Boolean reduction help

Discussion in 'Homework Help' started by PSU31, Feb 7, 2008.

  1. PSU31

    Thread Starter New Member

    Jan 31, 2008
    2
    0
    Im having trouble with these problems, any help is appreciated.

    h= ab'+bc'd'+abc'd+bc
    = ab'(c+c') + b(d+cd') + abc'd
    = ab' + bd + bc + abc'd
    = b(c+ac'd) + ab' + bd
    = bc + abd + ab' + bd
    = bc + ad + ab' + bd

    This is where Im stuck.
    should be 3 terms with 6 literals


    h = ab' + bc'd' + abc'd + bc
    = bc'(d' + ad) + ab' + bc
    = bc'd' + abc' + ab' + bc
    = bc + bd' + ab' + ac'

    should be 3 terms with 5 literals
     
  2. luck

    New Member

    Jan 25, 2008
    8
    0
    I don't really see what the difference is between the two equations. By mapping the first I get a+a'bc+a'bd' which could be rewritten as a+a'b(c+d')
     
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