GATE IT 2008 | Question: 37

Question

Consider the following state diagram and its realization by a JK flip flop

The combinational circuit generates J and K in terms of x, y and Q.
The Boolean expressions for J and K are :

• A. $\overline {x \oplus y}$ and $\overline {x \oplus y}$ 
• B. $\overline {x \oplus y}$ and $ {x \oplus y}$
• C. $ {x \oplus y}$ and $\overline {x \oplus y}$
• D. $ {x \oplus y}$ and $ {x \oplus y}$

Comments

Correct option is D

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Dont need to solve the complete pblm as it takes more time ...

1) current state value 0 on seeing input 00(values of (x,y)) should remain as 0 ... For this in JK-FF, I can give J=0,K=0 or J=0,K=1 ... 

       Now J = (x EXOR y)^c = (x EXNOR y) will make J = 1 ... which is wrong because J has to be given input as 0 ... So A) and B) are eliminated ...

2) similiarly current state value = 1 on seeing input 10 (values of (x,y)) should be modified as 0 .. For this I need to give J = 0,K=1 or J = 1,K=1

      Now  Now K = (x EXOR y)^c = (x EXNOR y) will make K = 0 ... which is wrong because k should be given value 1 ... So C) is eliminated ... 

Option D) is the answer ... 

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Detailed Video Solution with $\color{red}{\text{3 Different Methods to Solve}}$:

https://youtu.be/eDyeuTA0yMk  

Answer 1

From state diagram:$${\begin{array}{ccc|c|c|c}
\textbf{Q}&    \textbf{X}&  \textbf{Y}&\bf{ Q_{n+1}}& \textbf{J}& \textbf{K} \\\hline
0&0&0&0&0 &\text{X} \\ 0&0&1&1&1& \text{X} \\    0&1&0&1&1& \text{X} \\    0&1&1&0&0& \text{X} \\   1&0&0&1& \text{X}& 0 \\    1&0&1&0& \text{X}&1  \\    1&1&0&0& \text{X}& 1 \\   1&1&1&1& \text{X} &0\\  
 \end{array}}$$Excitation table of JK$${\begin{array}{cc|cc}\textbf{Q}&\bf{ Q_{n+1}}& \textbf{J}& \textbf{K} \\\hline
0&0&0&\text{d} \\ 0&1&1&\text{d} \\    1&0&\text{d}&1 \\   1&1&\text{d}&0       
 \end{array}}$$

Option D.

Comments

This question is merely conversion of Flip Flops. Here, XY Flip Flop is represented by state transition diagram. We have to realize it using JK flip flop.

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In conversion of Flip Flops, basic steps in state table to be filled:

1.Write the truth table(function table) of required flip flop(XY)

2.Write the excitation table of available JK Flip Flop and generate J and K in terms of X ,Y, Q using K-map method and here J=X$\oplus$Y, K=X$\oplus$Y.

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I think your favourite subject is DIgital, most of the PYQ are best answered by you

As usual, amazing answer.

Answer 2

all answers are awesome..may look at this too :

 

 

 

Comments

How do you say that the state diagram is corresponding to x,y and not j,k?

Answer 3

From the state diagram one can infer that Q_n+1 = Q_n, when x = y, and Q_n+1 = Q'_n, when x != y

For JK flip flop Q_n+1 = Q_n, if J=K=0 and

Q_n+1 = Q'_{n ,} if J=K=1

and as EX-OR gate is non-equivalence gate it satisfies for the above conditions of J and K when X and Y are taken as inputs.

Comments

@skan good explanation :)

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Thanks!