For the logic circuit shown,the output wavefo | vedclass

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

Question

(D) The given circuit consists of two $NAND$ gates at the input followed by another $NAND$ gate. Let the inputs be $A$ and $B$. The output of the firs

Vedclass · Accepted Answer

Vedclass · Answer

(D) The given circuit consists of two $NAND$ gates at the input followed by another $NAND$ gate. Let the inputs be $A$ and $B$. The output of the first two $NAND$ gates are $\overline{A}$ and $\overline{B}$ respectively (since they act as $NOT$ gates). The final $NAND$ gate takes $\overline{A}$ and $\overline{B}$ as inputs. The output $Y$ is given by $Y = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's theorem,$Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$. Thus,the circuit acts as an $OR$ gate. Analyzing the waveforms: For $t=0$ to $1$: $A=0, B=0 \implies Y = 0+0 = 0$. For $t=1$ to $2$: $A=0, B=1 \implies Y = 0+1 = 1$. For $t=2$ to $3$: $A=1, B=0 \implies Y = 1+0 = 1$. For $t=3$ to $4$: $A=1, B=1 \implies Y = 1+1 = 1$. Thus,the output $Y$ is $0$ from $t=0$ to $1$ and $1$ from $t=1$ to $4$.

For the logic circuit shown,the output waveform at $Y$ is

Similar Questions

The following configuration of gate is equivalent to

Which one of the following is the Boolean expression for $NOR$ gate?

The truth table for the circuit shown in the figure is

Choose the correct logic circuit for the given truth table having inputs $A$ and $B$.
Inputs Inputs Outputs
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

The real-time variation of input signals $A$ and $B$ are as shown below. If the inputs are fed into a $NAND$ gate,then select the output signal from the following.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Inputs	Inputs	Outputs
$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$