The waveforms $A$ and $B$ given below are given as input to a NAND gate. Then its logic output $Y$ is

Question

Aicharya · Accepted Answer

A NAND gate is made by connecting a NOT gate and a AND gate with each other. The symbol for NAND gate is given as

The Boolean expression for a NAND gate is given as
$Y=\overline{A \cdot B}=\bar{A}+\bar{B}$
Truth Table for a NAND gate is given as
$\begin{array}{|c|c|c|}
\hline \text { Input A } & \text { Input B } & \text { Output Y } \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}$
For $t_1$ to $t_2$; we have $\mathrm{A}=1$ and $\mathrm{B}=0$, using the Boolean algebra or the truth table for NAND gate we get $\mathrm{y}=1$
For $t_2$ to $t_3$; we have $\mathrm{A}=1$ and $\mathrm{B}=1$, using the Boolean algebra or the truth table for NAND gate we get $\mathrm{y}=0$
For $t_3$ to $t_4$; we have $\mathrm{A}=0$ and $\mathrm{B}=1$, using the Boolean algebra or the truth table for NAND gate we get $\mathrm{y}=1$
For $t_4$ to $t_5$; we have $\mathrm{A}=0$ and $\mathrm{B}=0$, using the Boolean algebra or the truth table for NAND gate we get $\mathrm{y}=1$
For $t_5$ to $t_6$; we have $\mathrm{A}=1$ and $\mathrm{B}=0$, using the Boolean algebra or the truth table for NAND gate we get $\mathrm{y}=1$
On the basis of the above results, the correct option is C from the given choices.

The waveforms $A$ and $B$ given below are given as input to a NAND gate. Then its logic output $Y$ is

Solution