Question 14.6: A coherent binary phase-shift-keyed (BPSK) transmitter opera......
A coherent binary phase-shift-keyed (BPSK) transmitter operates at a bit rate of 20 Mbps. For a probability of error P(e) of 10^{-4},
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. With BPSK, the minimum bandwidth is equal to the bit rate, 20 MHz. From Figure 24, the minimum C/N is 8.8 dB. Substituting into Equation 20 gives us
\frac{E_{b}}{N_{0}}=\frac{C}{N}+\frac{B}{f_{b}} (20)
= 8.8 dB + 10 log\frac{20 \,× \,10^{6}}{20 \,× \,10^{6}}
= 8.8 dB + 0 dB = 8.8 dB
Note: The minimum E_{b}/N_{0} equals the minimum C/N when the receiver noise bandwidth equals the bit rate which for BPSK also equals the minimum Nyquist bandwidth. The minimum E_{b}/N_{0} of 8.8 can be verified from Figure 25.
What effect does increasing the noise bandwidth have on the minimum C/N and E_{b}/N_{0} ratios?
The wideband carrier power is totally independent of the noise bandwidth. However, an increase in the bandwidth causes a corresponding increase in the noise power. Consequently, a decrease in C/N is realized that is directly proportional to the increase in the noise bandwidth. E_{b} is dependent on the wideband carrier power and the bit rate only. Therefore, E_{b} is unaffected by an increase in the noise bandwidth. N_{0} is the noise power normalized to a 1-Hz bandwidth and, consequently, is also unaffected by an increase in the noise bandwidth.
b. Because E_{b}/N_{0} is independent of bandwidth, measuring the C/N at a point in the receiver where the bandwidth is equal to twice the minimum Nyquist bandwidth has absolutely no effect on E_{b}/N_{0} . Therefore, E_{b}/N_{0} becomes the constant in Equation 20 and is used to solve for the new value of C/N. Rearranging Equation 20 and using the calculated E_{b}/N_{0} ratio, we have
\frac{E_{b}}{N_{0}}=\frac{C}{N}+\frac{B}{f_{b}} (20)
\frac{C}{N}=\frac{E_{b}}{N_{0}}-\frac{B}{f_{b}}
= 8.8 dB − 10 log\frac{40\, × \,10^{6}}{20\, × \,10^{6}}
= 8.8 dB − 10 log 2
= 8.8 dB − 3 dB = 5.8 dB
c. Measuring the C/N ratio at a point in the receiver where the bandwidth equals three times the minimum bandwidth yields the following results for C/N:
\frac{C}{N}=\frac{E_{b}}{N_{0}}− 10\, \log\frac{60 \,×\, 10^{6}}{20\, ×\, 10^{6}}
= 8.8 dB − 10 log 3 = 4.03 dB
The C/N ratios of 8.8, 5.8, and 4.03 dB indicate the C/N ratios that could be measured at the three specified points in the receiver and still achieve the desired minimum E_{b}/N_{0} and P(e).
Because E_{b}/N_{0} cannot be directly measured to determine the E_{b}/N_{0} ratio, the wideband carrier-to-noise ratio is measured and, then, substituted into Equation 20. Consequently, to accurately determine the E_{b}/N_{0} ratio, the noise bandwidth of the receiver must be known.
