Digital clock indicator logic: when should the PM indicator assert (go HIGH)? In a 12-hour digital clock, identify the exact time transition at which the PM indicator should change state from LOW to HIGH.
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ANever
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BGoing from 11:59:59 to 12:00:00
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CGoing from 12:59:59 to 01:00:00
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DOn the falling edge of the clock after enable goes high
Answer
Correct Answer: Going from 11:59:59 to 12:00:00
Explanation
Introduction / Context:A 12-hour clock toggles its AM/PM indicator at noon and midnight. Correctly placing the transition is important to avoid off-by-12-hours display bugs when minutes and hours roll over.
Given Data / Assumptions:
- AM corresponds to times 12:00:00 through 11:59:59 of the first half-day.
- PM corresponds to times 12:00:00 through 11:59:59 of the second half-day.
- We consider the exact second-level transition.
Concept / Approach:The PM indicator asserts at the instant the clock moves into the afternoon period: precisely at 12:00:00 noon (following 11:59:59 AM). It deasserts at midnight (12:00:00 AM) when the cycle restarts.
Step-by-Step Solution:
Track time through 11:59:59 AM.Increment seconds → 12:00:00 noon.Toggle AM/PM flip-flop → PM goes HIGH at this boundary.Next toggle at 11:59:59 PM → 12:00:00 AM (PM goes LOW).Verification / Alternative check:Truth table of AM/PM over a 24-hour period confirms toggles only at the two 12:00:00 transitions.
Why Other Options Are Wrong:
- A: The indicator certainly toggles twice per day.
- C: 12:59:59 to 01:00:00 does not change AM/PM.
- D: Refers to a signal edge rather than the time-state boundary.
Common Pitfalls:Incorrectly toggling at 11:59:59 or 12:00:01 introduces a one-second display error.
Final Answer:Going from 11:59:59 to 12:00:00