[Boolean algebra] Mismatch between algebra and karnaugh map?

Discussion in 'Homework Help' started by Josh Samman, Apr 5, 2013.

  1. Josh Samman

    Thread Starter New Member

    Mar 29, 2013
    11
    0
    Look at this equation:

    f = (a + c)*(a' + b)

    I solve it by distributing the terms:

    f = a'(a + c) + b(a + c)
    = aa' + a'c + ab + bc
    = a'c + ab + bc --> Resolution

    Then i plot it into the Karnaugh map:

    [​IMG]
    * Sorry for the bad drawing :rolleyes:

    And i get:

    f = a'c + ab(from the map)

    f
    = a'c + ab + bc --> (comparing to the resolution by algebra the "bc" term is missing).

    I look at the map and i know right away that the "bc" term gets simplified by ther other 2 (regarding that the bold 1's in the map are the "bc" term).

    Now i don't know how i can simplify f = a'c + ab + bc to f = a'c + ab using algebra only! :confused: How to do it? There's any rule i'm missing in there or what?
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,446
    3,361
    Expand each of the three terms a'c + ab + bc

    For example,

    a'c = a'b'c + a'bc

    Take it from there.
     
  3. Josh Samman

    Thread Starter New Member

    Mar 29, 2013
    11
    0
    F = (a'bc + a'b'c) + (abc + abc') + (abc + a'bc)
    = a'bc+ a'b'c + abc + abc';
    = [a'c(b+b')] + [ab(c'+c)]
    = a'c + ab

    \o/ - Thank you man, for the help ;)
     
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