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GATE | GATE-CS-2016 (Set 1) | Question 18

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  • Difficulty Level : Medium
  • Last Updated : 29 Sep, 2021
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We want to design a synchronous counter that counts the sequence 0-1-0-2-0-3 and then repeats. The minimum number of J-K flip-flops required to implement this counter is
Note : This question was asked as Numerical Answer Type.
(A) 1
(B) 2
(C) 4
(D) 5

Answer: (C)

Explanation: Total 4.

2 J-K flip- flops for synchronous counter + 2 J-K flip-flop to make 2 bit counter.

Actually, when we are repeating again then after 3 we don’t know that our Synchronous counter will go to which zero.
To make it work right, we need to move it to
1st zero after 3
2nd zero after 1
3rd zero after 2
i.e. 0 -> 1 -> 0 -> 2 -> 0 -> 3 (from here it again go to 1st zero).
In order to decide which zero to move on we use counter from 1 to 3, I have attached truth table of normal 2 bit synchronous counter using JK flip flop.


// This Explanation has been contributed by Mohit Gupta

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