Difficulty: Easy
Correct Answer: changes from LOW to HIGH
Explanation:
Introduction / Context: Synchronous sequential circuits rely on edge-triggered flip-flops to register data at precise instants. Understanding exactly when a positive edge-triggered flip-flop samples its inputs is crucial for proper timing closure, meeting setup/hold times, and avoiding metastability in digital systems.
Given Data / Assumptions:
Concept / Approach: By definition, a positive edge-triggered flip-flop samples its inputs at the instant the clock transitions from LOW to HIGH. Data present at the input must be stable for a window that starts tsu before the edge and continues through th after the edge. The output Q updates as a result of this sampling and remains stable until the next active edge (ignoring asynchronous overrides).
Step-by-Step Solution:
Recognize the sampling instant: the rising edge of the clock.Relate to timing: inputs must be valid during [edge − tsu, edge + th].Eliminate distractors: clock level (HIGH or LOW) does not define sampling for edge-triggered devices.Select correct completion: “changes from LOW to HIGH.”Verification / Alternative check:
Timing diagrams show Q updating at the clock’s rising edge, not throughout the HIGH level.Why Other Options Are Wrong:
is LOW / is HIGH: These describe levels (latch-like behavior), not edges.changes from HIGH to LOW: That is the negative edge condition, not positive.Common Pitfalls:
Confusing level-sensitive latches with edge-triggered flip-flops.Ignoring setup/hold margins near the active edge.Final Answer:
changes from LOW to HIGH
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