The logic gate equivalent to the given logic | vedclass

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(A) The given circuit consists of two $NOT$ gates connected to the inputs of a $NAND$ gate.Let the inputs be $A$ and $B$. The outputs of the $NOT$ gat

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$OR$

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(A) The given circuit consists of two $NOT$ gates connected to the inputs of a $NAND$ gate.Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$.These are the inputs to the $NAND$ gate,so the final output $Y$ is given by:$Y = \overline{\bar{A} \cdot \bar{B}}$Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.Thus,the output $Y = A + B$,which is the Boolean expression for an $OR$ gate.Truth table:$A$$B$$\bar{A}$$\bar{B}$$Y = \overline{\bar{A} \cdot \bar{B}}$$0$$0$$1$$1$$0$$0$$1$$1$$0$$1$$1$$0$$0$$1$$1$$1$$1$$0$$0$$1$Hence,the equivalent gate is an $OR$ gate.

The logic gate equivalent to the given logic circuit is

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