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During conversation, a colleague proposed that over-the-air television and radio broadcasters can determine the number of viewers or listeners based on the "load" on their signal. This seems to me like total bupkis, but he's piqued my curiosity and I've not been able to find a discernible answer when searching the web to prove him right or wrong.

Is such a thing even possible? Do the number of receivers within the broadcast range of a transmitter put any "load" on that signal? I always thought that the amount of power required for a transmitter simply determined the distance at which the signal could still be reliably received. AFAIK receiving a radio signal does not require any actual power at the listener's end, except to filter and amplify that signal into something useful, and that power is provided locally.

If this were true, it seemed plausible to me that one could place several signal monitors at a fixed radius from the transmitter and measure the signal strength at each. Monitors with weaker signal must have more receivers between that monitor and the transmitter, which could be used to extrapolate the number of receivers within that arc of the radius at, say, -3 dBm per receiver.

What I do know is that obstructions between the transmitter and receiver degrade the strength of the signal, so in that situation, one would have to account for buildings, trees, mountains, birds, precipitation, clouds, airplanes, helicopters, low-flying kayaks, large snowmen, and Santa Claus.

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    \$\begingroup\$ May be since in the near field, and especially in the reactive near field (λ/2π distance from antenna), there is no transmission as an electromagnetic wave, and may apply to either inductively or capacitively coupled. In this area E and H becomes really very complex. \$\endgroup\$ – GR Tech Aug 27 '15 at 17:28
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    \$\begingroup\$ For what it's worth, viewer statistics for radio and television are typically determined through listener studies, e.g. by Arbitron in the US. \$\endgroup\$ – duskwuff Aug 27 '15 at 20:35
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    \$\begingroup\$ It is just as possible as the number of plants or eyeballs having an influence on the sun's radiation output. All forms of electromagnetic energy become heat 100%. So, any sort of receiver, even a charcoal brick, is an equivalent "load" as the signal propagates away... To Infinity, And Beyond! \$\endgroup\$ – user56384 Aug 28 '15 at 14:27
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Actually, yes, a receiver can affect the transmitter. Passive RFID is based on this principle.

However, RFID only works at very close distances, where the receiver is absorbing something on the order of 10-4 to 10-5 of the transmitter's signal. In other words, the transmitter is sending out hundreds of milliwatts, while the receiver is absorbing a few microwatts. Such changes are just barely detectable at the transmitter with careful techniques.

However, for general broadcast radio, the transmitter is sending out tens to hundreds of kilowatts, while the receiver is absorbing tens to hundreds of femtowatts, which is a fraction on the order of 10-18. This is completely undetectable at the transmitter. Furthermore, receivers absorb signal regardless of whether they're turned on or not, so even if it were detectable, it would tell you nothing about how many people were actually listening.

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    \$\begingroup\$ Even there, small RFID tags work by actually taking the wireless power from the transmitter (RFID Reader) and transmitting back. longer distance tags (smart tolls, aircraft ID'd) use powered tags. If it weren't for the tag's transmission of data, the transmitter would know nothing. \$\endgroup\$ – R Drast Aug 27 '15 at 16:14
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    \$\begingroup\$ @RDrast: With passive RFID, the transmitter sends a continuous carrier, and the "transmitting back" by the tag is done on the same frequency. This is sensed on the transmitter by measuring small variations in the carrier amplitude at the antenna terminals. In effect, the tag is varying how much energy it is absorbing from the transmitter in a recognizable pattern. Active RFID uses completely different techniques. \$\endgroup\$ – Dave Tweed Aug 27 '15 at 17:01
  • \$\begingroup\$ If it is actually RFID, then the tag is actually transmitting back information after being powered up by the passive field. I know of no systems that do not include a a digital ID signature that is read by the reader, encoding information onto the returned data. \$\endgroup\$ – R Drast Aug 27 '15 at 17:04
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    \$\begingroup\$ @RDrast: Isn't that what I just said? \$\endgroup\$ – Dave Tweed Aug 27 '15 at 17:09
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    \$\begingroup\$ Further-furthermore: when talking about receivers from the point of view of the RF signal, large bodies of water, living meat (people, cats, cattle..) walls of a certain color or material and mountains made of certain types of rocks also count as receivers (they also absorb the signal). So any attempt to count receivers based on the load at the transmitter is merely characterizing the environment the transmitter is in. \$\endgroup\$ – slebetman Aug 27 '15 at 17:49
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It's technically possible to detect radio receivers if they are Superheterodyne receivers that use RF mixing to downmix the received signal to a well known intermediate frequency. You can scan for this frequency using a directional antenna and count the receivers around you.

Though this doesn't sound like what you're inferring since the transmitter can't detect the receiver based on signal "load" or other factors, it requires a special detector that's separate from the transmitter.

This is how radar detector detectors work. Also, some billboards use this technology to determine what radio station drivers are listening to so they can tailor the ads to the drivers' preferences:

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  • \$\begingroup\$ So if the radio company owned several such billboards in the area, connected to them somehow (cell radio perhaps)... I think they would be able to have a reasonable guess per the OP's question \$\endgroup\$ – user2813274 Aug 28 '15 at 16:44
  • \$\begingroup\$ It somewhat reminds this : en.wikipedia.org/wiki/The_Thing_%28listening_device%29 \$\endgroup\$ – TEMLIB Aug 28 '15 at 23:09
  • \$\begingroup\$ This is the correct answer. +1 \$\endgroup\$ – Deer Hunter Aug 29 '15 at 21:32
  • \$\begingroup\$ While this would work for relatively short distances, it seems highly unlikely this would be any help in the case given by the OP. \$\endgroup\$ – Mast Aug 31 '15 at 9:21
  • \$\begingroup\$ This works only if the mixer in the receiver is a very bad mixer that reflects some of the mixing product back to the antenna or leaks it somewhere else in such a large amount that it can detected from a distance AND if the exact IF is known. In addition this method could be easily undermined by using an unusual IF. Also there are dozens of other possible sources that would leak EM radiation of the same particular frequency (e.g. harmonics of µC-clocks). So practically there is NO WAY to use this to find radio receivers tuned to a particular station. \$\endgroup\$ – Curd Aug 31 '15 at 12:52
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No. There is no way for an AM or FM transmitter to determine how many people are listening. They provide exactly the same power output at carrier whether there are a million receivers within 1 mile or zero.

