GATE 2024 EC question paper PDF and answer key

The official GATE 2024 Electronics & Communication Engineering paper, organised by IISc Bengaluru: 65 questions for 100 marks in 3 hours. Download the official question paper and answer key as PDFs. Below is how the marks were split by topic, the complete official answer key and 5 questions solved step by step.

Source: the official paper and answer key published by IISc Bengaluru for GATE 2024 (official GATE site). Spotted a mistake? Email team@lemyte.com.

General Aptitude
10 · 15 marks
Electronics & Communication Engineering
55 · 85 marks
MCQ / MSQ / NAT
39 / 8 / 18
Marks to all
Q 14, 44, 48

Where the marks were

GATE 2024 EC topic-wise marks

Electronics & Communication Engineering questions only; General Aptitude adds 15 marks on top. The top three topics carried 39 of the 85 subject marks.

TopicQuestionsMarks
Networks, Signals and Systems1016
Electronic Devices712
Digital Circuits711
Communications710
Analog Circuits710
Control Systems58
Electromagnetics46
Calculus35
Linear Algebra23
Complex Variables12
Differential Equations11
Probability and Statistics11

Free solved questions

5 solved questions from GATE 2024 EC

Question text and figures as in the official paper, the answer from the official key, and a worked solution. The other 60 solutions are in your report after you take the paper.

Q2 · General Aptitude · MCQ · 1 mark

P, Q, R, S, and T have launched a new startup. Two of them are siblings. The office of the startup has just three rooms. All of them agree that the siblings should not share the same room.

If S and Q are single children, and the room allocations shown below are acceptable to all,

Figure for GATE 2024 EC question 2 (Analytical Aptitude)

then, which one of the given options is the siblings?

  • (A)
    P and T
  • (B)
    P and S
  • (C)
    T and Q
  • (D)
    T and R

Answer (official key): A

Solution

S and Q are single children, so the siblings are two of P, R and T. The pairs sharing a room cannot be siblings: P–R (first allocation) and R–T (second allocation). The only remaining pair is P and T.

Q10 · General Aptitude · MCQ · 2 marks

Two identical sheets A and B, of dimensions 24 cm × 16 cm, can be folded into half using two distinct operations, FO1 or FO2.

In FO1, the axis of folding remains parallel to the initial long edge, and in FO2, the axis of folding remains parallel to the initial short edge.

If sheet A is folded twice using FO1, and sheet B is folded twice using FO2, the ratio of the perimeters of the final shapes of A and B is

  • (A)
    14:11
  • (B)
    11:14
  • (C)
    18:11
  • (D)
    11:18

Answer (official key): A

Solution

  • FO1 halves the short side twice: 24×16→24×8→24×424 \times 16 \to 24 \times 8 \to 24 \times 4, perimeter 56 cm.
  • FO2 halves the long side twice: 24×16→12×16→6×1624 \times 16 \to 12 \times 16 \to 6 \times 16, perimeter 44 cm.

The ratio is 56:44=14:1156:44 = 14:11.

Q49 · Analog Circuits · MCQ · 2 marks

The opamps in the circuit shown are ideal, but have saturation voltages of ±10 V.

Figure for GATE 2024 EC question 49 (Analog Circuits)

Assume that the initial inductor current is 0 A. The input voltage (ViV_i) is a triangular signal with peak voltages of ±2 V and time period of 8 µs. Which one of the following statements is true?

  • (A)
    V01V_{01} is delayed by 2 µs relative to ViV_i, and V02V_{02} is a triangular waveform.
  • (B)
    V01V_{01} is not delayed relative to ViV_i, and V02V_{02} is a trapezoidal waveform.
  • (C)
    V01V_{01} is not delayed relative to ViV_i, and V02V_{02} is a triangular waveform.
  • (D)
    V01V_{01} is delayed by 1 µs relative to ViV_i, and V02V_{02} is a trapezoidal waveform.

Answer (official key): D

Solution

OPA1 is a non-inverting Schmitt trigger with thresholds ±10×10k100k=±1\pm 10 \times \frac{10\text{k}}{100\text{k}} = \pm 1 V. The input slope is 4 V4 μs=1\frac{4\text{ V}}{4\ \mu\text{s}} = 1 V/µs, so after each zero crossing it takes 1 µs to reach ±1\pm 1 V. V01V_{01} is therefore a ±10 V square wave delayed by 1 µs.

OPA2 with the series inductor integrates: dV02dt=−RLV01=∓107\frac{dV_{02}}{dt} = -\frac{R}{L}V_{01} = \mp 10^7 V/s =∓10= \mp 10 V/µs. It reaches ±10 V saturation in 2 µs of each 4 µs half-period, then holds, giving a trapezoidal waveform.

Q52 · Electronic Devices · MSQ · 2 marks

Which of the following statements is/are true for a BJT with respect to its DC current gain β\beta?

  • (A)
    Under high-level injection condition in forward active mode, β\beta will decrease with increase in the magnitude of collector current.
  • (B)
    Under low-level injection condition in forward active mode, where the current at the emitter-base junction is dominated by recombination-generation process, β\beta will decrease with increase in the magnitude of collector current.
  • (C)
    β\beta will be lower when the BJT is in saturation region compared to when it is in active region.
  • (D)
    A higher value of β\beta will lead to a lower value of the collector-to-emitter breakdown voltage.

Answer (official key): A, C, D

Solution

  • (A) True: high-level injection reduces emitter efficiency, so β\beta falls at high ICI_C.
  • (B) False: in the recombination-dominated low-current region, β\beta increases with ICI_C.
  • (C) True: in saturation, IC/IBI_C/I_B is forced below its active-region value.
  • (D) True: BVCEO≈BVCBO/β1/nBV_{CEO} \approx BV_{CBO}/\beta^{1/n}, so a higher β\beta gives a lower breakdown voltage.

