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I was reading about the reactive near field of an antenna here part of the relevant portion of which is quoted below:

"Because of this energy storage and return effect, if either of the inductive or electrostatic effects in the reactive near field transfer any field energy to electrons in a different (nearby) conductor, then this energy is lost to the primary antenna. When this happens, an extra drain is seen on the transmitter, resulting from the reactive near-field energy that is not returned. This effect shows up as a different impedance in the antenna, as seen by the transmitter."

I am unsure if I understand this correctly, or at least its implication in the real world. Does this mean that putting any sort of metal near a transmitting or receiving antenna will cause some extra loss in the form of inductive and electrostatic coupling? Would these losses come from the Faraday Effect and essentially using energy from the electric field to charge a nearby metal like charging a capacitor respectively? Does this mean that in the real world antenna and RF designers must ensure that there are no metals or any other conductors within the reactive near field of their antennas? Am I wrong in my understanding, or is perhaps the effect just too small to matter practically?

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    \$\begingroup\$ Do you understand that metal in the far field does not represent a loss or extra load on the transmitting antenna? \$\endgroup\$
    – Andy aka
    Feb 19, 2017 at 10:25
  • \$\begingroup\$ My understanding was that any metal outside the reactive near field, so anything in the radiative near field or far field, wouldn't represent an extra load and cause loss. I think this is because at those distances there is no longer any reactive returning/storing of energy? I guess the loss would only be a result of the faraday/electrostatic effects very close (reactive near field) to the antenna that you would be concerned with in almost any circuit design, not just RF or antennas. If this is incorrect please let me know! \$\endgroup\$
    – scuba
    Feb 19, 2017 at 10:58

3 Answers 3

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Do conductors in the reactive near field of an antenna cause loss?

Not necessarily. Consider a well-designed dipole antenna; you can place an array of "other elements" around it and turn the dipole into a Yagi-Uda antenna: -

enter image description here

The Yagi-Uda antenna uses "other elements" constructively to produce an EM emission directed towards a particular direction. These extra elements are in the near-field of the dipole-section of the antenna. The EM radiation becomes focussed like this: -

enter image description here

Attribution: By Chetvorno - Own work, CC0, https://commons.wikimedia.org/w/index.php?curid=54323935

If the placing of these elements isn't accurately controlled then you get alterations to the electrical impedance seen at the terminals of the antenna. In fact some Yagi-Uda designs utilize this and convert the natural 73 ohms of the dipole (the driven part of the antenna) into something radically different.

The implication of this is that ad hoc placement of perfectly conducting material around a dipole antenna will significantly change the electrical impedance. Basically, the antenna becomes detuned from its optimum frequency; the presense of conducting material lowers the electrical impedance and the dipole becomes what is known as "short". Consider the dipole and what happens when you operate it not at the perfect resonant point: -

enter image description here

When the length of the antenna corresponds to half a wavelength (nominal operating point for a dipole) the real impedance is 73 ohms and the reactive impedance is zero. If the antenna is "shortened" by the presence of conducting elements, the "real" part of the impedance falls rapidly towards zero ohms and the reactive part becomes capacitive, rising rapidly in impedance as length shortens.

Given that the electrical power delivery system to an antenna relies on impedance matching, you can see that an increase in power loss is inevitable. It's not irreconcilable; you could place a transformer and inductor at the dipole terminals to convert impedances and maintain the same power delivery but extra losses are inevitable. The biggest of these is the antenna conduction loss itself. Once the conduction losses of the antenna start to become a significant percentage of the electrical radiation resistance, you are on the downward slope.

Consider also the placement of a really big conductor close to a dipole. Let's call that really big conductor "earth". The graph below shows how the resistive impedance changes as the dipole is raised a distance above ground: -

enter image description here

If you placed the dipole only a small distance above ground (0.2 wavelengths or less) you can see that the impedance is significantly reduced and gets smaller as ground approaches.

The bottom line of what I'm trying to say is that the wiki article is correct but, it is secondary to the bigger picture that I've tried to outline above. Losses due to impedance mismatches (brought about by localized conductors/materials) are much more significant than the actual dielectric or conduction losses in those materials.

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  • \$\begingroup\$ Thank you, I certainly do understand now how impedance mismatch is very significant in terms of losses, I think you have quantified that well! While intuitively it makes sense that conduction/dielectric losses are not significant compared to impedance mismatch losses, I cannot say that I understand why this is quantitatively. Do you know of any resource that quantifies these losses so I can see with math/figures just how insignificant they are compared with the impedance mismatch effect? I wonder what percentage of total loss might be due to these effects in a typical antenna application? \$\endgroup\$
    – scuba
    Feb 19, 2017 at 21:46
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    \$\begingroup\$ I don't know of any resources other than to dig deep and dirty on google. A different approach would be to forget about the antenna being an em radiator and just concentrate on the near field phenomena of coupled coils as per near field tags that use only the magnetic bourne energy. It boils down to the rogue conductor being a shorted turn with resistive losses. You can model the antenna as a tuned parallel circuit and the rogue as a single turn with enough resistance so as not to destine the resonant frequency but low enough to take significant power \$\endgroup\$
    – Andy aka
    Feb 19, 2017 at 22:15
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    \$\begingroup\$ Of course this model doesn't produce far field em power but, importantly it wastes power in the rogue metal. This may be a more productive route to getting the idea. It's never easy when it comes to antennas; a simple equation can turn into a virtually unresolvable triple integral and, you may have noticed that there ain't any good visualisations of how disparate e and h fields (in the near field) magically align in the far field. So, forget about an antenna that converts 73 ohms to free space 377 ohms and concentrate on the mag field eddy current induction but not at the expense of detuning. \$\endgroup\$
    – Andy aka
    Feb 19, 2017 at 22:24
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    \$\begingroup\$ As soon as there is significant detuning you will be altering the antenna impedance in a complex way and this is where the tuned circuit analogy will fall down. \$\endgroup\$
    – Andy aka
    Feb 19, 2017 at 22:27
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Any sort of metal near the antenna will suck energy out of the RF field and induce eddy currents in the metal.

