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Homework Help: SOP vs POS

  1. Feb 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi my question is whether an SOP = POS for a given function F(A, B, C, D). Or is the SOP a complement of POS?

    Tasked to convert derive a simplfied SOP from a 4 var Kmap with dont cares.
    My answer was,
    F = C'.D' + A'.B'.C which is correct.

    Then for part B they asked for the simplified POS form.

    3. The attempt at a solution
    F = C'.D' + A'.B'.C
    F' = (C'.D' + A'.B'.C)' (Using Dmorg)
    F' = ((c+d)' + (a+b+c')')' (Using Dmorg)
    F' = (c+d)'.(a+b+c')'

    F = ((c+d)'.(a+b+c')')'
    F = (c+d) +(a+b+c') (Stuck here)

    If i used Kmap i got the answer which is

    I wanna know the boolean algebra way
    Last edited: Feb 6, 2015
  2. jcsd
  3. Feb 6, 2015 #2
    There is a modification to KMaps to get POS instead of SOP.
    It involves instead of circling the 1's you circle the 0's
    Prehaps look for it in your textbook? I only used it on the 1 assignment where we had to do it and then promptly forgot it.
    The other way is to use applications of DeMorgans to get there. But that seems like a lot of work :) the POS KMap was much easier
  4. Feb 6, 2015 #3
    Yea i got the KMap part. I want know the demorgans part
  5. Feb 7, 2015 #4
    I'm not sure what you actually want. Sum of minterms can be converted to product of maxterms and that is equivalent.
    The easy way to convert them is using a dual form (twice), so, you simply switch * and + (and 1 and 0) to obtain a dual function. Then multiply everything in order to again get a sum of minterms and use a dual.
    In your case: f = C'D' + A'B'C
    f_d = (C' + D') (A' + B' + C) = C'A' + C'B' + C'C + D'A' + D'B' + D'C = C'A' + C'B' + D'A' + D'B' + D'C
    (f_d)_d = f = (C' + A')(C' + B)(D' + A')(D' + B')(D' + C')
    If that's what you wanted. I don't know how you got: C'.D' + A'.B'.C == D.(B'+C').(A'+C'), because I can't get that. I can simplify my own form though, but I don't get your result.
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