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.
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ACorrect
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BIncorrect
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CApplies only to 7-bit ASCII
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DApplies only to odd parity systems
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EDepends on stop bits
Answer
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