GATE 2023 EC Q.23: solution and answer

GATE 2023 · ECQ.23 · 1 mark · MCQ

In the circuit shown below, P and Q are the inputs. The logical function realized by the circuit shown below is

GATE 2023 EC question 23: A single 2 × 1 MUX block with output Y on the right. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$Y = P Q$
(B)
$Y = P + Q$
(C)
$Y = \overline{P Q}$
(D)
$Y = \overline{P + Q}$

Answer and solution

ANSWER

The circuit realises the AND function, $Y = P Q$ : option (A).

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

Output equation of a $2 \times 1$ MUX with select line $S$ (complement written with ): $S = 0$ passes $I_{0}$ and $S = 1$ passes $I_{1}$ , so $Y = \overline{S} I_{0} + S I_{1}$
Read the connections off the figure: $S = Q$ , $I_{0} = 0$ (tied to ground), $I_{1} = P$ .
Substitute: $Y = \overline{Q} \cdot 0 + Q \cdot P = P Q$
Check with the truth table: when $Q = 0$ the MUX passes $I_{0} = 0$ , so $Y = 0$ whatever $P$ is. When $Q = 1$ it passes $I_{1} = P$ , so $Y = P$ . $Y = 1$ only for $P = Q = 1$ , which is the AND function.
Hence $Y = P Q$ , option (A).

KEY IDEA

A $2 \times 1$ MUX computes $Y = \overline{S} I_{0} + S I_{1}$ . Tie one data input to a constant and feed a variable into the other, and it becomes a basic gate. With $I_{0} = 0$ and $I_{1} = P$ , the output is $Y = S P$ . That is an AND gate between the select signal and $P$ .

WATCH OUT FOR

Getting the select convention backwards, as if $S = 1$ picked $I_{0}$ . That gives $Y = \overline{Q} P$ , which matches none of the options, and some students then guess OR or NAND. Another slip is reading the grounded input as logic 1 instead of 0.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 12 Combinational questions, with notes

GATE 2023 EC Q.23: solution and answer

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