Boolean Simplification

Discussion in 'Homework Help' started by Simca, Sep 25, 2009.

  1. Simca

    Thread Starter New Member

    Sep 25, 2009
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    0
    I have been working on the following Boolean simplifications but I seem to be missing a theorem in my approach

    1) AB + A'C + B'D + D'E + C'E Solution: AB + A'C + B'D + E

    and

    2) AD + BD + CD + A'B'C' Solution: D + A' B' C'

    I notice that it seems that the solution seems to include something to the effect that if you have one equation that contains the notted version of existing variable that you can get ride of the existing, keeping the common. I know that may not make sense, but im a little lost on how to do these 2.

    Thanks
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Simca,

    You are missing a Karnaugh map in your approach. The first problem is a 5 variable logic expression. That means there are 32 different possible minterms that need to be simplified. The K-map and the program below show this expression to have 26 minterms. That is too complicated to do using Boolean algebra and theorems. Once you K-map it, you can easily what minterms combine to make each simplified terms. That is the best way to do it.

    You can find the minterms easily by using this program. http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=boolean . It also is good for checking your results.

    Ratch
     
  3. Simca

    Thread Starter New Member

    Sep 25, 2009
    3
    0
    I appreciate your advise, but these ones have to be done using Algebra and then varified with the K Map. Thats how I got the solutions :)

    Thanks Anyway
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    simca,

    OK, then you have to expand out to the minterms of the problem in order to group and reduce them. Unless you are a logical Mozart, I suggest to use the K-map to see how to group the minterms. Once the terms are grouped correctly, you should be able to easily reduce each group for each term of the solution. You will have to duplicate some terms in order make the correct groups. Duplicate terms do not change the value of the expression. I will partially expand the first problem to show how it is done. You are in for a hell of a lot of work doing it manually.

    AB + A'C + B'D + D'E + C'E

    AB(C+C') + A'C(B+B' )+ (A+A')B'D + (A+A')D'E + (A+A')C'E

    ABC + ABC' + A'BC + A'B'C + AB'D + A'B'D + AD'E + A'D'E + AC'E + A'C'E

    ABC(D+D') + ABC'(D+D') + A'BC(D+D') + A'B'C(D+D') + AB'(C+C')D + A'B'(C+C')D + A(B+B')D'E + A'(B+B')D'E + A(B+B')C'E + A'(B+B')C'E

    ABCD + ABCD' + ABC'D + ABC'D' + A'BCD + A'BCD + A'BCD' + A'B'CD + A'B'CD' + ..... continue this expansion using (X+X') for the missing variable

    When you get done, you should have 26 terms as shown in your K-map.

    Ratch
     
  5. Simca

    Thread Starter New Member

    Sep 25, 2009
    3
    0
    Thanks, thats what I thought I may have to do, I tried to see if I could detect the ones that would show the solution, or check to see if i was missing something. I appreciate your help

    Simca
     
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