edited by
36,828 views
79 votes
79 votes
The minimum number of $\text{JK}$ flip-flops required to construct a synchronous counter with the count sequence $(0, 0, 1, 1, 2, 2, 3, 3, 0, 0, \ldots)$ is _______.
edited by

8 Answers

3 votes
3 votes

Found something interesting. Guys at MadeEasy were able to do this with 2 flip flops using an additional clock.

Although I would like to clarify that the answer in the official sheet is 3.

 MadeEasy answer

1 votes
1 votes
This series is equivalent to designing mod-8 counter and considering only the two most significant bits.

000, 001, 010, 011, 100, 101, 110, 111

Hence, 3 FFs are required.
0 votes
0 votes
2 should be correct because using some extra gates along with Flip-Flops we can count this sequence.
Answer:

Related questions

56 votes
56 votes
12 answers
2
go_editor asked Feb 12, 2015
19,309 views
The number of min-terms after minimizing the following Boolean expression is _______.$[D'+AB'+A'C+AC'D+A'C'D]'$
38 votes
38 votes
5 answers
3
36 votes
36 votes
6 answers
4
go_editor asked Feb 13, 2015
15,634 views
The number of states in the minimal deterministic finite automaton corresponding to the regular expression $(0+1)^* (10)$ is _____.