The following configuration of gates is equiv | vedclass

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(B) Let the inputs be $A$ and $B$.The output of the $OR$ gate $(G_1)$ is $(A+B)$.The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.The f

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$XOR$

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(B) Let the inputs be $A$ and $B$.The output of the $OR$ gate $(G_1)$ is $(A+B)$.The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.The final output $Y$ from the $AND$ gate $(G_3)$ is the product of the outputs of $G_1$ and $G_2$:$Y = (A+B) \cdot \overline{A \cdot B}$Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$:$Y = (A+B) \cdot (\overline{A} + \overline{B})$Expanding the expression:$Y = A \cdot \overline{A} + A \cdot \overline{B} + B \cdot \overline{A} + B \cdot \overline{B}$Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$:$Y = 0 + A \cdot \overline{B} + \overline{A} \cdot B + 0$$Y = A \cdot \overline{B} + \overline{A} \cdot B$This is the Boolean expression for an $XOR$ gate.

The following configuration of gates is equivalent to

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