Wikipedia says:

It can be shown that the aperture of a lossless isotropic antenna, which by definition has unity gain, is:

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

We also see this emerge as a $\lambda$ term in the Friis transmission equation:

$$ {\frac {P_{r}}{P_{t}}}=G_{t}G_{r}\left({\frac {\lambda }{4\pi R}}\right)^{2} $$

This would suggest that longer wavelength antennas are more efficient collectors of electromagnetic energy. Why is this? Wikipedia says it can be shown. Show me.

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    $\begingroup$ Physics.SE has a detailed mathematical analysis at Effective aperture of isotropic antenna. $\endgroup$ – Phil Frost - W8II Aug 8 '14 at 12:47
  • $\begingroup$ The first formula does not take gain into account. $\endgroup$ – Mike Waters May 26 '17 at 18:19
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    $\begingroup$ @MikeWaters Sure it does. The gain is 1, by definition of "isotropic antenna". $\endgroup$ – Phil Frost - W8II May 26 '17 at 18:27
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    $\begingroup$ Is the gain 1 or is it 0dB? I thought we recently had this discussion $\endgroup$ – Scott Earle May 28 '17 at 14:46
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    $\begingroup$ @ScottEarle 1 = 0dB (dBi if you want to be more specific) because $10^{0/10} = 1$. $\endgroup$ – Phil Frost - W8II May 28 '17 at 16:44

This is a topic that troubles most students and even finds it way into many technical papers and textbooks in the form of incorrect assertions and conclusions. While you will find some reasonable references to thermodynamic equivalencies in some texts, it seems the genesis of the isotropic effective aperture equation has been rarely published.

The answer to the question lies buried in the mechanisms of Fresnel (near field) and Fraunhoffer (far field) zones of antennas. The Fresnel zone is the specific area of interest. The non-radiating, non-dissipating (thus reactive) part of the Fresnel zone is generally considered to extend 1∕(2π) times the wavelength from the surface of the antenna. In a more general sense, this is considered the maximum distance from which EM (electromagnetic) waves can couple to a nearby object. This coupling is why the energy is not radiated or dissipated by a transmitting antenna.

Now we turn our attention to the elusive isotropic antenna. It is considered a point source - small enough in dimension compared to any other incorporated dimensions that it is dimensionless and infinitesimally small. By definition, the isotropic antenna radiates equally in all directions. By the theory of reciprocity, the isotropic antenna must then receive equally in all directions.

Now consider an EM plane wave approaching the isotropic antenna. The isotropic antenna cannot "look ahead" and see the plane coming when the plane wave is in its far field because the EM plane wave is not yet having any effect on the isotropic antenna. But as the EM plane wave gets very close to the isotropic antenna, it begins to cause current to flow in the isotropic antenna. How close does the EM plane need to be? The distance of the non-radiating Fresnel zone which, as stated previously, is considered to be 1∕(2π) times the wavelength of the frequency in question.

As the plane wave intersects the isotropic antenna (that is, the isotropic point is on the plane), the spherical receiving pattern of the isotropic antenna has the maximum possible coupling with the EM plane wave. Since the isotropic antenna is a point lying on the intersecting plane with its receive sphere bisected by the plane, the resulting pattern is a circle defined by a radius that originates at the isotropic antenna and extends for a radius of 1∕(2π) times the wavelength. This circle is the Ae (effective aperture) of the isotropic antenna. That is, it is generating current from EM waves within that radius.

But with any receiving antenna, we are more interested in determining the total power the receive antenna is able to make available to the receiver. This is a function of the radiative flux which is given in SI units of W/m2. While we do not know the radiative flux in this case, we can compute the normalize power received (the amount of power received if radiative flux = 1 watt/m2) by simply computing the area of the the Ae. Since the aperture is a circle and the area of a circle is given by:

$$ A_\text{circle} = \pi r^2 $$

we can substitute the radius of the isotropic $A_e$:

$$ A_e = \pi \left({ \lambda \over 2 \pi}\right)^2 $$

and simplify:

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

Thus emerges the standard definition of the Ae of an isotropic antenna, always with a gain of one. The dependency on λ2 is simply due to the minimum radius at which the isotropic antenna (or any antenna for that matter) can begin to receive or emit EM waves.

