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(D) The circuit consists of two $NOT$ gates followed by an $OR$ gate. This is equivalent to a $NOR$ gate.Let the inputs be $A$ and $B$. The outputs of

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(D) The circuit consists of two $NOT$ gates followed by an $OR$ gate. This is equivalent to a $NOR$ gate.Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$.The final output $Y$ of the $OR$ gate is $Y = \bar{A} + \bar{B}$.According to De Morgan's theorem,$\bar{A} + \bar{B} = \overline{A \cdot B}$.Thus,the circuit acts as a $NAND$ gate.The truth table for a $NAND$ gate is:$A=0, B=0 \implies Y=1$$A=0, B=1 \implies Y=1$$A=1, B=0 \implies Y=1$$A=1, B=1 \implies Y=0$Therefore,the output $Y$ is low only when both inputs $A$ and $B$ are high.

$A$	$B$	Output
$1$	$0$	$1$
$1$	$1$	$0$
$0$	$1$	$1$
$0$	$0$	$0$

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

In the given circuit as shown,the two input waveforms $A$ and $B$ are applied simultaneously. The resultant waveform $Y$ is:

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