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(C) The given circuit consists of two $NOT$ gates,one $NOR$ gate,and one $NOT$ gate at the output. Let the inputs be $A$ and $B$. The outputs of the f

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(C) The given circuit consists of two $NOT$ gates,one $NOR$ gate,and one $NOT$ gate at the output. Let the inputs be $A$ and $B$. The outputs of the first two $NOT$ gates are $\overline{A}$ and $\overline{B}$. These are fed into a $NOR$ gate,so the intermediate output $x$ is given by $x = \overline{\overline{A} + \overline{B}}$. By De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$. The final output $y$ is the inversion of $x$,so $y = \overline{x} = \overline{A \cdot B}$. This is the Boolean expression for a $NAND$ gate. The truth table for a $NAND$ gate is:$A$$B$$y$$0$$0$$1$$0$$1$$1$$1$$0$$1$$1$$1$$0$Based on the input waveforms,the output $y$ will be $0$ only when both $A$ and $B$ are $1$. Otherwise,it will be $1$.

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

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