1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simplifying output for a XOR gate using Boolean Algebra

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm trying to show that the output of this XOR circuit is ##F=A'B+AB'##, 480px-XOR_from_NOR.svg.png

    2. Relevant equations
    ##(A+B)'=A'\cdot B'##
    ##(A\cdot B)'=A'+B'##
    3. The attempt at a solution

    From the gates the output is ##[(A\cdot B)+(A+B)']'##, using De Morgan's laws this becomes ##[(A\cdot B)+(A+B)']'=(A\cdot B)'\cdot (A+B)=(A\cdot B)'\cdot (A+B)=(A'+B')\cdot (A+B)=0##? I cant seem to figure out what I'm doing wrong.
     
  2. jcsd
  3. Apr 7, 2016 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You did it right, but simplify the last expression applying the distributive law.
     
  4. Apr 7, 2016 #3
    yikes, can't believe I missed that one!
     
  5. Apr 7, 2016 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Never give up hope :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simplifying output for a XOR gate using Boolean Algebra
  1. Boolean Algebra (Replies: 5)

  2. Boolean algebra (Replies: 6)

  3. Electronics XOR gates (Replies: 11)

Loading...