Predicting the next state in a 4-bit Johnson (twisted-ring) counter Given on the 5th clock pulse: Q0 = 0, Q1 = 1, Q2 = 1, Q3 = 1. What is the state on the 6th clock pulse?

Introduction / Context:
A 4-bit Johnson counter (twisted-ring) is a shift register whose input is the inversion of its last stage. It generates a 2n-state sequence (for n flip-flops), useful for decoded timing signals and sequence control. Predicting the next state requires understanding the feedback and shift direction.

Given Data / Assumptions:

• 4-bit Johnson counter with stages Q0 (first) through Q3 (last).
• On the 5th clock: Q0=0, Q1=1, Q2=1, Q3=1.
• Feedback to the first stage is NOT(Q3).
• On each rising clock edge, bits shift toward higher-index stages.

Concept / Approach:
For a typical Johnson implementation, the new Q0 becomes the inverted previous Q3. Each stage Qi (i>0) takes the previous value of Qi-1. Thus, compute the inverted feedback for Q0, then shift prior values forward by one position.

Step-by-Step Solution:
Previous Q3 = 1 → new Q0 = NOT(1) = 0.New Q1 = old Q0 = 0.New Q2 = old Q1 = 1.New Q3 = old Q2 = 1.Therefore, next state: Q0=0, Q1=0, Q2=1, Q3=1.

Verification / Alternative check:
Listing the 8 states of a 4-bit Johnson sequence confirms the transition from 0 1 1 1 to 0 0 1 1 under the assumed shift direction and feedback polarity.

Why Other Options Are Wrong:

• Q0=1,1,1,0: Would imply Q0=NOT(Q3)=0 is violated.
• Q0=0,0,0,1 and Q0=1,0,0,0: Do not match the correct shift of old bits.

Common Pitfalls:

• Mixing up shift direction (left vs. right) and which stage feeds which.
• Forgetting the inversion in Johnson feedback.

Final Answer:
Q0 = 0, Q1 = 0, Q2 = 1, Q3 = 1