Boolean Algebra Problems

Discussion in 'Homework Help' started by molder22, Oct 7, 2006.

  1. molder22

    Thread Starter New Member

    Oct 7, 2006
    3
    0
    I am having trouble reducing the following probems.
    F = AB’CD’ + A’BCD’+AB’C’D+A’BC’D
    F = A’BC’+ A’B’C+ AC’D+ ACD’
    F = (A′ + BD′ + AD) (B + C′) (A+ B′C)

    Any help would be appreciated.
    Thanks in advance.
     
  2. cat

    New Member

    Sep 25, 2006
    8
    0
    You can't simplify that function further. If you've learned karnaugh maps, just draw a karnaugh map for the function and you'll see why.
    cat
     
  3. mik3ca

    Active Member

    Feb 11, 2007
    189
    0
    Here is my solution:

    using: F = AB’CD’ + A’BCD’+AB’C’D+A’BC’D

    F = C(AB'D + A'BD') + C'(AB'D + A'BD)
    F = C(AB'D + A'BD') + C'D(AB' + A'B)
    F = C(AB'D + A'BD') + C'D(A xor B)
     
  4. fxbit

    New Member

    Mar 15, 2007
    4
    0
    No advertising.
     
  5. Blue_turnip

    New Member

    Mar 5, 2007
    8
    0
    lol throwing in a xor doesn't really help simplify it, mate.;)
     
Loading...