As a rule of thumb, the bandwidth of a communication system with a complex signal must be equal to the bit-rate or symbol-rate?

Besides, after reading this post, it is not clear for me how a band-filter or low-filter signal can influence the maximum data rate.


1 Answer 1


In order to avoid inter symbol interference (ISI) the baseband transmit signal must meet Nyquist's first criterion. In frequency domain it can be formulated as follows. $$ \omega_\mathrm{N} = \frac{v_\mathrm{s}}{2} $$ With $v_\mathrm{s}$ the symbol rate and $\omega_\mathrm{N}$ the 3dB frequency of a low pass filter with symmetric flank. The flank must be point symmetric with regard to $\omega_\mathrm{N}$ and zero for $\omega > \omega_\mathrm{N}(1 + \alpha)$. Consequently, the baseband bandwidth $B=\omega_\mathrm{N}(1 + \alpha)$.

After modulation the bandwidth is doubled and the radio frequency bandwidth is $2B = 2\omega_\mathrm{N}(1 + \alpha) = v_\mathrm{s}(1 + \alpha)$. For the ideal case of a rectangular low pass filter $\alpha=0$ and $2B = v_\mathrm{s}$ which corresponds to the statement that the required bandwidth is equal to the symbol rate *). Similarly, if the bandwidth $B$ is given, the symbol rate is limited by $v_\mathrm{s} \leq \frac{2B}{1+\alpha}$

The required bandwidth indirectly depends on the bitrate, because a certain number of bits is transmitted per symbol.

The lowpass mentioned above is also referred to as impulse shaper and $\alpha$ as roll-off factor.

*) As an ideal lowpass isn't feasible, the required RF bandwidth in practice is always larger than the symbol rate.


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