In the following digital logic circuit, the output Y will be ' 1 ' for inputs
In the following digital logic circuit, the output Y will be ' 1 ' for inputs
$\mathrm{A}=0, \mathrm{~B}=0$
$\mathrm{A}=0, \mathrm{~B}=1$
$\mathrm{A}=1, \mathrm{~B}=0$
$\mathrm{A}=1, \mathrm{~B}=1$
Solution
$\therefore \quad$ There are two NOR gates, one NOT gate, and one NAND gate.
$\therefore \quad$ Output of NAND gate: $\overline{A \cdot B}$
Output of NOT and NOR Gate: $\overline{\overline{\mathrm{A}}+\mathrm{B}}$
Final output: $\overline{(\overline{\mathrm{A} \cdot \mathrm{B}})+(\overline{\overline{\mathrm{A}}+\mathrm{B}})}$
So, the output $\mathrm{Y}$ is 1 only if the input $\mathrm{A}$ and $\mathrm{B}$ is 1.
$\begin{aligned}
& \mathrm{A}=1 \\
& \mathrm{~B}=1 \\
& \mathrm{Y}=\overline{\overline{1 \cdot 1}+\overline{1}+1} \\
& \mathrm{Y}=1
\end{aligned}$
