Boolean Algebra - Simplification

Discussion in 'Homework Help' started by D0wnl04d3r, Dec 30, 2009.

  1. D0wnl04d3r

    Thread Starter New Member

    Dec 30, 2009
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    0
    Hi guys...

    I'm having real trouble simplifying the following equation, so that a minimal number of gates are used...
    I'm very new to this, I have no previous knowledge of electronics and have just started a degree...I just cannot understand this concept at all!

    here is the question...

    Simplify the following function, such that a minimal number of gates are used:
    (D.B'+A'C).(C+A')

    Do I need to create a truth table or Karnaugh graph from this? is it possible to simplify without therse? I've tried browsing through the site but find it all very confusing!

    thanks
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    D0wnl04d3r,

    A degree in what? How much was taught to you already? Is this part of a course you are taking? It find it worrisome that you are trying to simplify a Boolean expression, and you don't understand the concept.

    Did you see the post from just three days ago? http://forum.allaboutcircuits.com/showthread.php?t=32040 . What do you find confusing? We can answer specific questions, but we cannot start you from ground zero. You have to do some preliminary studying on your own.

    Ratch
     
  3. D0wnl04d3r

    Thread Starter New Member

    Dec 30, 2009
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    0
    it's a degree in IT but more than half of the course is electronics based. My lecturers so far have touched on Truth Tables & Karnaugh Graphs but there is no maths module as such within my course so the boolean algebra side I am finding very difficult...

    for staters I dont understand how to expand the current expression, so that the data can be input into the Karnaugh map
     
    Last edited: Dec 30, 2009
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
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    D0wnl04d3r,

    First of all, get rid of the dots between the logic variables and the terms. They are redundant and ridiculous.

    (D.B'+A'C).(C+A') = (DB' + A'C)(C +A')

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

    = B'CD + A'B'D + A'C ; Remove redundant term

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

    = A'B'CD + A'B'CD + A'B'C'D + A'B'CD + A'B'C + A'BC

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

    = A'B'CD + A'B'CD + A'B'C'D + A'B'CD + A'B'CD' + A'B'CD) + A'BCD' + A'BCD

    = minterms = Ʃ(3,11,1,11,2,3,6,7) = Ʃ(1,2,3,6,7,11) ; Remove redundant minterms

    From there you can use a Karnaugh map to simplify it.

    You can obtain the minterms a lot faster if you use the program from the following link.

    http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=logic+lovers

    Ratch
     
  5. D0wnl04d3r

    Thread Starter New Member

    Dec 30, 2009
    3
    0
    okay thankyou, i've done that now, need to have a more in depth look at the karnaugh map now, to see how I can group the 1's. thanks ever so much for your help!

    Well i've finally come up with an expression of :

    A'(B'+C)+CD

    is anyone able to verify whether or not this is correct?
     
    Last edited: Dec 31, 2009
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    D0wnl04d3r,

    No one should have to do so. You should be able to do it yourself. Did you download that program from the link, read the instructions, and plug in your K-map simplification? Did the program spit out the same minterms as you started with? If not, you did not intrepret the K-map correctly.

    Ratch
     
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