State count of a Johnson (twisted-ring) counter In a 6-bit Johnson counter, how many distinct states (bit patterns) are produced in the complete sequence?

Difficulty: Easy

Correct Answer: 12

Explanation:


Introduction / Context:
A Johnson counter (twisted-ring) is a shift register with complemented feedback. Its key property is that an n-stage Johnson counter cycles through 2n unique states, making it efficient for sequence generation and timing signals.


Given Data / Assumptions:

  • Number of stages n = 6.
  • Inverted feedback from last stage to input.
  • Clocking advances the pattern by one bit each cycle.


Concept / Approach:

Unlike a one-hot ring counter (n states), the Johnson counter gradually fills with 1s and then empties, creating two distinct sequences of length n each. Therefore, the total number of unique states is 2n.


Step-by-Step Solution:

Compute total states = 2 * n.For n = 6, total = 2 * 6 = 12.These 12 states repeat cyclically with no duplicates within one cycle.


Verification / Alternative check:

Enumerating states for a smaller n (e.g., n = 3) yields 6 unique patterns, reinforcing the 2n rule; scaling to n = 6 yields 12.


Why Other Options Are Wrong:

  • 2 and 6: too few; confuse with ring or stage count.
  • 24: would correspond to 4n or 2^(n+1), which is not a Johnson property.


Common Pitfalls:

  • Mixing Johnson (2n states) with regular ring (n states) or binary counters (2^n states).


Final Answer:

12

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