IIRC 

 

IIRC, most of the basic principles were explained, with illustrations, in
Albert Smith’s 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 80’s (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 I’ve now got an up and running spreadsheet.
I’ve 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,

 

I’m 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 I’m 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.

 

I’ve tried Googling but maybe I’m not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

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