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I'm trying to design a low frequency dipole interferometer antenna array. I am realizing how little I know about radio antennae. I'm confused by the concept of effective area $A_e$ and how it applies to arrays. I haven't been able to get a clear answer on that.

Let's say I'm using my array to receive 10MHz signals. That's $\lambda = 30\,\text{m}$, so is my effective area $A_e \approx \lambda^2/4\pi \approx 72\,\text{m}^2$? Is that for each dipole antenna? How does effective area change when I combine them?

What factors do I need (e.g. dipole length, spacing, baseline) to calculate the effective collecting area for the array as a whole?

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  • $\begingroup$ Krauss' Antennas doesn't show an equation for $A_e$ of this form. Please cite your reference. $\endgroup$
    – Brian K1LI
    Apr 15, 2019 at 14:16
  • $\begingroup$ Here's one: cv.nrao.edu/course/astr534/AntennaTheory.html $\endgroup$
    – Spuds
    Apr 15, 2019 at 14:18
  • $\begingroup$ The citation says, "Any antenna, from a short dipole to the 100-m diameter Green Bank Telescope, has the same average collecting area $A_e$ that depends only on wavelength." This seems counterintuituve. Krauss describes effective aperture, $A_e$, as the ratio between the power delivered to a load and the power incident on the antenna, which seems more intuitive. $\endgroup$
    – Brian K1LI
    Apr 15, 2019 at 15:19
  • $\begingroup$ Correction: Krauss writes that the maximum effective aperture, $A_{em}=\frac{\lambda^2}{\Omega_A}$, which comports with your citation. Krauss goes on, though, to detail effective apertures for several different antennas, which do depend on the antenna's physical properties. $\endgroup$
    – Brian K1LI
    Apr 15, 2019 at 15:32

2 Answers 2

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For some context, the relevant part of the cited source:

The average collecting area is defined by

$$ \langle A_e \rangle \equiv {\int_{4\pi} A_e d \over \int_{4\pi}d}.$$

The effective collecting area of a receiving antenna is independent of its radiation environment, so this result applies for any type of radiation, not just blackbody radiation. Without using Maxwell's equations we have obtained the remarkable result true for all lossless antennas:

$$ \langle A_e \rangle = {\lambda^2 \over 4 \pi} \tag{3A6} $$

  • Any antenna, from a short dipole to the 100-m diameter Green Bank Telescope, has the same average collecting area $\langle A_e \rangle$ that depends only on wavelength.

In the case of an isotropic antenna, the effective collecting area in any direction equals the average collecting area:

$$ A_e(\theta, \phi) = \langle A_e \rangle = {\lambda^2 \over 4 \pi} $$

The description follows after a description about thermodynamic equilibrium and this illustration:

enter image description here

A cavity in thermodynamic equilibrium at temperature $T$ containing a resistor $R$ is coupled to an antenna, also at temperature $T$, through a filter passing frequencies in the range $\nu$ to $\nu+d \nu$.

The discussion involves the thermodynamic equilibrium between the resistor and the antenna. This "average collecting area" is describing what should be fairly intuitive: this equilibrium holds regardless of what kind of antenna is put in the left cavity.

In other words, the average collecting area has something to do with how the antenna responds, on average, over all directions. It must be the same for all lossless antennas because anything else would violate conservation of energy.

Note the average collecting area is $\langle A_e \rangle$. Just $A_e$ on its own is effective aperture, which has an additional term $G$ added for gain:

$$ A_e = {\lambda^2 G \over 4 \pi} $$

Gain is often given as a single number, just the maximum gain for the antenna. But more completely, gain is defined as a function over direction and polarization. Since the average collecting area is the same for any lossless antenna design, and the effective aperture is the average collecting area multiplied by gain, it follows that if gain is to be increased in some direction, it must necessarily be decreased in some other direction if no laws of thermodynamics are to be broken.

For an array of $N$ antennas, the effective aperture is at best $N A_e$ where $A_e$ is the effective aperture of a single antenna. In practice it's less, due to either limited performance of the correlators, or due to overlap between the fields of the antennas.

If the antennas are several wavelengths apart, then the interactions between them will be negligible and performance will be dominated by the correlators.

If the antennas are closer than that, mutual coupling between them must be considered. The interactions are not simple or amenable to a closed form solution, so usually some antenna modelling software is used to solve the equations numerically.

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  • $\begingroup$ For now though, I am more concerned about how an array of multiple dipoles affects overall effective area, if at all. I am looking for a rough estimate at this point. $\endgroup$
    – Spuds
    Apr 15, 2019 at 18:39
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    $\begingroup$ @Spuds I believe at best, the effective area of an array is equal to that of a single dipole, multiplied by the number of dipoles in the array. But it can be less, depending on array configuration. But if you're building an interferometer then you might also be concerned with angular resolution, which is something else entirely. $\endgroup$ Apr 15, 2019 at 18:44
  • $\begingroup$ That is my intuition as well. I can size for interferometry, but I would love a source for sizing for total collecting area in different configuration. $\endgroup$
    – Spuds
    Apr 15, 2019 at 18:46
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    $\begingroup$ @Spuds If the antennas are far enough apart that their near fields don't overlap, I think your sensitivity will be limited by the correlators rather than the antennas. And if they do overlap, I think antenna modelling software is your best bet. $\endgroup$ Apr 15, 2019 at 18:48
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    $\begingroup$ @Spuds I added a bit more to the answer. $\endgroup$ Apr 15, 2019 at 19:19
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The effective aperture can be expressed in terms of the antenna's gain: $$A_e=\frac{\lambda^2 G}{4\pi}$$ where G is the "linear" value of gain, not the logarithm of a ratio expressed in decibels. Assuming you are able to obtain the gain of your dipole array, this may be a more workable expression.

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  • $\begingroup$ For now though, I am more concerned about how an array of multiple dipoles affects overall effective area, if at all. I am looking for a rough estimate at this point. $\endgroup$
    – Spuds
    Apr 15, 2019 at 18:39
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    $\begingroup$ The gain of a dipole array is readily obtained with modeling software. If you describe the array, I may be able to reply with a gain value. Need number, size (length and diameter) and dispositions of dipole elements. $\endgroup$
    – Brian K1LI
    Apr 15, 2019 at 18:50
  • $\begingroup$ Is there a decent free tool you could recommend, maybe via python? This is a small part of my project, one which I am really only looking to estimate. $\endgroup$
    – Spuds
    Apr 15, 2019 at 18:57
  • $\begingroup$ I recommend 4nec2; AFAIK, it only runs under Windows. $\endgroup$
    – Brian K1LI
    Apr 15, 2019 at 19:12

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