Johnson counter fact check: An n-bit Johnson (twisted-ring) counter produces 2n unique states before repeating. For an 8-bit Johnson counter, is the sequence length eight states?

Introduction / Context:Johnson counters are widely used for sequence generation and timing because they visit a set of unique states with simple feedback. Knowing the exact sequence length is essential for designing decoders and detecting end-of-sequence conditions.

Given Data / Assumptions:

• In a Johnson counter, the inverted output of the last flip-flop feeds the input of the first.
• There are n flip-flops in the chain.
• All flip-flops shift simultaneously on the clock edge.

Concept / Approach:A Johnson counter cycles through 2n states, not n. For n = 8, the sequence length is 16 states. This differs from a simple ring counter, which has exactly n states (one-hot patterns). Understanding the distinction prevents incorrect assumptions about decoding and resource allocation.

Step-by-Step Solution:1) Start from a valid seed state.2) On each clock, shift and feed back the inverted last bit to the first.3) Count the number of distinct states before repetition; algebra and examples show 2n unique states.4) For n = 8, obtain 16 unique states.

Verification / Alternative check:Enumerate states for a 4-bit Johnson counter (2n = 8) to build intuition, then generalize to 8-bit for 16 states.

Why Other Options Are Wrong:“Correct” confuses Johnson with ring counters. Other qualifiers do not alter the fundamental 2n property.

Common Pitfalls:Misidentifying counter type; assuming one-hot behavior for Johnson sequences.

Final Answer:Incorrect