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.


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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