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Boolean Algebra Simplification

  1. Apr 8, 2017 #1
    1. The problem statement, all variables and given/known data

    How to simplife it? I would like to see a process of simplification.

    Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

    The ' denotes a bar over the previous letter.


    My second question is:

    Y = BC + AB'D = (uploaded picture)

    Y = A'D + B'D + A'BC + ABC'D' = ???????


    2. Relevant equations

    Simplification Rules


    3. The attempt at a solution

    I tried it but can't get to the correct result.

    The correct result should be (made through Karnaugh map):

    Y = A'D + B'D + A'BC + ABC'D'


    Thank you for any advices and I'm sorry for my english.
     
  2. jcsd
  3. Apr 8, 2017 #2
    uploaded pic
     

    Attached Files:

    • asd.PNG
      asd.PNG
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  4. Apr 9, 2017 #3

    NascentOxygen

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    Staff: Mentor

    Hi Rossy. :welcome:

    Break it down into manageable parts, e.g., look at "taking out a common factor"

    A'B'C'D + A'B'CD = A'B'D(C' + C) = ????

    Can you complete this simplification?
     
  5. Apr 9, 2017 #4
    Hi NascantOxygen,

    it's A'B'D

    Could you look at the image I uploaded, it's my attempt.. as you can see I can't get to the right solution.
     

    Attached Files:

  6. Apr 9, 2017 #5

    NascentOxygen

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    Staff: Mentor

    The most obvious mistake is that you have ignored one term in the given expression, viz.,

    + A'BC'D +
     
  7. Apr 9, 2017 #6

    NascentOxygen

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    Staff: Mentor

    In Boolean algebra you may encounter an expression such as (C + C' D) and wish to simplify it. One way is to AND the C term with something having the value TRUE, and I'll choose (1 + D) which has the value 1 because anything ORed with TRUE is TRUE.

    C(1 + D) + C' D

    = C.1 + CD + C'D

    = C + D(C + C')

    = C + D
     
  8. Apr 10, 2017 #7
    So if I take this expression A'BC'D + ABC'D' I can simplyfy it like this BC'(A + D)(A' + D') but how do I continue?
     
  9. Apr 10, 2017 #8

    NascentOxygen

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    Staff: Mentor

    You can't go further with that. You have extracted the factor BC' and that's it.
     
  10. Apr 11, 2017 #9
    So you can't get from this expression

    Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

    to this expression by Boolean Algebra

    Y = A'D + B'D + A'BC + ABC'D'
     
  11. Apr 11, 2017 #10

    NascentOxygen

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    Staff: Mentor

    Yes, it can be simplified to that.
     
  12. Apr 11, 2017 #11
    Can't get further...
     

    Attached Files:

  13. Apr 11, 2017 #12

    NascentOxygen

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    Staff: Mentor

    Take two of these terms (shown in green) and simplify them using the help shown in post #6

    A'BC + B'D + A'BC'D + ABC'D'
     
  14. Apr 12, 2017 #13
    Is it right?
     

    Attached Files:

  15. Apr 12, 2017 #14

    NascentOxygen

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    Staff: Mentor

    That is right.
     
  16. Apr 12, 2017 #15
    Thank you for your time and patience with me!! :) The problem is solved.
     
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