GATE 2020 EC question paper PDF and answer key

The official GATE 2020 Electronics & Communication Engineering paper, organised by IIT Delhi: 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 IIT Delhi for GATE 2020 (official GATE site). Spotted a mistake? Email team@lemyte.com.

General Aptitude
10 · 15 marks
Electronics & Communication Engineering
55 · 85 marks
MCQ / MSQ / NAT
42 / 0 / 23
Marks to all
Q 41, 43

Where the marks were

GATE 2020 EC topic-wise marks

Electronics & Communication Engineering questions only; General Aptitude adds 15 marks on top. The top three topics carried 38 of the 85 subject marks.

TopicQuestionsMarks
Networks, Signals and Systems1015
Analog Circuits813
Electronic Devices610
Digital Circuits69
Communications69
Control Systems58
Electromagnetics58
Calculus34
Linear Algebra23
Differential Equations23
Probability and Statistics23

Free solved questions

5 solved questions from GATE 2020 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.

Q5 · General Aptitude · MCQ · 1 mark

A superadditive function f(⋅)f(\cdot) satisfies the following property

f(x1+x2)≥f(x1)+f(x2)f(x_1 + x_2) \ge f(x_1) + f(x_2)

Which of the following functions is a superadditive function for x>1x > 1?

  • (A)
    exe^x
  • (B)
    x\sqrt{x}
  • (C)
    1/x1/x
  • (D)
    e−xe^{-x}

Answer (official key): A

Solution

For x1,x2>1x_1, x_2 > 1: ex1+x2=ex1ex2≥ex1+ex2e^{x_1 + x_2} = e^{x_1}e^{x_2} \ge e^{x_1} + e^{x_2} (since ab≥a+bab \ge a + b when a,b≥2a, b \ge 2). The other three functions are concave or decreasing, so they are subadditive.

Q9 · General Aptitude · MCQ · 2 marks

a,b,ca, b, c are real numbers. The quadratic equation ax2−bx+c=0ax^2 - bx + c = 0 has equal roots, which is β\beta, then

  • (A)
    β=b/a\beta = b/a
  • (B)
    β2=ac\beta^2 = ac
  • (C)
    β3=bc/(2a2)\beta^3 = bc/(2a^2)
  • (D)
    b2≠4acb^2 \ne 4ac

Answer (official key): C

Solution

With equal roots, the sum is 2β=ba2\beta = \frac{b}{a} and the product is β2=ca\beta^2 = \frac{c}{a}. Multiplying: β3=b2a⋅ca=bc2a2\beta^3 = \frac{b}{2a}\cdot\frac{c}{a} = \frac{bc}{2a^2}.

Q38 · Networks, Signals and Systems · MCQ · 2 marks

The current I in the given network is

Figure for GATE 2020 EC question 38 (Networks, Signals and Systems)

  • (A)
    0 A.
  • (B)
    2.38∠−96.37∘2.38\angle -96.37^\circ A.
  • (C)
    2.38∠143.63∘2.38\angle 143.63^\circ A.
  • (D)
    2.38∠−23.63∘2.38\angle -23.63^\circ A.

Answer (official key): C

Solution

Take the junction between the sources (and, through the wire carrying I, the node between the two Z's) as reference. Then the top node is at 120∠−90∘=−j120120\angle -90^\circ = -j120 and the bottom node at 120∠−30∘=103.92−j60120\angle -30^\circ = 103.92 - j60. The current I entering the middle node leaves through both Z's: I=0−VTZ+0−VBZ=−103.92−j18080−j35=207.8∠120∘87.3∠−23.63∘=2.38∠143.63∘ AI = \frac{0 - V_T}{Z} + \frac{0 - V_B}{Z} = -\frac{103.92 - j180}{80 - j35} = \frac{207.8\angle 120^\circ}{87.3\angle -23.63^\circ} = 2.38\angle 143.63^\circ\ \text{A}

Q39 · Networks, Signals and Systems · MCQ · 2 marks

A finite duration discrete-time signal x[n]x[n] is obtained by sampling the continuous-time signal x(t)=cos⁡(200πt)x(t) = \cos(200\pi t) at sampling instants t=n/400t = n/400, n=0,1,⋯ ,7n = 0, 1, \cdots, 7. The 8-point discrete Fourier transform (DFT) of x[n]x[n] is defined as

X[k]=∑n=07x[n]e−jπkn4,k=0,1,⋯ ,7.X[k] = \sum_{n=0}^{7} x[n]e^{-j\frac{\pi kn}{4}}, \quad k = 0, 1, \cdots, 7.

Which one of the following statements is TRUE?

  • (A)
    All X[k]X[k] are non-zero.
  • (B)
    Only X[4]X[4] is non-zero.
  • (C)
    Only X[2]X[2] and X[6]X[6] are non-zero.
  • (D)
    Only X[3]X[3] and X[5]X[5] are non-zero.

Answer (official key): C

Solution

x[n]=cos⁡(πn2)=cos⁡(2π⋅2n8)x[n] = \cos\left(\frac{\pi n}{2}\right) = \cos\left(\frac{2\pi \cdot 2n}{8}\right), which is exactly the k=2k = 2 bin (and its mirror k=6k = 6). All other bins are zero.

