The output of the given combination of logic | vedclass

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(B) $1$. The first two gates are $NOT$ gates (since inputs $A$ and $B$ are connected to both terminals of the $NOR$ gates). Thus,the outputs are $\bar

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$NAND$ Gate

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(B) $1$. The first two gates are $NOT$ gates (since inputs $A$ and $B$ are connected to both terminals of the $NOR$ gates). Thus,the outputs are $\bar{A}$ and $\bar{B}$.$2$. These outputs are fed into a $NOR$ gate. The output of this gate is $\overline{\bar{A} + \bar{B}}$.$3$. By De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.$4$. This result $(A \cdot B)$ is then fed into a final $NOT$ gate (a $NOR$ gate with shorted inputs). The final output $Y$ is $\overline{A \cdot B}$.$5$. The expression $\overline{A \cdot B}$ represents a $NAND$ gate.

The output of the given combination of logic gates represents:

Similar Questions

Give the truth table for the circuit shown in the figure.

If $X$ and $Y$ are the inputs, the given circuit works as . . . . . . .

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

To get an output $Y = 1$ in the given circuit,which of the following inputs will be correct?

The given circuit works as:

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