The output (Y) of the given logic gate is sim | vedclass

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(C) Let the inputs to the first $NAND$ gate be $A$ and $A$. The output is $Y_1 = \overline{A \cdot A} = \bar{A}$.Let the inputs to the $NOR$ gate be $

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$AND$ gate

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(C) Let the inputs to the first $NAND$ gate be $A$ and $A$. The output is $Y_1 = \overline{A \cdot A} = \bar{A}$.Let the inputs to the $NOR$ gate be $B$ and $B$. The output is $Y_2 = \overline{B+B} = \bar{B}$.The final gate is a $NOR$ gate with inputs $Y_1$ and $Y_2$. The output $Y$ is given by:$Y = \overline{Y_1 + Y_2}$Substituting the values of $Y_1$ and $Y_2$:$Y = \overline{\bar{A} + \bar{B}}$Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.Thus,the output $Y = A \cdot B$,which is the output of an $AND$ gate.

The output $(Y)$ of the given logic gate is similar to the output of an/a

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