Even parity rule in digital communications: “In even parity, the sum (count) of 1-bits in the code group must be even.” Determine if this definition is accurate and complete.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Parity bits provide a simple, low-cost method to detect single-bit errors during storage or transmission. The question checks knowledge of the precise rule that defines even parity and how the parity bit relates to the data bits.


Given Data / Assumptions:

  • A “code group” is the set of data bits plus, when present, a parity bit.
  • Even parity aims for an even count of logic 1s across the group after adding the parity bit.
  • Scope includes serial links (UART), memory systems, and simple buses where parity may be used.


Concept / Approach:
Parity enforces a constraint on the count of 1s. With even parity, the transmitter sets the parity bit so that the total number of 1s in data + parity is even. At the receiver, the same count is performed; a mismatch indicates an odd number of flipped bits (most notably a single-bit error).


Step-by-Step Solution:

Count the 1s in the data bits.If the count is already even, set parity bit to 0; if odd, set parity bit to 1.Transmit data plus parity; receiver repeats the count check.If total 1s are odd at the receiver, flag a parity error.


Verification / Alternative check:
Worked example: data 1011001 has four 1s (even). Even parity bit = 0; total remains even. If noise flips one bit, total becomes odd; error is detected. If two bits flip, parity may miss it (even number of errors).


Why Other Options Are Wrong:

  • Incorrect: The definition provided is the standard rule.
  • Applies only to 7-bit ASCII or odd parity: Parity rules are independent of the particular payload width and parity selection (odd/even).
  • Depends on stop bits: Stop bits are UART framing parameters separate from parity logic.


Common Pitfalls:
Confusing parity with checksums/CRCs; assuming parity guarantees multi-bit error detection (it does not); mixing up the direction (even vs. odd).


Final Answer:
Correct

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