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

 

As the question just speaks about synchronous counters, I thought of using a Johnson Counter in which with n flip-flops we could count mod(2n). Considering that, the question is counting 6 bits therefore we require 3 flip-flops. 

 

Answer 2

0

1

0

2

0

3

It means

00

01

00

10

00

11

but to differentiate

000

001

100

110

000

011

Still need one more bit

0000

0001

0100

0110

1000

1011

 

0----->1----->4---->6---->8---->11---->0

The counter is working like this.

So 4 flip flops

Answer 3

…………………………………...

Answer 4

1. Total States: The counter needs to represent 6 distinct states: $0, 1, 0, 2, 0, 3$ the repeat..
2. Required Flip-Flops:
   • Each flip-flop can represent $2$ states (0 or 1).
   • To represent 6 states, we need $2^n >= 6$, where n is the number of flip-flops.
   • The smallest value of n that satisfies this condition is $n = 4$.
   • The answer could be 3 also as he condition $2^n >= 6$ is satisfied for $n=3$ as well.

Therefore, $4$ or $3$ J-K flip-flops are required to implement this counter.