My question is related to the performance of 2-FSK modulation scheme against M-FSK (M > 2). From the SNR vs BER curve it seems that as the number of tones (frequencies) increases the BER decrease for the same SNR compared to 2-FSK(pics attached). (I assume maintaining the same symbol duration)

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As the number of tones increases the bandwidth increases but i can't figure out why the performance of MFSK also increases. At first glance i would say that the resilience to the noise should be the same as the 2-FSK.

Which is the benefit of using 2-FSK instead of M-FSK (M>2) given a fixed available bandwidth?



1 Answer 1


(I assume maintaining the same symbol duration)

nope! That is usually not the case, because it would be unfair – if you send the same number of symbols per second, you'd get twice as many bits per second using 4-FSK rather than 2-FSK.

Therefore, the comparison is typically done with identical Bit Energy rather than identical Symbol duration (which would be, for constant envelope signals, be equal to identical symbol energy).

Hence, with higher modulation order, you get less symbols per bit, and hence, for constant $E_b$ more $E_{sym}$.

Now, errors happen on a symbol level – and a Symbol Error Rate of $r$ doesn't imply a BER of $r$ – in fact, a wrong symbol has at most its worth in bits in wrong bits, but usually less (see: Gray code), so $\text{BER}\le\text{SER}$ generally holds, and one can in an uncorrelated noise scenario be almost certain that $\text{BER}<\text{SER}$.

  • $\begingroup$ Thanks Marcus, i assumed the same symbol duration and i couldn't figure out why. But if i understood well the evaluation is done within the same bit energy which means that the symbol duration for the case of 4-FSK is twice of 2-FSK, maintaining in this case the same bit rate. In such case the BER is lower as the symbol duration is longer. $\endgroup$
    – peter
    Sep 2, 2016 at 19:50
  • $\begingroup$ jep! And because if, for example, one fourth of all symbols is wrong for 4-PSK, then less than one fourth of the bits is, whereas for 2-FSK SER = BER! $\endgroup$ Sep 2, 2016 at 20:48

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