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46 votes

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 :

  1. $\overline {x \oplus y}$ and $\overline {x \oplus y}$ 
  2. $\overline {x \oplus y}$ and $ {x \oplus y}$
  3. $ {x \oplus y}$ and $\overline {x \oplus y}$
  4. $ {x \oplus y}$ and $ {x \oplus y}$
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3 Answers

Best answer
76 votes
76 votes

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.

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12 votes
12 votes

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


 

 

 

8 votes
8 votes

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

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

Qn+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.

Answer:

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