Exclusive-OR binary operation can be represented as

Exclusive-OR binary operation can be represented as

$ar{A} cdot B + A cdot ar{B}$
$A cdot ar{B} + ar{A} cdot B$
$A cdot B + ar{A} cdot ar{B}$
$(ar{A} + ar{B}) cdot (A + B)$
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UPSC CISF-AC-EXE – 2020
The Exclusive-OR (XOR) binary operation of two inputs A and B is true (1) if and only if the inputs are different. Its truth table is:
A | B | A XOR B
–|—|——–
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
The Sum of Products (SOP) representation for this function includes minterms where the output is 1. These are when (A=0 and B=1) or (A=1 and B=0).
– A=0 and B=1 is represented as $\bar{A} \cdot B$.
– A=1 and B=0 is represented as $A \cdot \bar{B}$.
Combining these with an OR operator gives the SOP form: $\bar{A} \cdot B + A \cdot \bar{B}$.
Option A is $\bar{A} \cdot B + A \cdot \bar{B}$, which directly matches the standard SOP form of XOR.
– XOR outputs 1 when inputs are different.
– The standard SOP form of XOR(A, B) is $\bar{A}B + A\bar{B}$.
– Boolean algebra allows representing logic functions using AND (`.`), OR (`+`), and NOT (`bar` or prime).
Option B is identical to A due to the commutativity of addition. Option C represents XNOR ($A \cdot B + \bar{A} \cdot \bar{B}$), which is the complement of XOR. Option D, $(A+B)(\bar{A}+\bar{B})$, is the Product of Sums (POS) canonical form for XOR, also a correct representation. However, option A is the standard SOP form.