The logic gate equivalent to the combination | vedclass

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(A) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate. Let the inputs be $A$ and $B$. The outputs of the two $NOT$ gates are $\ba

Vedclass · Accepted Answer

$AND$

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(A) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate. Let the inputs be $A$ and $B$. The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$. These are the inputs to the $NOR$ gate. The output $y$ of the $NOR$ gate is given by $y = \overline{\bar{A} + \bar{B}}$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$. The expression $y = A \cdot B$ represents an $AND$ gate. Therefore,the combination is equivalent to an $AND$ gate.

The logic gate equivalent to the combination of logic gates shown in the figure is

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