Digital communications terminology check: is a ‘‘baud’’ always equivalent to a fixed number of bits (for example, one bit)? Choose the best equivalence from the options below.

Difficulty: Easy

Correct Answer: None of the above

Explanation:


Introduction / Context:
Students often confuse ‘‘baud’’ and ‘‘bit rate’’. Baud is a unit of symbol rate (signals per second), while bit rate measures bits per second. Depending on the modulation scheme, each symbol (one baud) can represent one or more bits—or even carry coding overhead—so there is no single fixed equivalence in bits for one baud.


Given Data / Assumptions:

  • We are choosing a definition-equivalence, not performing arithmetic.
  • Options suggest specific bit counts per baud.
  • We must decide whether any fixed mapping is always true.


Concept / Approach:
Baud = symbols per second. Bit rate = bits per second. For binary signaling (2-level), one symbol encodes 1 bit, so baud = bit/s in that special case. For higher-order modulation (e.g., 4-level, 8-level, 16-QAM), bits per symbol = log2(M), where M is the number of signal levels/states. Thus, 1 baud may equal 2, 3, 4, or more bits depending on M, and with coding/line codes the mapping can deviate further. Therefore, there is no universal, ‘‘always’’ equivalence to a fixed bit count.


Step-by-Step Solution:

Recognize that baud measures symbol changes per second.Relate bits per symbol to log2(M) for M-ary modulation.Conclude that none of the fixed values listed are ‘‘always’’ correct.


Verification / Alternative check:
Examples: NRZ binary → 1 bit/symbol; 16-QAM → 4 bits/symbol; 64-QAM → 6 bits/symbol. This variability disproves any single fixed mapping.


Why Other Options Are Wrong:

a byte: fixed at 8 bits; unrelated to symbol rate.a bit: true only for binary modulation, not ‘‘always’’.100 bits / 16 bits: arbitrary fixed counts with no general basis.


Common Pitfalls:
Using ‘‘baud’’ and ‘‘bps’’ interchangeably; forgetting modulation order and coding change bits per symbol.


Final Answer:
None of the above.

More Questions from Networking

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion