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Antenna Gain vs Effective Area

  1. Oct 31, 2004 #1
    I have been looking for info on the physical basis for this antenna equation :

    Gain = 4 * Pi * (Effective Area ) / Wavelength^2

    I have found some tutorials that say it is based on reciprocity, and more specifically, on thermodynamic balance between a resistive load connected to an antenna, and a black body "target" surface at the same temperature as the load.

    Briefly, it seems that the wavelength^2 term comes in because the power radiated towards the antenna from the black surface contains this factor (a la the Rayleigh-Jeans eqn).

    My question is, what happens to the above equation as we move towards visible wavelengths? If thermodynamic balance is to be preserved, then the gain vs effective area equation would have to follow the Planck equation. Does this mean the radiation pattern would change in a different way from the classically expected one?

    I ask this because I have found that many discussions on optical communications assume that the classical antenna equations can be used to find the antenna gain.
  2. jcsd
  3. Feb 19, 2005 #2
    Hi! I also posted a query on this topic after you. Unfortunately there were no responses. However I have done some reading of my own and have come to a an (imperfect) explanation of the wavelength dependence of the effective area of an antenna. If you are still interested in the topic, please let me know. Then I will post my views on the issue. And I am also very interested in your explanation invoking the Rayleigh Jeans formula.
  4. Feb 26, 2005 #3
    Last edited by a moderator: Apr 21, 2017
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