A digital link uses Phase Shift Keying (PSK) at a bit rate of 9600 bps with 16 distinct phase levels (16-PSK). What is the line signaling speed, i.e., the modulation rate in baud (symbols per second)?

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:

  • Bit rate (R_b) = 9600 bits per second.
  • Modulation = 16-PSK (M = 16 distinct phases).
  • Bits per symbol (k) = log2(M) = log2(16) = 4 bits per symbol.
  • No coding or framing overhead is included in the basic calculation.


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:

Compute k = log2(16) = 4 bits/symbol. Compute baud = 9600 bps / 4 bits/symbol = 2400 symbols/second. Express the modulation rate: 2400 baud. Select the closest matching option exactly: 2400 baud.


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:

  • 1200 baud: would correspond to k = 8 bits/symbol at 9600 bps (not 16-PSK).
  • 4800 baud: would correspond to QPSK (k=2) at 9600 bps.
  • 9600 baud: true only for binary schemes (k=1) at 9600 bps.
  • 600 baud: does not match any common M for 9600 bps without coding/overhead assumptions.


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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