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

Both 3 and 4 were given the correct answer to this question in GATE this year

Answer 2

both 3 and 4 are correct in the official key

Comments

Actually only one can be correct.. :P
IISC people are clever.. they just increased the error allowed in the answer - 3.0 to 4.0 :)

Answer 3

Can we think like this ?

First we build a circuit for the sequence ; 0 - 1 - 4 - 6 - 8 - 11

To build this we need 4 FF, and after doing this, just connect $Q_1$ and $Q_0$ terminals to the decoder.

Comments

What is advantage of it?

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Can we just ignore the two most significant bit and take only the least 2 significant bit. No need of decoder. Can we ?

Answer 4

2no of flip flops needed >= mod of the circuit 

in this question mod of counter is 6 and let no of flip flops be X;

2x >= 6

answer = 3 

Comments

Answer is either 3 or 4 as per IISC.But i think this approach is right also.