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Using 3 input LUT's to Implement a Boolean Function

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the function f(w1,w2,w3,w4) = SUM( m(0, 1, 3, 6, 8, 9, 14, 15)). Derive an implementation using the minimum possible number of three-input LUTs.


    2. Relevant equations
    Where SUM(m(0, 1, 3, 6, 8, 9, 14, 15)) is the sum of the min terms, a Sum of Products that from a K map is (~w1*~w2*~w3)+(~w1*~w2*w4)+(w1*w2*~w3)+(w1*~w2*w3)+(~w1*w2*w3*~w4) Where ~w is "not w".

    3. The attempt at a solution
    Thus far all I can do is get the minimum cost solution from the K map, which I gave above. I guess I'm just a little confused as where to start considering a 3 input LUT has 8 stored values and 3 different inputs and one output. Any help will be beneficial! Thanks,

    Matt
     
  2. jcsd
  3. Oct 27, 2009 #2
    Noone have any ideas? I'm not looking for the answer just a push in the right direction. Thanks, Matt
     
  4. Nov 7, 2009 #3
    Use shannon's expansion theorem
     
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