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I understand why the effective area of an antenna is related to the wavelength. I'm aware of the below equation:

$$G = \frac{4\pi A_{eff}}{\lambda^2}$$

However this suggests to me that one of either gain or effective aperture should remain fixed. I believe that both gain and effective aperture vary with wavelength, so this seems to suggest this equation is incorrect (or at least oversimplified)? Can anyone help with my understanding?

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    \$\begingroup\$ The equation is right and I can't understand why it should do what you say it suggests to you. \$\endgroup\$
    – Andy aka
    Commented May 5, 2023 at 15:44
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    \$\begingroup\$ You'll get different answers for what "depends on wavelength" depending on whether that means you change the operating frequency while leaving the antenna the same, change the antenna dimensions while leaving the frequency the same, or change both in proportion. But that equation is correct in any case, if you carry out the computation with the correct values. \$\endgroup\$
    – hobbs
    Commented May 5, 2023 at 15:51
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    \$\begingroup\$ I don't understand the vote to close. The question is valid, just shows some misunderstanding on the part of the poster. \$\endgroup\$
    – SteveSh
    Commented May 5, 2023 at 16:52
  • \$\begingroup\$ Thank you to everyone for their comments, it has helped my understanding. I guess at this stage what I'm really interested in would be what a graph of both Aeff and Gain against wavelength looks like for a single antenna \$\endgroup\$
    – Christian
    Commented May 6, 2023 at 9:13

2 Answers 2

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Welcome to the site, Christian!

The initial assumption in your question, saying you "understand why the effective area of an antenna is related to the wavelength", is not strictly correct. For aperture-style antennas that are many wavelengths in size (e.g. parabolic dishes, horns) the effective aperture is roughly equal to the physical aperture size, and doesn't vary with frequency or wavelength. With that understanding, the gain of a given antenna being inversely proportional to wavelength should make sense.

I believe the source of your confusion about aperture comes at the small antenna end of the scale. For example, dipoles made of thin metallic elements have an effective aperture that is much greater than their actual physical area (if they are carefully tuned and matched). For small antennas, their theoretical or measured directivity or gain should be used as the basis for calculations.

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  • \$\begingroup\$ With small antennas == small relative to the wavelength of interest. \$\endgroup\$
    – SteveSh
    Commented May 5, 2023 at 16:49
  • \$\begingroup\$ Thank you for your answer mark. Based on your answer, I'm taking it that it is the gain that varies more for large antennas (relative to wavelength), and the effective aperture varies more for antennas that are small in regards to wavelength. Is this correct? \$\endgroup\$
    – Christian
    Commented May 5, 2023 at 18:31
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This formula is a general antenna law. It can be used to calculate G from Ae at certain wavelength. It assumes that Ae is known just at the same wavelength which is in the denominator of the formula.

The same antenna behaves differently at different wavelengths. If you for ex. have an antenna which has Ae = 1 square meter at 10 cm wavelength, you can use this formula and that 1 sq.meter only at 10 cm wavelength to calculate G. This formula is useless for estimating G for ex. at 20 cm wavelength if you know only Ae at 10 cm.

Maybe useful: The formula is used in radio and radar theory. It's needed in deriving formulas for how much a radar receiver or a radio receiver might theoretically get signal power from the antenna. People want to have in these formulas antenna gain G, not Ae. Ae is used to calculate how much power an antenna catches from certain rf power density (watts per square meter), but in the final formula Ae is presented with G by reversing your G vs. Ae formula.

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  • \$\begingroup\$ "The same antenna behaves differently at different wavelengths." Yes, but the question is by how much? You can design antenna elements that have fairly uniform performance over a 10:1 or even 20:1 frequency range. If you then make a large array of those elements, then the effective aperture is the same across that frequency range, and so the (aperture) gain of the array is given by the the equation in OP's question. \$\endgroup\$
    – SteveSh
    Commented May 5, 2023 at 23:45
  • \$\begingroup\$ So building on your answer, if you were to plot Ae vs wavelength, you are saying there would be no trend that described that graph, such that you could infer the Ae at a wavelength for which you have no data? \$\endgroup\$
    – Christian
    Commented May 6, 2023 at 9:21

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