GATE 2024 EC Q.41: solution and answer

GATE 2024 · ECQ.41 · 2 marks · MCQ

The propagation delay of the $2 \times 1$ MUX shown in the circuit is $10 ns$ . Consider the propagation delay of the inverter as $0 ns$ .

[Figure]

If S is set to 1 then the output Y is ________.

GATE 2024 EC question 41: A 2 X 1 MUX block with data inputs labelled 0 (upper) and 1 (lower), select input S at the bottom, and output Y on the… — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
a square wave of frequency 100 MHz
(B)
a square wave of frequency 50 MHz
(C)
constant at 0
(D)
constant at 1

Answer and solution

ANSWER

With S = 1 the output is fed back to itself through one inversion and the 10 ns MUX delay, so Y toggles every 10 ns. That is a square wave with a 20 ns period, frequency 50 MHz: option (B).

Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Read the wiring from the figure: input 0 is tied straight to the output, $I_{0} = Y$ , and input 1 comes from the output through the inverter, $I_{1} = \overline{Y}$ (overline means complement).
MUX equation with a propagation delay of $t_{p d} = 10 ns$ : $Y (t) = \overline{S} I_{0} (t - t_{p d}) + S I_{1} (t - t_{p d})$ .
With $S = 1$ the MUX passes only input 1: $Y (t) = I_{1} (t - 10 ns) = \overline{Y} (t - 10 ns)$ . The inverter adds no delay, so the whole loop delay is the MUX's 10 ns.
So the output always equals the complement of what it was 10 ns earlier. The loop is a single inversion fed back through a delay, which has no steady state. A constant $Y = 0$ would need $0 = \overline{0} = 1$ , which is impossible, and a constant $Y = 1$ would need $1 = \overline{1} = 0$ , also impossible. That rules out (C) and (D).
Apply the relation twice: $Y (t) = \overline{Y} (t - 10 ns) = \overline{\overline{Y}} (t - 20 ns) = Y (t - 20 ns)$ . The output repeats every 20 ns, and in each 20 ns period the second 10 ns is the complement of the first. So Y toggles every 10 ns with $T_{hi g h} = T_{l o w} = 10 ns$ , which is a square wave.
Period $T = 2 \times t_{p d} = 20 ns$ , so $f = \frac{1}{T} = \frac{1}{20 ns} = 50 MHz$ . The answer is (B).

KEY IDEA

With S = 1 the MUX connects Y to its own complement, delayed by the MUX propagation delay: $Y (t) = \overline{Y} (t - t_{p d})$ . One inversion in a feedback loop gives an oscillator (a ring oscillator). The output toggles once every $t_{p d}$ , so a full cycle takes $2 t_{p d}$ and $f = \frac{1}{2 t _{p d}}$ .

WATCH OUT FOR

Taking the period as one propagation delay ( $T = 10 ns$ ) and choosing 100 MHz (A). One delay is only half a cycle: the output needs one 10 ns delay to go from 0 to 1 and another to go back to 0, so $T = 20 ns$ . Another mistake is to treat the circuit as static and pick a constant output (C or D), which ignores the inverting feedback. The input-0 path does not matter when S = 1.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 12 Combinational questions, with notes

GATE 2024 EC Q.41: solution and answer

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