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Consider a $4$-bit Johnson counter with an initial value of $0000.$ The counting sequence of this counter is 

  1. $0, 1, 3, 7, 15, 14, 12, 8, 0$
  2. $0, 1, 3, 5, 7, 9, 11, 13, 15, 0$
  3. $0, 2, 4, 6, 8, 10, 12, 14, 0$
  4. $0, 8, 12, 14, 15, 7, 3, 1, 0$
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Johnson counter /twisted ring counter :-

"N" bit then there will be "2N" states 

Option B easily eliminated as it has 9 state form rest we will have to check out.

Answer :- D 

 

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  1. Structure: It consists of 4 flip-flops connected in a ring, with the inverted output of the last flip-flop feeding back to the input of the first flip-flop.
  2. Initial State: Starts with all flip-flops at $0 (0000)$.
  3. Counting:
    • With each clock pulse, the bits shift one position to the right.
    • The inverted output of the last flip-flop determines the new bit shifted into the first flip-flop.

Counting Sequence:

  1. $0000$ (initial state)
  2. $1000$ (inverted $0$ from last flip-flop shifts in)
  3. $1100$
  4. $1110$
  5. $1111$
  6. $0111$ (inverted $1$ from last flip-flop shifts in)
  7. $0011$
  8. $0001$
  9. $0000$ (repeats)
Answer:

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