The Boolean expression for a two-input Ex-OR | vedclass

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Question

(C) The truth table for an Ex-$OR$ gate is as follows:| Input $A$ | Input $B$ | Output $Y = A \oplus B$ || :--- | :--- | :--- || $0$ | $0$ | $0$ || $0

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$Y=(\bar{A} \cdot B)+(A \cdot \bar{B})$

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(C) The truth table for an Ex-$OR$ gate is as follows:| Input $A$ | Input $B$ | Output $Y = A \oplus B$ || :--- | :--- | :--- || $0$ | $0$ | $0$ || $0$ | $1$ | $1$ || $1$ | $0$ | $1$ || $1$ | $1$ | $0$ |The output equation for an Ex-$OR$ gate is $Y = A \oplus B = (\bar{A} \cdot B) + (A \cdot \bar{B})$.Key points to remember:$(1)$ The output is low $(0)$ when both inputs are the same.$(2)$ The output is high $(1)$ when both inputs are different.

The Boolean expression for a two-input Ex-$OR$ gate is (where $A$ and $B$ are inputs,and $Y$ is the output):

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