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The minimum number of $\text{JK}$ flip-flops required to construct a synchronous counter with the count sequence $(0, 0, 1, 1, 2, 2, 3, 3, 0, 0, \ldots)$ is _______.
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176 votes
176 votes
First, lets design a counter for $0, 1, 2, 3$. It is a MOD - $4$ counter. Hence, number of Flip Flops required will be two. Count sequence will be:

$00 \to 01 \to 10 \to 11$
 
Count sequence mentioned in question is:

$00 \to 00 \to 01 \to 01 \to 10 \to 10 \to 11 \to 11$
 
Now, two flip flops won't suffice. Since we are confronted with repeated sequence, we may add another bit to the above sequence:

$000 \to 100 \to 001 \to 101 \to 010 \to 110 \to 011 \to 111$
 
Now each and every count is unique, occurring only once. Meanwhile, our machine has been extended to a MOD - $8$ counter. Hence, three Flip Flops suffice.

Just neglect the MSB flip flop output and take the o/p of other two only. So, we have :

$0,0,1,1,2,2,3,3,\ldots$

So, correct answer: $3.$
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37 votes
37 votes
People are asking for explaination , I am not giving full solution but an approach how to handle these kind of questions .

First for creating  counter for any sequence we need to identify the states .

Lets say we are taking a sequence of 0-1-2-3 then the states of flip flop (counter design) will be (ff1-ff0) 00->01->10->11-> , only 2 flipflop will be sufficient to give output in 2 bits.

but for the above sequence as you see there are two digits of same kind  0,0,1,1,2,2,3,3 , for which binary values will be 00,00,01,01,10,10,11,11 we cant diffrenciate first 00 from 2nd 00 or third 01 from fourth 01 using only 2 bits so we will add 3rd bit as follow

000,100,001,101,010,110,011,111 here if you see i have just added one extra bit and we can now differenciate between each number seperately , if we made a counter for this sequence and take output of only  first two flipflops (ff1,ff0) from (ff2,ff1,ff0) then our sequence is realized.

 

(PS for futher clearance on the topic refer Morris Mano book )
10 votes
10 votes

Here it appears to have 8 discrete states. That requires 3 bits, therefore 3 flip-flops. From inspection, there are 4 groups of 2 identical bits per group. I'd break that down to into a divide by two (1 flip-flop) followed by a divide by four (2 flip-flops).

4 votes
4 votes
question is slightly ambigus.the thing is we can get sequence 0,0,1,1,2,2,3,3, by 2 FF.

how?

the o/p of the FF will be sampled at twice the i/p clock frequency.

e.g. let my FF retains a state for 2 seconds and i sample o/p at 1 second interval.

and this is quite possible.

and other cases users have already discussed
Answer:

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