Beginner question about logic simplification

Discussion in 'Homework Help' started by Mcftdhorapusswrtrvocm, Sep 27, 2014.

  1. Mcftdhorapusswrtrvocm

    Thread Starter New Member

    Sep 27, 2014
    1
    0
    Hello there!
    I need some help with simplification. We were given an equation and told to build the circuit out of it, but I can't simplify the function any more and it seems like way too much to build as is.

    Ʃ m(0,1,3,5,6,9,10,15) + d(8,14)

    and the truth table from that is:
    D C B A - Z
    0 0 0 0 - 1
    0 0 0 1 - 1
    0 0 1 0 - 0
    0 0 1 1 - 1
    0 1 0 0 - 0
    0 1 0 1 - 1
    0 1 1 0 - 1
    0 1 1 1 - 0
    1 0 0 0 - x
    1 0 0 1 - 1
    1 0 1 0 - 1
    1 0 1 1 - 0
    1 1 0 0 - 0
    1 1 0 1 - 0
    1 1 1 0 - x
    1 1 1 1 - 1

    I simplified the equation to:

    A ( B C + B C ) + D ( B C + B C ) + C ( A XOR D ) + A B D

    So that final function leaves me with 12 gates, and that seems like way too much for my first project, and I feel like there must be some way to simplify it but everything I've tried doesn't work. Once I was told that ( B C + B C ) could become ( B XNOR C) but that doesn't work or make sense.
    here's a drawing:
    DLD1.png

    And if you have any tips on how to put this all into an actual circuit, any help would be much appreciated. It just doesn't make any sense to me yet. I'll probably end up going to the teacher for help on Monday, but that's when the class is and I'm supposed to have it finished for the class.
    Thanks for any help you can offer.

     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,378
    494
    In one of my textbooks it says:
    A+A=1
    and
    A1=A

    So your A(BC+BC)=A(1)=A
     
  3. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    It seem to me that A ( BC + BC ) + D ( BC + BC ) would be equivalent to ( A + D )( BC + BC ). In which case, you could use 9 gates.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    Looking at the truth table you observe that there eight products that produce a 1 and six products that produce a 0 and two that produce a don't care. You have implemented a canonical SOP (Sum of Products), but you could have chosen to implement the equally valid POS (Product of Sums)
     
  5. arivvu

    New Member

    Sep 7, 2012
    9
    0
    @shteii01 and @LDC3
    i think he (thread starter) mentioned ([B' and C'] or [B and C]) it is not ([B and C]' or [B and C]). both will give different result, if it is first case result will not be 1 always.
     
  6. jjw

    Member

    Dec 24, 2013
    173
    31
    Have you tried Karnaugh maps?
    It might be easier to simplify with them.
     
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