The resultant gate and its Boolean expression | vedclass

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(B) The circuit consists of two $NOT$ gates ($G1$ and $G2$) followed by a $NOR$ gate. The inputs to the $NOT$ gates are $A$ and $B$. Thus,the outputs

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$AND, A \cdot B$

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(B) The circuit consists of two $NOT$ gates ($G1$ and $G2$) followed by a $NOR$ gate. The inputs to the $NOT$ gates are $A$ and $B$. Thus,the outputs of the $NOT$ gates are $C = \overline{A}$ and $D = \overline{B}$. These are fed into a $NOR$ gate. The output $Y$ of the $NOR$ gate is given by $Y = \overline{C+D}$. Substituting the values of $C$ and $D$,we get $Y = \overline{\overline{A} + \overline{B}}$. By De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$. This is the Boolean expression for an $AND$ gate. The truth table is as follows: $A$$B$$C=\overline{A}$$D=\overline{B}$$Y=\overline{C+D}$$0$$0$$1$$1$$0$$0$$1$$1$$0$$0$$1$$0$$0$$1$$0$$1$$1$$0$$0$$1$

The resultant gate and its Boolean expression in the given circuit is

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