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 _______. Digital Logic gatecse-2015-set2 digital-logic digital-counter normal numerical-answers + – go_editor asked Feb 12, 2015 • edited Jun 21, 2021 by Lakshman Bhaiya go_editor 36.8k views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments mehul vaidya commented Jan 5, 2020 reply Follow Share I have basic doubt here what is sequence here ? is it 0, 0, 1, 1, 2, 2, 3, 3, 0, 0 (in this case 4 FF needed ) or 0, 0, 1, 1, 2, 2, 3, 3 (in this case three FF needed) 1 votes 1 votes codingo1234 commented Jan 12, 2020 reply Follow Share @mehul vaidya same doubt occuring here, 0 votes 0 votes taran97 commented Jan 15, 2021 reply Follow Share For 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3,... we will require 4 FF. Right? 0 votes 0 votes Please log in or register to add a comment.
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. gmrishikumar answered Sep 18, 2018 gmrishikumar comment Share Follow See all 4 Comments See all 4 4 Comments reply Asim Siddiqui 4 commented Oct 28, 2018 reply Follow Share that's why writers on GATE OVERFLOW are much better than made easy/ace/any other coaching institute's writers. 7 votes 7 votes Nitesh Singh 2 commented Aug 20, 2019 reply Follow Share This is not a classic example of a synchronous counter. Although we could do that but it will become advance level. 0 votes 0 votes palashbehra5 commented Nov 23, 2021 reply Follow Share gmrishikumar Couldn't deny it, this is actually very interesting. 0 votes 0 votes shikhar500 commented Dec 25, 2022 reply Follow Share i am not able to understand this solution can anyone plz explain me this ? 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes The patter can be generated by a mod 8 counter so the answer is 3. Gowthaman Arumugam answered Feb 8, 2015 Gowthaman Arumugam comment Share Follow See all 2 Comments See all 2 2 Comments reply gate2016 commented Apr 14, 2015 reply Follow Share how?? plz explain 1 votes 1 votes richa07 commented Jul 17, 2015 reply Follow Share Pls explain it... 0 votes 0 votes Please log in or register to add a comment.
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. heena singh answered Nov 27, 2017 heena singh comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 2 should be correct because using some extra gates along with Flip-Flops we can count this sequence. Akash Rathor answered Feb 22, 2015 Akash Rathor comment Share Follow See all 2 Comments See all 2 2 Comments reply kshirsagar1992 commented Feb 24, 2015 reply Follow Share Can you kindly elaborate what kind of ckt can be used along with flip-flops to obtain above sequence ? As per my understanding, a frequency divider ckt (divide by 2) can be used, but then it will require one more flipflop along with two output flipflops. Hence 3 flipflops in total. Is there any other method which uses less than 3 flipflops? 1 votes 1 votes kakkar commented Mar 2, 2015 reply Follow Share I think as GATE AUTHORITY mentioned the counter as synchronous counter then we can go for the Clock frequency sampling policy(changing its period length).. By this we can accomplish the task with 2 flip flop... BUT MAY BE ITS WRONG. 0 votes 0 votes Please log in or register to add a comment.