I know that FSK occupies a larger bandwidth compared other keying schemes,

$$ BW= f_1 -f_2 +2R_b $$

where $f_1$ and $f_2$ are carrier frequencies and $R_b$ is the bit rate.

For achieving a particular bit rate say 8 Kbps, how can I find the minimum bandwidth if I am free to choose the carrier frequencies?

  • 1
    $\begingroup$ Your expression for the bandwidth of a binary (orthogonal) FSK system is not quite correct. Generally speaking, the difference $|f_!-f_2|$ between the two carriers needs to be a multiple of the bit rate $R_b$ so that the Mark and Space signals are orthogonal. The minimum bandwidth is thus $3R_b$. Alternatively, you can choose two vastly different carrier frequencies and get two (spectrally non-overlapping) OOK signals occupying $2R_b$ Hz each for a total bandwidth of $4R_b$ Hz but with lots of unused frequency spectrum separating them. $\endgroup$ Dec 7, 2015 at 20:54
  • $\begingroup$ @DilipSarwate: how do you define BW occupancy to get those $3R_b$? As far as I know, the rule of thumb for M-ary FSK is $B=M/2T$. $\endgroup$
    – vaz
    Dec 7, 2015 at 21:10
  • 1
    $\begingroup$ @vaz $3R_b$ is basically the "Carson's rule" bandwidth with $f_1$ and $f_2$ both being multiples of $R_b$ and differing by $R_b$. With rectangular baseband pulses, the Mark and Space signals have sinc spectra centered at $f_1$ and $f_2$ respectively with the first null of each spectrum coinciding with the center frequency of the other. The first null of the sinc spectrum on the other side is also $R_b$ Hz away, giving $3R_b$ as the bandwidth: from $f_1-R_b$ to $f_2+R_b$ where $f_2 = f_1+R_b$. See Chapter 6.6 of T.T.Ha, Theory and Design of Digital Communication Systems. $\endgroup$ Dec 7, 2015 at 22:07


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