Boolean Algebra Help

Discussion in 'Homework Help' started by 5kWatz, Jun 8, 2012.

  1. 5kWatz

    Thread Starter New Member

    Jun 8, 2012
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    We are learning Boolean Algebra Simplification in my Electronics class and I have a fairly good understanding of how to apply the rules, but I ran into one problem that I can't get past. I have checked it and I know that they are equivalent.

    How to get from
    A'B' + A'C' + AB + BC'
    to A'B' + AB + BC'
    using the Boolean rules?

    Is there a rule that I am missing. This seems like it should be so easy.

    Thanks in advance for any help!
    5kWatz
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Have you done Karnaugh mapping as yet?
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    The steps needed to solve this problem are very simple and, once you get there, are immediately obvious why they work. But seeing them in the first place is not very obvious because it uses a property of Boolean algebra that doesn't have a counterpart in the normal algebra you have spent years learning to think in.

    Q1) Which term do you need to make go away?

    Q2) Which other terms have factors in common with the one you need to make go away?

    Q3) Let's say that my three variables are XYZ and I need to make the term XY go away (and, by go away, I mean that, just like this problem, we have reason to believe that it is an unnecessary term). How would I write this term so that it used all three variable? Hint: What expression that is only dependent on Z could I AND with XY while not changing it's behavior?

    Q4) Given the answer to Q3, can you expand it into two terms and then absorb each term into one of the other existing terms.

    Take a shot at answering these questions and, if needed, I'll throw out some more explicit hints.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    If boolean equations can be simplified it is because there are redundancies.
    Sometimes the redundancy is not obvious by just looking that the equation.
    When you learn to use Karnaugh maps the picture becomes much clearer.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    I agree. Using bubble logic to apply DeMorgan's theorems is also usually a lot more quick and efficient. But in either case you are using a tool that works because of the underlying Boolean algebra properties and there is merit in learning to work with those properties directly. And it's not an either-or situation. As long as Karnaugh maps and bubble logic are presented from a Boolean algebra foundation, a period in which all three are used in conjunction leads to much better insite into logic relationships than would likely occur if each were presented as isolated topics (which, sadly, is effectively how they are often treated).

    This particular problem strikes me as one that is intended to let the student practice applying the 'nontraditional' properties of Boolean algebra. Once the solution is found and understood based on the direct properties, then examining how it is arrived at using a Karnaugh map underscores not only the fact that such maps help you to visualize and remove redundancies, but established a link between which specific Boolean algebra properies are at play to make that visualization and removal possible.
     
  6. 5kWatz

    Thread Starter New Member

    Jun 8, 2012
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    Okay, I believe that you are implying that I do ....

    A'B' + A'C'(B+B') + AB + BC'
    = A'B' + A'BC' + A'B'C' + AB + BC'
    = A'B'(1+C) + AB + (A'+1)BC'
    = A'B' + AB + BC'

    So, it was a simple as expanding a term into two terms that would each cancel out. Very nice! Thank you for the help!


    We haven't done Karnaugh, but are using Multisim to simplify ... so I knew that they were equivalent. We are still learning the basics of Boolean simplification techniques.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Yes, that is EXACTLY what I was trying to steer you to. I usually try to keep my questions such that each one is a single concept question that needs just one concept added to where the prior questions should have gotten you. In this case, I was concerned that I wasn't able to phrase the question well enough to keep it from being too confusing in it's own right. I'm glad you were able to extract the points I was trying to describe. Good Job!

    The thing to understand from this exercise is that it is possible for a term to be redundant not because it can be combined with another term, but because it can be broken up into multiple pieces and as long as each and every one of those pieces can be absorbed, individually, into another term then the original term is not necessary. The terms that do the absorbing can be completely independent and unrelated to each other. This is what makes it hard to spot algebraically.

    What you will discover when you get to Karnaugh maps is that they visualize precisely these kinds of interactions between terms to reveal redundant combinations that can be removed. What should also be covered at some point is that, while these redundant terms, known as "concensus terms", are not necessary on paper, they are not always undesireable and, in fact, they can be critical to avoiding what are known as "timing hazards". Consider that just a teaser that will hopefully make the material more interesting to look forward to.
     
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