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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 _____________.

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+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

+1
no of states is 6, but out of those 3 should correspond to 0..
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do u have any link that supports the theory behind ur answer ?? coz i just followed standard procedure !!
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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..
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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
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@Shaik Masthan

plz explain this question!

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@ why 3 can not be the correct answer?

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in the official key, they provided 3 or 4
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@Shaik Masthan

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

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@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.

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)

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)

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)

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)