Q57 · Communications · Numerical · 2 marks

SPM(t)S_{PM}(t) and SFM(t)S_{FM}(t) as defined below, are the phase modulated and the frequency modulated waveforms, respectively, corresponding to the message signal m(t)m(t) shown in the figure.

SPM(t)=cos⁡(1000πt+Kpm(t))S_{PM}(t) = \cos\left(1000\pi t + K_pm(t)\right)

and

SFM(t)=cos⁡(1000πt+Kf∫−∞tm(τ)dτ)S_{FM}(t) = \cos\left(1000\pi t + K_f\int_{-\infty}^{t}m(\tau)d\tau\right)

where KpK_p is the phase deviation constant in radians/volt and KfK_f is the frequency deviation constant in radians/second/volt. If the highest instantaneous frequencies of SPM(t)S_{PM}(t) and SFM(t)S_{FM}(t) are same, then the value of the ratio KpKf\frac{K_p}{K_f} is ______ seconds.

Figure for GATE 2020 EC question 57 (Communications)

Answer (official key): 2

Solution

The highest PM frequency uses the largest positive slope of m(t)m(t), 102=5\frac{10}{2} = 5 V/s; the highest FM frequency uses the peak value, 10 V. Equating 5Kp=10Kf5K_p = 10K_f gives KpKf=2\frac{K_p}{K_f} = 2 s.

The other 60 questions are solved in your report when you take GATE 2020 EC as a 3-hour test.

Take GATE 2020 EC as a test

Official answer key

GATE 2020 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 2020 EC answer key (PDF)
QTopicTypeMarksAnswer
1Verbal AptitudeMCQ1A
2Verbal AptitudeMCQ1B
3Verbal AptitudeMCQ1B
4Verbal AptitudeMCQ1C
5Quantitative AptitudeMCQ1A
6Analytical AptitudeMCQ2A
7Quantitative AptitudeMCQ2B
8Spatial AptitudeMCQ2C
9Quantitative AptitudeMCQ2C
10Analytical AptitudeMCQ2B
11Linear AlgebraMCQ1C
12CalculusMCQ1C
13CalculusMCQ1B
14Differential EquationsMCQ1C
15Networks, Signals and SystemsMCQ1C
16Electronic DevicesMCQ1B
17Electronic DevicesMCQ1B
18Analog CircuitsMCQ1D
19Networks, Signals and SystemsMCQ1C
20Digital CircuitsMCQ1A
21Control SystemsMCQ1B
22CommunicationsMCQ1D
23ElectromagneticsMCQ1A
24Networks, Signals and SystemsMCQ1A
25Networks, Signals and SystemsNumerical148
26Networks, Signals and SystemsNumerical12.8 to 2.85
27Analog CircuitsNumerical1644 to 657
28Analog CircuitsNumerical16
29Digital CircuitsNumerical114
30Digital CircuitsNumerical13.05 to 3.08
31ElectromagneticsNumerical16.25
32CommunicationsNumerical10.5
33Control SystemsNumerical1160
34CommunicationsNumerical16
35Probability and StatisticsNumerical10.25
36Linear AlgebraMCQ2A
37Differential EquationsMCQ2A
38Networks, Signals and SystemsMCQ2C
39Networks, Signals and SystemsMCQ2C
40Networks, Signals and SystemsMCQ2A
41Electronic DevicesMCQ2Marks to all
42Electronic DevicesMCQ2A
43Electronic DevicesMCQ2Marks to all
44Electronic DevicesMCQ2A
45Analog CircuitsMCQ2A
46Analog CircuitsMCQ2C
47Analog CircuitsMCQ2B
48Digital CircuitsMCQ2A
49Digital CircuitsMCQ2A
50Control SystemsMCQ2A
51Analog CircuitsMCQ2D
52CommunicationsMCQ2B
53ElectromagneticsMCQ2A
54Analog CircuitsNumerical2800
55ElectromagneticsNumerical21
56ElectromagneticsNumerical20.8
57CommunicationsNumerical22
58CommunicationsNumerical23
59Control SystemsNumerical23.95 to 4.05
60Digital CircuitsNumerical276 to 77
61CalculusNumerical22.25
62Networks, Signals and SystemsNumerical258.5 to 58.8
63Networks, Signals and SystemsNumerical2-2
64Probability and StatisticsNumerical20.299 to 0.301
65Control SystemsNumerical230

Questions about GATE 2020 EC

How many questions are in the GATE 2020 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 42 MCQs, 23 numerical answer (NAT) questions. The paper lasted 3 hours.

Is there negative marking in GATE 2020 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 2020 EC?

Outside General Aptitude, the biggest topics were Networks, Signals and Systems (15 marks), Analog Circuits (13 marks), Electronic Devices (10 marks). The full topic-wise split is in the table on this page.

Were any GATE 2020 EC questions awarded marks to all?

Yes. The official key awarded full marks to everyone for questions 41, 43.

Where does this GATE 2020 EC answer key come from?

From the official answer key published by IIT Delhi, which organised GATE 2020. 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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