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I put this into a k-map and it has 5 terms! is somthing wrong?

  1. Oct 17, 2005 #1
    Okay i have 4 inputs, A,B,C,D, and when i put it into a K-map i got 5 terms, its a prime number circuit. It will output 1 if the number is prime, example...
    2,3,5,7,11,13. So 1101 == 13 would output a 1.
    A = 1, B = 1, C = 0, D = 1; Output = 1
    From suming up the min terms on my truth table i got the following boolean expression:
    Note: 'A means A complemented

    'A'BC'D + 'A'BCD + 'AB'CD + 'ABCD + A'BCD + AB'CD

    I put it into a K-map and got the following:
    'A'BC'D + 'ACD + 'ABD + 'BCD + B'CD

    Does that look right to you? I already wasted an hour implemented the wrong diagram because i screwed up the k-map and i want to make sure i didn't screw it up again. Thanks.
     
  2. jcsd
  3. Oct 17, 2005 #2

    ranger

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    Gold Member

    I'm getting four terms for the k-map:
    Code (Text):
    B'CD+ 'ACD + 'A'BC'D + A'BCD
    [​IMG]
     
    Last edited: Oct 17, 2005
  4. Oct 17, 2005 #3
    Thanks for the responce...but ]shouldn't u loop some more? Here is what I got...but it still isn't right because when i create a circuit out of it using MAX PLUS II, it says A is not needed!! which makes no sense to me...
    Here is my k-map...
    http://img137.imageshack.us/img137/5189/lastscan6nq.jpg
     
  5. Oct 18, 2005 #4

    ranger

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    Gold Member

    Ah yes. I see where I forgot to loop.
    I think you make some mistakes in the k-map
    Code (Text):
    'A'BC'D + 'ACD + 'ABD + 'BCD + B'CD
    Thats what you have. I'm look at the k-map and I cant see 'A'BC'D in the k-map. I see A'BCD. Why is it that you put the upper right most 1 by itself. Just looping it with the 1 to the left of it is good enough. Looking at the last row I see that you have 'BCD. How did you end up with that?
    Here is what I got
    Code (Text):
    'A'BC + 'ACD + 'ABD + B'CD + A'BCD
    I'm also taking this course right now, I know how confusing this topic can get.
     
    Last edited: Oct 18, 2005
  6. Oct 21, 2005 #5
    That is wrong, but thanks! I figured it out finally!
     
  7. Oct 30, 2005 #6
    For whatever it is worth, there are four terms, each of three variables. In other words, each loop encompasses two cells.

    I have a couple of suggestions (in real life - - - if your instructor does it the other way, you are stuck with it in class). First, try using the apostrophe after the variable to show a "negation", this is a bit more familiar to most of us. (It is unfortunate that most of our typing program packages don't provide for an "overline".) Second, make "A" your low-order variable. Then, A'B'CD' would be "0100". This makes it a little easier to put it into an ordered K-Map (I keep preaching the use of ordered maps - - - but most still don't bother.) Third, try to learn and use the ordered K-Map somewhere down the line. It makes the operation far more automatic and gives more capability.

    KM
     
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