In almost all long distance communication AM is preferred over FM. Do type of modulation has anything to do with this or it is just depending on transmitting frequency?

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    \$\begingroup\$ I think you need to justify your assertion. \$\endgroup\$ – Andy aka Aug 28 '14 at 7:11
  • \$\begingroup\$ I second that. May that be that the traditionally allocated slots for AM are in the f range where the EM waves follow the earth/bounce on the ionosphere? AM is usually less energy efficient than FM but the receiver is trivial to build... Nothing to do with distance though. \$\endgroup\$ – Vladimir Cravero Aug 28 '14 at 7:24
  • \$\begingroup\$ Are you linking carrier frequency and modulation scheme and asserting its the modulation scheme that is more attractive over long distances? \$\endgroup\$ – JonRB Aug 28 '14 at 7:25
  • \$\begingroup\$ Related: What is the basic difference between AM and FM radio? \$\endgroup\$ – JYelton Aug 28 '14 at 15:57

I don't think so. Voyager 1 is a great example of this - it uses phase modulation (similar to frequency modulation) and it has been able to transmit messages to earth that are ridiculously long distances away.

As always, the "link loss" between transmitter and receiver is paramount in understanding how much attenuation occurs between transmitter and receiver. This is: -

link loss (dB) = = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.

What you may be influencing your question is that ionospheric bounce occurs quite well in the lower frequencies that are dominated by public AM transmission such as a local radio station. AM is used not because it gets further but because the thousands of radio receivers tuned-in have a simpler job demodulating the transmission i.e. it is cost driven - non-suppressed carrier AM is easy and cheap to decode, hence the extremely simple crystal set radios of the last century.


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