Difficulty: Easy
Correct Answer: clock is HIGH
Explanation:
Introduction / Context:Master–slave flip-flops use two level-sensitive latches in sequence. The master latch is typically enabled when the clock is HIGH (for a positive-logic design), and the slave latch is enabled when the clock is LOW, resulting in an effective edge-controlled transfer of data. Understanding when inputs must be stable avoids unintended toggling and races.
Given Data / Assumptions:
Concept / Approach:Because the master latch is transparent when the clock is HIGH, any change on J or K during that interval can alter what the master captures, leading to incorrect or unpredictable outcomes. Therefore, J and K must be stable throughout the HIGH phase so that the master captures a single, well-defined state to transfer to the slave on the subsequent LOW phase (or edge-equivalent moment).
Step-by-Step Solution:
1) Identify transparency window: master is transparent while CLK = HIGH.2) Requirement: keep J and K steady during this interval.3) Slave transfers when CLK = LOW; by then, the master state is fixed.4) Result: predictable output transition without double toggling.Verification / Alternative check:Refer to timing diagrams; when J and K vary during clock HIGH, the master can respond multiple times, producing errors once the slave samples.
Why Other Options Are Wrong:
Common Pitfalls:Forgetting that level-sensitive windows exist within master–slave designs; assuming they behave like ideal edge-triggered devices.
Final Answer:clock is HIGH
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