JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Semiconductors

Each input first passes through a NAND gate whose two inputs are tied together, so the upper gate produces $\overline{A}$ and the lower gate produces $\overline{B}$. These two outputs feed the final NAND gate, giving $Y=\overline{\overline{A}\cdot \overline{B}}=A+B$, which is the OR operation. From the waveforms, $A=0$ for $0<t<2$ and $A=1$ for $2<t<4$, while $B=1$ only during $1<t<2$ and $3<t<4$. Applying $Y=A+B$: during $0<t<1$ both inputs are low so $Y=0$; during $1<t<2$ the input $B$ is high so $Y=1$; and during $2<t<4$ the input $A$ is high so $Y=1$. Hence the output stays low from $t=0$ to $t=1$ and remains high from $t=1$ to $t=4$.