The truth table for the following logic circu | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given logic circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. The output expression is derived as follows:$1$. The upper branch gives the output $\overline{A} \cdot B$.$2$. The lower branch gives the output $A \cdot \overline{B}$.$3$. The final $OR$ gate combines these to give $Y = \overline{A}B + A\overline{B}$,which is the Boolean expression for an $XOR$ gate.The truth table is:$A$$B$$\overline{A}$$\overline{B}$$\overline{A} \cdot B$$A \cdot \overline{B}$$Y = \overline{A}B + A\overline{B}$$0$$0$$1$$1$$0$$0$$0$$0$$1$$1$$0$$1$$0$$1$$1$$0$$0$$1$$0$$1$$1$$1$$1$$0$$0$$0$$0$$0$

The truth table for the following logic circuit is:

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

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

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

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