Boolean Algebra Help

Discussion in 'Homework Help' started by Joepiper5, Aug 3, 2011.

  1. Joepiper5

    Thread Starter New Member

    Aug 3, 2011
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    Hey guys, Im struggling getting my head around this Boolean algebra stuff

    Im stuck on where to start with what seems to be a simple problem:

    f=(ab).c+d There is a bar over each variable plus a bar over the whole equation.

    Any help is greatly appreciated
    Thanks
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
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    If you don't tell us where you're trying to go, any road will get you there... :p

    I assume you want to simplify the expression -- but your notation is a bit strange. Is "ab" a single symbol or does it signify multiplication? I assume the period means multiplication. I'll assume the overbars denote negation, so look up De Morgan's Laws for some help to get started.
     
  3. Joepiper5

    Thread Starter New Member

    Aug 3, 2011
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    sorry about that, yes i need to simplify it so that the final expression is a Sum of Product expression with no bar appearing in the expression. I'll keep looking into it, the textbook been used brushes over the topic with only 1 example which seems to be much easier. sooo frustrating
     
  4. TBayBoy

    Member

    May 25, 2011
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    Mostly Harmless :)
     
  5. Joepiper5

    Thread Starter New Member

    Aug 3, 2011
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    lol. So i watched some videos i found on youtube and had a attempt,

    I'll represent the bar with ' and double bar is ''

    so f=(ab)' *c'+d' with a bar over the whole thing
    -- =(ab)''+c''+d''
    =(a''+b'')+c''+d'' since a''=a
    =(a+b)+c+d

    What do you think??
     
  6. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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  7. Joepiper5

    Thread Starter New Member

    Aug 3, 2011
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    cant figure out how to enter it correctly into the calculator, i want to see it on paper too so i can see the process
     
  8. Zazoo

    Member

    Jul 27, 2011
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    Hello, you made a small mistake in your first step:

    When you break apart the top-most negation bar the + d' becomes * d'', like this:

    = [(ab)''+c'']*d''

    (Note the brackets to maintain the correct grouping of d'' with the remaining expression).

    Also, in step 2: (ab)" = (ab), instead of (a"+b"). You can treat everything in the parentheses as a unit, so negating the negation just gives you back the original expression.

    See if you can work it through again.
    ^Mike^
     
    Last edited: Aug 3, 2011
  9. Joepiper5

    Thread Starter New Member

    Aug 3, 2011
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    Thank you so much mike. I appreciate the help
     
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