The output of the given combination of logic | vedclass

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(B) $1$. The first two gates are $NOT$ gates (since inputs $A$ and $B$ are connected to both terminals of the $NOR$ gates). Thus,the outputs are $\bar

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$NAND$ Gate

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(B) $1$. The first two gates are $NOT$ gates (since inputs $A$ and $B$ are connected to both terminals of the $NOR$ gates). Thus,the outputs are $\bar{A}$ and $\bar{B}$.$2$. These outputs are fed into a $NOR$ gate. The output of this gate is $\overline{\bar{A} + \bar{B}}$.$3$. By De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.$4$. This result $(A \cdot B)$ is then fed into a final $NOT$ gate (a $NOR$ gate with shorted inputs). The final output $Y$ is $\overline{A \cdot B}$.$5$. The expression $\overline{A \cdot B}$ represents a $NAND$ gate.

The output of the given combination of logic gates represents:

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