First, find the truth table for this combination of gates:

Second, show that the following combination is equivalent to the preceding combination of gates:

A'+B | A+C | ||||
---|---|---|---|---|---|

A | B | C | D | E | D^{.}E |

0 | 0 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 |

1 | 1 | 0 | 1 | 1 | 1 |

0 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 1 | 1 | 1 | 1 |

1 | 0 | 1 | 0 | 1 | 0 |

1 | 1 | 1 | 1 | 1 | 1 |

A^{.}B |
A'^{.}C |
||||
---|---|---|---|---|---|

A | B | C | F | G | F+G |

0 | 0 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 |

1 | 1 | 0 | 1 | 0 | 1 |

0 | 0 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 1 | 1 |

1 | 0 | 1 | 0 | 0 | 0 |

1 | 1 | 1 | 1 | 0 | 1 |

Note that the Karnaugh map for the truth table is:

C | 0 | 1 | |
---|---|---|---|

AB\ | |||

00 | 0 | 1 | |

01 | 0 | 1 | |

11 | 1 | 1 | |

10 | 0 | 0 |

The second gate combination may be found directly by boxing the ones with two 1x2 boxes. The first gate combination may be found by boxing the zeros with two 1x2 boxes and then NORing rather than ORing. I derived the alternative combination of gates (in the first part) for the purpose of this quiz by application of DeMorgan's theorems to the second Karnaugh approach.

"http://www.physics.udel.edu/~watson/phys345/quiz/7soln.html"

Last updated November 6, 1998.

Copyright George Watson, Univ. of Delaware, 1998.