Sop=pos

Discussion in 'Math' started by dhamphire, Aug 2, 2008.

  1. dhamphire

    Thread Starter New Member

    Aug 2, 2008
    3
    0
    Please help me..
    Prove that:
    BCD+ACD+ABD+ABC=(CD)(B+C)(A+C)(A)(A'B+D)

    or i wast just wrong with the Kmapping?

    thanks....
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    dhamphire,

    Expanding out the SOP, we get the 5 logic values of 7,11,13,14,15 . The complement of the SOP are the missing 11 logic values which are 0,1,2,3,4,5,6,8,9,10,12 . Using the Quine-McCluskey method to simplify the 11 logic values we get A'B' + A'C' + B'C' + A'D' + B'D' + C'D' . This represents the complement of the given original expression. To reverse the complement, use DeMorgan's theorem to get (A+B)(A+C)(B+C)(A+D)(B+D)(C+D) in POS format.

    Ratch
     
  3. dhamphire

    Thread Starter New Member

    Aug 2, 2008
    3
    0
    Thanks Ratch... :)
     
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