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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 $\text{J-K}$ flip-flops required to implement this counter is _____________.

in Digital Logic by Loyal (7.2k points)
retagged by | 17.8k views
+3

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;

2>= 6

answer = 3 

+1
no of states is 6, but out of those 3 should correspond to 0..
0
do u have any link that supports the theory behind ur answer ?? coz i just followed standard procedure !!
0
no i dont have any link , but i m sure.. lets see in the official key.
And its not a standard question, so you cant  follow standard procedure..
0
for that type ques ans is not clear .someone plz explain in detail ....
+1
we need four JK flipflops..
0->1->0->2->0->3 They are in non determensitic manner

0000->0001->0100->0010->1000->0011

There are 6 states and 3 of them correspond to same state.. ie non-deterministic
to differentiate between 0,1,2,3 we need 2 bits.

 
to differentiate between 3 0's we need 2 bits..
So total 4 bits=4 FF
0

@Shaik Masthan

plz explain this question!

0
read the best answer, i can't explain more than that answer !
0

@ why 3 can not be the correct answer?

0
in the official key, they provided 3 or 4
0

@Shaik Masthan

The extra bits which we are putting to eliminate duplicate can be of any order ?

0

@Shaik Masthan

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

0

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

12 Answers

0 votes

Sequence 0-1-0-2-0-3

we need 2 bits for (0-1-2-3) sequence and then to differentiate between 3 0's we have to calculate no. of flip flops needed

  1. 0 after 3 

  2. 0 after 2

  3. 0 after 1

To differentiate between these 3 0's we need 2 FFs as 

Q0

Q1

Q0N

Q1N

J0

K0

J1

K1

0

0

0

1

0

X

1

X

0

1

0

0

0

X

X

1

1

0

0

0

X

1

0

X

1

1

0

0

X

1

X

1

So total , 2+2 = 4FFs are needed

by Active (1k points)
0 votes

Answer is 3 not 4.

we can arrange flip flops output in following manner.

A(O/P of FF1) B(O/P of FF2) C(O/P of FF3) X(sequence MSB) Y((sequence MSB))
0 0 0 0 0
0 0 1 0 1
0 1 0 0 0
0 1 1 1 0
1 0 0 0 0
1 0 1 1 1

So,

X = ~BC and

Y = C(~BA + ~AB)

Hence, it is mod-8 counter with 110 as reset state and 111 as invalid state

3 Flip Flops are required. Please do comments if anything is wrong.

by (193 points)
0 votes

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.

by (221 points)
0 votes
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
by Junior (977 points)
Answer:

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