Question

In a coherent BPSK modulation scheme, if the input bit rate is doubled while keeping the symbol energy as constant, what is the effect on the transmission bandwidth ($BW$) and the probability of bit error ($P_e$)?

• A. $BW$ doubles, $P_e$ increases
• B. $BW$ doubles, $P_e$ remains same
• C. $BW$ remains same, $P_e$ increases
• D. $BW$ doubles, $P_e$ decreases

Hint:

Be alert to the wording: if they say power is kept constant while bit rate doubles, bit energy $E_b = \frac{P}{R_b}$ would be halved, which would increase $P_e$. But since they specified energy is kept constant, $P_e$ remains completely unchanged!

Answer 1

The Correct Option is B

Concept: For a Binary Phase Shift Keying (BPSK) modulation scheme:
• The transmission bandwidth required for a main-lobe transmission of a rectangular pulse is proportional to the bit rate ($R_b$). Specifically, for binary systems, $BW = 2R_b$ (or $BW = R_b$ for baseband/minimum bandwidth allocations depending on pulse shaping definitions; either way, it scales linearly with $R_b$).
• The probability of bit error ($P_e$) for coherent BPSK depends solely on the ratio of bit energy ($E_b$) to noise power spectral density ($N_0$): $$P_e = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$$ Step-by-step Analysis of the Given Conditions:
• Condition 1: Input bit rate is doubled. Let the initial bit rate be $R_{b1}$ and the new bit rate be $R_{b2} = 2R_{b1}$. Since the transmission bandwidth $BW$ is directly proportional to the bit rate ($BW \propto R_b$): $$BW_2 = 2 \times BW_1$$ Hence, the transmission bandwidth doubles.
• Condition 2: Symbol energy is kept constant. In binary modulation configurations such as BPSK, one symbol encapsulates exactly one bit of data information. Therefore, the symbol energy ($E_s$) is strictly equal to the bit energy ($E_b$): $$E_b = E_s$$ The question explicitly dictates that the symbol energy remains completely constant ($E_{s2} = E_{s1}$). This directly implies that the bit energy remains invariant as well ($E_{b2} = E_{b1}$).
• Since $P_e$ is determined solely by $E_b$ and $N_0$, and neither of these parameters is modified: $$P_{e2} = Q\left(\sqrt{\frac{2E_{b2}}{N_0}}\right) = Q\left(\sqrt{\frac{2E_{b1}}{N_0}}\right) = P_{e1}$$ Therefore, the probability of bit error remains identical/same.