The correct truth table for the given logic g | vedclass

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(B) The logic circuit consists of an $AND$ gate $(A, B)$, followed by a $NOT$ gate, an $OR$ gate $(C, D)$, an $AND$ gate, and a final $NOT$ gate.Let t

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Inputs $(A, B, C, D)$Output $(Y)$$1, 1, 0, 1$$1$$0, 0, 1, 1$$0$$1, 0, 1, 0$$0$$1, 1, 1, 1$$1$

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(B) The logic circuit consists of an $AND$ gate $(A, B)$, followed by a $NOT$ gate, an $OR$ gate $(C, D)$, an $AND$ gate, and a final $NOT$ gate.Let the output of the first $AND$ gate be $A \cdot B$. After the $NOT$ gate, it becomes $\overline{A \cdot B}$.The output of the $OR$ gate is $C + D$.These two signals are fed into an $AND$ gate, giving $(\overline{A \cdot B}) \cdot (C + D)$.The final $NOT$ gate gives the output $Y = \overline{(\overline{A \cdot B}) \cdot (C + D)}$.Using De Morgan's Law, $Y = \overline{(\overline{A \cdot B})} + \overline{(C + D)} = (A \cdot B) + (\overline{C + D})$.Evaluating for the given inputs:$1$. $A=1, B=1, C=0, D=1$: $Y = (1 \cdot 1) + \overline{(0+1)} = 1 + 0 = 1$.$2$. $A=0, B=0, C=1, D=1$: $Y = (0 \cdot 0) + \overline{(1+1)} = 0 + 0 = 0$.$3$. $A=1, B=0, C=1, D=0$: $Y = (1 \cdot 0) + \overline{(1+0)} = 0 + 0 = 0$.$4$. $A=1, B=1, C=1, D=1$: $Y = (1 \cdot 1) + \overline{(1+1)} = 1 + 0 = 1$.Comparing with the options, option $B$ is correct.

The correct truth table for the given logic gate circuit is:

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Inputs $(A, B, C, D)$	Output $(Y)$
$1, 1, 0, 1$	$1$
$0, 0, 1, 1$	$0$
$1, 0, 1, 0$	$1$
$1, 1, 1, 1$	$0$