Question

The following configuration of gate is equivalent to

• A. NAND
• B. XOR
• C. OR
• D. none of these

Answer 1

Answer: (B)

$Y=(A+B)\cdot \overline{AB}$
The given output equation can also be written as

$Y=(A+B)\cdot (\bar{A}+\bar{B})$ (De Morgan's theorem)
$=A\bar{A}+A\bar{B}+B\bar{A}+B\bar{B}$
$=0+A\bar{B}+\bar{A}B+0=\bar{A}B+A\bar{B}$
This is the expression for XOR gate.