What is the relationship between the reach of a RF signal and its frequency?
What I mean is: If power is kept constant, should I use high or low frequency waves to get a better reach? Why?
\$\begingroup\$
\$\endgroup\$
1
-
1\$\begingroup\$ What have you already read or done do find out? \$\endgroup\$– user16324Commented Mar 24, 2013 at 22:24
Add a comment
|
4 Answers
\$\begingroup\$
\$\endgroup\$
11
In free space, it doesn't matter. The power per incident area of a propagating wave is inversely proportional to the square of the distance from the transmitter. This is true regardless of frequency.
Certain frequencies are reflected, refracted, absorbed, and scattered differently by different materials. There is no single monotonic relationship until you get to really high energies, like gamma rays and beyond. At these really high energies (high frequencies), the waves basically just blast thru any material in their way, with higher energies passing thru material with less attenutation. Up to below Xray frequencies, there is no single answer, and it depends on the material between the transmitter and receiver.
Diffraction effects can make low fequencies (long wavelengths) appear to bend around objects, but this actually occurs at all wavelengths. The "near" layer where diffraction effects occur scales with wavelength, so it appears to us at a fixed human scale that long wavelengths go "around" objects where short wavelengths don't, but that is due to our perception scale. On the scale of the earth, commercial AM radio frequencies around 1 MHz are low enough to diffract around the curvature of the earth to some extent making over the horizon AM reception possible. Commercial FM radio, being 100x shorter wavelength, exhibits this effect much less for the same size earth, so FM radio appears to us to be mostly occluded by the horizon.
answered Mar 24, 2013 at 22:57
-
-
\$\begingroup\$ The type of modulation isn't relevant. \$\endgroup\$ Commented Mar 24, 2013 at 23:40
-
\$\begingroup\$ For an absolute fixed distance and an absolute fixed receiving antenna aperture, the higher frequency benefits from receiving antenna gain. But if the receiving antenna is sized by wavelength (such as both are 1/4 wave), the lower frequency has the advantage due to size (but that's the same as being "closer in wavelengths"). If everything (even distance) is sized in wavelengths, it should come out even. But in real life distance in wavelengths is never the situation. \$\endgroup\$– SkaperenCommented Mar 25, 2013 at 7:43
-
\$\begingroup\$ @Skape: Remember that antenna "gain" is really directionality. Yes you can form a more directional beam with a larger antenna relative to the wavelength, but that is not what the OP asked about. \$\endgroup\$ Commented Mar 25, 2013 at 11:44
-
\$\begingroup\$ @LeonHeller I believe Olin was using "AM radio frequencies" to refer to the MF broadcast band between roughly 500 kHz and 1700 kHz by a name that is more commonly understood. \$\endgroup\$– userCommented Mar 25, 2013 at 12:25
\$\begingroup\$
\$\endgroup\$
You are looking at a quite deep topic called radio wave propagation.
From what I understand, low frequencies can follow the curvature of the earth (ground wave) and can get around obstacles more easily due to their larger wavelength. Higher frequencies are often restricted to line-of-sight propagation.
\$\begingroup\$
\$\endgroup\$
2
"Free-space propagation of a waveform between antennas is completely independent of frequency"
However, apart from the above mentioned effect of low-frequencies following the curvature of the earth, another common reason why lower frequency signals get better communication distances is that their quarter-wave (or other type) antennas are physically larger, giving a larger "effective area" in the antenna equations, and better coupling between transmitter and receiver.
-
\$\begingroup\$ Damn I was gonna say that!!! \$\endgroup\$– Andy akaCommented Mar 24, 2013 at 23:33
-
1\$\begingroup\$ When you start heading into the centimetric wavelengths, super-high-gain antennas become practical in terms of physical size. In the HF range (3-30 MHz), you're lucky to get 10 dBi (dB above an isotropic antenna) before physical antenna size becomes a severely limiting factor. In the SHF range (3-30 GHz), 10 dBi is a walk in the park as far as the antenna goes (though you'll have other concerns). (A simple dipole at proper height gives you 2.15 dBi, if I don't recall wrong.) This doesn't necessarily invalidate your point, but there's a limit to how physically large antennas are practical. \$\endgroup\$– userCommented Mar 25, 2013 at 12:20
\$\begingroup\$
\$\endgroup\$
Besides from what has been said by the others (in free space frequency doesn't matter, if there is not free space absorbtion, refraction, diffraction, scattering etc. has to be considered and those dependend very much on frequency and the matter inbetween) it will also be important how easy it is the construct effective antennas for the frequency you use. You need to get the power somehow into the space and out of it.
E.g. if you want to use a rather low frequency in order to penetrate into matter (e.g. below earth or sea level) but it is very difficult to build antennas for such low frequencies (several km wavelength).
