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I'm stuck on boolean algebra, it has 4 variables!

  1. Sep 24, 2005 #1
    Hello everyone, i'm designing a IC that will result in a 1 output if the number is > 9. So I wrote out the truth table and or'ed all the min terms and got the following:
    Picture of truth table and boolean expression I'm stuck now! any help would be great!
     
  2. jcsd
  3. Sep 24, 2005 #2
    AND the MSB with each of the two preceeding bits and then OR that.

    (AB)+(AC) will yield a 1 when the output is greater than 9.
     
  4. Sep 24, 2005 #3
    Thanks for the reply but i finally figured out how k-maps work, and when i threw that big mess in the k-map it reduced down to:
    z = AB + AB'C
    note; B' = B complimented
    here is a picture:
    http://img43.imageshack.us/img43/3353/w0t1bp.jpg
     
  5. Sep 24, 2005 #4
    You don't need the B! though. You're adding complexity by having it.
     
  6. Sep 24, 2005 #5
    I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?
     
  7. Sep 24, 2005 #6
    You have it backward. It's because you didn't use all possible overlaps that you have redundancies. Go back this time and group your "1's" so that each group is as large as possible, taking every overlap you can get, but the smallest number of groupings possible. (In this case there are obviously only two groupings needed, so there will be two 'AND' terms.) When you do that you will see the answer.

    KM

    By the way: your drawing was hard to see.
     
  8. Sep 24, 2005 #7
    You do get the right answer if you arrange things logically:

    Code (Text):

    ______________________________
    |  \cd                        |
    |ab \ ________________________|
    |     | cd  | cd! | c!d | c!d!|
    |______________________________
    |  ab |  1  |  1  |  1  |  1  |
    |______________________________
    | ab! |  1  |  1  |  0  |  0  |
    |______________________________
    | a!b |  0  |  0  |  0  |  0  |
    |______________________________
    |a!b! |  0  |  0  |  0  |  0  |
    _______________________________
     
    You see AB is one part of the solution because it works for all combinations of CD. If you look in the columns you will see AC is the other solution because the solution AC does not depend on B at all.
     
  9. Sep 24, 2005 #8
    Thanks so much guys, my professor so told us that we should not overlap, infact we should avoid overlapping, what the heck is he talking about! I ended up with AB+AC like u guys said!
     
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