Physics - Semiconductors Question with Solution | TestHub

From the given logic circuit, we can identify the gates and their outputs:

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

The bottom gate is an OR gate with inputs $A$ and $B$. Its output is $A+B$.

The final gate is an AND gate, which takes the outputs of the NAND and OR gates as its inputs. Therefore, the final output $Y$ is given by:

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

Using De Morgan's laws, we can expand $\overline{A\cdot B}$ as $\overline{A}+\overline{B}$:

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

Expanding the expression, we get:

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

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

$Y=0+\overline{A}\cdot B+A\cdot \overline{B}+0$

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

This is the standard Boolean expression for an Exclusive-OR (X-OR) gate.

Answer: X-OR gate.