Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:
Edge-triggered flip-flops sample inputs at a specific clock transition and then update outputs after a propagation delay referenced to that same edge. The idea that the output waits for the opposite clock edge mixes up edge-triggered behavior with two-phase master–slave latch operation.
Given Data / Assumptions:
Concept / Approach:
At the active edge, the flip-flop captures inputs and begins internal transitions. The output reflects the new state after a short delay (nanoseconds), not at the opposite clock edge. Master–slave latches internally use two non-overlapping phases but still present a single-edge interface; the public Q does not wait for the trailing edge in true edge-triggered implementations.
Step-by-Step Solution:
Identify the specified active edge (e.g., rising edge).Ensure D/J/K inputs are stable for tSU/tH around this edge.Observe that Q updates after the edge by tPLH/tPHL, completing well before the next opposite edge.Therefore, the statement claiming change only at the trailing edge is incorrect.
Verification / Alternative check:
Datasheet timing diagrams show clock-to-Q referenced to the same edge. Simulation with a standard D or J–K flip-flop confirms immediate (delayed) response after the active edge.
Why Other Options Are Wrong:
“Correct” contradicts standard timing. “True only for master–slave” and “true only with gated clocks” add conditions that still do not defer Q until the opposite edge in edge-triggered devices.
Common Pitfalls:
Confusing transparent latching intervals with edge capture; ignoring propagation delay and thus expecting instantaneous change at the edge.
Final Answer:
Incorrect
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