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+12 votes

Which one of the following circuits is NOT equivalent to a $2$-input $XNOR$ (exclusive $NOR$) gate?

asked in Digital Logic by Veteran (101k points)
edited by | 1k views

3 Answers

+15 votes
Best answer
A) $(AB'+A'B)'=A\odot B$

B) $(A'(B')'+(A')'B')'=(A\oplus B)'=A\odot B$

C) $A'B'+(A')'B=A\odot B$

D) $((AB')'.(A+B'))'=(AB')+(A+B')'=AB'+A'B=A\oplus B$

So, Answer is D)
answered by Veteran (96.3k points)
edited by
$((A.B')'.(A+B'))' = A.B' + (A+B')' = A.B' + A'.B = A \oplus B$

I think this should be the expression.  Please re-check your option D)
@ srestha
+5 votes
All options except D produce XNOR. See following image (Source:




Source :GeeksforGeeks
answered by Active (2.1k points)
edited by

Correction : All options except D produce XNOR.

Corrected, thank you.
+4 votes
ans d)
answered by Loyal (5.2k points)

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