#### Question

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 |

clickto share

#### Solution

Canonical POS expression for a Boolean function G

`=(X + Y + Z)(X + Y + bar(Z))(X + bar(Y) + bar(Z))(bar(X) + bar(Y) + Z)`

Is there an error in this question or solution?

#### APPEARS IN

#### Related Questions VIEW ALL [4]

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 |

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 |

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 |

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 |

Solution for question: Derive a Canonical Pos Expression for a Boolean Function G, Represented by the Following Truth Table: concept: Obtaining Sum of Product (SOP) and Product of Sum (POS) Form the Truth Table. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science), CBSE (Science)