The truth table for the following logic circu | vedclass

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(A) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. The output expression is derived as follows:$1$. The inputs to th

Vedclass · Accepted Answer

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

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(A) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. The output expression is derived as follows:$1$. The inputs to the top $AND$ gate are $\overline{A}$ and $B$,so its output is $\overline{A} \cdot B$.$2$. The inputs to the bottom $AND$ gate are $A$ and $\overline{B}$,so its output is $A \cdot \overline{B}$.$3$. These two outputs are fed into an $OR$ gate,resulting in the final output $Y = \overline{A} \cdot B + A \cdot \overline{B}$.This is the expression for an $XOR$ gate. The truth table is:$A$$B$$Y$$0$$0$$0$$0$$1$$1$$1$$0$$1$$1$$1$$0$

The truth table for the following logic circuit is

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

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

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

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