Yes, you may (at your choice) add or subtract 180 degrees for the reflected 
vertically polarised wave only!

I think Brent meant to say the same, but with another viewpoint....

 

Gert

 

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Thursday 6 August 2015 11:13
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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