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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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20 Answers

1 votes
1 votes

PS NS

00 01

01 00

00 10

10 00

00 11

11 00

In total there are 6 states. 0,1,2,3 are represented by 2 bits while the three 00's are represented by 2 bits. So in total 4 bits are required.

1 votes
1 votes

Given the sequence is 0-1-0-2-0-3.

There are three transitions from 0.

Hence, the number of bits has to be added to existing states is 2-bits to represent 4  distinct states and eliminate non-determinism. 

Please find the below answer useful. Also, let me know if it needs correction. 

 

0 votes
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

0 votes
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.

Answer:

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