For a noiseless channel of bandwidth 3 kHz transmitting binary level signals, what is the maximum theoretical data rate according to the Nyquist criterion?

Introduction / Context:
The Nyquist criterion provides an upper bound on the data rate for a noiseless channel given a limited bandwidth and a specified number of discrete signaling levels. This question asks you to apply the Nyquist formula to a 3 kHz channel using binary signaling (two levels) to find the maximum bit rate in bits per second (bps).

Given Data / Assumptions:

• Channel bandwidth B = 3 kHz (that is, 3000 Hz).
• Number of signal levels M = 2 (binary).
• The channel is idealized as noiseless; intersymbol interference is neglected under Nyquist sampling.

Concept / Approach:
Nyquist maximum data rate for a noiseless channel is given by: R_max = 2 * B * log2(M) For binary signaling (M = 2), log2(2) = 1, so the formula simplifies to: R_max = 2 * B Substituting B = 3000 Hz yields 6000 bps, which is 6 Kbps.

Step-by-Step Solution:

Verification / Alternative check:
Compare with higher-level modulation: if M = 4 (quaternary), R_max would double to 12 Kbps for the same bandwidth, confirming that binary (M = 2) gives 6 Kbps at 3 kHz.

Why Other Options Are Wrong:

• 3 Kbps: Misses the factor of 2 in the Nyquist formula for baseband signaling.
• 12 Kbps / 24 Kbps: These would require higher M (more levels) or additional assumptions; they exceed the binary case.
• None of the above: Incorrect because 6 Kbps is exactly the Nyquist limit here.

Common Pitfalls:
Confusing Nyquist's noiseless limit with Shannon's noisy-channel capacity, or forgetting that binary signaling (M = 2) makes log2(M) equal to 1.

Final Answer:
6 Kbps