I have a lowcostmobile Rad5G-Ultra that I've been used for years with my LTE router.

Now I have to use this couple of antennas for 5G through a ZTE MC801a (3500 MHz).

Please, can you tell me the optimal distance between the two antennas? I asked lowcostmobile and they told me 25 cm. I knew instead it had to be two times the wavelength. Could anyone provide e definitive answer eventually with a link to the math involved? Lowcostmobile Rad5G-Ultra antenna data

I don't know if it matters but the BTS I have to use is at 2.7 km.

  • $\begingroup$ Genuinely curious re "I knew instead it had to be two times the wavelength." Why did you bother asking them? $\endgroup$
    – webmarc
    Commented Oct 26, 2023 at 13:41

1 Answer 1


25 cm seems more than reasonable. The two linearly polarizing antennas are orthogonal, so by simple arguments of geometry, their far fields are orthogonal, and 25cm is sufficiently above a wavelength, so that we don't have to consider near field reactive coupling.

The exact formula why orthogonal fields do not couple is actually vector space math, so I'd go find a text book for higher math for engineers, first semester. That answers why if you believe me that from orthogonal geometry yield orthogonal field components, and from orthogonal field follows no interaction physically, you're all set.

If you want in on how the physics care about vector spaces and orthogonality, then get the text book from the same uni that goes into static fields and waves. Finally, you would want to arrive at Maxwell's equations and the bath needed for basics of finite element modeling to be able to simulate antennas, but you'd need to go through the math basics first. For an antenna as complex as this, you need to understand what you simulate, and simulate, to figure out what the near field is, as there's no closed form formula for complex geometric systems that you could reasonably calculate by hand.

It's also the number that the company that designed and simulated the antenna gave you - so I'd honestly just trust them on that. If the computer said "strength of the near field at the reflecting end at more than 12.5cm negligible", it is.

I knew instead it had to be two times the wavelength

That sounds like one of these rules of thumbs that are very rarely accurate. Never heard of that for such antenna constellations, so unless that $2\lambda$-rule comes with a very solid explanation and well-reasoned boundaries to what it applies to, I'd ignore that piece of "knowledge".

  • $\begingroup$ Thank you man: ) $\endgroup$
    – Netsurfer
    Commented Oct 25, 2023 at 4:22
  • $\begingroup$ A larger separation would be wanted if the two antennas had the same polarization and you wanted to get increased gain (smaller vertical beamwidth) by stacking them. In that case 2λ could be right (but you would have to model it, it depends on the pattern of the individual antennas). $\endgroup$ Commented Oct 25, 2023 at 21:06
  • $\begingroup$ @hobbs-KC2G to be fair, a wave that's 45° offset to the polarization of one antenna will be offset by -45° to the other, and vice versa. For these, you do get a directive gain. However, the reason to install two orthogonally polarized antennas to begin with is not the directive gain your array antenna gets, but the diversity gain that your MIMO system gets :) The reason is that these two polarizations should see different multipath channels. Or, if the base station actually uses cross-polarized antennas itself, you directly get two independent channels in a true line-of-sight-only scenario $\endgroup$ Commented Oct 25, 2023 at 22:39
  • $\begingroup$ The more relevant problem is that these bands are 500 MHz (n78) to 900 MHz (n77) wide – and suddenly your extreme wavelengths are 15% different, and that can quite drastically change your pattern if you're optimizing for 2λ of the lower frequency, then the antenna distance is 2.6λ at the upper frequency, and that puts pretty ugly sidelobes and nulls where you don't want them. If you build a directive system, you need to either be smarter than have constant physical vertical phase center offset, or you need to correct the phase in the signal processing chain – which is exactly what happens in $\endgroup$ Commented Oct 25, 2023 at 22:56
  • $\begingroup$ … MIMO processing anyway. So, either, this 2λ rule is dangerous, because it acts as if the band was narrow relative to the center frequency (and 500 MHz is not "much much less" than 3500 MHz), or it's a bit useless, because the phase of the emitted signal isn't fixed, but adjusted by the two transceivers anyways, and all you want is "sufficient decorrelation", not a specific distance. $\endgroup$ Commented Oct 25, 2023 at 22:58

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