Boolean Algebra

Discussion in 'Homework Help' started by Kash88, Nov 27, 2014.

  1. Kash88

    Thread Starter New Member

    Nov 27, 2014
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    CW = ¬A.¬B.¬C + A.B.¬C + A.¬B.C + A.B.C

    Simplify the expression and mention which laws used and where they were used..

    ¬ = NOT
    + = OR
    . = AND
     
  2. Kash88

    Thread Starter New Member

    Nov 27, 2014
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    my attempt -
    A(B(C+¬C) + ¬B.C) + ¬A.¬B.¬C
    A(C(B.¬B) + ¬A.¬B.¬C
    A (C+B) + ¬A.¬B.¬C
    A.B + A.C + ¬A.¬B.¬C

    What now? :confused:
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    That looks like about it to me, though you were supposed to indicate which laws you used at each step. To be more rigorous when asked to show the laws being used, apply one law in one place on each line.

    Be sure to consider the validity of your result -- does it really match the original expression.
     
  4. Kash88

    Thread Starter New Member

    Nov 27, 2014
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    I'm meant to simplify it and end up with an XOR in there some how?:confused:

    Classmates have managed to get to ¬A XOR (B.C) no idea how they've done that..
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    This brings up the point of what it means to "simplify" something. What is the metric by which one expression is deemed "simpler" than another expression? Fewest terms? Fewest gates? What constraints must be met.

    The usual implication when "simplifying" a Boolean expression is for the end result to be either a SOP or a POS with each term/factor having as few variables as possible. The use of XOR is generally not allowed as it is not a basic Boolean operator. Are you sure that the use of an XOR is (1) allowed, and (2) is "simpler"?

    Are you sure that what they ended up with is even correct? What does their solution yield for (ABC) = (001)? Does that agree with the original expression?

    Why do you say that you are meant to simplify it and end up with an XOR in the result? Is that part of the problem statement?
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    If you really want to get an XOR into the expression, look at the truth table and see if you can break it into two halves such that the output in one half is the opposite of the output in the other half.

    I get an expression involving one NOR gate and one XOR gate.
     
  7. Kash88

    Thread Starter New Member

    Nov 27, 2014
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    It's not part of the problem we've just been told by our tutor that we should end up with an XOR gate
     
  8. Kash88

    Thread Starter New Member

    Nov 27, 2014
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    I'm more of a visual learner if you could come up with an example/similar expression and go through it please, I'd be able to understand it better.
    Thanks
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    First, consider the Truth Table for the basic XOR function:

    Y = A XOR B

    Code (Text):
    1.  
    2. A B Y
    3. 0 0 0
    4. 0 0 1
    5. 0 1 1
    6. 1 1 0
    7.  
    Let's reorganize things as follows:

    Code (Text):
    1.  
    2. A B=0 B=1
    3. 0  0   1
    4. 1  1   0
    5.  
    In this table, the middle column gives me Y when B=0 and the last column gives me Y when B=1.

    Do you see how the two output columns are opposites of each other? That's the key.

    The middle column is some function of A, let's call it X, and the last column is therefore X'.

    When B=0, Y=X and when B=1, Y=X'. This can be written in Boolean algebra as

    Y = B'·X + B·X'

    Well, that's the definition of

    Y = B XOR X

    In our case, X=A, so we have

    Y = B XOR A

    which shouldn't come as any surprise.

    So let's just make up a function in three variables that has this property:

    Code (Text):
    1.  
    2. AC B=0 B=1
    3. 00  1   0
    4. 01  0   1
    5. 10  1   0
    6. 11  1   0
    7.  
    In this case, our X is a function of A and C and our result will again be of the form:

    Y = B XOR X

    We just need to figure out what X is. By inspection, we can see that it is

    X = A'·C

    so we have

    Y = B XOR (A'·C)
     
    Kash88 likes this.
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