Give the truth table for the circuit shown in | vedclass

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(A) Let the output of the $NAND$ gate be $C$. The boolean expression for $C$ is $C = \overline{A \cdot B}$.Next, the top $AND$ gate takes inputs $A$ a

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$A, B, X$$0, 0, 0$$0, 1, 1$$1, 0, 1$$1, 1, 0$

Vedclass · Answer

(A) Let the output of the $NAND$ gate be $C$. The boolean expression for $C$ is $C = \overline{A \cdot B}$.Next, the top $AND$ gate takes inputs $A$ and $C$, giving output $D = A \cdot C = A \cdot (\overline{A \cdot B})$.Similarly, the bottom $AND$ gate takes inputs $B$ and $C$, giving output $E = B \cdot C = B \cdot (\overline{A \cdot B})$.The final output $X$ is the $OR$ of $D$ and $E$, so $X = D + E = A \cdot (\overline{A \cdot B}) + B \cdot (\overline{A \cdot B}) = (A + B) \cdot (\overline{A \cdot B})$.Using the truth table:$A$$B$$C = \overline{A \cdot B}$$D = A \cdot C$$E = B \cdot C$$X = D + E$$0$$0$$1$$0$$0$$0$$0$$1$$1$$0$$1$$1$$1$$0$$1$$1$$0$$1$$1$$1$$0$$0$$0$$0$This truth table corresponds to the $XOR$ gate.

Give the truth table for the circuit shown in the figure.

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$A$ $B$ $C$ $Y$
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To obtain the truth-table shown, from the following logic circuit, the gate $G$ should be
$A$ $B$ $Y$
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$A$	$B$	$C$	$Y$
$0$	$0$	$0$	$1$
$0$	$0$	$1$	$1$
$0$	$1$	$0$	$0$
$0$	$1$	$1$	$0$
$1$	$0$	$0$	$1$
$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$
$1$	$1$	$1$	$0$

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