The truth table for the two-input Ex-OR gate | vedclass

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(D) The Boolean expression for an Exclusive-$OR$ (Ex-$OR$) gate is given by:$Y = A \oplus B = \bar{A} \cdot B + A \cdot \bar{B}$The output of an Ex-$O

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Table $(S)$

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(D) The Boolean expression for an Exclusive-$OR$ (Ex-$OR$) gate is given by:$Y = A \oplus B = \bar{A} \cdot B + A \cdot \bar{B}$The output of an Ex-$OR$ gate is $HIGH$ $(1)$ only when the inputs are at different logic levels. If both inputs are the same ($0,0$ or $1,1$),the output is $LOW$ $(0)$.Evaluating the truth table:$1$. For $A=0, B=0$: $Y = 0 \oplus 0 = 0$$2$. For $A=0, B=1$: $Y = 0 \oplus 1 = 1$$3$. For $A=1, B=0$: $Y = 1 \oplus 0 = 1$$4$. For $A=1, B=1$: $Y = 1 \oplus 1 = 0$Comparing this with the given tables,Table $(S)$ matches this truth table.

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

The truth table for the two-input Ex-$OR$ gate is (where $A$ and $B$ are the inputs and $Y$ is the output).
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

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The truth table for the two-input Ex-$OR$ gate is (where $A$ and $B$ are the inputs and $Y$ is the output).$A$$B$$Y$$0$$0$$0$$0$$1$$1$$1$$0$$1$$1$$1$$0$