Semiconducting Devices Question 73

Question: The truth table shown in figure is for [Pb. CET 1998]
A 0 0 1 1
B 0 1 0 1
Y 1 0 0 1

Options:

A) XOR

B) AND

C) XNOR

D) OR

Show Answer

Answer:

Correct Answer: C

Solution:

For XNOR gate $ Y=\bar{A}\bar{B}+AB $ i.e.

$ \bar{0}.\bar{0}+0.0=1.1+0.0=1+0=1 $

$ \bar{0}.\bar{1}+0.1=1.0+0.1=0+0=0 $

$ \bar{1}.\bar{0}+1.0=0.1+1.0=0+0=0 $

$ \bar{1}.\bar{1}+1.1=0.0+1.1=0+1=1 $



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