GATE CSE 2026 | Set 2 | Question: 49

Question

Consider the digital circuit shown below with two input lines $\text{A}$ and $\text{B}$, two select lines $\text{S0}$ and $\text{S1}$, and an output line $\text{Y}$. The blocks $\text{Q}$ and $\text{M}$ represent active high $2:4$ decoder and $4$-to-$1$ multiplexer, respectively. Out of $16$ possible input combinations, the number of combinations that produce $\mathrm{Y}=1$ is $\_\_\_\_\_\_$ (answer in integer)

Note: One input combination is an instance of [$\text{A B S1 S0}$].

Comments

Nice Quality Question ✅

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Easy explanation https://gateoverflow.in/523097/gate-cse-2026-set-2-question-49?show=543318#a543318

Answer 1

Easiest and simplest way to solve these kind of Question is this 

Comments

bro kaise fill kiya 1 in kmap  table??

 

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Pramod,  A'B'S1'S0'+S1S0'+S1S0AB ko expend krlo jese A'B'S1'S0' = 0000 = 0 mtlb cell 0 me 1 age like that sabko expend krlo

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ha got it bro thanks

 

Answer 2

$D0 = A'B' \quad \&  \quad D3=AB$

$So : Y= \bar A \bar B \bar S_1 \bar S_0 + 0 \bar S_1 S_0 +  \underbrace {1S_1 \bar S_0}_{\text{This gives 4 more outputs}}+  A B S_1 S_0$

             $ = \bar A \bar B \bar S_1 \bar S_0 + \begin{aligned}
\bar A \bar B S_1 \bar{S}_0 \\
\bar A B S_1 \bar{S}_0 \\
A \bar B S_1 \bar{S}_0 \\
AB S_1  \bar S_0
\end{aligned} + A B S_1 S_0 $

$So, answer$ $is$ $6$

Answer 3

$$\text{The Truth Table Approach}$$

 

Answer 4

Since its an active high decoder it will make exactly one of its output lines  1 (high) for any given input

A	B	S1	S0	Y
0	0	0	0	D0
0	0	0	1	0
0	0	1	0	1
0	0	1	1	D3
0	1	0	0	D0
0	1	0	1	0
0	1	1	0	1
0	1	1	1	D3
1	0	0	0	D0
1	0	0	1	0
1	0	1	0	1
1	0	1	1	D3
1	1	0	0	D0
1	1	0	1	0
1	1	1	0	1
1	1	1	1	D3

 

the given circuit gives 1 for input combination

0 0 0 0 ( D0 will be in active high state)

0 0 1 0 ( 1 )

0 1 1 0 ( 1 )

1 0 1 0 ( 1 )

1 1 1 0 ( 1 )

1 1 1 1 ( D3 will bw in active high state)

therefore the answer is 6

Answer 5

The answer is $\boxed{6}$.

$A, B, S1, S0$ is the function.

Now, we have to get $D0$ to get input $0$ in the multiplexer. So, we have only one choice for $AB=00$ and $S1S0=00$. For this combination, the output is $1$.

$ABS1S0=0000$

So, there is only $1$ choice.

Now, we have to get $D3$ to get input $3$ in the multiplexer. So, we have only one choice for $AB=11$ and $S1S0=11$. For this combination, the output is $1$.

$ABS1S0=1111$

So, there is only $1$ choice.

Now, as we can see in the MUX, when $S1S0=10$, the output is $1$. For this, we don't care what $AB$ is. For $A$, there are two choices, and for $B$, there are two choices.

$ABS1S0=XX10$

For this combination, the total number of choices is $2\times2=4$.

Therefore, to get $Y=1$, we have a total of $1+1+4=\boxed{6}$.

\[
\begin{array}{|c|c|c|c|}
\hline
S1S0 & \text{MUX input} & \text{Condition on }AB & \text{Combinations}\\
\hline
00 & D0 & AB=00 & 1\\
\hline
01 & 0 & - & 0\\
\hline
10 & 1 & AB=XX & 4\\
\hline
11 & D3 & AB=11 & 1\\
\hline
\text{Total} & & & \boxed{6}\\
\hline
\end{array}
\]