Nanding an expression ?

Discussion in 'Homework Help' started by Gingerbreadman, May 31, 2011.

  1. Gingerbreadman

    Thread Starter New Member

    May 23, 2011
    3
    0
    Hi, i'm trying to implement the expression: F = ((B.A).(D+C))+(D.C).(B+A)
    using only NAND gates.

    How would i convert the function using de-morgan's theorem? I know it should not be hard yet every time i start i get stuck with the wrong answer.

    Please can you help?

    If you could show me step by step so i know how it gets there that would be great.

    Thanks for reading,

    Dave
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    I don't have the time to take you step by step right now, but here are some tips:

    (A+B)=(A'B')'
    A'=A NAND 1
    (AB)=((AB)')'

    You could also convert the expression to a truth table, find the Sum of Products and convert it to a NAND network. That doesn't use DeMorgan's theorem but it's more algorithmic.
     
  3. Gingerbreadman

    Thread Starter New Member

    May 23, 2011
    3
    0
    Hey, thanks for answering, and quickly too !

    i've just about got it, i'm just thinking about (d' . c') . (d' + c')

    (where . means AND, + means OR and ' means not of course.)

    I can get this !
     
  4. Gingerbreadman

    Thread Starter New Member

    May 23, 2011
    3
    0
    ok so i may have that bit wrong, or it just won't tell you what i'm thinking.

    But thanks becuase i just noticed the NOT law you posted.

    A' = A Nand 1

    Thanks !
     
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