Question

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

Choose one option
Official answer and solution
Official answer
D
From
Official GATE 2011 answer key

Solution

Derive the Boolean transfer function for each circuit and compare it with the 2-input XNOR function.

  1. The 2-input XNOR operation on inputs AA and BB produces output A⊙B=A⊕B‾=AB+A‾ B‾A \odot B = \overline{A \oplus B} = A B + \overline{A}\,\overline{B}.
  1. Option A: An XOR gate followed by an inverter produces A⊕B‾=A⊙B\overline{A \oplus B} = A \odot B (XNOR).
  1. Option B: Inverting both inputs prior to an XOR gate followed by an output inverter gives:
A‾⊕B‾‾=A⊕B‾=A⊙B\overline{\overline{A} \oplus \overline{B}} = \overline{A \oplus B} = A \odot B
  1. Option C: Inverting a single input to an XOR gate gives A‾⊕B=A‾ B‾+A‾‾ B=A‾ B‾+AB=A⊙B\overline{A} \oplus B = \overline{A}\,\overline{B} + \overline{\overline{A}}\,B = \overline{A}\,\overline{B} + A B = A \odot B (XNOR).
  1. Option D: The top NAND gate produces A⋅B‾‾=A‾+B\overline{A \cdot \overline{B}} = \overline{A} + B. The bottom OR gate produces A+B‾A + \overline{B}. Feeding these into the final NAND gate gives:
(A‾+B)(A+B‾)‾=AB+A‾ B‾‾=A⊕B\overline{(\overline{A} + B)(A + \overline{B})} = \overline{A B + \overline{A}\,\overline{B}} = A \oplus B
  1. Circuit D produces the XOR function, not XNOR. Therefore, circuit D is NOT equivalent to an XNOR gate.
Final answer

D

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2011 Computer Science and Information Technology (CS)
Question
13 of 65, printed as CS Q.13
Section
Computer Science and Information Technology

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