4-Bit Johnson Counter — If the counter is initially cleared (all zeros) and after the first clock pulse the output is 0001, what is the output after the next clock pulse (standard Johnson feedback)?

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:

• 4 flip-flops, synchronously clocked.
• Initial: 0000; after first clock: 0001 (given).
• Feedback: insert NOT(Q_last) into the vacated end on each shift; shift direction consistent with the given transition.

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:

Verification / Alternative check:

Why Other Options Are Wrong:

Common Pitfalls:

Final Answer: