**Q.1**Exam - CBSE12D-2017/C06D/3

Reduce the following Boolean Expression to its simplest form using K-Map:

F(X,Y,Z,W)= Σ (0,1,2,3,4,5,10,11,14)

Reduce the following Boolean Expression to its simplest form using K-Map:

F(X,Y,Z,W)= Σ (0,1,2,3,4,5,10,11,14)

Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table:

U | V | W | F(U,V,W) |
---|---|---|---|

0 | 0 | 0 | 1 |

0 | 0 | 1 | 0 |

0 | 1 | 0 | 1 |

0 | 1 | 1 | 1 |

1 | 0 | 0 | 0 |

1 | 0 | 1 | 0 |

1 | 1 | 0 | 1 |

1 | 1 | 1 | 0 |

Reduce the following Boolean Expression to its simplest form using K-Map :

F(X,Y,Z,W) = ∑(0,1,4,5,6,7,8,9,11,15)

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 |

Reduce the following Boolean expression to its simplest form using K-Map:

E(U,V,Z,W)= Ó (2,3,6,8,9,10,11,12,13)

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 |

Reduce the following Boolean Expression to its simplest form using K-Map:

G(U,V,W,Z) = Σ (3,5,6,7,11,12,13,15)

Derive a Canonical POS expression for a Boolean function FN, represented by the following 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 |

Reduce the following Boolean Expression to its simplest form using K-Map:

F(P,Q,R,S)= Σ(0,4,5,8,9,10,11,12,13,15)

Derive a Canonical SOP expression for a Boolean function G, represented by the following truth table:

A | B | C | G(A,B,C) |
---|---|---|---|

0 | 0 | 0 | 1 |

0 | 0 | 1 | 0 |

0 | 1 | 0 | 1 |

1 | 0 | 0 | 0 |

1 | 0 | 1 | 0 |

1 | 1 | 0 | 1 |

1 | 1 | 1 | 1 |

Verify the following using Boolean Laws.

X’+ Y’Z = X’.Y’.Z’+ X’.Y.Z’+ X’Y.Z+ X’.Y’.Z+ X.Y’.Z

LHS X’ + Y’.Z = X’.(Y + Y’).(Z + Z’) + (X + X’).Y’.Z