IIRC
IIRC, most of the basic principles were explained, with illustrations, in Albert Smiths early IEE papers published in the 1980s. I only had a paper copy of those when I started experimenting with his swept-frequency site calibration methods in the late 80s (at HP) and I assume that the principles have been expounded and expanded many times over the years so can anyone point to some decent downloadable documents which illustrate the issues? John Allen W.London, UK From: Pawson, James [mailto:james.paw...@echostar.com] Sent: 10 August 2015 09:56 To: EMC-PSTC@LISTSERV.IEEE.ORG Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC Very helpful, thanks Brent! From: Brent DeWitt [mailto:bdew...@ix.netcom.com] Sent: 08 August 2015 01:19 To: Pawson, James; EMC-PSTC@LISTSERV.IEEE.ORG Subject: RE: [PSES] Calculating Reflection Angles on OATS/SAC Hi James, The image concept again is useful. By definition, the ground reference plane is at zero potential. For that to be true, charges on the real antenna and its image must be equal and opposite. Put a plus on one end of a dipole and a minus on the other and look at them. If they are vertical, and the bottom of the real dipole has the minus sign, the top of the image must be plus for the charges to cancel. For the horizontal example, if the left end is plus the same end of the image must be minus for the same reason. In the extreme thought experiment, if you lowered the vertical dipole so its center point were at the ground plane (now a monopole), its image would complete the dipole. The same extreme applied to the horizontal dipole would have the two cancelling each other out entirely. We can see this in reality, since the vertical polarization with the antenna at one meter height is usually the strongest emission at low frequencies where the path length in wavelengths is small. The first maximum from the horizontal dipole occurs when there is a 180 degree path length difference between the real antenna and its image. Does that help any? Brent From: Pawson, James [mailto:james.paw...@echostar.com] Sent: Thursday, August 06, 2015 5:13 AM To: EMC-PSTC@LISTSERV.IEEE.ORG Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC Many thanks for all of the replies on this topic. The conceptual key I lacked was the image of the receiver below the ground plane which made the calculations a lot simpler and Ive now got an up and running spreadsheet. Ive also been introduced to things like cotangents and arctangents which are new to me. The only thing I still remain confused about is the phase of the reflection from the ground plane. Gert wrote: Note that vertical waves invert in polarity on reflection with the ground plane, where horizontal polarized waves do not. Brent wrote: and take the difference for phase, remembering that the horizontally polarized image is 180 degrees out of phase to start with while the vertical image is in phase. I might be misunderstanding but these statements seem to contradict each other. I can kind of see how a vertically polarised wave would be reflected inverted. If this was the case, could this be compensated for by subtracting 180° from the reflected ground ray to ensure the phases added/subtracted correctly at the RX antenna? Thanks again James _____________________________________________ From: Pawson, James Sent: 31 July 2015 15:59 To: EMC-PSTC@LISTSERV.IEEE.ORG Subject: Calculating Reflection Angles on OATS/SAC Hi, Im trying to calculate the distances/angles at which a maximum (in phase) or minimum (anti-phase) signal would occur on an OATS/SAC. I can do this simply when the TX and RX antennae are the same height above the reflecting surface as the point of reflection lies halfway between the two antennae, Distance_tx = Distance_rx. The direct and reflected paths can be calculated using simple geometry and the wavelength is given by lambda = c / f. However when the height of the RX antenna is different to the height of the TX antenna then the horizontal distance to the reflection point is no longer equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx / Distance_rx remains the same because the angle of reflection is the same. But Im left with two unknown Distance terms in the equation. Is there a standard equation for calculating the reflection angle on an OATS/SAC with a varying height antenna? Or can someone give me some pointers to help me figure it out myself? I was so distracted thinking about this that I missed my turnoff whilst cycling home the other day. Ive tried Googling but maybe Im not putting in the right search term. Any assistance gratefully received. Thanks and regards, James - ---------------------------------------------------------------- This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. 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