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XOR gate to XNOR gate boolean algebra

  1. Aug 23, 2010 #1
    Xor gate with negated input and negated output.

    The expected output is an XNOR gate. I can't get it with boolean algebra.

    My initial formula is:
    ((AB')' + (A'B)')'

    The expected output is:
    AB + (AB)' right?

    Please help me.
  2. jcsd
  3. Aug 23, 2010 #2
    Is this your homework?
  4. Aug 23, 2010 #3
    No, this was my seatwork and I didn't get the answer.
  5. Aug 24, 2010 #4
    must you show it with boolean algebra? I'd recommend, since this has only 2 inputs and 1 output, simply running through the 4 possible inputs and calculating the 4 possible outputs 1 at a time in a truth table.
  6. Aug 24, 2010 #5


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    Neither of those expressions looks right to me.
  7. Sep 18, 2010 #6
    I'm pretty sure that we're talking about inverting only one of the inputs to the XOR gate.
    So, instead of A XOR B, we would have A' XOR B

    This would, in fact have the result as A XNOR B
  8. Jul 30, 2011 #7

    By De Morgan's Theorem
    (AB')' • (A'B)'

    Again by De Morgan's Theorem
    (A'+B'') • (A''+B')

    (A'+B) • (A+B')


    *AA' and BB' are equal to 0

    Therefore [(AB')+(A'B)]' = A'B'+AB
    Hope this helps
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