Official previous-year papers are the closest practice to the real GATE ECE exam. This page covers where the Electronics and Communication Engineering (EC) papers and answer keys are published, the current pattern, how to use the papers by section, and 12 original practice questions with worked solutions. For structured practice, see the Myndaq GATE EC course.
The GATE EC paper pattern
As published by IIT Madras on the official pattern page and in the information brochure, read on 16 September 2026:
- 65 questions, 100 marks, 3 hours - computer-based, in a forenoon (9:30 AM to 12:30 PM) or afternoon (2:30 PM to 5:30 PM) session
- Marks split for EC - General Aptitude 15 marks from 10 questions; Engineering Mathematics 13 marks and EC subject questions 72 marks from the other 55
- Question types - each worth 1 or 2 marks: MCQ (four options, one correct), MSQ (four options, one or more correct) and NAT (a number typed on a virtual keypad to the decimal places the question states)
- Negative marking - 1/3 mark for a wrong 1-mark MCQ and 2/3 for a wrong 2-mark MCQ; none on MSQ or NAT, and no partial marks on MSQ
- Calculator - the on-screen virtual calculator only
- Score validity - three years from the announcement of results
See also the GATE marking scheme guide and NAT and virtual calculator strategy.
Where the official EC papers and answer keys are
- Official GATE Downloads page (IIT Madras) - EC question papers with answer keys for each year from 2021 to 2026, plus a bulk download of papers from 2007 to 2026.
- GATE 2026 papers and keys page (IIT Guwahati) - the EC master question paper and answer key, with a note that on-screen question and option order may differ from the master paper, so read each key with its own paper.
This page reproduces no official question. Two checks first:
- Match old questions to the current syllabus. IIT Madras has published revised syllabi, so set aside any question whose topic the official EC syllabus PDF no longer lists.
- Trust the official key. Check any other worked solution against it before you learn from it.
Score full papers against the key with the GATE marks calculator.
How to use previous-year questions, section by section
Solve one section at a time untimed first, then sit full three-hour papers and log each lost mark by section, question type and cause. GATE publishes no marks split across the EC subject sections, so count it from recent papers, as the GATE EC syllabus guide explains.
- Engineering Mathematics - finish every calculation; later sections reuse these methods.
- Networks, Signals and Systems - redraw each circuit and name the transform you use.
- Electronic Devices - keep a constants sheet; powers of ten cost marks.
- Analog Circuits - bias point and region first, then the small-signal model.
- Digital Circuits - redo minimisation and timing by hand; the syllabus also lists memories and computer organisation.
- Control Systems - drill Routh tables, root locus and Bode sketches.
- Communications - list the power, bandwidth and capacity formulas you got wrong.
- Electromagnetics - practise reflection, matching and wave numericals often.
Practice questions in the GATE EC style
The 12 questions below are original practice questions written by Myndaq in the official GATE style. They are not official past-paper questions.
Q1 - Engineering Mathematics, MCQ, 1 mark
C is the circle |z| = 2, traversed anticlockwise. The integral of f(z) = (z + 1) / (z(z − 3)) around C is:
- (A) −2πi/3
- (B) 2πi/3
- (C) 2πi
- (D) 0
Answer: (A). Of the poles z = 0 and z = 3, only z = 0 lies inside C. Its residue is (0 + 1)/(0 − 3) = −1/3, so the integral is 2πi × (−1/3) = −2πi/3. Option (C) also counts z = 3.
Q2 - Networks, Signals and Systems, NAT, 2 marks
A 12 V DC source in series with a 4 Ω resistor is connected across a 12 Ω resistor, and a variable load R is connected across the 12 Ω resistor. What is the maximum power, in watts, that R can receive? Give the answer rounded off to two decimal places.
Answer: 6.75. The Thevenin voltage is 12 × 12/(4 + 12) = 9 V, and the Thevenin resistance is 4 Ω in parallel with 12 Ω, which is 3 Ω. Maximum power occurs at R = 3 Ω and equals 9²/(4 × 3) = 81/12 = 6.75 W.
Q3 - Networks, Signals and Systems, MSQ, 2 marks
With input x(t) and output y(t), which systems are both linear and time-invariant?
- (A) y(t) = x(t − 2)
- (B) y(t) = t · x(t)
- (C) y(t) = x(t) + 1
- (D) y(t) = the integral of x(τ) from −∞ to t
Answer: (A) and (D). A delay and a running integrator are both LTI. (B) is linear but time-varying: a shifted input gives t · x(t − t₀), but the shifted output is (t − t₀) · x(t − t₀). (C) is not linear, because a zero input gives an output of 1.
Q4 - Electronic Devices, NAT, 2 marks
Silicon at 300 K is uniformly doped with 10¹⁶ donors per cm³, all ionised. Take nᵢ = 10¹⁰ cm⁻³, electron mobility 1250 cm²/V·s and q = 1.6 × 10⁻¹⁹ C. What is the resistivity in Ω·cm? Give the answer rounded off to two decimal places.
Answer: 0.50. n = 10¹⁶ cm⁻³ and p = nᵢ²/n = 10⁴ cm⁻³, which is negligible. The conductivity is qnμₙ = 1.6 × 10⁻¹⁹ × 10¹⁶ × 1250 = 2.0 S/cm, so the resistivity is 1/2.0 = 0.50 Ω·cm.
