Question

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, the ratio \(\frac{C}{N} = 8.8~dB\), the minimum theoretical \(\frac{{{E_b}}}{{{N_0}}}\) ratio for a receiver bandwidth equal to the minimum double-sided Nyquist bandwidth will be
1. 4.8 dB
2. 6.4 dB
3. 8.8 dB
4. 10.4 dB

Answer 1

Correct Answer - Option 3 : 8.8 dB

The bit energy is given by:

E_b = C × T_b

C = Carrier Power

\(E_b = \frac{C}{R_b}\)

The total noise power is given by:

N = N₀ × B

\(N_0 = \frac{N}{B}\)

∴ The required ratio is given by:

\(\frac{E_b}{N_0} = \frac{C}{R_b}\times\frac{B}{N}\)

For Nyquist 'or' sinc pulses, the BPSK bandwidth is equal to the bit rate Rb, i.e.

\(B = R_b\)

\(\therefore\frac{E_b}{N_0} = \frac{C}{R_b}\times\frac{R_b}{N}\)

\(\frac{E_b}{N_0} = \frac{C}{N}\)

Given \(\frac{C}{N}=8.8~dB\)

\(\frac{E_b}{N_0} = \frac{C}{N}=8.8~dB\)