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+12 votes
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Which one of the following circuits is NOT equivalent to a $2$-input $XNOR$ (exclusive $NOR$) gate?

asked in Digital Logic by Veteran (106k points)
edited by | 1.1k views

3 Answers

+15 votes
Best answer
  1. $(AB'+A'B)'=A\odot B$
  2. $(A'(B')'+(A')'B')'=(A\oplus B)'=A\odot B$
  3. $A'B'+(A')'B=A\odot B$
  4. $((AB')'.(A+B'))'=(AB')+(A+B')'=AB'+A'B=A\oplus B$


So, Answer is (D)

answered by Veteran (101k points)
edited by
0
$((A.B')'.(A+B'))' = A.B' + (A+B')' = A.B' + A'.B = A \oplus B$

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: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html)

 

 

 

Source :GeeksforGeeks
answered by Active (2.2k points)
edited by
0

Correction : All options except D produce XNOR.

+1
Corrected, thank you.
+4 votes
ans d)
answered by Loyal (5.3k points)
Answer:

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