The logic circuit in the figure is equivalent | vedclass

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(D) Let the inputs be $A$ and $B$. The first gate is a $NAND$ gate,so its output is $Y_1 = \overline{A \cdot B}$.This output $Y_1$ is fed into both in

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$NAND$ gate

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(D) Let the inputs be $A$ and $B$. The first gate is a $NAND$ gate,so its output is $Y_1 = \overline{A \cdot B}$.This output $Y_1$ is fed into both inputs of a $NOR$ gate. The output of a $NOR$ gate with inputs $Y_1$ and $Y_1$ is $Y_2 = \overline{Y_1 + Y_1} = \overline{Y_1}$.Substituting $Y_1$,we get $Y_2 = \overline{\overline{A \cdot B}} = A \cdot B$.This output $Y_2$ is then passed through a $NOT$ gate. The final output is $Y = \overline{Y_2} = \overline{A \cdot B}$.Since the final output is $\overline{A \cdot B}$,the given logic circuit is equivalent to a $NAND$ gate.

The logic circuit in the figure is equivalent to

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