Physics - Semiconductors Question with Solution | TestHub

The signal $A$ is given to a NOR gate and $B$ is given to an OR gate. The output of these two signals is given to a NAND gate. Using the laws of Boolean algebra,

$Y=\overline{\overline{A} ·B} =A+\overline{B}$.

Analysing from the input graphs, one can see,

$$\begin{gathered} t_1-t_2: A=0, \overline{B}=0, Y=0 \\ {t}_2-t_3: A=1, \overline{B}=0, Y=1 \\ {t}_3-t_4: A=0, \overline{B}=1, Y=1 \\ {t}_4-t_5: A=1, \overline{B}=0, Y=1 \\ {t}_5-t_6: A=1, \overline{B}=1, Y=1 \end{gathered}$$