Difficulty: Easy
Correct Answer: 0011
Explanation:
Introduction: A Johnson counter (twisted-ring) feeds back the inverted last stage into the first stage, producing a sequence of two-hot patterns before returning to all zeros. The exact bit ordering in questions varies by textbook convention; here the second state is given as 0001, and you must identify the next state under standard Johnson feedback with right-shifted insertion of the inverted last bit.Given Data / Assumptions:
Concept / Approach: From 0001 to the next state under Johnson operation, the register shifts the pattern and inserts the inverted last bit. With the given convention (moving the “1” toward the MSB while inserting NOT of the dropping bit), the two-hot run grows: 0001 → 0011, then 0111, then 1111, etc., before shrinking back toward 0000.Step-by-Step Solution:
State S1 = 0001 (given).Invert last bit (1 → 0) and shift according to the stated convention.Result S2 = 0011, extending the run of ones.Verification / Alternative check:
Enumerate the classic 4-bit Johnson sequence under this bit order: 0000 → 0001 → 0011 → 0111 → 1111 → 1110 → 1100 → 1000 → 0000.Why Other Options Are Wrong:
0010: Would shrink, not extend, the run; not Johnson behavior here.1000: Belongs to the later part of the cycle in this ordering.0110: Does not follow directly from 0001 under standard Johnson feedback.Common Pitfalls:
Mixing left vs right shift conventions; follow the one implied by the given step.Confusing Johnson with simple ring sequences.Final Answer:
0011
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