0 votes 0 votes In a Johnson’s counter LSB is complemented and a circular right shift operation has to be done to get the next state. For ring counter also a shortcut exists ? Digital Logic digital-counter digital-logic + – amitqy asked Jan 4, 2019 amitqy 427 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Shaik Masthan commented Jan 4, 2019 reply Follow Share In a Johnson’s counter LSB is complemented and a circular right shift operation has to be done to get the next state. Did you ever analyses why it is happen ? then automatically you will answer to your question by yourself ! ( Hint :- Clearly observe the circuit diagram of Johnson counter diagram ) Answer to your question is :- " with out complement( i.e., keep the same value ), just perform circular right shift ! " Take care with LSB or MSB bits, according to the question you have to name it ! 1 votes 1 votes amitqy commented Jan 4, 2019 reply Follow Share I never analysed it before. I just did this type of question by the long way by making tables and all. I got your point. Since in ring counter input of the first flip-flop is not complement of the output of the last flip-flop so it's just a right shift without a compliment.Thank you 0 votes 0 votes Please log in or register to add a comment.
