The following figure shows the variation of i | vedclass

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

Question

(B) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.Based on the input waveforms:For $t = 0-2 \ s$: $A=1, B=0 \implies A \cdot B = 0 \im

Vedclass · Accepted Answer

Vedclass · Answer

(B) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.Based on the input waveforms:For $t = 0-2 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 2-4 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 4-6 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 6-8 \ s$: $A=1, B=1 \implies A \cdot B = 1 \implies Y = \overline{1} = 0$.For $t > 8 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.Thus,the output $Y$ is $1$ for $0-6 \ s$,$0$ for $6-8 \ s$,and $1$ for $t > 8 \ s$. This matches the waveform shown in option $B$.

The following figure shows the variation of inputs $A$ and $B$ with time. If these inputs are given to a $NAND$ gate,select the correct output signal $Y$ from the given options.

Similar Questions

The output $Y$ when all the three inputs $A, B, C$ are first low and then high will be

Identify the logic operation carried out by the given circuit.

The logic circuit is shown below:
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $0$
$1$ $0$ $0$
$1$ $1$ $1$

To obtain the given truth table,which logic gate should be placed at $G$?

The logic behind the $NOR$ gate is that it gives

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module