The truth table for the system of four NAND g | vedclass

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(B) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.This $Y_1$ is fed into the next two $NAND$ gates.The output of the second

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

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(B) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.This $Y_1$ is fed into the next two $NAND$ gates.The output of the second $NAND$ gate is $Y_2 = \overline{A \cdot Y_1} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A} + (A \cdot B) = \overline{A} + B$.The output of the third $NAND$ gate is $Y_3 = \overline{B \cdot Y_1} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B} + (A \cdot B) = \overline{B} + A$.The final output $Y$ is the $NAND$ of $Y_2$ and $Y_3$:$Y = \overline{Y_2 \cdot Y_3} = \overline{(\overline{A} + B) \cdot (\overline{B} + A)} = \overline{(\overline{A} \cdot \overline{B} + \overline{A} \cdot A + B \cdot \overline{B} + B \cdot A)} = \overline{(\overline{A} \cdot \overline{B} + 0 + 0 + A \cdot B)} = \overline{\overline{A} \cdot \overline{B}} + \overline{A \cdot B} = (A + B) \cdot (\overline{A} + \overline{B}) = A \cdot \overline{B} + B \cdot \overline{A}$.This is the Boolean expression for an $XOR$ gate.The truth table for an $XOR$ gate is:If $A=0, B=0$, then $Y=0$.If $A=0, B=1$, then $Y=1$.If $A=1, B=0$, then $Y=1$.If $A=1, B=1$, then $Y=0$.This matches option $B$.

The truth table for the system of four $NAND$ gates as shown in the figure is:

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

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

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

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