antenna-theory.com

[requiem]

Good day everyone,

I'm having difficulties with one of my projects, main reason - lack of experience and knowledge... I'm still working on it.

I have to install channel selective GSM900 repeater on watertower (20m high), but I suspect there will be problem with antenna isolation, because the distance between the antennas is about 2.5 meters. Thing that comes in handy is that the tower is built of metal, which will help with antenna isolation problem, maybe add some 15 dB.
In order to operate repeater at its maximum gain (90 dB) isolation requirements are Gain + 15 dB = 105 dB.
For service and donor antenna I'm planning to use Kathrein 730 376 Vpol 900MHz Antennas.

I have tried to google antenna isolation formulas, but all formulas I have found are related to antennas with Isotropic or Omni directional radiation patterns.

Antennas will be placed with horizontal separation about 2.5 meters far from each other, no vertical separation. Angle between the antennas is about 97 degrees.

Planned antenna placement
Antenna pattern

However - maybe I should use X polar instead of V polar antennas - that way the polarisation difference would be 90 degrees (then I will use only one of two RF cable connectors for each antenna), that should improve isolation.

Can anyone help me with the answer - how can I achieve 105 dB isolation?

[Schubert]

105 dB sounds like a very ambitious isolation target.

Here's what influences isolation when you're about 7.5 wavelengths away (as in your case):

1) Distance
2) Direction of antenna(s) peak radiation relative to each other
3) Polarization

Now, it sounds like your distance is fixed. The first thing to try would be to make one of your antennas horizontally polarized. This would give you polarization isolation which can be significant if you have somewhat pure linear polarization coming off each antenna.

Other than that you can experiment with shielding - placing metal between the antennas to redirect the energy.

There's no simple formula in this case.