Digital modulation – baud rate with M-ary PSK A 9600 bit/s digital stream is transmitted using 8-PSK (8-level phase shift keying). What is the resulting modulation rate in baud (symbols per second)?

Introduction / Context:
Modulation schemes like M-ary PSK map multiple bits onto one symbol. The baud rate (symbols per second) is therefore lower than the bit rate when each symbol represents more than one bit. This question checks your ability to relate bit rate, constellation size, and baud rate for M-ary modulation.

Given Data / Assumptions:

• Bit rate (Rb) = 9600 bit/s.
• Modulation = 8-PSK, which means M = 8 constellation points.
• Each symbol carries log2(M) bits.

Concept / Approach:
For M-ary modulation, bits per symbol = log2(M). Baud rate (Rs) = Rb / (bits per symbol). For 8-PSK, bits per symbol = log2(8) = 3. Therefore, Rs = 9600 / 3 = 3200 symbols/second (3200 baud).

Step-by-Step Solution:
Compute bits per symbol: log2(8) = 3.Relate baud and bit rate: Rs = Rb / (bits per symbol).Insert numbers: Rs = 9600 / 3.Calculate result: Rs = 3200 baud.

Verification / Alternative check:
If the scheme were BPSK (M=2), bits per symbol = 1, giving 9600 baud. For QPSK (M=4), bits per symbol = 2, giving 4800 baud. These comparisons validate that 8-PSK should yield 3200 baud.

Why Other Options Are Wrong:
1200 baud: would imply 8 bits per symbol, which is not 8-PSK.4800 baud: corresponds to QPSK (2 bits/symbol), not 8-PSK.9600 baud: corresponds to BPSK (1 bit/symbol), not 8-PSK.

Common Pitfalls:

• Confusing bit rate with baud rate; they are equal only when 1 bit per symbol.
• Using log10 instead of log2 for bits per symbol.
• Ignoring coding/overhead; in basic calculations we assume ideal mapping without FEC overhead.

Final Answer:
3200 baud