Physics - Semiconductors Question with Solution | TestHub

The Boolean expression for the output $Y$ can be determined from the logic circuit.

The top gate is a NAND gate with inputs $A$ and $B$. Its output is $Y_1=\overline{A\cdot B}$.

The middle gate is a NOT gate with input $A$. Its output is $\overline{A}$.

The bottom gate is a NOR gate with inputs $\overline{A}$ and $B$. Its output is $Y_2=\overline{\overline{A}+B}$.

The final gate is a NOR gate with inputs $Y_1$ and $Y_2$. Its output is $Y=\overline{Y_1+Y_2}$.

Substituting the expressions for $Y_1$ and $Y_2$ into the final output equation:

$Y=\overline{\overline{A\cdot B}+\overline{\overline{A}+B}}$

Using De Morgan's laws, the expression simplifies to:

$Y=\overline{(\overline{A\cdot B})}\cdot \overline{(\overline{\overline{A}+B})}$

$Y=(A\cdot B)\cdot (\overline{A}+B)$

Expanding the expression:

$Y=A\cdot B\cdot \overline{A}+A\cdot B\cdot B$

Since $A\cdot \overline{A}=0$ and $B\cdot B=B$, the expression becomes:

$Y=0+A\cdot B$

$Y=A\cdot B$

Given that the output $Y=1$:

$A\cdot B=1$

This condition is satisfied only when both inputs are $1$.

Therefore, $A=1$ and $B=1$.

Answer: $A=1,B=1$