GATE CSE 2016 Set 1 | Question: 8

Question

We want to design a synchronous counter that counts the sequence $0-1-0-2-0-3$ and then repeats. The minimum number of $\text{J-K}$ flip-flops required to implement this counter is _____________.

Comments

@Shaik Masthan

If the sequence was 0 - 1 - 0 - 2 - 3 .... then we would require 3 FFs right????

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Please update this. The answer is 3 or 4 as per the official key.

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anwer is 4 ,  the sequence has 3 distinct numbers 1,2,3 to store that 2 bit is required and the zeros are 3 distinct zeros as 0-1,0-2,0-3 they are not the same , therefore for 3 distinct zeros you need 2 more bits to store them .

final answer is 2+2 = 4 bits means 4 FF .

as 1 ff stores 1 but of info

Answer 1

Answer is 3.

Question paper: https://engineering.careers360.com/sites/default/files/GATE-2016-S5-CS.pdf

Answer key: https://engineering.careers360.com/sites/default/files/GATE-2016-official-CS-1-AnsKey.pdf

I don't understand why some people are complicating the answer by saying we need to differentiate between the 3 0s. No we don't. The clock cycles are there to do that for us. In brief, the circuit would know in which clock cycle the FFs are. The 3 0s are occurring at cycle 1, cycle 3 and cycle 5. They're implicitly different.

Answer 2

Answer : 3

counts sequenc = 0−1−0−2−0−3

so , Mod of Counter = 6

Min No. of Flip-Flop required (x) =  ceil(log2(Mod of Counter))

x= ceil(log2(6))

x= 3

Answer 3

since the MOD of this problem is 6 because the nos of states this counter will sequence before coming to back original first state.

0-1-0-2-0-3-----sequence

1-2-3-4-5-6  MOD counting

hence the nos of FFs wil be 2^x>=6  which gives x>=3.

Hence​​​​​​​ MINIMUM nos of J-K flipflops required is 3.

Answer 4

What if we want to design the same using T FF and D FF.

How many minimum FFs are required?