The output waveform of the given logical circ | vedclass

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(D) The circuit consists of two $NAND$ gates as $NOT$ gates (since both inputs are tied together) followed by a $NAND$ gate.$1$. The first $NAND$ gate

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(D) The circuit consists of two $NAND$ gates as $NOT$ gates (since both inputs are tied together) followed by a $NAND$ gate.$1$. The first $NAND$ gate with inputs $A$ and $A$ gives output $Y_1 = \overline{A \cdot A} = \overline{A}$.$2$. The second $NAND$ gate with inputs $B$ and $B$ gives output $Y_2 = \overline{B \cdot B} = \overline{B}$.$3$. The final $NAND$ gate takes $Y_1$ and $Y_2$ as inputs,so the final output $Y = \overline{Y_1 \cdot Y_2} = \overline{\overline{A} \cdot \overline{B}}$.$4$. Using De Morgan's theorem,$\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$.$5$. Thus,the circuit acts as an $OR$ gate.$6$. Analyzing the input waveforms:- For $t - For $t_1 - For $t_2 - For $t_3 - For $t_4 - For $t_5 - For $t > t_6$: $A=0, B=0 \implies Y = 0+0 = 0$.Comparing this with the given options,the waveform matches option $D$.

The output waveform of the given logical circuit for the following inputs $A$ and $B$ as shown below,is:

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