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Homework Help: Boolean Algebra, Logic Diagram, K-Map, Nor gates HELP

  1. Feb 2, 2007 #1
    Boolean Algebra, Logic Diagram, K-Map, Nor gates... HELP!!!!

    1. The problem statement, all variables and given/known data
    F(A,B,C,D) = Sigma(2,4,6,10,12)
    d(A,B,C,D) = Sigma(0,8,9,13) [Dont Care Functions]

    Implement the function using no more than 2 NOR gates.
    2. Relevant equations

    3. The attempt at a solution
    First of all, I am wondering if this is PHYSICALLY POSSIBLE.

    Anyways, if you draw the K-map for it, you get
    d 0 0 1
    1 0 0 1
    1 0 0 0
    d d 1 1

    As for the grouping, I grouped the Top 4 0s (0100, 1100, 0101, 1101), mid 4 0s (0101, 1101, 0111, 1111) and 2 0s on the right and middle (1011, 1111)

    If you're doing Nor implementation.. it's usually easier to group the 0s so
    F' = BD + A'D + ABC

    (As for the notation, ' are inversion, + is or and * is And EG : AB is A and B)

    But we want F.. so we apply the demorgan's theorem

    F = (B'D')(AD')(A'B'C')

    and that is our function.

    I looked at this function and I thought there is no way to solve this problem only by using NOR gates.

    Please help!
    Last edited: Feb 2, 2007
  2. jcsd
  3. Feb 2, 2007 #2
    I think you got the table wrongly. There are 5 minterms in the problem statement but you have 6 in your table; and there should be 4 dont cares while you have only 3 in the table.
  4. Feb 2, 2007 #3
    On a related note, De Morgan does not give you the transformation above. Read up on De Morgan again.
  5. Feb 2, 2007 #4
    Oops, I meant to put a d instead of that 1. It's fixed now.

    And yeah, F is supposed to be

    F = (B'+D')(A+D')(A'+B'+C')

    But this still does not answer my question. It is utterly impossible to represent this just using two nor gates.
  6. Feb 2, 2007 #5
    With this K-map, shouldn't there be only two groups of 0's?
  7. Feb 2, 2007 #6

    ? I can't see how you can have only two groupings of 0's and cover all 0's here.

    Can you please explain how that might be done?
  8. Feb 2, 2007 #7


    Now I can do it from here.

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