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Johnson Counter States — For a Johnson (twisted-ring) counter constructed with N flip-flops, how many unique states are produced per full sequence?

Difficulty: Easy

Correct Answer: 2N

Explanation:


Introduction:
A Johnson counter (also called a twisted-ring counter) is formed by feeding the inverted output of the last flip-flop back to the input of the first. Unlike a normal ring counter, which has N states for N flip-flops, the Johnson counter produces more unique patterns for the same number of stages.

Given Data / Assumptions:

  • N flip-flops clocked synchronously.
  • Feedback: inverted last stage to first stage.
  • Sequence repeats after a fixed number of clocks.


Concept / Approach:
The inverted feedback creates a sequence where runs of 1s and 0s expand and contract through the register. The resulting cycle length is 2N states, offering easier decoding (many “two-hot” combinations) and glitch-resistant outputs for one-of-K control signals.

Step-by-Step Solution:

Start with all zeros; each clock injects a 1 at one end while previous bits shift.After N clocks, the register is all ones; subsequent clocks inject zeros and shrink the run of ones.Total distinct patterns until returning to the start = 2N.


Verification / Alternative check:

Simulate a small case (N=4): states = 0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 (8 = 2N states).


Why Other Options Are Wrong:

N: That is the count for a simple ring counter, not a Johnson counter.2^N: That is the maximum for a generic N-bit register, not this constrained sequence.N + 1: Underestimates the Johnson sequence length.


Common Pitfalls:

Confusing ring counters (N states) with Johnson counters (2N states).Counting “transitions” instead of “distinct states.”


Final Answer:

2N
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