Q5 - Analog Circuits, NAT, 1 mark
An ideal op-amp inverting summer has a 100 kΩ feedback resistor. V₁ = 0.2 V is applied through 10 kΩ and V₂ = −0.5 V through 20 kΩ. What is the output in volts? Give the answer rounded off to one decimal place.
Answer: 0.5. Vₒ = −100 kΩ × (0.2/10 kΩ − 0.5/20 kΩ) = −100 kΩ × (20 µA − 25 µA) = +0.5 V.
Q6 - Analog Circuits, MSQ, 2 marks
A common-emitter BJT amplifier is biased at I_C = 1 mA, with β = 100, thermal voltage 25 mV, a 5 kΩ collector resistor, a fully bypassed emitter resistor, no Early effect and no load. Which statements are correct?
- (A) The transconductance is 40 mA/V
- (B) r_π is 2.5 kΩ
- (C) The midband gain from base to collector is −200
- (D) r_e is about 2.5 kΩ
Answer: (A), (B) and (C). g_m = 1 mA/25 mV = 40 mA/V; r_π = β/g_m = 100/(40 mA/V) = 2.5 kΩ; the gain is −g_m R_C = −40 mA/V × 5 kΩ = −200. (D) is false: r_e = V_T/I_E, and with I_E about 1.01 mA, r_e is about 25 Ω.
Q7 - Digital Circuits, MCQ, 2 marks
The 8-bit two's complement numbers 01111010 and 00001100 are added. The result, read as two's complement, and the overflow status are:
- (A) +134, no overflow
- (B) −122, overflow
- (C) −122, no overflow
- (D) +6, overflow
Answer: (B). The operands are +122 and +12, and their sum is 10000110. Two positive operands have given sign bit 1, so overflow has occurred. 10000110 is −(01111001 + 1) = −01111010 = −122; the true sum, +134, exceeds +127.
Q8 - Control Systems, NAT, 2 marks
A unity negative-feedback system has G(s) = K / (s(s + 4)). For what value of K is the closed-loop damping ratio 0.5? Give the answer as an integer.
Answer: 16. The characteristic equation is s² + 4s + K = 0. Matching it with s² + 2ζωₙs + ωₙ² gives 2ζωₙ = 4, so with ζ = 0.5, ωₙ = 4 rad/s and K = ωₙ² = 16.
Q9 - Communications, MCQ, 1 mark
An additive white Gaussian noise channel has 3 kHz bandwidth and a signal-to-noise power ratio of 255. Its Shannon capacity is:
- (A) 8 kbps
- (B) 12 kbps
- (C) 24 kbps
- (D) 48 kbps
Answer: (C). C = B log₂(1 + S/N) = 3000 × log₂256 = 3000 × 8 = 24,000 bits per second.
Q10 - Communications, MSQ, 2 marks
s(t) = 10[1 + 0.5 cos(2π × 10³ t)] cos(2π × 10⁶ t) volts. Which statements are correct?
- (A) The modulation index is 0.5
- (B) The transmission bandwidth is 1 kHz
- (C) Each sideband component has amplitude 2.5 V
- (D) The sidebands together carry one-ninth of the total power
Answer: (A), (C) and (D). μ = 0.5, and each sideband has amplitude μA_c/2 = 2.5 V, at 999 kHz and 1001 kHz, so the bandwidth is 2 kHz and (B) is false. In a 1 Ω load the carrier power is 10²/2 = 50 W and each sideband carries 2.5²/2 = 3.125 W, so the sideband share is 6.25/56.25 = 1/9.
Q11 - Electromagnetics, NAT, 2 marks
A lossless 50 Ω line feeds a 150 Ω resistive load. What characteristic impedance, in ohms, must a quarter-wave matching transformer have? Give the answer rounded off to two decimal places.
Answer: 86.60. A quarter-wave section presents Z_T²/Z_L, so matching needs Z_T = √(50 × 150) = √7500 = 86.60 Ω.
Q12 - Electromagnetics, MCQ, 1 mark
A 300 MHz uniform plane wave travels in a lossless non-magnetic medium of relative permittivity 4. With c = 3 × 10⁸ m/s, its wavelength in the medium is:
- (A) 0.25 m
- (B) 0.5 m
- (C) 1 m
- (D) 2 m
Answer: (B). The phase velocity is c/√4 = 1.5 × 10⁸ m/s, so λ = 1.5 × 10⁸ / (3 × 10⁸) = 0.5 m. Option (C) is the free-space value.
Quick answers
Where can I download GATE ECE previous year papers officially?
From the Downloads page of the official GATE website (EC papers and keys for 2021 to 2026, plus a bulk download of papers from 2007 to 2026) and, for 2026, from IIT Guwahati's GATE 2026 master question papers and answer keys page.
Are these practice questions from real GATE papers?
No. They are original Myndaq practice questions in the GATE style.
How many marks does Engineering Mathematics carry in GATE EC?
13 marks, per the official pattern page.
Is there negative marking on MSQ and NAT questions?
No. Only wrong MCQ answers lose marks, and MSQs get no partial marks.
