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blog:2025-02-03:antenna_isolation [2025/02/03 19:44] – 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 for example, if two antennas used around 147 MHz are 3 m apart (feedline to feedline), the isolation will be about:+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} \$   * \$ \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} \$   * \$ \text{Isolation} = 28 + 40 \log(\frac{3 \text{ m}}{2.04 \text{ m}}) \approx 35 \text{ dB} \$
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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.1738640681.txt.gz · Last modified: by va7fi