Question 14.7: A coherent 8-PSK transmitter operates at a bit rate of 90 Mb......
A coherent 8-PSK transmitter operates at a bit rate of 90 Mbps. For a probability of error of 10^{-5},
a. Determine the minimum theoretical C/N and E_{b}/N_{0} ratios for a receiver bandwidth equal to the minimum double-sided Nyquist bandwidth.
b. Determine the C/N if the noise is measured at a point prior to the bandpass filter where the bandwidth is equal to twice the Nyquist bandwidth.
c. Determine the C/N if the noise is measured at a point prior to the bandpass filter where the bandwidth is equal to three times the Nyquist bandwidth.
a. 8-PSK has a bandwidth efficiency of 3 bps/Hz and, consequently, requires a minimum bandwidth of one-third the bit rate, or 30 MHz. From Figure 24, the minimum C/N is 18.5 dB. Substituting into Equation 20, we obtain
\frac{E_{b}}{N_{0}}=\frac{C}{N}+\frac{B}{f_{b}} (20)
\frac{E_{b}}{N_{0}}= 18.5\, dB + 10\log\frac{30\,MHz}{90\,Mbps}
= 18.5 dB + (−4.8 dB) = 13.7 dB
b. Rearranging Equation 20 and substituting for E_{b}/N_{0} yields
\frac{C}{N}=13.7\,dB\, -\, 10\log\frac{60\,MHz}{90\,Mbps}
=13.7 dB − (−1.77 dB) = 15.47 dB
c. Again, rearranging Equation 20 and substituting for E_{b}/N_{0} gives us
\frac{C}{N}=13.7\,dB \,-\, 10\log\frac{90\,MHz}{90\,Mbps}
= 13.7 dB (dB) = 13.7 dB
