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I'm kind of confused about which is the most correct unit when it comes to the measure of wireless signal's strength. It seems that the correct unit is V/m (usually expessed in dBuV/m), however it seems that most of the time W are used (in the form of dBm).

From what I can tell dBm actually expresses the power that arrives at the receiving antenna, which is related to the actual signal strength and it will vary according to an inverse square law.

What has been puzzling me is that I found this article (which I'm unsure wether is correct or not) that states:

The magnitude of the E component of a radio wave varies inversely with the distance from the transmitter in a free-space, line-of-sight link. If the distance is doubled, the E-field intensity is cut in half; if the distance increases by a factor of 10, the E-field intensity becomes 1/10 (0.1 times) as great. The E component of an EM field is measured in a single dimension, so the intensity-versus-distance relation is a straight inverse rule, not the inverse-square law.

Could someone please clarify this statement and shed some light on how dBm and dBuV/m are related?

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    \$\begingroup\$ dBuV/m is measure electric field strength which only makes sense if it is static. dBm is the amount of power (referenced to 1mW) induced in the receiver. That power is an expression of a voltage/current that is used by the receiver electronics. \$\endgroup\$ – Tom Carpenter Feb 20 '16 at 19:51
  • \$\begingroup\$ Static in what sense? I was thinking about a static point in space. For example 20m away from the antenna. \$\endgroup\$ – petersaints Feb 20 '16 at 20:07
  • \$\begingroup\$ Static in the sense of non-time-varying. \$\endgroup\$ – Tom Carpenter Feb 20 '16 at 20:36
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    \$\begingroup\$ @tom dBuV/m is an RMS value for radio waves just as dBm is an average power of the time varying E and H fields when an antenna liberates power into it's correct terminating resistance. \$\endgroup\$ – Andy aka Feb 20 '16 at 22:21
  • \$\begingroup\$ @TomCarpenter Ok. I'm starting to see where your point might be. Maybe I'm just confused because I'm trying to apply Coulomb's law electric field derived notions (i.e., en.wikipedia.org/wiki/Coulomb%27s_law#Electric_field), in which the field intensity-versus-distance relation is an inverse square-law, to propagating waves. This comes from the fact that in my mind they are not that different. They are both fields (electrical ones) that are expressed in N/C or V/m (those units are equivalent). So I may have wrongly assumed they had more in common than what they actually have. \$\endgroup\$ – petersaints Feb 20 '16 at 22:24
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The E field has dimensions of volts per metre and the H field is amps per metre. For a far-field radio wave, the ratio of E to H is 377 (120\$\pi\$). Basically, if the E field received is X the H field is 377 times lower than X. 377 ohms is the impedance of free space and is measured in ohms because the "per metre" parts cancel out. The 377 ohms comes from this equation: -

enter image description here

The power is E x H so strictly speaking it's watts per square metre. The antenna has a certain "aperture size" measured in square metres so, a certain watts per sq metre hitting an antenna of a certain area liberates true watts into the terminating impedance. Imagine it like a sheet of paper being illuminated by a light bulb - the power received falls away with distance squared and this implies the individual E and H fields reduce linearly with distance.

You can use dBm or dBuV/m or dBI/m if you are talking about radio waves - whichever you use leads to the same conclusion and cross references the other units perfectly. However, you cannot sensibly state dBm without implying an antenna is present. You could state dBm per sq metre of course because that is the power illuminating a notional 1m surface at some distance from the transmit antenna.

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Incited voltage decreases linearly as you increase distance between transmitter and reciever. Received V(dist'X') = V(dist'0') x (dist'0'/dist'x').

Incited current also decreases linearly with distance, so rec'd I(dist'X') = I(dist'0') x (dist'0'/dist'x').

Power equals voltage x current, so putting it all together, you get: P(dist'X') = P(dist'0') x (dist'0'/dist'X')^2

This is also why we generally use linear, rather than logarithmic, values when calculating dBV(voltage) and dBI(current) values, but use logarithmic calculations for dBa(sound pressure), dBi(gain) and dBm(power) values.

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  • \$\begingroup\$ I would agree if this was V. But would this conclusion still hold for V/m? \$\endgroup\$ – petersaints Feb 20 '16 at 20:21
  • \$\begingroup\$ V/m is a measurement of how many volts (or partial/micro-volts) of potential can be 'collected' on a 1m long piece of conductor, so it's still just a voltage measurement. \$\endgroup\$ – Robherc KV5ROB Feb 20 '16 at 21:12
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    \$\begingroup\$ @RobhercKV5ROB be careful about this - you cannot liberate a volt on a 1m long wire in an E field of 1V/m. The wire will short out the E field! \$\endgroup\$ – Andy aka Feb 20 '16 at 21:37
  • \$\begingroup\$ @Andyaka Yes, I meant the theoretical basis for the value. I understand actually harvesting the energy would require specific adjustments based on lambda, etc. -- Just wasn't sure how to succinctly word that in the comment :S \$\endgroup\$ – Robherc KV5ROB Feb 20 '16 at 21:44

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