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According to Wikipedia, audio frequency-shift keying (AFSK) is done by using binary FSK, i.e. 2 tones of different frequencies, a tone for the 1 bit and a tone for the 0 bit.

  • But isn't it possible to choose more than 2 frequencies (any frequency between 0 and half the sampling rate) to transmit more than binary data?
  • Why don't we use say 10 tones of different frequencies, to transmit in decimal, or say 256 tones to transmit in chars? Wouldn't that make transmission much more efficient?
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The three main reasons appear to be these:

1) In AFSK, it's not just jumping back and forth between frequencies. The tones must also be continuous phase. In other words, when the tone changes, there can't be any jump in phase. For example, if you're sending a 1200 hz tone and the waveform is at its peak when you switch to 2200 hz, the waveform is still at its peak, so it can't be just any two frequencies. It has to be at the peak. It can't start back at zero, or halfway through; it can't start anywhere but at the peak. Additionally, the length of each tone must be the same. Because of these two points, the amount of wavelengths that share compatible periods is finite. For instance, if you use 1000 hz, your next compatible is something like 1500 hz (my apologies if my numbers are off; I'm worn out).

2) As a follow-on from 1, since AFSK is audio, the range of compatible frequencies is not that high.

3) The amount of bandwidth required for the transmission increases for each additional tone, making data efficiency greater, but power and bandwidth efficiency less.

That doesn't meant that multiple frequencies can't be used; in fact MFSK has been used since the mid-60s. However, MFSK has several drawbacks that make it untenable in some situations. Delay spread, when broadcasting to multiple locations, and fading make it less desirable for some applications.

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    $\begingroup$ sorry I'm a bit new to this, but why is it necessary for the tones to be continuous phase (in point 1)? $\endgroup$ – Rufus Sep 28 '16 at 7:27
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    $\begingroup$ Discontinuities contain high frequency content, which may not fit within the allowed bandwidth. $\endgroup$ – hotpaw2 Sep 28 '16 at 8:02
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There's a time vs. frequency trade-off that can work against you. At a given signal-to-noise ratio, it takes longer (and often more compute power) to reliably determine which of two frequencies are being sent when they are close together than when they are far apart or orthogonal within the time interval. So if you use 10 times more frequencies within the same bandwidth, but have to slow down the rate at which you change frequencies by 10X to get the same statistical reliability of detection of each one apart from all the others, you can end up with a slower data rate (in bits per second).

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