Four NOR gates are connected as shown in the | vedclass

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(B) Let the output of the first $NOR$ gate be $C = \overline{A+B}$.This signal $C$ is fed into the next two $NOR$ gates along with inputs $A$ and $B$

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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 $NOR$ gate be $C = \overline{A+B}$.This signal $C$ is fed into the next two $NOR$ gates along with inputs $A$ and $B$ respectively.The outputs of these two gates are $D = \overline{A+C} = \overline{A+\overline{A+B}}$ and $E = \overline{B+C} = \overline{B+\overline{A+B}}$.Using De Morgan's theorem,$D = \overline{A} \cdot (A+B) = \overline{A}A + \overline{A}B = 0 + \overline{A}B = \overline{A}B$.Similarly,$E = \overline{B} \cdot (A+B) = \overline{B}A + \overline{B}B = A\overline{B} + 0 = A\overline{B}$.The final output $Y$ is the $NOR$ of $D$ and $E$: $Y = \overline{D+E} = \overline{\overline{A}B + A\overline{B}}$.This is the expression for an $XOR$ gate,which is $A \oplus B$.The truth table for $A \oplus B$ is:$A, B$$Y$$0, 0$$0$$0, 1$$1$$1, 0$$1$$1, 1$$0$

Four $NOR$ gates are connected as shown in the figure. The truth table for the given figure is:

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

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

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

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