The output Y of the given logic circuit is: | vedclass

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(A) Let the output of the first $NAND$ gate be $X = \overline{AB} = \bar{A} + \bar{B}$.The upper $NAND$ gate has inputs $A$ and $X$. Its output is $Y_

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$A \cdot \bar{B}$

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(A) Let the output of the first $NAND$ gate be $X = \overline{AB} = \bar{A} + \bar{B}$.The upper $NAND$ gate has inputs $A$ and $X$. Its output is $Y_1 = \overline{A \cdot X} = \overline{A \cdot (\bar{A} + \bar{B})} = \overline{A\bar{A} + A\bar{B}} = \overline{0 + A\bar{B}} = \overline{A\bar{B}} = \bar{A} + B$.The lower $OR$ gate has inputs $X$ and $B$. Its output is $Y_2 = X + B = (\bar{A} + \bar{B}) + B = \bar{A} + (\bar{B} + B) = \bar{A} + 1 = 1$.The final $NAND$ gate has inputs $Y_1$ and $Y_2$. Its output is $Y = \overline{Y_1 \cdot Y_2} = \overline{(\bar{A} + B) \cdot 1} = \overline{\bar{A} + B} = A \cdot \bar{B}$.

The output $Y$ of the given logic circuit is:

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