Q57 · Communications · Numerical · 2 marks

Let X(t)=Acos⁡(2πf0t+θ)X(t) = A\cos(2\pi f_0t + \theta) be a random process, where amplitude AA and phase θ\theta are independent of each other, and are uniformly distributed in the intervals [−2,2][-2, 2] and [0,2π][0, 2\pi], respectively. X(t)X(t) is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of X(t)X(t) is RX(τ)=23cos⁡(2πf0τ)R_X(\tau) = \frac{2}{3}\cos(2\pi f_0\tau), the signal to quantization noise ratio (in dB, rounded off to two decimal places) at the output of the quantizer is __________.

Answer (official key): 45 to 45.3

Solution

Signal power is RX(0)=23R_X(0) = \frac{2}{3}. The quantizer covers [−2,2][-2, 2] with 282^8 levels, so Δ=4256=164\Delta = \frac{4}{256} = \frac{1}{64} and the noise power is Δ212=149152\frac{\Delta^2}{12} = \frac{1}{49152}. SQNR=2/31/49152=32768⇒45.15 dB\text{SQNR} = \frac{2/3}{1/49152} = 32768 \Rightarrow 45.15\ \text{dB}

The other 60 questions are solved in your report when you take GATE 2024 EC as a 3-hour test.

Take GATE 2024 EC as a test

Official answer key

GATE 2024 EC answer key

All 65 answers from the official key. Numerical answers are ranges; “or” means the key accepts either answer.

Download the official GATE 2024 EC answer key (PDF)
QTopicTypeMarksAnswer
1Verbal AptitudeMCQ1A
2Analytical AptitudeMCQ1A
3Quantitative AptitudeMCQ1B
4Quantitative AptitudeMCQ1C
5Quantitative AptitudeMCQ1A
6Verbal AptitudeMCQ2C
7Spatial AptitudeMCQ2D
8Analytical AptitudeMCQ2C
9Spatial AptitudeMCQ2A
10Analytical AptitudeMCQ2A
11Differential EquationsMCQ1A
12Control SystemsMCQ1A
13Control SystemsMCQ1D
14CommunicationsMCQ1Marks to all
15ElectromagneticsMCQ1A
16ElectromagneticsMCQ1B
17Analog CircuitsMCQ1A
18Analog CircuitsMCQ1A
19Digital CircuitsMCQ1A
20CommunicationsMCQ1A
21Networks, Signals and SystemsMCQ1A
22Electronic DevicesMCQ1A
23Analog CircuitsMCQ1A
24Networks, Signals and SystemsMSQ1A, C, D
25CalculusMSQ1A, C
26Electronic DevicesMSQ1A, B, C
27Digital CircuitsNumerical12047
28CommunicationsNumerical12.49 to 2.51
29Digital CircuitsNumerical111
30Linear AlgebraNumerical15
31Networks, Signals and SystemsNumerical12
32Networks, Signals and SystemsNumerical10.75
33Probability and StatisticsNumerical10.5
34CommunicationsNumerical14
35Analog CircuitsNumerical10.25
36CalculusMCQ2A
37Control SystemsMCQ2A
38Control SystemsMCQ2B
39ElectromagneticsMCQ2A
40Digital CircuitsMCQ2B
41Digital CircuitsMCQ2B
42Digital CircuitsMCQ2B
43Complex VariablesMCQ2D
44Networks, Signals and SystemsMCQ2Marks to all
45CommunicationsMCQ2B
46Digital CircuitsMCQ2B
47CommunicationsMCQ2A
48Networks, Signals and SystemsMCQ2Marks to all
49Analog CircuitsMCQ2D
50Analog CircuitsMCQ2B
51Analog CircuitsMCQ2B
52Electronic DevicesMSQ2A, C, D
53Control SystemsMSQ2A, C
54CalculusMSQ2A, B, D
55Linear AlgebraMSQ2A, B
56Networks, Signals and SystemsMSQ2A, D or A, C, D
57CommunicationsNumerical245 to 45.3
58ElectromagneticsNumerical2200
59Networks, Signals and SystemsNumerical2112
60Networks, Signals and SystemsNumerical21.5
61Electronic DevicesNumerical24.1 to 4.5
62Networks, Signals and SystemsNumerical22.5
63Electronic DevicesNumerical20.17 to 0.19
64Electronic DevicesNumerical252.4 to 52.6
65Electronic DevicesNumerical24 to 4.26

Questions about GATE 2024 EC

How many questions are in the GATE 2024 EC paper?

65 questions for 100 marks: 10 General Aptitude questions worth 15 marks and 55 Electronics & Communication Engineering questions worth 85 marks. By type, there were 39 MCQs, 8 MSQs, 18 numerical answer (NAT) questions. The paper lasted 3 hours.

Is there negative marking in GATE 2024 EC?

Yes, for MCQs only. A wrong MCQ answer costs one-third of its marks (−⅓ for a 1-mark question, −⅔ for a 2-mark question). MSQ and numerical (NAT) questions have no negative marking, and an MSQ earns marks only when every correct option is chosen.

Which topics carried the most marks in GATE 2024 EC?

Outside General Aptitude, the biggest topics were Networks, Signals and Systems (16 marks), Electronic Devices (12 marks), Digital Circuits (11 marks). The full topic-wise split is in the table on this page.

Were any GATE 2024 EC questions awarded marks to all?

Yes. The official key awarded full marks to everyone for questions 14, 44, 48.

Where does this GATE 2024 EC answer key come from?

From the official answer key published by IISc Bengaluru, which organised GATE 2024. Range answers for numerical questions and questions with more than one accepted answer are kept exactly as the key gives them.

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