Determine the minimum bandwidth required to achieve a P(e) of 10^{-7} for an 8-PSK system operating at 10 Mbps with a carrier-to-noise power ratio of 11.7 dB.
From Figure 48, the minimum E_{b}/N_{0} ratio to achieve a P(e) of 10^{-7} for an 8-PSK system is 14.7 dB. The minimum bandwidth is found by rearranging Equation 44:
\frac{E_{b}}{N_{0}}=\frac{C}{N}×\frac{B}{f_{b}} (44)
\frac{B}{f_{b}}=\frac{E_{b}}{N_{0}}-\frac{C}{N}= 14.7 dB – 11.7 dB = 3 dB
\frac{B}{f_{b}} = antilog 3 = 2
B = 2 × 10 Mbps = 20 MHz