I am trying to understand how to add the background noise into the equation. I am trying to work out the pathloss of a radio signal over a certain distance.

We have the Friis Formula for that for example. Now the Friis Formula has 2 parameters for the gains of the receiver and transmitter antenna. I don't necessarily know whether the antenna gain variable contains or is influenced by the atmoshperic noise. I think it's just the internal noise inside the antenna, like how a not perfect conductor will generate heat and that loss can interfere with the signal.

But even then to just be safe I like to use the maximum gain formula, as in the fundamental limit on the antenna gain for a theoretically perfect antenna, which I have talked about and asked some questions about here.

Ok so I read this paper:

  • "Atmospheric Magnetic Noise Measurements in Urban Areas" (2014) - Christian Schlegel, Matt Mallay, and Chris Touesnard

Where they break down the background noise, in this case the magnetic noise into 2 parts:

  • Thermal ( the heat energy in the air)
  • Atmospheric (micropulsation of the Earth's magnetic field)

They both disrupt the signal of the antenna, and basically at lower frequencies you have more atmospheric noise ,and in higher frequencies you get more thermal or even man made noise, at very high frequencies you get cosmic noise.

  • David Gibson also wrote an interesting paper on this: "Channel Characterisation and System Design for Sub-Surface Communications" (2003)

To keep it short, the formula is basically this:

$$F_a(f) \approx 294.15 - 36 \log_{10}f $$

I have checked it, it matches the data of both Gibson's research and that of the 2014 paper, which has experimental evidence, although there is a 20 db difference between indoor and outdoor noise. Gibson's research relies on the 1968 CCIR study, which might be obsolete now as the electromagnetic spectrum with all the new technologies affecting it have changed.


The question is this: Could I consider this atmospheric noise as a fundamental minimum in dB that a signal has to have in order to be able to be detected by an antenna.

For example this formula says that a 1 Hz signal will have ~ 294.15 dB level noise interference, so would this mean that the receiver antenna has to receive higher than 294.15 dBW signal, probably way higher, in order to establish a communication channel. I mean the level of the noise would totally overwhelm any signal that would have a power level lower than this and it would be impossible to send information through.

So going with the Nearfield formula that I've talked about in my previous question:

  • A 1 Hz magnetic signal, going between 2 magnetic antennas placed 50 centimeter from eachother with a boundary sphere of 20 cm has a path loss of roughly 16 dB
  • Now inserting the atmospheric noise component, which is 294.15 dB (probably +20 dB higher in a building)
  • Would this mean that the total attenuation of the signal in this case is 310.15 dB? And note that this is the minimum attenuation, since we used the maximum possible gain for the antennas, and in reality the gain will be much lower, so the attenuation would be much higher. Also this would be the attenuation of the signal, the antenna would still receive the energy, it's just that you could not decode it due to the overwhelming noise interference.
  • \$\begingroup\$ How can a "1 Hz signal will have ~ 294.15 dB noise". Think about what you are saying. \$\endgroup\$
    – Andy aka
    Jul 27, 2017 at 7:54
  • \$\begingroup\$ @Andyaka sorry I have clarified it now. \$\endgroup\$
    – David K.
    Jul 28, 2017 at 6:58
  • \$\begingroup\$ Post a shot of the document that covers that errant formula please. It's too long for me to find it. \$\endgroup\$
    – Andy aka
    Jul 28, 2017 at 7:35
  • \$\begingroup\$ @Andyaka it's on page 4, right above the 19th formula in the "Atmospheric Magnetic Noise Measurements in Urban Areas" (2014) paper. It's a crude model approximation of Gibson's values, but it's actually an underestimation, according to their measurements the noise is much higher. \$\endgroup\$
    – David K.
    Jul 28, 2017 at 9:03
  • \$\begingroup\$ Listen, I don't want to gripe at you but that 1st paper (Atmospheric Magnetic Noise Measurements in Urban Areas) needs IEEE passwords and logins. So, please do as I recommend and paste in the page that mentions the formula. \$\endgroup\$
    – Andy aka
    Jul 28, 2017 at 11:06

1 Answer 1


The question is this: Could I consider this atmospheric noise as a fundamental minimum in dB that a signal has to have in order to be able to be detected by an antenna.

If any of the desired signal is present within the capture area of the antenna, the signal is received regardless of the noise that is present. That is to say that the noise is simply another signal that the antenna receives (with the same gain) in addition to the desired signal. The receive antenna system will also generate a limited amount of thermal noise.

The Friis formula allows you to calculate, for a given frequency and antenna system, the ratio of received power to transmitted power. That is its primary function.

The question of what is the minimum signal to noise ratio to carry out communications requires additional calculations. It is dependent upon the receiver, modulation method, the receive bandwidth, and signalling rate. There are communication methods that allow a -20 dB signal to noise ratio at the receiver and communications can still be carried out reliably, for example.

Your posting has several issues that you need to address. A dB is always an expression of a power ratio. When you say:

so would this mean that the receiver antenna has to receive higher than 294.15 dBm signal

the m following the dB means you are referencing this to a milliwatt. The absolute power level you are quoting is 2.5 x 1026 watts. Not likely.

In other cases you use the term dB with no normative reference which makes your statements ambiguous. The proper use of dB is a fundamental requirement to communicate effectively in this discipline.

Then there is this:

A 1 Hz magnetic signal, going between 2 magnetic antennas

There is no such thing as a magnetic (only) signal nor is there a magnetic (only) antenna. In RF and antenna engineering the signals are all electromagnetic (EM) signals. They always have an E and H field. There are antennas called magnetic antennas but this is simply a classification of a loop antenna that is still a conventional EM antenna.

I have seen from your past posting that you tend to apply boundary sphere conditions and you have done it here again:

with a boundary sphere of 20 cm has a path loss of roughly 16 dB

I encourage you to drop this line of thinking if you are interested in antenna engineering - it is sending down an errant path. The proper reference is the dBi gain of an antenna, not its boundary sphere.

  • \$\begingroup\$ I see, I made a few errors, I should have said dBW, because we should use SI values to avoid confusion in such long calculations. I think I have explained my question wrongly, should I delete this question and ask it again more precisely? \$\endgroup\$
    – David K.
    Jul 28, 2017 at 9:05
  • \$\begingroup\$ No you don't understand the boundary sphere. It is used to compute the maximum gain of the antenna that it physically can have, so that we can eliminate the gain variable and focus on the other variables. If we apply the theoretical maximum gain, then we can compute the other stuff with more certainty, not having to worry about the other variables, thus we assume 2 ideal antennas. \$\endgroup\$
    – David K.
    Jul 28, 2017 at 9:09
  • 1
    \$\begingroup\$ I would encourage you to edit your question to get it into the best condition possible by your judgement. I will be happy to edit my response accordingly. \$\endgroup\$
    – Glenn W9IQ
    Jul 28, 2017 at 10:58
  • \$\begingroup\$ You have edited your question and now reference a 294.15 dBW ratio. This is worse! That now 1000 times the previous value. As I said, spend some time getting very comfortable as to how dB's work. Then you will also feel comfortable getting away from boundary spheres in order to solve real world examples. \$\endgroup\$
    – Glenn W9IQ
    Jul 29, 2017 at 12:09

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