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

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

@Abhrajyoti00 Yes Brother, I was also thinking same but Pranav’s point is also valid saying only shift register arises question in mind that which type of shift register..? Is it ring or johson or something else…? I was also in confusion because question specifically says “function performed by arrangement”. If instead of above line if they asked “function performed by arrangement is an application of” then definitely option A is right

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@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!

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

In the table, you shows sequence of states, along with sequence complemented outputs (states) , with 3 FF we can have any (shift) register of 3 bits only. (not as shifting 000111)

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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?