In the following digital circuit,what will be | vedclass

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(C) Let the output of the $NAND$ gate be $P = \overline{AB}$.Let the output of the $OR$ gate be $Q = A+B$.The $AND$ gate takes $P$ and $Q$ as inputs,s

Vedclass · Accepted Answer

$0, 0, 1, 0$

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(C) Let the output of the $NAND$ gate be $P = \overline{AB}$.Let the output of the $OR$ gate be $Q = A+B$.The $AND$ gate takes $P$ and $Q$ as inputs,so its output is $R = P \cdot Q = (\overline{AB}) \cdot (A+B)$.Using Boolean algebra: $R = (\bar{A} + \bar{B}) \cdot (A+B) = \bar{A}A + \bar{A}B + \bar{B}A + \bar{B}B = 0 + \bar{A}B + A\bar{B} + 0 = A \oplus B$ ($XOR$ operation).The final $NOR$ gate takes $P$ and $R$ as inputs,so $Z = \overline{P+R} = \overline{(\overline{AB}) + (A \oplus B)}$.Let's evaluate for each input $(A, B)$:$1$. $(1, 0): P = \overline{1 \cdot 0} = 1, Q = 1+0 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.$2$. $(0, 0): P = \overline{0 \cdot 0} = 1, Q = 0+0 = 0, R = 1 \cdot 0 = 0. Z = \overline{1+0} = 0$.$3$. $(1, 1): P = \overline{1 \cdot 1} = 0, Q = 1+1 = 1, R = 0 \cdot 1 = 0. Z = \overline{0+0} = 1$.$4$. $(0, 1): P = \overline{0 \cdot 1} = 1, Q = 0+1 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.The outputs are $0, 0, 1, 0$.

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

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

In the following digital circuit,what will be the output at $Z$ when the inputs $(A, B)$ are $(1,0), (0,0), (1,1), (0,1)$?

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Which of the following represents the $NAND$ gate?

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$A$ $B$ $Y$
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Which logic gate is represented by the following truth table?

$A$ $B$ $Y$

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To get an output of $1$ from the circuit shown in the figure,the input must be:

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