radiation occurs due to current at the far ends. Like two short dipoles spaced a distance (1/2 wavelength with a resonant slot) apart. So at horizontal slot has the E-field vertical and the H-field horizontal. The polarization is vertical for a horizontal slot. I have fooled around with NEC simulations but the mesh that represents the sheet becomes huge and cumbersome when the grid size is kept small (like 1/16 wavelength). I have had some success using the automatic ground plane generation feature and sticking several planes together. Dave From: C N [mailto:abx...@hotmail.com] Sent: Thursday, July 10, 2003 8:29 PM To: emc-p...@majordomo.ieee.org Subject: RE: apertures I usually have a problem with these equations. They give no reference to the angle at which the wave is impinging upon the sheet. The way slots work is to put amaximum differential between the longest sides. If the impinging wave puts a circulating current laterally across the slot, the slot will radiate at a maximum. If the impinging wave puts a circulating current longitudinally along the slot, the slot will radiate at a minimum. It's just that there's a lot of other characteristics which define how a slot radiates. Or are we assuming that everything is perfectly lined up for maximum performance? Regards, Doug McKean _ STOP MORE SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
RE: apertures
I thought a slot is driven with a potential across the closest sides. Radiation occurs due to current at the far ends. Like two short dipoles spaced a distance (1/2 wavelength with a resonant slot) apart. So at horizontal slot has the E-field vertical and the H-field horizontal. The polarization is vertical for a horizontal slot. I have fooled around with NEC simulations but the mesh that represents the sheet becomes huge and cumbersome when the grid size is kept small (like 1/16 wavelength). I have had some success using the automatic ground plane generation feature and sticking several planes together. Dave From: C N [mailto:abx...@hotmail.com] Sent: Thursday, July 10, 2003 8:29 PM To: emc-p...@majordomo.ieee.org Subject: RE: apertures I usually have a problem with these equations. They give no reference to the angle at which the wave is impinging upon the sheet. The way slots work is to put amaximum differential between the longest sides. If the impinging wave puts a circulating current laterally across the slot, the slot will radiate at a maximum. If the impinging wave puts a circulating current longitudinally along the slot, the slot will radiate at a minimum. It's just that there's a lot of other characteristics which define how a slot radiates. Or are we assuming that everything is perfectly lined up for maximum performance? Regards, Doug McKean _ STOP MORE SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
Re: apertures
On Thu, 10 Jul 2003 12:58:49 -0600, drcuthb...@micron.com 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? 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
RE: apertures
I usually have a problem with these equations. They give no reference to the angle at which the wave is impinging upon the sheet. The way slots work is to put amaximum differential between the longest sides. If the impinging wave puts a circulating current laterally across the slot, the slot will radiate at a maximum. If the impinging wave puts a circulating current longitudinally along the slot, the slot will radiate at a minimum. It's just that there's a lot of other characteristics which define how a slot radiates. Or are we assuming that everything is perfectly lined up for maximum performance? Regards, Doug McKean _ STOP MORE SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
RE: apertures
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
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
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
RE: apertures
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: 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 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: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: emc_p...@symbol.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org 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
