The truth table for the given circuit of four | vedclass

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Question

(B) Let the inputs be $A$ and $B$.$1$. The first $NAND$ gate produces output $C = \overline{A \cdot B}$.$2$. The second $NAND$ gate (top) takes inputs

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the inputs be $A$ and $B$.$1$. The first $NAND$ gate produces output $C = \overline{A \cdot B}$.$2$. The second $NAND$ gate (top) takes inputs $A$ and $C$, producing $D = \overline{A \cdot C} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A \cdot (\overline{A} + \overline{B})} = \overline{A \cdot \overline{A} + A \cdot \overline{B}} = \overline{0 + A \cdot \overline{B}} = \overline{A \cdot \overline{B}} = \overline{A} + B$.$3$. The third $NAND$ gate (bottom) takes inputs $B$ and $C$, producing $E = \overline{B \cdot C} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B \cdot (\overline{A} + \overline{B})} = \overline{B \cdot \overline{A} + B \cdot \overline{B}} = \overline{B \cdot \overline{A} + 0} = \overline{B \cdot \overline{A}} = \overline{B} + A$.$4$. The final $NAND$ gate takes inputs $D$ and $E$, producing $Y = \overline{D \cdot E} = \overline{(\overline{A} + B) \cdot (A + \overline{B})} = \overline{\overline{A} \cdot A + \overline{A} \cdot \overline{B} + B \cdot A + B \cdot \overline{B}} = \overline{0 + \overline{A} \cdot \overline{B} + A \cdot B + 0} = \overline{\overline{A} \cdot \overline{B} + A \cdot B} = A \oplus B$ ($XOR$ gate).The correct table is $B$.

The truth table for the given circuit of four $NAND$ gates is:

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

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

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

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