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(D) The circuit consists of a $NOR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of the $N

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$NOR$

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(D) The circuit consists of a $NOR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of the $NOR$ gate is $Y_1 = \overline{A+B}$.The output of the $NAND$ gate is $Y_2 = \overline{A \cdot B}$.The final output $Y$ of the $AND$ gate is $Y = Y_1 \cdot Y_2 = \overline{A+B} \cdot \overline{A \cdot B}$.Using De Morgan's laws,$\overline{A+B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$.So,$Y = (\overline{A} \cdot \overline{B}) \cdot (\overline{A} + \overline{B})$.Expanding this,$Y = (\overline{A} \cdot \overline{B} \cdot \overline{A}) + (\overline{A} \cdot \overline{B} \cdot \overline{B})$.Since $\overline{A} \cdot \overline{A} = \overline{A}$ and $\overline{B} \cdot \overline{B} = \overline{B}$,we get $Y = (\overline{A} \cdot \overline{B}) + (\overline{A} \cdot \overline{B})$.Thus,$Y = \overline{A} \cdot \overline{B} = \overline{A+B}$.This is the Boolean expression for a $NOR$ gate.

The output $(Y)$ of the given logic implementation is similar to the output of a . . . . . . gate.

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