Question

The Boolean expression for \(\text{X-OR}\) gate \(C = (A \oplus B)\) is equivalent to

• A. \((\overline{A}\cdot B) + (A\cdot \overline{B})\)
• B. \(A + (\overline{A}\cdot \overline{B})\)
• C. \((A\cdot B) + \overline{B}\)
• D. \((A\cdot B) + (\overline{A}\cdot \overline{B})\)

Hint:

Remember: XOR = Different → \((\overline{A}B) + (A\overline{B})\)
XNOR = Same → \((AB) + (\overline{A}\overline{B})\)

Answer 1

The Correct Option is A

Concept: XOR (Exclusive OR)
An XOR gate gives output:
• 1 when inputs are different
• 0 when inputs are same

Step 1: Truth Table
\[ \begin{array}{|c|c|c|} \hline A & B & A \oplus B \hline 0 & 0 & 0 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \hline \end{array} \]

Step 2: Form Boolean expression
Output is 1 for:
• \(A=0, B=1 \Rightarrow \overline{A}B\)
• \(A=1, B=0 \Rightarrow A\overline{B}\) So, \[ A \oplus B = (\overline{A}B) + (A\overline{B}) \]

Step 3: Compare options

• (A) matches exactly \checkmark
• (B) simplifies to OR-like expression $\times$
• (C) incorrect $\times$
• (D) represents XNOR $\times$ Final Answer: Option (A)