I am new to radar and curious about the geometric spreading in free space. I set up an air-coupled antenna about a half meter above the ground, with a large PEC on the ground. Then I adjusted the height of the antenna and recorded the corresponding reflected amplitude. I expected the reflected amplitude should be proportional to 1/r because there is only free space loss. However, I got 1/r^0.2 for one antenna (1Ghz, I am not sure about the type of it) and 1/r^0.5 for another one (a 2Ghz horn antenna). Am I understanding the geometric spreading wrong? Or the wave is not always spherical spreading for the different antennas? Any help will be really appreciated.

Two antennas are both ground penetrating radar. The set-up is shown as follows (PEC should be on the ground, which is missing in this photo):

enter image description here

  • \$\begingroup\$ Hint: Method of images. The operating wavelength, and particularly the ratio \$\lambda/h\$ (where \$h\$ is the height above ground) are likely to be key parameters in this problem. \$\endgroup\$ – The Photon Nov 18 '19 at 17:59
  • \$\begingroup\$ A picture (or two) is worth a thousand (or two) words. Pictures of the antennas and your setup? \$\endgroup\$ – Aaron Nov 18 '19 at 21:30
  • \$\begingroup\$ @ThePhoton Thank you for your comment. I added the central frequency of antennas I used in the question. But if it is a near-field problem, will the amplitude fall off faster, usually in 1/r^2 or even 1/r^3? Am I misunderstanding you? Thank you \$\endgroup\$ – Juliet Nov 19 '19 at 3:51
  • \$\begingroup\$ @Aaron Thanks for your reminder. I added the antenna setup here. \$\endgroup\$ – Juliet Nov 19 '19 at 3:51
  • \$\begingroup\$ Where did you get this "PEC"? I have a shielding application that could really use some of that. \$\endgroup\$ – The Photon Nov 19 '19 at 4:42

Bob Jacobsen is correct -- the most likely problem is that there is not enough distance in your setup.

Here is an overview of near and far field in radar. Figure 1 in that article provides some nice intuition on the near field and the far field:

  • In the far field, we can see that the wave fronts all align with each other in the center line of the antenna pattern, so the signal at this centerline behaves the same as a signal we'd get from a point source. In this region spherical spreading loss applies and power will decrease as \$1 \over r^2\$ (amplitude will decrease as \$1 \over r\$).
  • In the near field (very close to the antenna) we can see that things are quite a mess. To my (limited) knowledge there is no useful, simple rule about power-versus-distance in this case.

From the above webpage, we see some approximations for where the near field and far field begin. You mention frequencies of 1GHz and 2GHz, which means wavelengths of roughly 0.3m and 0.15m. I'm guessing that your antennas shown in your picture have a dimension of roughly that size (i.e. your antenna aperture is a single wavelength or less, rather than multiple wavelengths large). So, from that page, the far field begins when \$r \gg \lambda\$.

In your setup you state that you are about 0.5m above the ground, which I would say doesn't qualify as "much greater than" your wavelength here. (Your experiment was performed with a distance of 1.6 and then 3.2 wavelengths. It is interesting to note that the 3.2 wavelengths case ended up closer to the expected answer.)

If possible, I would recommend repeating your experiment with maybe 5-10 wavelengths of distance. The tough part here is that you'd need that 5-10 wavelengths of distance in all directions -- I doubt the antenna pattern is very focused so you need to be sure that you're not getting coupling through side lobes that could be corrupting your measurement results.

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  • \$\begingroup\$ Thank you for your help. I used to refer to Wiki page of the near and far field. It says the amplitude will drop faster in the near field., while in my case it drops slower. Is this possibly because in their case they assumed an omnidirectional antenna, and the waveguide in real life will affect the radiation? \$\endgroup\$ – Juliet Nov 19 '19 at 17:42
  • \$\begingroup\$ If you have a true omnidirectional antenna (i.e. a point source) then there isn't any near field at all. I would guess that the problem is just that the near field is very complex -- there are a lot of different interactions between the various wave fronts, and per the Wiki page there can be inductive effects as well. So I would say that even if there was some rule for the relationship between received power and distance in the near field, I would doubt that you could ever reasonably measure that relationship in a reliable, repeatable fashion. \$\endgroup\$ – Mr. Snrub Nov 19 '19 at 19:22

Far enough away that a source looks like a point, the amplitude drop off is like \$1/r\$ (power drops off like \$1/r^2\$).

But consider a point very close to an infinite plane source. There, the fields don’t fall off at all: sufficiently close to a planar antenna, there’s no drop off.

In general, you need to be quite few antenna-sizes away from an antenna before you reliably see the \$1/r\$ decrease in amplitude.

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  • \$\begingroup\$ Thank you for your help. I used to think the amplitude will fall off faster in near field. \$\endgroup\$ – Juliet Nov 19 '19 at 17:26

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