It is commonly known that greater antenna heights usually improve the performance of these point-point terrestrial paths. Here's a NEC study showing the relationship of receive antenna height AGL to its output signal.

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    $\begingroup$ "Here's a NEC study showing the relationship of receive antenna height AGL to its output signal." --> where? $\endgroup$ Jul 24 '19 at 9:45
  • $\begingroup$ It is included In Answer 1 (below). $\endgroup$ Jul 24 '19 at 11:27

Below is a graphic showing that an approximately linear relationship applies. That is, doubling the receive antenna height above ground ~doubles the r-f voltage existing across the output terminals of the receive antenna (other things equal).

Elevating the receive antenna sufficiently may provide more improvement than that, if it reduces the propagation losses resulting from obstructed paths such as possible from nearby buildings, hills, etc.

enter image description here

  • $\begingroup$ Did you simulate this yourself, or are you quoting someone else? $\endgroup$
    – rclocher3
    Jul 22 '19 at 14:29
  • $\begingroup$ I did the study myself using 4nec2 (with a NEC4.2 engine). I will post the .nec file for it if anyone is interested. Right now I'm wondering why this Q&A deserved the loss of 2 rep points, in someone's opinion. $\endgroup$ Jul 22 '19 at 16:40
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    $\begingroup$ I'd guess because the "question" isn't really a question. Also I suspect the answer is overgeneralized. $\endgroup$ Jul 22 '19 at 17:10
  • $\begingroup$ What ground type did you use, and what values? I don't think NEC can account for earth curvature. It looks like your graph shows the expected field extinction at a S-N ground. $\endgroup$
    – tomnexus
    Jul 23 '19 at 2:44
  • $\begingroup$ The ground plane in this S-N study was set to 5 mS/m, d.c. 12. NEC assumes a flat ground plane, but for the 30-mile path length I chose, the resulting error is insignificant for the purpose of the study. A space-wave e-m field has a very low relative value at zero elevation above the ground plane for all path lengths, because of the loss of Fresnel clearance. Thanks for asking for clarification before coming to, and posting a conclusion. $\endgroup$ Jul 23 '19 at 12:53

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