It's not like charging a capacitor. That would mean that the electric energy just moved somewhere else but is still there. Instead the eddy currents will turn the energy into heat.

The transmitter will 'see' this energy loss. One way to look at this is that the impedance of the antenna and matching network will go lower.

E.g the transmitter will see a larger load.

As an RF designer you'll usually try to minimizie metals near the RF field, but that is not always possible. One way to deal with it is to guide the field away from the metal by using ferite material.

Also it's the RF designers job to make sure that - no matter how much metal you put near the antenna - the impedance never gets so low that the transmitter overheats (remember, impedance goes low, so the more metal you put near the antenna, the more the antenna looks like a short circuit).

You can do this by using the fact that metal near the antenna has a detuning effect and the antenna system will quickly going out of tune for the target RF frequency. This already limits the amount of energy transfered from the transmitter to the antenna somewhat.

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  • \$\begingroup\$ I understand the eddy current part of "sucking energy out of the RF field." That is pretty much Faraday's law and caused by the magnetic field and is nothing like charging a capacitor like you said. But is there also not a second effect caused by the electric field, in addition to this first effect caused by the magnetic field? The electric field portion of the EM wave in the near field won't induce some voltage or charge on the metal near the antenna? \$\endgroup\$
    – scuba
    Feb 19, 2017 at 8:35
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    \$\begingroup\$ "Any sort of metal near the antenna will suck energy out of the RF field and induce eddy currents in the metal." While that's of course technically true for any real lossy metal, is it practically an important factor here? Often, we can design antennas using a perfectly electrically conducting (PEC) approximation to whatever actual metal is used. PECs are lossless. If that approximation is valid inside an antenna, why would it not be valid in the nearfield outside the antenna? I think most of what you're saying is really about detuning an antenna, not energy loss. \$\endgroup\$
    – LedHead
    Feb 19, 2017 at 9:24
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Bits of metal near antenna may or may not have induced current, but needn't cause signal loss. They can just as well cause beam steering, or (if, for example, the metal is a dish) can cause significant antenna gain (not power gain, 'antenna gain' means focusing, or direction limiting to form a beam).

A well-tuned antenna might become less effective due to alteration of the driven impedance, however. The antenna will typically have a tuning network that is adjusted for transmitter efficiency, and that adjustment determines losses in transmission. Adding or deleting a metal object would equally likely cause losses due to poor tuning.

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  • \$\begingroup\$ For the first part of your answer, what would be a situation in which they would not have induced current? Doesn't Faraday's Law dictate that some current must be induced from the changing magnetic field normal to the conductor surface, unless you are assuming the antenna is highly directional and there just isn't radiation where the bits of metal are? Are you saying that in a dish antenna, the dish does not have induced current from Faraday's Law? And do you also imply that even if there is induced current there might not be signal loss? How could this be? \$\endgroup\$
    – scuba
    Feb 19, 2017 at 11:06
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    \$\begingroup\$ A dipole antenna makes an E field that's vertical; so, it won't induce currents in a thin horizontal piece of metal: the direction is wrong. It's normal for antennae to have both driven, and unconnected (or grounded) passive elements, i.e. 'nearby metal bits'. Magnetic field makes CIRCULAR currents, but a long thin element... doesn't complete such a circuit. As for 'signal loss', that depends on direction: you don't CARE if signal doesn't penetrate the ground, after all. Loss happens regardless. \$\endgroup\$
    – Whit3rd
    Feb 19, 2017 at 12:31
  • \$\begingroup\$ Thank you, I feel that I understand the answers to all of my questions except the last one and that is because I did not ask it properly. I did not mean to say "signal loss" but rather "antenna loss" in that induced currents (assuming the metal geometry supports circular currents, which as you have shown might not be the case) in metals in the reactive-near field would store/return energy(reactive) to the antenna rather than allowing it to be radiated out into the far field, that energy would be "lost." That would happen, right? \$\endgroup\$
    – scuba
    Feb 19, 2017 at 21:50
  • \$\begingroup\$ Loss of energy, i.e. resistive heating, can certainly occur (but try putting a few pieces of wire on a paper plate in a microwave oven: only the resonant lengths make scorchmarks). Scattering of energy into unuseful directions is more likely. \$\endgroup\$
    – Whit3rd
    Feb 20, 2017 at 7:33

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