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प्रश्न
Derive a Canonical POS expression for a Boolean function FN, represented by the following truth truth table:
| X | y | z | FN (X, Y, Z) |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
उत्तर
Output: `(barX + barY + barZ)(X + barY + barZ)(barX + barY + barZ)(barX + barY + barZ)`
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Obtaining Sum of Product (SOP) and Product of Sum (POS) Form the Truth Table
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संबंधित प्रश्न
Derive a Canonical POS expression for a Boolean function G, represented by the following truth table:
| X | Y | Z | G(X, Y, Z) |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Derive a Canonical POS expression for a Boolean function F, represented by the following truth table
| P | Q | R | F(P,Q,R) |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