If you now consider the effect of gain by any other type of antenna, you can see that it is simply increasing the area of the Ae of the isotropic antenna by the magnitude of the gain. Since the Ae is multiplied by the radiative flux and now the gain of the antenna, the received or transmitted power is scaled proportionally. It should be noted however, that the gain of an antenna does not physically extend the radius at which an EM wave can generate current in the antenna as this boundary condition is immutable. The gain and pattern of most amateur radio antennas is determined by the current vector patterns of the antenna.

If you find this explanation to be helpful and wish to requote it, I ask that you kindly give attribution to me, Glenn Schulz W9IQ.

Footnote: The Hairy Ball Theorem has been mentioned in this thread. The Theorem states that given a ball completely covered in hair, you cannot comb the hair in such a way as to have no partline. I can prove the theorem wrong: take a comb and do a 'fro on the ball. Hair is combed - no partline. Quod erat demonstrandum.

  • $\begingroup$ Yes, but a 'fro isn't a tangent hairstyle :) $\endgroup$ – Phil Frost - W8II Jun 1 '17 at 18:33
  • $\begingroup$ I was thinking about making a corollary to an isotropic, but I will let your review stand. LOL! $\endgroup$ – Glenn W9IQ Jun 1 '17 at 19:10
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    $\begingroup$ Again, super great answer. I formatted the math, and I think corrected some missing braces. Hope I didn't mess anything up. $\endgroup$ – Phil Frost - W8II Jun 7 '17 at 0:17
  • $\begingroup$ @Phil - Thanks for the bounty points! That was very kind. $\endgroup$ – Glenn W9IQ Jun 21 '17 at 20:32

This would suggest that longer wavelength antennas are more efficient collectors of electromagnetic energy. Why is this?

I think of it like in optics. A larger lens will collect more light because more light is incident on the lens. More photons hit the lens.

A radio antenna is a lens for a different wavelength of light, i.e. radio waves, not visible light. Therefore, the (electrically) larger the antenna, the more photons are incident on the conductor, the more energy will be collected.

Said differently: If you had 2 antennas of the same length, you would collect twice as much energy as a single antenna. If you had two lenses of the same size, you would collect twice as much light as a single lens. Instead of 2 antennas, you could have a single antenna of twice the length and collect twice the energy of the original antenna. Same for the optical lens.

  • $\begingroup$ So this is useful, but not quite the answer I'm looking for. Say we compare two resonant dipoles for different wavelengths: it's apparent the longer wavelength antenna is longer, and bigger, and so has a larger effective aperture. Yet, it's gain is the same. Why is there a frequency term in the relationship between effective aperture and wavelength? $\endgroup$ – Phil Frost - W8II Sep 21 '16 at 12:52
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    $\begingroup$ Trying to understand your question: is it the same as in your other comment, "Why would increasing the frequency on a dish antenna increase the antenna gain?"? $\endgroup$ – Jacob Davis Sep 21 '16 at 17:31
  • $\begingroup$ Well I think the root of the problem is understanding why the relationship between gain and effective aperture has a frequency term. But understanding why dish antennas change gain with frequency is a way of getting there. $\endgroup$ – Phil Frost - W8II Sep 21 '16 at 20:43
  • $\begingroup$ I wonder if this is analogous to the way different frequencies (colors) of light refract differently through a prism. Seems similar to a lens, which is similar to an antenna. If a prism is an optical antenna, then a radio antenna would refract (focus?) light differently by frequency. So perhaps if we know why light refracts differently based on frequency this would answer this root problem. $\endgroup$ – Jacob Davis Sep 21 '16 at 20:57
  • $\begingroup$ That does seem reasonable, but I haven't been able to follow that line of thought to a good conclusion. Imagine an ideal isotropic antenna: the gain is a constant 1 by definition, but as the frequency goes down, the effective aperture increases. $\endgroup$ – Phil Frost - W8II Sep 22 '16 at 14:47

This is simpler than the Friis equation makes it out to be.