Digital transmissions that require a subscription can on the other hand possibly know how many receivers there are, if there is a two way verification link. Or like WiFi, each 'receiver' is actually interacting with the transmitter, but in neither case does it affect the output power of the transmitter, or is it able to be sensed by monitoring the output power.

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Sounds like complete and utter cr*p for all practical purposes. The actual energy extracted by a receiver is microscopic.

Though there is a story of a farmer who built a big tuned loop in order to extract free power from a nearby radio transmitter. Sufficient to distort the field pattern and be detected.

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  • \$\begingroup\$ I heard that near radio towers usage of neon lights is prohibited because it lights up from the strong signal. But thats just heard. \$\endgroup\$ – akaltar Aug 27 '15 at 21:43
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    \$\begingroup\$ @akaltar If you hold a flourescent tube light under those big transmission lines with one end closer to the lines, it will light up. I'm pretty sure that our two situations are based off the same principle. \$\endgroup\$ – Zach Mierzejewski Aug 28 '15 at 3:11
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Assuming the the field in question is the electromagnetic field, and all interactions are in the 'far field', then the question is 100% no, no you cannot sense increased load.

RF is just the production of light, albeit a frequency much lower than visible (WiFi runs at 2.4 GHz. Red light is ~400 THz).

Does a star experience more 'drain' because its light is being absorbed by my eye? Or a piece of silicon? Or a spherical cow?

Does a lightbulb experience more 'drain' because its light is being absorbed by my office walls?

The is answer is absolutely no, once the antenna has produced photons, the energy is gone and all drain on that device to produce that photon has already occurred.

...

The answer is different if you consider near field - where inductive reactance dominates. This is how purely-passive, non-transitting RFID tags mentioned in the comments work - they have an inductive circuit that is tuned to the frequency of the inductor that makes up the antenna, like a big open-air transformer. Here the antenna/transformer/inductor does actually sense increased load, because it is being coupled to the RFID's inductor.

Near field, however, only operates within about 1 wavelength from the transmitter. This is why near-field purely-passive non-transmitting RFID tags must use low frequencies, so that they can have reasonable operating distances.

A good reference is the following paper by two IEEE RF scientists: http://www.ee.washington.edu/faculty/nikitin_pavel/papers/RFID_2007.pdf

To quote:

Low frequency (LF, 125-134 KHz) and high frequency (HF, 13.56 MHz) RFID systems are short range systems based on inductive coupling between the reader and the tag antennas through a magnetic field. Ultra-high frequency (UHF, 860-960 MHz) and microwave (2.4 GHz and 5.8 GHz) RFID systems are long-range systems which use electromagnetic waves propagating between reader and tag antennas

Some wavelength calculations for those above frequencies, for the curious:

  • 125 KHz == 2398.34 meters
  • 13.56 MHz == 22.11 meters
  • 2.4 GHz == 0.125 meters
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This is explained here in detail:

So, in the optimum case half of the power absorbed by the antenna is immediately re-radiated. Clearly, an antenna which is receiving electromagnetic radiation is also emitting it. This is how the BBC catch people who do not pay their television license fee in England. They have vans which can detect the radiation emitted by a TV aerial whilst it is in use (they can even tell which channel you are watching!).

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    \$\begingroup\$ FTR, this is quite different from, as in the question, the transmission tower detecting it. Also purely for the record, in the vast majority of cases the authorities in the UK simply track purchase of TV sets at electronics stores, and align those records. (Which is quite chilling!) \$\endgroup\$ – Fattie Aug 28 '15 at 14:57
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There is no way to detect the number of receivers from transmitting point. Once the EM wave leaves the near field of the antenna the wave becomes transversal electromagnetic wave and it has no influence on transmitter. However there is an interaction between sorrounding antennas at close distance (near field - half wavelength), but this is barely detectable.

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Technically, it could be estimated. A known power level source will transmit to a certain distance before losing signal strength to half power (-3db). Every antenna and receiver between the source and that -3db distance will tap some of the power from the signal. If you have a receiver sensitive enough at the -3db distance, the number of interfering listeners in between could be estimated. Now, do that process in a circular pattern around the source and you could estimate the number of signal interceptors between the source and the known power level perimeter. A similar process can be used in cable transmission by determining the amount of signal power needed to maintain that -3db level at the end of the transmission line. (ie, each receiver requires 5 milliwatts to show the signal to its receiver, the end of the line will see a minus 5 milliwatts for each customer watching that channel between the source and the line end. If the end of the line experiences a half watt signal strength loss (500 milliwatts) it means 100 people are tuned to that channel.

That is doable physics. Whether the radio stations or cable providers do it is not known.

http://en.wikipedia.org/wiki/Transmission_(telecommunications)

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    \$\begingroup\$ 1. Receiving antennas in the path will extract the same power from the signal whether they are being listened to or not. 2. So would (maybe not quite so efficiently) every building, tree, vehicle, cow, telephone pole, etc., \$\endgroup\$ – The Photon Aug 27 '15 at 16:51
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    \$\begingroup\$ Those trees and cows need to pay their TV fee! \$\endgroup\$ – user56384 Aug 28 '15 at 14:31

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