Simplification of Boolean Expression

Discussion in 'Homework Help' started by mathsfailure, Nov 28, 2009.

  1. mathsfailure

    Thread Starter New Member

    Nov 28, 2009
    4
    0
    Hi guys,
    I need help with my homework regarding simplification of boolean algebra.

    X = A.B + C . (A.C +B)

    I'm not sure how to do the compliment as in the bar above the letter.
     
  2. mathsfailure

    Thread Starter New Member

    Nov 28, 2009
    4
    0
    is it correct to do it this way

    (A+B+C) * (A+C+B)
    AA+AC+AB+BA+BC+BB+CA+CC+CB


    I'm kinda lost.
     
  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    I cant see bar letters in the equation. If you want to write bar letters write them like this: A'.
     
  4. mathsfailure

    Thread Starter New Member

    Nov 28, 2009
    4
    0
    X = ]A.B' + C]' . (A'.C' +B')


    yup here it is.
     
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    I will give you a hint:

    [A.B'+C]'=(A'+B).C'

    Try to simply it more.
     
  6. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    de morgans theorem:

    A' + B' = (AB)'

    and

    A'B' = (A+B)'
     
  7. mathsfailure

    Thread Starter New Member

    Nov 28, 2009
    4
    0
    ive simplified the equation to somewhat where i got confused

    (A'+B+C')(A'+C'+B')
    A'A' + A'C' + A'B' + BA' +BC'+BB'+C'A'+C'C'+C'B'
    A'+A'C'+A'B'+BA'+BC'+C'A'+C'+C'B'

    Confused
     
  8. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    You can continue by taking common factors as to make some terms zero. However, I think it is easier to proceed like this:

    (AB'+C')'(A'C'+B')=(A'+B)C(A'C'+B')=(A'C+BC)(A'C'+B')=A'CA'C'+A'B'C+BCA'C'+BB'C

    Then apply the rule:

    AA'=0

    to simplify the final expression.
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    mathsfailure,

    You will find the program in the link below to be useful. It gives the minterms of any Boolean expression. If the minterms of the problem match your solution or your intermediate simplifications, then you know you did not make a mistake with the algebra. Once you know the minterms, you can simplify the Boolean expression with a Karnaugh map. Also search for threads in this forum for "logic".

    Ratch

    http://forum.allaboutcircuits.com/showthread.php?p=135582#post135582
     
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