The truth table for the given logic circuit i | vedclass

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Question

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the $NOR$ gate is $\overline{A+B}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the $NOR$ gate is $\overline{A+B}$.The output of the $AND$ gate is $A \cdot B$.These two outputs are fed into a $NAND$ gate.Therefore,the final output $Y$ is given by:$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$Using De Morgan's Law,$\overline{X \cdot Z} = \overline{X} + \overline{Z}$:$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$$Y = (A+B) + (\overline{A} + \overline{B})$$Y = (A + \overline{A}) + (B + \overline{B})$Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:$Y = 1 + 1 = 1$Thus,for any combination of inputs $(A, B)$,the output $Y$ is always $1$.

The truth table for the given logic circuit is:

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

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

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

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