XOR gate to XNOR gate boolean algebra

Ogakor

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.

I_am_learning

Is this your homework?

Ogakor

Quote by thecritic

Is this your homework?

No, this was my seatwork and I didn't get the answer.

*xcvxcvvc*

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.

vela

Quote by Ogakor

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.

zgozvrm

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

RonnieRongz

[(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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Ogakor	XOR gate to XNOR gate boolean algebra Tweet 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.

I_am_learning Recognitions: Gold Member	Is this your homework?

xcvxcvvc	XOR gate to XNOR gate boolean algebra 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.

zgozvrm	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

RonnieRongz	[(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