The logic performed by the circuit shown in t | vedclass

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Question

(A) The circuit consists of two $NOR$ gates used as $NOT$ gates at the input,followed by a $NOR$ gate at the output.Input $a$ is fed into a $NOR$ gate

Vedclass · Accepted Answer

$AND$

Vedclass · Answer

(A) The circuit consists of two $NOR$ gates used as $NOT$ gates at the input,followed by a $NOR$ gate at the output.Input $a$ is fed into a $NOR$ gate with both inputs tied together,resulting in output $\bar{a}$.Input $b$ is fed into a $NOR$ gate with both inputs tied together,resulting in output $\bar{b}$.These outputs $\bar{a}$ and $\bar{b}$ are then fed into a final $NOR$ gate.The final output $Y$ is given by $Y = \overline{\bar{a} + \bar{b}}$.Using De Morgan's theorem,$\overline{\bar{a} + \bar{b}} = \overline{\bar{a}} \cdot \overline{\bar{b}} = a \cdot b$.This is the Boolean expression for an $AND$ gate.Thus,the circuit is equivalent to an $AND$ gate.

The logic performed by the circuit shown in the figure is equivalent to:

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