A digital system uses a 6-MHz channel and 4-level signaling (M = 4). Assuming the ideal Nyquist criterion for a noiseless baseband channel, what is the maximum symbol rate achievable?

Introduction / Context:
The Nyquist criterion sets a theoretical upper bound on symbol rate for intersymbol-interference-free transmission over an ideal baseband channel with limited bandwidth. Separating symbol rate (baud) from bit rate (bit/s) is essential, especially when multi-level signaling is used.

Given Data / Assumptions:

• Channel bandwidth B = 6 MHz.
• Signal levels M = 4 (2 bits per symbol if used for bit rate computation).
• Assume ideal, noiseless baseband with perfect Nyquist pulses.
• Question asks explicitly for symbol rate (baud), not bit rate.

Concept / Approach:
For an ideal baseband channel, the Nyquist maximum symbol rate is R_s,max = 2 * B symbols per second. With B = 6e6 Hz, the maximum symbol rate is 12e6 symbols/s, i.e., 12 Mbaud/s. If we were asked for bit rate, we would multiply by log2(M): R_b,max = R_s,max * log2(4) = 12e6 * 2 = 24 Mb/s.

Step-by-Step Solution:
Write Nyquist formula for baud: R_s,max = 2 * B.Substitute B = 6 MHz → R_s,max = 12 MHz in baud.Match to options: 12 Mbaud/s.

Verification / Alternative check:
Cross-check using bit rate notion: with M = 4, each symbol carries 2 bits, giving 24 Mb/s maximum in the same conditions—consistent with 12 Mbaud/s symbol rate.

Why Other Options Are Wrong:
1.5 or 6 Mbaud/s: Below the Nyquist limit for 6 MHz.

24 Mbaud/s: Would require >12 MHz bandwidth for ideal Nyquist signaling.

None of the above: Incorrect because 12 Mbaud/s fits.

Common Pitfalls:
Confusing bit rate with baud; multi-level signaling increases bits per symbol but not the maximum symbol rate allowed by Nyquist for a given bandwidth.

Final Answer:
12 Mbaud/s