Bandwidth of a communication system
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 $\omega &gt; \omega_\mathrm{N}(1 + \alpha)$. 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.