Boolean Alg Help

Discussion in 'Homework Help' started by 20FinalKonekt12, Feb 3, 2009.

  1. 20FinalKonekt12

    Thread Starter New Member

    Feb 3, 2009
    4
    0
    Its been a while since I did Boolean simplification... I know tisk tisk

    Good thing I came across this sick forum.

    OK, so here's the expression I got from a simple circuit:

    X = ((A'B')*(C'D'))+((C'D')*(B+D))'

    If someone could show me a step by step that would be great, I'm mostly getting stuck on ((C'D')*(B+D))' half of the expression.
    Not sure how the simplification rules handle this? DeMorgans? Idk, but any help is greatly appreciated

    Also, if someone could show me how to fill this into a kmap that would be nice too. I know how kmaps work once its all filled out, just not sure how to apply this expression to the map.

    I appreciate any help, thank you
     
  2. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Hello 20Finalkonekt12,

    you can use Demorgan's theorems for the second part[(AB)' = A'+B']
    ((C'D')*(B+D))' = ((C'D')' + (B+D)')

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

    For karnaugh mapping refer this page
    http://www.allaboutcircuits.com/vol_4/chpt_8/6.html
     
  3. 20FinalKonekt12

    Thread Starter New Member

    Feb 3, 2009
    4
    0
    Ok cool,

    so does the first half, ((A'B')*(C'D')), stop in this form:
    (A'+B')(C'+D') ?
     
  4. Alexei Smirnov

    Active Member

    Jan 7, 2009
    43
    1
    (A'B')(C'D') = (A+B)'(C+D)' ?
     
  5. 20FinalKonekt12

    Thread Starter New Member

    Feb 3, 2009
    4
    0
    ...yeh that would be the question
     
  6. Alexei Smirnov

    Active Member

    Jan 7, 2009
    43
    1
    It's not a question, it's true: A'B' = (A+B)'

    The question was why
    (A'B')*(C'D') stop in this form (A'+B')(C'+D')
     
  7. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Alexei Smirnov,

    You do not have to use the asterisk in your logic expressions.

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

    Ratch
     
  8. Alexei Smirnov

    Active Member

    Jan 7, 2009
    43
    1
    Shame on me... I've missed the last step:
    (A'B')(C'D') = (A+B)'(C+D)' = ((A+B)+(C+D))' = (A+B+C+D)'

    Still not (A'+B')(C'+D') though.
     
  9. 20FinalKonekt12

    Thread Starter New Member

    Feb 3, 2009
    4
    0
    Awesome, thanks for all of your help so far

    So what im getting from combined posts is the the simplified function is:
    (A+B+C+D)'+C+D+(BD)'

    And then to applying this to a kmap:
    [​IMG]

    Does this look right?
     
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