In the following digital circuit, the output | vedclass

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(D) Let the output of the $NAND$ gate be $Y_1 = \overline{A \cdot B}$.Let the output of the $NOT$ gate be $A' = \overline{A}$.Let the output of the $N

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$A = 1, B = 1$

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(D) Let the output of the $NAND$ gate be $Y_1 = \overline{A \cdot B}$.Let the output of the $NOT$ gate be $A' = \overline{A}$.Let the output of the $NOR$ gate be $Y_2 = \overline{A' + B} = \overline{\overline{A} + B} = A \cdot \overline{B}$.The final output $Y$ is the output of a $NOR$ gate with inputs $Y_1$ and $Y_2$, so $Y = \overline{Y_1 + Y_2} = \overline{\overline{A \cdot B} + A \cdot \overline{B}}$.Using De Morgan's law, $Y = (A \cdot B) \cdot (\overline{A \cdot \overline{B}}) = (A \cdot B) \cdot (\overline{A} + B) = (A \cdot B \cdot \overline{A}) + (A \cdot B \cdot B) = 0 + (A \cdot B) = A \cdot B$.For $Y = 1$, we must have $A = 1$ and $B = 1$.

In the following digital circuit, the output '$Y$' will be '$1$' for inputs '$A$' and '$B$' having values:

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