GATE CSE 1993 | Question: 6-3

Question

For the initial state of $000$, the function performed by the arrangement of the $\text{J-K}$ flip-flops in figure is:

• A. Shift Register
• B. $\text{Mod- 3}$ Counter
• C. $\text{Mod- 6}$ Counter
• D. $\text{Mod- 2}$ Counter
• E. None of the above

Comments

sahi toh hai @Prakash_

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@ayush18288

transition state from 111 to 011 is missing in your diagram

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Correct. @Amjad. Thanks for pointing it out .

Answer 1

Circuit behaves as shift register and $\mod6$ counter

$$\begin{array}{c|c}\hline \textbf{Clock Cycle}  &  \textbf{Output} \\\hline \text{1} & \text{100}\\ \text{2} & \text{110}\\ \text{3} & \text{111}\\ \text{4} & \text{011}\\ \text{5} & \text{001}\\ \text{6} & \text{000}\\\hline \end{array}$$

This is Johnson counter which is an application of Shift Register. And Johnson counter is mod 2N counter.

Comments

@pratik2404 I understand. But see:-

Shift register → Johnson Or Ring (Can’t say)

But Johnson → Shift Register (Guaranteed). Hence option A is right as the given counter is Johnson Counter and thus also a Shift Register

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@Abhrajyoti00 Totally Convinced. Thanks man!

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@Pooja PalodI think ,  it is not a johnson counter , but it is twisted ring counter ..... by adding  AND gates to twisted ring counter  ( i.e for n i/p twistd counter  we need 2n  fan-in-2 AND gates required  , in our question  3 i/p twisted ring counter need   6 fan-in-2 AND gates required ) it become a  johnson counter  

 

knowledge from  @Deepak Poonia sir

Answer 2

According to me, I think that A, C and D are the answer. Pleas, find the attached screen shot.

Comments

@Raju Kalagoni, here I didn't show the part of find JK, (its my bad) since, in this case it will be same. But, better approach is to find the JK part for each J0K0, J1K1, and J2K2, otherwise we may not get the right answer. To be safer side we should find the JK part as well.

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Thankyou @ manojsahu @ shashank023

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Initial state given is 000 so option D) is not possible!

Answer 3

1st approach : Start from 000 and use conventional method (i.e. Truth table) to find modulus.

2nd approach : see, given circuit is nothing but Jonson Counter . Number of States in Jonson counter will be 2n where n is number of flip flops. here n is 3 so number of states are 6 . mod 6 counter..

Comments

Yes, It's twisted ring counter i.e Johnson Counter which does mod 2N counting.

Here, n=3 so, mod 6 counter. 

Johnson counters are application of Shift Registers.

Answer 4

A. Shift Register 
C. Mod-6 Counter

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Given circuit diagram is for a Mod-6 Johnson Counter.

Comments

Sir i think answer will be both shift register as well as the mod 6 counter. Right?

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why not mod 2 counter?

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@Abhineet Singh it will be mod2 as well as mod6 but the thing is we need to go with the optimal solution that's why it is mod 6 counter

Answer 5

Ans : A and C 

Shift Register & mod 6 counter. 

Comments

@ananya_23 is this screenshot from any lecture ?

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No.. It is solved by me.. Solution based on Deepak sir's explanation.

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okay

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Thanks, ma'am for this explanation.

Answer 6

Short method:

Johnson’s counter is Mod 2N counter and ring is Mod N counter.

Detailed  Explanation:

If you look closely at the question, you will notice that the input of J and K are complements of each other and if you recall, if the input values are the complement of each other, JK FF behaves as D FF. 

Here in the diagram we can see how D FF is formed from JK FF.

So this means the JK in our question will behave as D FF. You can also replace the JK FF with D FF if you want for easier understanding.

So now we can equate the values of J with Q,

Hence we can get the state diagram. And we will have 6 states since it was mentioned that it will start from 000 thereby eliminating the possibility of MOD 2 counter.

Also the shift register option is correct.