GATE ECE 2022 | Question: 9

Question

Consider the $ \text{2–bit}$ multiplexer $\text{(MUX)}$ shown in the figure. For $\text{OUTPUT}$ to be the $\text{XOR}$ of $\text{C}$ and $\text{D},$ the values for $A_{0}, A_{1}, A_{2},$ and $A_{3}$ are _______________.

• A. $A_{0} = 0, A_{1} = 0, A_{2} = 1, A_{3} = 1$
• B. $A_{0} = 1, A_{1} = 0, A_{2} = 1, A_{3} = 0$
• C. $A_{0} = 0, A_{1} = 1, A_{2} = 1, A_{3} = 0$
• D. $A_{0} = 1, A_{1} = 1, A_{2} = 0, A_{3} = 0$

Answer 1

Detailed Video Solution: https://youtu.be/52mf21k7k6E 

We want $OUTPUT = C \oplus D.$

Note: Boolean variable $C$ is connected to MSB select line $S_1$ & boolean variable $D$ is connected to LSB select line $S_0.$

So, when $S_1 = C = 0 $ & $S_0 = D = 0,$ OUTPUT will be $I_0 = A_0$ which should be $0 \oplus 0 $ i.e. $0.$ So, $A_0 = 0.$

when $S_1 = C = 0 $ & $S_0 = D = 1,$ OUTPUT will be $I_1 = A_1$ which should be $0 \oplus 1 $ i.e. $1.$ So, $A_1 = 1.$

when $S_1 = C = 1 $ & $S_0 = D = 0,$ OUTPUT will be $I_2 = A_2$ which should be $1 \oplus 0 $ i.e. $1.$ So, $A_2 = 1.$

when $S_1 = C = 1 $ & $S_0 = D = 1,$ OUTPUT will be $I_3 = A_3$ which should be $1 \oplus 1 $ i.e. $0.$ So, $A_3 = 0.$

So, answer is Option C.

Detailed Video Solution: https://youtu.be/52mf21k7k6E