Draw a circuit diagram to the following boolean expression ((A'C)')(B + C') and show the behavior of the truth table. Any help?

Answer 1

I'm not sure how this is going to look, as I 'drew' it with a fixed width font, and I can't seem to figure out how to change the font for this answer. If the image is a jumble, copy-and-paste it into a text editor, where you can set the font to some fixed font (e.g. Courier New). -+- Sid

A +-----+ A' +-----+

A >----->+ not +---->+ |

+-----+ | and |

| | A'& +-----+ (A'&C) +-----+

+--------------->+ +----->+ not +-------->+ |

| +-----+ +-----+ | | ((A'&C)')&(B+C')

| | and +-------------------->

| B +-----+ B+C | |

B >----------------->+ +--------------------->+ |

| | or | +-----+

| C +-----+ C' | |

C >-+--->+ not +---->+ |

+-----+ +-----+

0 1 2 3 4 5 6 7

-----------------+---+---+---+---+---+---+---+---+

Inputs A | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |

B | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |

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

-----------------+---+---+---+---+---+---+---+---+

A' | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |

A'&C | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |

(A'&C)' | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |

C' | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

B+C' | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |

((A'&C)')&(B+C') | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |

-----------------+---+---+---+---+---+---+---+---+