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Horizontal Dipole - zero degrees elevation
On Sat, 20 Jun 2009 17:33:19 -0700, Roy Lewallen
wrote: snip Richard and Dave have answered your question -- the reason for the zero field at zero elevation angle EZNEC result is that the model ground is perfectly flat and infinite in extent, and the observation point is essentially at an infinite distance. This is useful for evaluating sky wave propagation at long distances, but not for line-of-sight propagation where Earth curvature can be a factor. Free space analysis is better for this. Programs which calculate reflection coefficients for various ground shapes aren't very useful at VHF and above, because local features, especially in an urban environment, cause reflections that are often much more significant than the ones calculated by the program. All you can do with these programs in a situation like yours is to get a general idea of what's happening. I think I understand now. snip but is it true that there should be no signal if everything was ideal? If by ideal you mean that the ground is perfectly flat and infinite in extent, yes. Got it snip The EZNEC plot looks more like it is showing additive and subtractive combining of the signal. That's exactly what EZNEC is doing. It calculates the sum of two "rays" -- idealized straight line radiation from the antenna. One is the "direct ray", which goes directly from the antenna to the observation point. (Actually, the fields from all segments are individually calculated and summed.) The second is the "reflected ray", which reflects from the ground plane to arrive at the same distant point. Some geometric investigation will show that the reflected ray at an elevation angle of -x degrees will add to the direct ray at an elevation angle of +x degrees at the distant point. The reflection coefficient of the ground, which is determined by the frequency, wave polarization, and ground conductivity and permittivity, is used to determine the amplitude and phase of the reflected ray. The reflection coefficient of low angle horizontally polarized waves is nearly -1 for any reasonable ground characteristics, so the low angle pattern is nearly the same for a horizontally polarized antenna over the Earth as for one over a perfectly conducting ground. If you look the the distance from the antenna to a distant point via the direct route and compare it to the distance to the same point via the reflected route, you'll see that the difference between the two routes (that is, the distances traversed by the two "rays") monotonically increases with increasing elevation angle. At zero elevation angle, the two distances are the same. Since the reflected ray undergoes a phase reversal (as Richard explained), expressed as a -1 reflection coefficient, the direct and reflected rays cancel at the distant point, resulting in zero field strength. As the elevation angle increases, the difference between the two two ray paths increases, resulting in imperfect cancellation. At some elevation angle, assuming the antenna is high enough off the ground, the difference in path lengths will be exactly a half wavelength. At this angle, the two rays will reinforce at the distant point. At a higher angle, they'll be exactly one wavelength different, and the rays will cancel again. And so forth. The path distance difference goes from zero at zero elevation angle to a maximum straight up of twice the antenna height above ground, so the number of "flower petals" you see depends on the height, in wavelengths, of the antenna above ground. I noticed that. It makes sense now. snip Hope this helps. Roy Lewallen, W7EL It did. Richards and the others helped a lot, but your explanation made the light bulb turn on ;-) It also explains why I saw a similar result with a virtical dipole. Thanks again to all who helped. 73, Pat |
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