The output of an OR gate is connected to both | vedclass

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(D) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y = A + B$.This output $Y$ is connected to both inputs of a $NAND$

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$NOR$ gate

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(D) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y = A + B$.This output $Y$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1$ and $X_2$,where $X_1 = X_2 = Y = A + B$.The output of a $NAND$ gate with inputs $X_1$ and $X_2$ is given by $\overline{X_1 \cdot X_2}$.Substituting $X_1 = X_2 = A + B$,the final output $Y^{\prime}$ is $\overline{(A + B) \cdot (A + B)}$.Using the Boolean identity $X \cdot X = X$,we get $Y^{\prime} = \overline{A + B}$.The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.Therefore,the combination acts as a $NOR$ gate.

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$

The output of an $OR$ gate is connected to both the inputs of a $NAND$ gate. The combination will serve as

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To get the truth table shown, from the following logic circuit, the Gate $G$ should be
$A$ $B$ $Y$
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The output $F$ of the logic circuit given below is

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