Q. The truth table shown in figure is for [table] A | 0 | 0 | 1 | 1 ; B | 0 | 1 | 0 | 1 ; Y | 1 | 0 | 0 | 1 [/table]
A) XOR
B) AND
C) XNOR
D) OR
Explanation
(c) For ‘XNOR’ gate $Y = \overline{A}\ \overline{B} + AB$
$i.e.,\ \overline{0}.\overline{0} + 0.0 = 1.1 + 0.0 = 1 + 0 = 1$
$\overline{0}.\overline{1} + 0.1 = 1.0 + 0.1 = 0 + 0 = 0$
$\overline{1}.\overline{0} + 1.0 = 0.1 + 1.0 = 0 + 0 = 0$
$\overline{1}.\overline{1} + 1.1 = 0.0 = 1.1 = 0 + 1 = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.