For the following logic circuit,the truth tab | vedclass

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Question

(C) The given circuit consists of two $NOT$ gates followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The outputs of the $NOT$ gates are $\bar{A}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given circuit consists of two $NOT$ gates followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$.These are the inputs to the $NAND$ gate.Therefore,the output $Y = \overline{\bar{A} \cdot \bar{B}}$.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.The truth table for an $OR$ gate is:$A$$B$$Y$$0$$0$$0$$0$ and $1$$1$$1$$1$ and $0$$0$$1$$1$ and $1$$1$$1$

For the following logic circuit,the truth table is:

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

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

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

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