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Karnaugh map

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  1. Nov 18, 2015 #1
    Hi All,

    I am given a canonical expression with active low output Y*

    Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)

    I would like to use a karnaugh map to obtain the Product of Sum.

    However, I am confused because of the terms "active low".

    Is my truth table correct?

    A B C D Y*
    0 0 0 0 1
    0 0 0 1 0
    0 0 1 0 1
    0 0 1 1 0
    0 1 0 0 0
    0 1 0 1 1
    0 1 1 0 0
    0 1 1 1 1
    1 0 0 0 1
    1 0 0 1 0
    1 0 1 0 1
    1 0 1 1 0
    1 1 0 0 1
    1 1 0 1 1
    1 1 1 0 0
    1 1 1 1 1

    Thank You!
     
  2. jcsd
  3. Nov 18, 2015 #2
    Active low means the same thing as logical NOT. It is the same thing as if you put a line over Y.

    However, for the case:
    Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)
    your answer is opposite because they are asking you do define Y*

    If the question asked for Y then you would have to invert your output.
     
  4. Nov 18, 2015 #3
    Hi,

    Thanks for your reply.
    just to confirmed soo my answer should be:
    A B C D Y*
    0 0 0 0 0
    0 0 0 1 1
    0 0 1 0 0
    0 0 1 1 1
    0 1 0 0 1
    0 1 0 1 0
    0 1 1 0 1
    0 1 1 1 0
    1 0 0 0 0
    1 0 0 1 1
    1 0 1 0 0
    1 0 1 1 1
    1 1 0 0 0
    1 1 0 1 0
    1 1 1 0 1
    1 1 1 1 0

    All my output should be inverted. Am I correct?
     
  5. Nov 18, 2015 #4
    Oops, sorry I was looking at POS instead of SOP. (It's been a few years) I think you were right the first time.

    Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)


    ABCDY* Y
    0 0 0 0 0 1
    0 0 0 1 1 0
    0 0 1 0 0 1
    0 0 1 1 1 0
    0 1 0 0 1 0
    0 1 0 1 0 1
    0 1 1 0 1 0
    0 1 1 1 0 1
    1 0 0 0 0 1
    1 0 0 1 1 0
    1 0 1 0 0 1
    1 0 1 1 1 0
    1 1 0 0 0 1
    1 1 0 1 0 1
    1 1 1 0 1 0
    1 1 1 1 0 1

    Does the term "Active Low" make sense though? You are going to see that a lot. In this case you can ignore it because your truth table has a star on it. But in general its important.
     
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