XOR gate to XNOR gate boolean algebra

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Discussion Overview

The discussion centers on the transformation of a XOR gate into an XNOR gate using boolean algebra. Participants explore the algebraic expressions and methods to achieve the expected output, while also considering alternative approaches such as truth tables.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents an initial formula: ((AB')' + (A'B)')', seeking assistance in deriving the expected output of AB + (AB)'.
  • Another participant questions whether this is a homework problem, and later clarifies it is related to seatwork.
  • A suggestion is made to use a truth table to evaluate the outputs instead of relying solely on boolean algebra.
  • A participant proposes that inverting only one input of the XOR gate (A' XOR B) results in an XNOR gate output.
  • Another participant provides a detailed algebraic manipulation using De Morgan's Theorem, ultimately simplifying the expression to A'B' + AB.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial formulas presented, with no consensus reached on the best approach to derive the XNOR gate output.

Contextual Notes

Some expressions presented by participants are challenged, and there is uncertainty regarding the validity of the initial formulas. The discussion also reflects varying preferences for algebraic versus truth table methods.

Ogakor
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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.
 
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Is this your homework?
 
thecritic said:
Is this your homework?

No, this was my seatwork and I didn't get the answer.
 
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.
 
Ogakor said:
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.
Neither of those expressions looks right to me.
 
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
 
[(AB')+(A'B)]'

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

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

Simplify
(A'+B) • (A+B')

Multiply
A'A+A'B'+AB+BB'

*AA' and BB' are equal to 0

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

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