The logic circuit below has the truth table, | vedclass

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Question

(B) The given logic circuit consists of two $NOT$ gates at the inputs of a $NOR$ gate, followed by another $NOT$ gate at the output.Let the inputs be

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$NAND$ gate

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(B) The given logic circuit consists of two $NOT$ gates at the inputs of a $NOR$ gate, followed by another $NOT$ gate at the output.Let the inputs be $A$ and $B$. The outputs of the first two $NOT$ gates are $\bar{A}$ and $\bar{B}$.These are fed into a $NOR$ gate, so the intermediate output $Y_1 = \overline{\bar{A} + \bar{B}}$.By De Morgan's theorem, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.Finally, this output $Y_1$ passes through another $NOT$ gate, so the final output $Y = \bar{Y_1} = \overline{A \cdot B}$.The Boolean expression $Y = \overline{A \cdot B}$ represents a $NAND$ gate.Thus, the truth table of the given circuit is the same as that of a $NAND$ gate.

The logic circuit below has the truth table, same as that of

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