Boolean Algebra help...

Discussion in 'Homework Help' started by avenger09123, Feb 5, 2009.

  1. avenger09123

    Thread Starter New Member

    Feb 5, 2009
    1
    0
    Ok here's the problem....
    (A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

    And here's my work....
    (A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

    [A'D+A(B+C)] [A'B+B(A+D')] (A+C+D')
    -----Did the stuff inside the brackets now------

    (A'D+AB+AC)(A'B+AB+D'B)(A+C+D')

    (A'D+AB+AC)(B+A'BC+A'BD'+AB+ABC+ABD'+ABD'+BCD'+B)
    ^^^^^^^^eliminate uneccessary doubles
    (A'D+AB+AC)(AB+A'BD'+BC+ABD'+B+BCD')

    OK WHERE DO I GO FROM HERE? OR DID I SCREW UP AND WHERE??????
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    avenger09123,

    I will let you decide that.

    F = (A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

    F' = ABC' + A'D + AB'D + A'B' + A'C'D by DeMorgan's theorem

    Plot the F' terms on a Karnaugh map with a "x" mark.

    Read the 6 terms on the K-map which have not been marked. These are the simplified F terms.

    F = A'BD' + AB'C + AB'D' which is your answer.

    Ratch
     
  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Better to use a K-map to minimize such quite big problems as you will get the optimum solution.
     
  4. photonxyz

    New Member

    Feb 13, 2009
    3
    0
    original problem

    (A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

    Simplifies to the following;

    A'BD'+AB'D'+ABC
     
  5. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    photonxyz,

    Yes, you are right. I should not have put a "'" on the B variable of the ABC term. I wonder why the OP did not catch that mistake.

    Ratch
     
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