GATE CSE 2024 | Set 1 | Question: 54

Question

Consider a digital logic circuit consisting of three $2$-to-$1$ multiplexers $\text{M1, M2}$, and $\text{M3}$ as shown below. $\mathrm{X} 1$ and $\mathrm{X} 2$ are inputs of $\mathrm{M} 1$. $\text{X3}$ and $\text{X4}$ are inputs of $\text{M2}$. $\text{A, B}$, and $\text{C}$ are select lines of $\text{M1, M2}$, and $\text{M3}$, respectively.

For an instance of inputs $\mathbf{X} \mathbf{1}=\mathbf{1}, \mathbf{X} \mathbf{2}=\mathbf{1}, \mathbf{X} \mathbf{3}=\mathbf{0}$, and $\mathbf{X} \mathbf{4}=\mathbf{0}$, the number of combinations of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ that give the output $\mathbf{Y}=\mathbf{1}$ is ____________.

Answer 1

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As we know that the output equations of any $2\times1$ multiplexer is $Y_{out}=\bar AI_0+AI_1$

 where A is the selection line and $I_0,I_1$ act as input to the multiplexer.

The output of the first multiplexer: $Q_1=\bar AX_1+AX_2=\bar A+A=1$

Similarly the output of the second multiplexer: $Q_2=\bar BX_3+BX_4=0+0=0$

Now this output is act as input to the third multiplexer. so output equations is:

$Y=\bar C.1+C.0=\bar C$

As we know with the three inputs total eight combinations are possible, since output $Y$ is independent from $A,B$ and dependent on $C$. it gives output as logic one when $C=0$

so the correct combinations are as follows:

A	B	C	Y
0	0	0	1
0	1	0	1
1	0	0	1
1	1	0	1

Correct answer is $4$

Answer 2

Y = C'Q1 + CQ2

Let's find Q1 and Q2

Q1 = A'X1 + AX2

Q2 = B'X3 + B X4

Now replace value of X1,X2,X3 and X4

Q1 = A'.1 + A.1 = A' + A = 1

Q2 = B' 0 +B.0 = 0 + 0 = 0

Q1 = 1 AND Q2 = 0

Replacing the value of Q1 and Q2 in the first equation we got at the beginning.

Y = C'.1 +C.0 =C'

We get Y =C'

for three variables A, B, and C there are a total of 8 combinations and among them, there are four combinations where C = 0

when  C = 0 then Y =1 hence for 4 Combinations Y =1.

Answer 4

Answer 3

Answer: 4