Johnson counter sequencing (4-bit) On the third clock pulse, a 4-bit Johnson sequence is Q0 = 1, Q1 = 1, Q2 = 1, Q3 = 0. What are Q0..Q3 on the fourth clock pulse?

Introduction / Context:
A Johnson (twisted-ring) counter feeds back the inverted last stage into the first stage. It produces a 2N-state sequence using N flip-flops and is widely used for timing and divide-by counters.

Given Data / Assumptions:

• 4-bit Johnson counter with feedback = NOT(Q3) into the first stage.
• Third pulse state: Q0 Q1 Q2 Q3 = 1 1 1 0.
• Positive-edge shifting to the right (conventional tabulations).

Concept / Approach:
On each clock, every Q shifts toward the MSB, and the inverted last stage enters the first stage. Given Q3 = 0, NOT(Q3) = 1 becomes the new incoming bit, so the next state turns on the remaining stage.

Step-by-Step Solution:
Present state: 1 1 1 0.Feedback bit = NOT(Q3) = 1.Next state after clock: 1 1 1 1.

Verification / Alternative check:
Enumerate a typical 4-bit Johnson sequence: 0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 … The transition 1110 → 1111 matches the given step.

Why Other Options Are Wrong:
Other states do not follow from feedback = 1 at this point or belong to earlier/later steps in the cycle.

Common Pitfalls:
Confusing ring vs. Johnson counters, or forgetting to invert the last stage when computing the incoming bit.

Final Answer:
Q0 = 1, Q1 = 1, Q2 = 1, Q3 = 1