Difficulty: Medium
Correct Answer: 2400 baud
Explanation:
Introduction / Context:
Digital modulation conveys bits by changing a signal’s properties (amplitude, frequency, or phase). With M-ary schemes like 16-PSK, each symbol carries log2(M) bits. The modulation rate in baud equals symbols per second, which may be lower than the bit rate when multiple bits are packed into each symbol. Calculating baud from bps is a frequent exam task because it reinforces the distinction between bit rate and symbol rate.
Given Data / Assumptions:
Concept / Approach:
Relationship: R_b = (symbols per second) * (bits per symbol). Therefore, symbols per second (baud) = R_b / k. For 16-PSK, k = 4, which reduces the symbol rate compared to binary schemes. This is the standard approach across M-ary modulation families (M-PSK, M-QAM, etc.).
Step-by-Step Solution:
Verification / Alternative check:
As a cross-check, if the system were binary PSK (M=2), k=1 and baud would equal 9600. For 4-PSK (QPSK), k=2 → 4800 baud. For 8-PSK, k=3 → 3200 baud. These comparisons align with the 16-PSK result of 2400 baud, confirming correctness.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing “baud” and “bps”; forgetting to take log2(M); ignoring overhead (which is excluded unless explicitly stated); assuming baud equals bps for all modulations, which is false for M-ary schemes.
Final Answer:
2400 baud
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