The logic circuit shown has the input wavefor | vedclass

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(C) The given circuit consists of two $NOT$ gates at the inputs,followed by a $NOR$ gate,and finally another $NOT$ gate at the output.Let the inputs b

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(C) The given circuit consists of two $NOT$ gates at the inputs,followed by a $NOR$ gate,and finally another $NOT$ gate at the output.Let the inputs be $A$ and $B$.The outputs of the first two $NOT$ gates are $\bar{A}$ and $\bar{B}$.These are fed into a $NOR$ gate,so its output is $\overline{\bar{A} + \bar{B}}$.By De Morgan's Law,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.This output is then passed through a final $NOT$ gate,so the final output $Y = \overline{A \cdot B}$.This is the Boolean expression for a $NAND$ gate.$A$ $NAND$ gate gives a low output $(0)$ only when both inputs are high $(1)$. In all other cases,the output is high $(1)$.Comparing this logic with the given input waveforms,the output waveform corresponds to option $C$.

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

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