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3-input XNOR in SOP form

Discussion in 'Electronic Design' started by KL, Oct 28, 2006.

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  1. KL

    KL Guest

    I need to know how to express a 3-input XNOR gate in SOP form....I am
    kinda lost, missed a lecture and trying to figure this out on my own
    just ain't cutting in. Anyone able to help?
     
  2. I never thought about that one. What's the truth table for a 3-input
    XNOR gate?
     
  3. KL

    KL Guest

    Wouldn't it be:

    a|b|c|f
    0|0|0|1
    0|0|1|0
    0|1|0|0
    0|1|1|0
    1|0|0|0
    1|0|1|0
    1|1|0|0
    1|1|1|1

    Because the output is '1' only if the inputs are the same, right?
    --

    KL

    ______________________________00
    _______________________0000000000
    ______________________00___00___00
    __________________________00____00
    _________________________00_____00
    ________________________00______00
    ________________000____00_______00
    ___________________00000000000000000
    _____________________00_________00_000
    ____________________00__________00
    ____________________00__________00
    ___________________00___________00
    _______00________00_____________00
    _________00______00______________00___
    ___________000000_________________00000

    ********ROLL TIDE*********
     
  4. Arlet

    Arlet Guest

    An N-input XNOR gate outputs a '1', for an even number of '1' inputs.
     
  5. KL

    KL Guest

    Why is that? In our notes, for a 2 input XNOR gate it said that the
    output is '1' only if the inputs are the same...which would hold true
    the way you say it, but I just want to be clear on this....if that is
    even possible.
    --

    KL

    ______________________________00
    _______________________0000000000
    ______________________00___00___00
    __________________________00____00
    _________________________00_____00
    ________________________00______00
    ________________000____00_______00
    ___________________00000000000000000
    _____________________00_________00_000
    ____________________00__________00
    ____________________00__________00
    ___________________00___________00
    _______00________00_____________00
    _________00______00______________00___
    ___________000000_________________00000

    ********ROLL TIDE*********
     
  6. Arlet

    Arlet Guest

    It's the definition that makes the most sense, and that is commonly
    used. It is equivalent to cascading multiple 2-input XNOR gates, so
    XNOR(a,b,c) = a XNOR b XNOR c (write it out using truth tables to
    verify it's the same as the 'even' criterium).
     
  7. Arlet

    Arlet Guest

    Oops. That's not true. What I meant to say was that you can cascade
    multiple XOR gates to create a multiple input XOR. You can then invert
    the output to create a multiple input XNOR.


    ..
     
  8. KL

    KL Guest

    OK so when it is XNOR(a,b) it is one when a=0 and b=0 because there is
    an 'even' number of 1's (ie 0) is that right?

    so my table would be reversed?
    --

    KL

    ______________________________00
    _______________________0000000000
    ______________________00___00___00
    __________________________00____00
    _________________________00_____00
    ________________________00______00
    ________________000____00_______00
    ___________________00000000000000000
    _____________________00_________00_000
    ____________________00__________00
    ____________________00__________00
    ___________________00___________00
    _______00________00_____________00
    _________00______00______________00___
    ___________000000_________________00000

    ********ROLL TIDE*********
     
  9. Arlet

    Arlet Guest

    I'm not sure what you mean by "reversed', but if you just count the
    ones, and see if the number is even/odd, you'll get the right answer.
     
  10. KL

    KL Guest

    Well my original truth table looked like:
    a|b|c|f
    0|0|0|1
    0|0|1|0
    0|1|0|0
    0|1|1|0
    1|0|0|0
    1|0|1|0
    1|1|0|0
    1|1|1|1

    rethinking this...would the correct table look like:
    a|b|c|f
    0|0|0|1
    0|0|1|0
    0|1|0|0
    0|1|1|1
    1|0|0|0
    1|0|1|1
    1|1|0|1
    1|1|1|0
    --

    KL

    ______________________________00
    _______________________0000000000
    ______________________00___00___00
    __________________________00____00
    _________________________00_____00
    ________________________00______00
    ________________000____00_______00
    ___________________00000000000000000
    _____________________00_________00_000
    ____________________00__________00
    ____________________00__________00
    ___________________00___________00
    _______00________00_____________00
    _________00______00______________00___
    ___________000000_________________00000

    ********ROLL TIDE*********
     
  11. Rich Grise

    Rich Grise Guest


    What happens if you draw a truth table for
    (NOT (A XOR B) XOR C)) ?

    I'm too lazy to do it myself at the moment, and if I did, you wouldn't
    learn anything.

    I like that odd/even distinction - it brings back the parity bit! :)

    Cheers!
    Rich
     
  12. I didn't know it was missing.
     
  13. Tam/WB2TT

    Tam/WB2TT Guest

    I believe that conceptually, you want this: Given signals A,B,C, and 2 input
    XNOR gates 1 and 2, connect A and B to inputs of XNOR1. Connect the output
    of XNOR1 and C to inputs of XNOR2. The XNOR2 output will be the desired
    signal.

    Tam
     
  14. Slurp

    Slurp Guest

    A 3 input XNOR does not make any sense. An exclusive-or is a logical
    operation between two operands which produces an true output only when the
    inputs are different.
    You can XOR multiple bit words together, but the operation is only performed
    on bit pairs of the same binary significance to produce a result of the same
    binary significance.
    It performs the same as an add function without a carry.

    The XOR or XNOR is a 2 input function.

    Slurp
     
  15. Arlet

    Arlet Guest

    It makes perfect sense. An XOR performs the same function as an add
    without carry, or a modulo-2 addition. For an N-input XOR you can
    simply extend the definition by using a modulo-2 addition of N bits.
    Since the XOR operation is associative: (a XOR b) XOR c == a XOR (b XOR
    c), and commutative: a XOR b == b XOR a, you can simply implement the
    N input XOR by cascading N-1 2 input XORs in any combination.

    An XNOR can be made by inverting the output of an equivalent XOR.

    Of course, it's all a matter of definition, but this is the one that
    makes the most sense, and is commonly used.
     
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