Electronic Devices Question 271
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 $
