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Boolean Algebra Simplification

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey there, I'm having trouble simplifying a boolean expression using XOR and XNOR functions.
    The final goal is to draw a logic circuit for the expression using NAND and XOR gates only.


    2. Relevant equations

    Assuming
    W' = Not W


    W' X' Y' Z' + W' X' Y Z + W' X Y' Z + W' X Y Z' + W X Y' Z' + W X' Y' Z


    3. The attempt at a solution


    W' X' (Y⊕Z)' + W' X (Y⊕Z) + ....

    So far if I'm working correctly (?) I can simplify the first four expressions, It's just the last two that get me as they have no like terms ?

    Once I understand this I should have no troubles drawing the logic circuit.

    Thankyou for any responses
     
  2. jcsd
  3. Apr 30, 2013 #2

    jedishrfu

    Staff: Mentor

    you could try drawing a truth table for the terms and see if you spot a pattern that matches XOR or NAND

    like X xor Y'
     
  4. Apr 30, 2013 #3
    Hmm okay thanks,

    I've come up with this

    W' X' (Y⊕Z)' + W' X (Y⊕Z) + W Y' (X⊕Z)

    Is this on the right track..before I simplify further.
     
  5. Apr 30, 2013 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    W x y' z' + w x' y' z = w y'❲ .......... ❳

    wow, PF has such a nanny editor that it converted my upper case to lower!
     
  6. Apr 30, 2013 #5

    jedishrfu

    Staff: Mentor

    That seems right. Once complete you could do a truth table on it and see if it agrees with original.
     
  7. Apr 30, 2013 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    You can see how to simplify this further? If not, I suggest that you construct a 4 variable Truth Table. Also construct the Truth Table for your simplified expression, and with any luck there may be some correspondence.

    I haven't tried it.
     
  8. May 1, 2013 #7
    For future reference (im not sure why anyone would care lol, but anyway), the equation can be simplified down to:

    W' (X⊕Y⊕Z)' + Y' (W⊕X⊕Z)' .....where (A⊕B⊕C)' represents a XNOR gate.

    This was obtained using karnough maps but Im sure there are easier ways...
     
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