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(A) $1$. The input $A$ passes through a $NOT$ gate,resulting in $\overline{ A }$.$2$. The input $B$ passes through a $NOT$ gate,resulting in $\overlin

Vedclass · Accepted Answer

$\overline{ A } \cdot \overline{ B } + A \cdot B$

Vedclass · Answer

(A) $1$. The input $A$ passes through a $NOT$ gate,resulting in $\overline{ A }$.$2$. The input $B$ passes through a $NOT$ gate,resulting in $\overline{ B }$.$3$. These two outputs ($\overline{ A }$ and $\overline{ B }$) are fed into an $AND$ gate,producing $\overline{ A } \cdot \overline{ B }$.$4$. The original inputs $A$ and $B$ are fed directly into another $AND$ gate,producing $A \cdot B$.$5$. Finally,the outputs of these two $AND$ gates are fed into an $OR$ gate.$6$. Therefore,the final output expression is $Y = \overline{ A } \cdot \overline{ B } + A \cdot B$.

Write down the boolean expression for output $Y$ of the system shown in the figure.

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