Difficulty: Easy
Correct Answer: presetting one flip-flop and clearing all the others.
Explanation:
Introduction / Context:
A ring counter circulates a single 1 (or single 0) through a chain of flip-flops. For deterministic behavior, exactly one stage must start in the opposite state to the others. If the counter powers up with all zeros (or multiple ones), it may stick or behave unpredictably. Proper initialization ensures the intended cyclic sequence begins immediately.
Given Data / Assumptions:
Concept / Approach:
To create the single circulating 1, one flip-flop must be preset to 1 while all others are cleared to 0. On each clock, the 1 shifts to the next stage, producing a one-hot sequence of period n. Without this initial condition, the ring can land in the all-zero state (no token to move) or in a multi-one state (generating erroneous patterns).
Step-by-Step Solution:
Verification / Alternative check:
Scope the outputs Q0..Qn−1: a single HIGH should appear and step through each stage in turn. If the ring stalls, verify that precisely one stage was preset and the rest cleared.
Why Other Options Are Wrong:
Clearing all stages leaves no 1 to circulate (dead state).
Presetting all stages yields many 1s, not a one-hot pattern.
“Clearing one and presetting the others” inverts the required token convention and would not match a 1-hot ring unless the logic were inverted everywhere.
Common Pitfalls:
Relying on random power-up states; neglecting reset synchronization; or mis-wiring preset/clear polarities so the token is not created reliably.
Final Answer:
presetting one flip-flop and clearing all the others.
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