Introduction / Context:
The maximum error-free symbol rate in a noiseless baseband channel is governed by the Nyquist criterion for zero intersymbol interference. For binary signaling (two levels), the bit rate equals twice the bandwidth when ideal conditions are met. This question assesses your ability to apply the Nyquist formula to compute theoretical maximum bit rate.
Given Data / Assumptions:
- Noiseless channel, bandwidth B = 3 kHz.
- Binary signaling (M = 2), ideal ISI-free (Nyquist) pulses.
- No coding constraints beyond binary PAM.
Concept / Approach:
Nyquist criterion: the maximum symbol rate Rs in a baseband channel without ISI is Rs = 2B symbols/s. With binary symbols (1 bit per symbol), the maximum bit rate Rb equals Rs.
Step-by-Step Solution:
Start with Nyquist: Rs = 2B.Compute symbol rate: Rs = 2 × 3000 = 6000 symbols/s.Binary signaling gives 1 bit per symbol → Rb = 6000 bps.
Verification / Alternative check:
Shannon capacity C = B log2(1 + SNR) exceeds Nyquist rate for sufficiently high SNR, but in the noiseless case Nyquist limit for binary pulses yields 2B bps; increasing constellation size M could raise bits per symbol.
Why Other Options Are Wrong:
12,000 and 10,000 bps: These would require higher-order modulation (M > 2) or passband assumptions not stated.3,000 bps: This equals B and ignores the factor 2.1,500 bps: Far below the theoretical limit in a noiseless scenario.
Common Pitfalls:
Confusing baseband Nyquist limit with passband constraints, or misapplying Shannon capacity without SNR details.
Final Answer:
6,000 bps
Discussion & Comments