Difficulty: Medium
Correct Answer: burst
Explanation:
Introduction / Context: Error classifications help diagnose channel issues and select suitable detection/correction schemes. When a transmitted character turns into another, the difference in their binary codes reveals whether the corruption affected one bit, several bits scattered, or a contiguous run (burst).
Given Data / Assumptions:
Concept / Approach: Compare the binary patterns. G = 0100 0111 and D = 0100 0100 differ in the two least-significant bits: 11 -> 00. Two adjacent bit positions changed, which is characteristic of a short burst error (contiguous bits corrupted).
Step-by-Step Solution: Write binary: G = 0100 0111.Write binary: D = 0100 0100.Identify differences: bit0 and bit1 (LSBs) flipped from 1 to 0.Adjacent flips imply a burst of length 2 rather than an isolated single-bit error.
Verification / Alternative check: Hamming distance between G and D is 2. Because the flipped bits are contiguous, it aligns with “burst” classification (a run of bits altered). If the two flipped bits were far apart, it would be multiple-bit non-burst.
Why Other Options Are Wrong: single-bit: Only one bit differs; here two do.
multiple-bit: True that multiple bits changed, but “burst” is a more precise classification given adjacency. recoverable: Not a standard error type; recoverability depends on coding. framing: Concerns start/stop/flag alignment, not symbol bit flips.Common Pitfalls: Equating any 2-bit error with generic “multiple-bit.” When adjacency exists, “burst” conveys cause (e.g., short noise spike).
Final Answer: burst
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