The combination of logic gates shown below be | vedclass

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(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A

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$OR$ gate

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(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the first $NAND$ gate (acting as $NOT$) is $\bar{A}$.The output of the second $NAND$ gate (acting as $NOT$) is $\bar{B}$.These two outputs are fed into a third $NAND$ gate.The final output $Y$ is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.This is the Boolean expression for an $OR$ gate.Therefore,the combination acts as an $OR$ gate.

The combination of logic gates shown below becomes:

Similar Questions

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

The values of $Y$ and $Z$ in the given logic circuit are

In which of the logic gate the following statement is true? "The output is high when either of the inputs is high but not if both inputs are high"

What is the output $Y$ of the logic circuit shown below?

Find the truth table for the function $Y$ with inputs $A$ and $B$ represented in the following figure.

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