For the given Boolean expression for one truth table of an unknown function.
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| A |
B |
C |
F |
| 0 |
0 |
0 |
1 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
0 |
0 |
| 0 |
1 |
1 |
0 |
| 1 |
0 |
0 |
1 |
| 1 |
0 |
1 |
1 |
| 1 |
1 |
0 |
0 |
| 1 |
1 |
1 |
0 |
The output recycles after the last state. If the above function is implemented using J-K flip flop then,
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$ J_A = B(\bar AC + A\bar C), K_A = A + \bar A\bar C$
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$ J_A = \bar B(A\bar C +\bar AC), K_A = A + \bar A\bar B$
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$ J_A = B(\bar AC + A\bar C), K_A = B + AC$
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$ J_A = \bar C(\bar AB + B\bar A), K_A = A + \bar A\bar C$
A
Correct answer
Explanation
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|
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| A |
B |
C |
F |
JA |
KA |
D |
| 0 |
0 |
0 |
1 |
X |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
X |
0 |
| 0 |
1 |
0 |
0 |
0 |
X |
0 |
| 0 |
1 |
1 |
0 |
1 |
X |
1 |
| 1 |
0 |
0 |
1 |
X |
0 |
1 |
| 1 |
0 |
1 |
1 |
X |
1 |
0 |
| 1 |
1 |
0 |
0 |
0 |
X |
0 |
| 1 |
1 |
1 |
0 |
1 |
x |
1 |
$J_A = \bar ABC + AB\bar C = B (\bar AC + A\bar C)$
$K_A = A + \bar A\bar C $
$D = A\bar B\bar C + BC$