The output y of the given logic gate network | vedclass

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(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P

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(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P = \overline{\bar A \cdot B} = A + \bar B$.The lower $NAND$ gate has inputs $A$ and $\bar B$. Its output is $Q = \overline{A \cdot \bar B} = \bar A + B$.The final gate is a $NOR$ gate,so the output $y = \overline{P + Q}$.Substituting the values of $P$ and $Q$:$y = \overline{(A + \bar B) + (\bar A + B)}$$y = \overline{(A + \bar A) + (B + \bar B)}$Since $A + \bar A = 1$ and $B + \bar B = 1$:$y = \overline{1 + 1} = \overline{1} = 0$.Thus,the output $y = 0$.

The output $y$ of the given logic gate network is:

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