Simplifying Boolean Expression

Discussion in 'Homework Help' started by ramla, Dec 22, 2008.

  1. ramla

    Thread Starter New Member

    Dec 22, 2008
    2
    0
    Hi,

    I have an expression of nine literals as follows:

    BC+AC'+AB+BCD.

    I have to reduce it to four literals.

    This is the way I proceeded.

    BC+AC'+AB+BCD = BC(1+D)+AC'+AB
    = BC+AC'+AB

    I have reduced it to six literals. But I am not able to proceed further.

    Please Help!!!


    Thanks and Regards,
    Ramla
     
  2. chadchoud

    Member

    Dec 18, 2008
    11
    0
    It cannot be simplified more than that, I think. But wait for a confirmation from an expert :)
     
  3. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    I think using K-map we will get a more simplified answer.
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    ramla,

    chadchoud,

    vvkannan,

    Correct, a K-map or one of the tabulation methods is the way to go. Doing it with Boolean algebra is the pits. Once you have done it with the K-map or any other closed method, you will easily be able see what terms make up the final solution, and then back compute your Boolean algebraic method to get the minimal solution.

    Behold:

    BC+AC'+AB
    BC+AC'+AB(C+C')
    BC+AC'+ABC+ABC'
    BC(1+A)+AC'(1+B)
    BC+AC'

    Ratch
     
    Last edited: Dec 22, 2008
  5. ramla

    Thread Starter New Member

    Dec 22, 2008
    2
    0
    Thanks a lot ratch..
     
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