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blog:2025-02-03:antenna_isolation [2025/02/03 19:35] – created va7fiblog:2025-02-03:antenna_isolation [2025/02/03 19:51] (current) – va7fi
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 Suppose you want to mount two (vertically polarized) antennas for two different radios that will use similar frequencies.  How far apart should you mount them so that the radios don't interfere with each other? Suppose you want to mount two (vertically polarized) antennas for two different radios that will use similar frequencies.  How far apart should you mount them so that the radios don't interfere with each other?
  
-The radiation pattern of dipole looks like a doughnut, so it makes sense that for a vertically polarized antenna, the "quietest" spots would be right above and below.+The radiation pattern of a dipole looks like a doughnut, so it makes sense that for a vertically polarized antenna, the "quietest" spots would be right above and below.
  
 The isolation formula for vertical separation is((See: https://calc.commscope.com/qvisolation.aspx))((and: https://ijarcce.com/upload/2013/july/66-o-gechug2002-antenna%20isolation%20technique%20for%20interference.pdf)): The isolation formula for vertical separation is((See: https://calc.commscope.com/qvisolation.aspx))((and: https://ijarcce.com/upload/2013/july/66-o-gechug2002-antenna%20isolation%20technique%20for%20interference.pdf)):
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 </WRAP> </WRAP>
  
-So from this, we see that if we wanted 70 dB of isolation, we'd need the two antennas separated by 11 wavelengths.  At 147 MHz, that would be: \$11 \times \frac{300}{147} \approx 22 \text{ m}\$.+So for example, if two antennas used around 147 MHz are 3 m apart (feedpoint to feedpoint), the isolation will be about: 
 +  * \$ \lambda = \frac{300}{147} = 2.04 \text{ m} \$ 
 +  * \$ \text{Isolation} = 28 + 40 \log(\frac{3 \text{ m}}{2.04 \text{ m}}) \approx 35 \text{ dB} \$ 
 + 
 +And if we wanted 70 dB of isolation, from the table, we see that we'd need the two antennas separated by 11 wavelengths, which at 147 MHz would be: \$11 \times \frac{300}{147} \approx 22 \text{ m}\$.
  
 The first thing to note is that there's a sort of "diminishing returns" thing going on: the greater the separation, the greater the isolation.  But each additional meter of separation doesn't give the same isolation.  But there's a neat trick to it.  Here are a few specially chosen numbers from the table again. See if you can spot a pattern: The first thing to note is that there's a sort of "diminishing returns" thing going on: the greater the separation, the greater the isolation.  But each additional meter of separation doesn't give the same isolation.  But there's a neat trick to it.  Here are a few specially chosen numbers from the table again. See if you can spot a pattern:
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 \$$ \text{Isolation} =28 + 40 \cdot \log \left(\frac{d}{\lambda}\right) \qquad \text{or} \qquad  \text{Isolation} = 28 + 12 \cdot \log_2 \left(\frac{d}{\lambda}\right) \$$ \$$ \text{Isolation} =28 + 40 \cdot \log \left(\frac{d}{\lambda}\right) \qquad \text{or} \qquad  \text{Isolation} = 28 + 12 \cdot \log_2 \left(\frac{d}{\lambda}\right) \$$
  
-They both say: when the separation is one wavelength, the isolation will be 28.+They both say: when the separation is one wavelength, the isolation will be 28 dB.
   * But then the first one says: every time you multiply that separation by 10 (1, 10, 100, 1000, etc), you'll add 40 dB of isolation.   * But then the first one says: every time you multiply that separation by 10 (1, 10, 100, 1000, etc), you'll add 40 dB of isolation.
   * And the second one says: every time you multiply that separation by 2 (1, 2, 4, 8, 16, etc), you'll add 12 dB of isolation.   * And the second one says: every time you multiply that separation by 2 (1, 2, 4, 8, 16, etc), you'll add 12 dB of isolation.
blog/2025-02-03/antenna_isolation.1738640110.txt.gz · Last modified: 2025/02/03 19:35 by va7fi