The longer wavelength isotropic antenna is bigger, and aperture measures the effective electrical size of the antenna as a radiator. It is no surprise that the electrical size of an antenna that is physically larger is also larger.

Half-wave Dipoles have a constant gain, but are similarly larger at longer wavelengths.

The Friis equation is a particular algebraic expression of a conservation-of-energy equation, in a theory where all radiated energy is either received or radiated into space in a sphere.

The Gains in the Friis equation are generally lambda-dependent. Once again, it is common radio wisdom that smaller wavelengths, higher frequencies, require smaller elements for yagis or smaller reflectors for dishes to have the same gain. Thus if the antenna size stayed constant then as frequency increases, and lambda decreases, the gain should rise. This suggests gain depends on some inverse or perhaps inverse power of lambda.

For example, a the gain of a parabolic reflector dish of, say 3m diameter, will depend on the frequency and thus lambda. Wikipedia:Parabolic Reflector notes that the gain scales as 1/lambda^2. The Friis equation needs the gain of both antennas (receiving and transmitting), and so with dishes the transmit power received would scale as lambda^-2 * lambda^-2 * lambda^2 = lambda^-2 which provides a very different impression of what frequencies are best.

It is important to realize the Friis equation is a free space theory where the signal can spread out in a sphere.

It neglects the Earth, and with it all absorption or reflection by terrain, multipath reflections which can interfere constructively or destructively resulting in hot spots and dead zones, changes of media (e.g. trees, clouds), and various engineering requirements which may play a role at very low or high frequencies.

For instance, in the parabolic reflector example, we could move from microwaves to lasers to reduce lambda and increase power received according to the Friis equation. But this may not work out. Why? Aiming is much more critical with the laser, which relateds to the change in gain with a misaligned receive or transmit antenna. The parabolic antenna must now be a parabolic mirror of good quality instead of, say, stressed wire screening. And, a solid object such as a tree or bird is known to block a laser beam while a microwave might still be received.

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    $\begingroup$ "The longer wavelength isotropic antenna is bigger" why? Aren't ideal isotropic antennas point sources? $\endgroup$ – Phil Frost - W8II Feb 27 '14 at 12:11
  • $\begingroup$ Well as a practical matter I was imagining a shortened deranged dipole as an imperfect isotropic radiator. Mathematically, the point sources are a convenience that also can't exist due to electromagnetism's vector fields. See en.wikipedia.org/wiki/Isotropic_radiator and en.wikipedia.org/wiki/Hairy_ball_theorem $\endgroup$ – Paul Feb 27 '14 at 20:43
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    $\begingroup$ The hairy ball theorem implies that we can't have an isotropic radiator, but we can still have point sources. Still, this is tangential to the question. How do you show that $A_\mathit{eff} = \lambda^2 / (4\pi)$ ? Why would increasing the frequency on a dish antenna increase the antenna gain? $\endgroup$ – Phil Frost - W8II Mar 4 '14 at 16:36

"For example, a λ/2 wave long dipole in free space has a capture area of about 0.13λ²." What matters is the radiation pattern, not the physical size. A very short loss-less dipole (supra-conducting coils) will have a dipole pattern and therefore the same capture area 0.13λ². It would however have a VERY small bandwidth. The bandwidth is closely related to the Q value, the ratio between stored energy and flowing energy. With Q=100 and a bandwidth of about 1%, the matched load at the feed-point absorbs energy from about 100 subsequent oscillations of the field and they extend over a 100 times larger volume of space than a single oscillation. Not much of a proof, but perhaps some food for thought. /Leif

