The Boolean expression for two-input Ex-OR gate is (A, B are inputs, Y is output)
- $Y=(\bar{A}+B) \cdot(A+\bar{B})$
- $\mathrm{Y}=(\mathrm{A}, \mathrm{B})+(\overline{\mathrm{A}}+\overline{\mathrm{B}})$
- $\mathrm{Y}=(\overline{\mathrm{A}} \cdot \mathrm{B})+(\mathrm{A} \cdot \overline{\mathrm{B}})$
- $\mathrm{Y}=(\mathrm{A} \cdot \mathrm{B})+(\overline{\mathrm{A}}+\mathrm{B})$
Solution
Output equation: $\mathrm{Y}=\mathrm{A} \oplus \mathrm{B}=(\overline{\mathrm{A}} \cdot \mathrm{B})+(\mathrm{A} \cdot \overline{\mathrm{B}})$
Key points to remember:
(1) the output is low when both the inputs are the same
(2) the output is high when both the inputs are different
.Asked in: MHT CET 2022 (08 Aug Shift 2)