3-input XNOR in SOP form

I never thought about that one. What's the truth table for a 3-input XNOR gate?

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?

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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*********

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

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.

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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*********

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).

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.

.

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*********

Correct.

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.

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*********

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

I didn't know it was missing.

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

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

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.