  • $\begingroup$ So when I build that tiny dipole with unobtanium which is 100% and has similar gain to a dipole, why does the aperture increase as the frequency decreases? $\endgroup$ – Phil Frost - W8II May 31 '17 at 12:51
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    $\begingroup$ When the frequency decreases, the power you can extract from that tiny dipole becomes bigger (in proportion to wavelength squared.) $\endgroup$ – sm5bsz Jun 6 '17 at 3:14
  • $\begingroup$ I know that. It's in the formula at the top of the question. Why? $\endgroup$ – Phil Frost - W8II Jun 6 '17 at 12:38
  • $\begingroup$ Look at the antenna as a rx antenna. The incoming wave will generate a current in a matched load. The current creates a wave that partly cancels the incoming wave. If the feed-point is shorted, twice as much current flows, and the antenna reflects the energy from an area twice as large as the capture area. Any antenna that absorbs power generates a re-radiated field with a size proportional to wl/2 or bigger. Free fields can not exist with a smaller size. From that follows that absorbed fields have a size proportional to wl/2. (The above is no proof, maybe inspiration for thinking...) $\endgroup$ – sm5bsz Jun 6 '17 at 23:58
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    $\begingroup$ You can edit your answer instead of using comments. $\endgroup$ – Phil Frost - W8II Jun 7 '17 at 0:20

In aperture antennas, like dishes, the effective aperture is not frequency dependant. It depends on the physical dimensions and the efficiency.

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    $\begingroup$ References? btw, Welcome to Ham Rado SE. Please take the tour at ham.stackexchange.com/Tour $\endgroup$ – SDsolar May 27 '17 at 0:38
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    $\begingroup$ Yes, I'd like to see some references too. "Depends on the physical dimensions" is very hand-wavy $\endgroup$ – Scott Earle May 28 '17 at 14:44
  • $\begingroup$ I'll buy it if you change "effective" to "physical". But even if we assume no variations in aperture efficiency, and thus the effective aperture does remain constant with wavelength, the gain will change with frequency. Why is that? $\endgroup$ – Phil Frost - W8II May 30 '17 at 23:07

Effective aperture (Ae) is a direct function of antenna gain and operating wavelength.

From W8JI and Jasik's book, at www.w8ji.com/capture_area_ae_effective_aperture.htm.

Ae (sometimes referred to as capture area) is determined by the voltage available across a load matching the antenna feed impedance for a given electromagnetic field strength density. In simple terms if the antenna is placed in a electromagnetic field of a certain intensity, a certain amount of power will appear in the load at the antenna terminals. The area of space around the antenna that provided this amount of power is the effective aperture.

Many people confuse physical area, or Ap, with effective aperture. They are not the same. Physical size only determines effective aperture as physical size might affect gain of an antenna. Gain and wavelength determines capture area, but capture area itself has nothing to do with actual physical size or physical area of the antenna.

For example, a λ/2 wave long dipole in freespace has a capture area of about 0.13λ². This means a lossless freespace dipole has an Ae of approximately 0.13 square wavelengths. This effective aperture is about 100 times larger than the actual physical area of a thin wire dipole antenna. Energy is extracted from an elliptically shaped area slightly longer than the dipole and about λ/4 diameter at the center. This is why increasing conductor diameter or using a cage of wires will not increase electrical aperture or capture area. As a matter of fact if we built a lossless or very low loss small dipole, perhaps λ/20 long, capture area or Ae would be within a few percent of a full size dipole!

Length itself has very little effect unless the change in length significantly affects antenna gain. We must have a change in gain to change Ae (effective electrical aperture). Physical aperture (Ap) changes do not affect Ae unless the gain changes.

  • $\begingroup$ While this is an enlightening diatribe on what effective aperture is, it doesn't answer the question, which is why the relationship between gain and aperture include wavelength. $\endgroup$ – Phil Frost - W8II May 26 '17 at 18:29
  • $\begingroup$ I don't like the article because it's incomplete ("it can be shown"), but so far I haven't any reason to believe it's wrong. $\endgroup$ – Phil Frost - W8II May 26 '17 at 18:49

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