Boolean Identities question

Discussion in 'Homework Help' started by mario_wins, Nov 3, 2008.

  1. mario_wins

    Thread Starter New Member

    Nov 3, 2008
    3
    0
    Hi,

    I just starts learning about combinational circuits and boolean identities. Is there an easy way to remeber the laws in AND & OR form inlcuding:

    Null (or Dominance) Law
    Idempotent Law
    Inverse Law
    Commutative Law
    Associative Law
    Distributive Law
    Absorption Law
    DeMorgan’s Law
    Double Complement Law
    Identity Law

    For exmaple i need to simplify x'yz + xz, although i don't know which laws to use or how to remember them.

    Thanks.

     
  2. Dave

    Retired Moderator

    Nov 17, 2003
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  3. Ratch

    New Member

    Mar 20, 2007
    1,068
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    mario_wins,

    Well, I have done a lot of Boolean simplifications in my time, so I feel qualified to answer you. I say learn the principles that these laws represent, but don't try to use them to algebraically to reduce a Boolean expression. Not unless the expression so simple that it can readily be seen to have a simpler solution. Lets take your above example. x'zy + xz = x'yz + xz(y +y') = x'yz + xyz + xy'z = minterm(3,5,7). From these three minterms, you can easily plot them on a Karnaugh map, or use the Quine-McCluskey tabulation method. Both methods will give you xz + yz for an answer. Now that it has been reduced, and you know the answer, you can go back and do it by the algebraic method. x'yz + xyz + xy'z + xy'z = (x + x')yz + x(y+y')z = yz + xz . So the algebraic method is hit and miss, but either K-map or Quine-McCluskey method will surely give you the simplest answer, or another alternate simplest answer if it exists. The algebraic method is the worst possible way to simplify a Boolean expression because it requires you to be able to see the pattern of simplification. That is sometimes hard to do, especially if the Boolean expression is complicated. The other two methods mentioned above will always converge on the solution or multiple solutions if more than one solution exists. Besides the two methods require no intuition. It is just plug and chug similar to an arithmetic operation between two numbers.

    Ratch
     
    Last edited: Dec 28, 2009
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