The output y of the given logic gate network | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P = \overline{\bar A \cdot B} = A + \bar B$.The lower $NAND$ gate has inputs $A$ and $\bar B$. Its output is $Q = \overline{A \cdot \bar B} = \bar A + B$.The final gate is a $NOR$ gate,so the output $y = \overline{P + Q}$.Substituting the values of $P$ and $Q$:$y = \overline{(A + \bar B) + (\bar A + B)}$$y = \overline{(A + \bar A) + (B + \bar B)}$Since $A + \bar A = 1$ and $B + \bar B = 1$:$y = \overline{1 + 1} = \overline{1} = 0$.Thus,the output $y = 0$.

Entry	$A$	$B$	$Y$
$1$	$0$	$1$	$0$
$2$	$1$	$0$	$0$
$3$	$1$	$1$	$1$
$4$	$0$	$0$	$1$

The output $y$ of the given logic gate network is:

Similar Questions

The output $Y$ of the combination of gates shown is equal to :-

$A$ typical integrated circuit $(IC)$ consisting of logic gates $\leq 1000$ is termed as

What will be the input of $A$ and $B$ for the Boolean expression $(\overline{A + B}) \cdot (\overline{A \cdot B}) = 1$?

For a two-input $AND$ gate,the four entries are shown in the truth table. Identify the correct ones out of these ($A, B =$ input,$Y =$ output).
Entry $A$ $B$ $Y$
$1$ $0$ $1$ $0$
$2$ $1$ $0$ $0$
$3$ $1$ $1$ $1$
$4$ $0$ $0$ $1$

In the logic circuit shown in the figure,if inputs $A$ and $B$ are $0$ and $1$ respectively,the output at $Y$ is $x$. The value of $x$ is...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module