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(D) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate. Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}

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(D) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate. Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$. These are fed into a $NOR$ gate,so the final output $Y$ is: $Y = \overline{\bar{A} + \bar{B}}$ Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$. Thus,the circuit acts as an $AND$ gate. The truth table for an $AND$ gate is: $A$$B$$Y$$0$$0$$0$$0$$1$$0$$1$$0$$0$$1$$1$$1$ Comparing the input waveforms $A$ and $B$ with the $AND$ gate logic,the output $Y$ is high $(1)$ only when both $A$ and $B$ are high $(1)$. Looking at the provided waveforms,this corresponds to option $D$.

The logic circuit shown below has the input waveforms $A$ and $B$ as shown. Pick out the correct output waveform.

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