GATE EC 2024 Previous Year Question Paper with Solutions

 

GATE 2024 Electronics and Communication Engineering (EC) Previous Year Paper

Original PYQs • Answer Checking • Step-by-Step Solutions

Introduction to GATE EC 2024 Previous Year Paper

If you are preparing for the GATE Electronics and Communication Engineering (EC) examination, practicing previous year question papers is an important part of your preparation. This post provides the GATE 2024 Electronics and Communication Engineering (EC) Previous Year Question Paper for practice. These questions are taken from the actual GATE examination and will help you understand the exam pattern, difficulty level, question format, and important concepts.

GATE 2024 was organized by IISc Bengaluru. The EC paper covers important subjects such as Engineering Mathematics, Networks, Signals and Systems, Electronic Devices, Analog Circuits, Digital Circuits, Control Systems, Electromagnetics, and Communication Systems. Practicing this paper will help you understand how theoretical concepts are applied to numerical and conceptual questions in the examination.

The purpose of this page is not only to provide previous year questions but also to offer an effective practice experience. Students can attempt the questions independently, check their answers, and understand their mistakes with the help of detailed solutions. Regular practice can improve conceptual clarity, problem-solving skills, calculation speed, and confidence.

GATE 2024 EC Examination Details

100 Marks Total Marks

3 Hours Exam Duration

15 Marks General Aptitude

85 Marks EC Subject Section

Important: The GATE 2024 EC paper included Multiple Choice Questions (MCQ), Multiple Select Questions (MSQ), and Numerical Answer Type (NAT) questions. Questions could carry either 1 mark or 2 marks. Negative marking applies to MCQs, while MSQ and NAT questions do not have negative marking.

How to Practice This Previous Year Paper

1. Read each question carefully and identify the important information provided.
2. Apply the appropriate formula, concept, or problem-solving method to solve the question.
3. For Numerical Answer Type (NAT) questions, enter your numerical answer and click the Check Answer button.
4. For MCQ and MSQ questions, select the correct option. Remember that MSQ questions may have more than one correct option.
5. If your answer is incorrect, open the solution and understand whether the mistake occurred in calculation or concept.
6. After completing the paper, identify the topics that need more practice and revise them accordingly.

Benefits of Practicing GATE EC Previous Year Questions

Regular practice of previous year questions helps students improve their problem-solving ability, calculation speed, and conceptual understanding. It also provides a better understanding of the examination pattern and helps students develop effective time-management skills.

Instead of memorizing answers, focus on understanding the concepts behind every question. Review the solutions after attempting the questions and practice additional problems related to the same topics. Consistent practice can help you prepare more effectively for the GATE EC examination.

Strengthen Your GATE EC Preparation!
Attempt the GATE 2024 EC Previous Year Paper, check your answers, and learn from every question.

Question 1

If ‘→’ denotes increasing order of intensity, then the meaning of the words [charm → enamor → bewitch] is analogous to [bored → _______ → weary]. Which one of the given options is appropriate to fill the blank?

Correct Answer: (A) jaded

Solution: “Jaded” means tired or bored, especially after too much of something. It fits the progression from “bored” toward “weary.”

Question 2

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,

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

Correct Answer: (A) P and T

Solution: Q and S are single children, so neither can be part of the sibling pair. In both room arrangements, P and T occupy different rooms. Therefore, P and T are the siblings.

Question 3

Five years ago, the ratio of Aman’s age to his father’s age was 1:4, and five years from now, the ratio will be 2:5. What was his father’s age when Aman was born?

Correct Answer: (B) 30 years

Solution: Let Aman’s present age be A and his father’s present age be F.

Five years ago:

(A − 5)/(F − 5) = 1/4

Therefore, F = 4A − 15.

Five years from now:

(A + 5)/(F + 5) = 2/5

Therefore, 5A + 25 = 2F + 10.

Solving both equations gives A = 15 and F = 45.

Father’s age when Aman was born = 45 − 15 = 30 years.

Question 4

For a real number x > 1,

1/log₂x + 1/log₃x + 1/log₄x = 1

The value of x is

Correct Answer: (C) 24

Solution: Using the identity 1/logₐx = logₓa, the given equation becomes:

logₓ2 + logₓ3 + logₓ4 = 1

Using the logarithm property:

logₓ(2 × 3 × 4) = 1

logₓ24 = 1

Therefore, x = 24.

Question 5

The greatest prime factor of (3¹⁹⁹ − 3¹⁹⁶) is

Correct Answer: (A) 13

Solution: Factor out the common term 3¹⁹⁶:

3¹⁹⁹ − 3¹⁹⁶ = 3¹⁹⁶(3³ − 1)

= 3¹⁹⁶(27 − 1)

= 3¹⁹⁶ × 26

= 3¹⁹⁶ × 2 × 13

The prime factors are 2, 3, and 13. The greatest prime factor is 13.

Question 6

Sequence the following sentences (P, Q, R, S) in a coherent passage:

P: Shifu’s student exclaimed, “Why do you run since the bull is an illusion?”

Q: Shifu said, “Surely my running away from the bull is also an illusion.”

R: Shifu once proclaimed that all life is illusion.

S: One day, when a bull gave him chase, Shifu began running for his life.

Correct Answer: (C) RSPQ

Solution:

First, Shifu proclaims that all life is an illusion (R). Then, a bull chases him and he runs (S). His student questions him (P), and Shifu replies (Q).

Therefore, the coherent sequence is RSPQ.

Question 7

Four identical cylindrical chalk-sticks, each of radius 𝑟 = 0.5 cm and length 𝑙 = 10 cm, are bound tightly together using a duct tape as shown in the following figure.

The width of the duct tape is equal to the length of the chalk-stick. The area (in cm²) of the duct tape required to wrap the bundle of chalk-sticks once, is

Correct Answer: (D) 10 (4 + π)

Solution:

The four identical cylindrical chalk-sticks are tightly bound together.

The perimeter of the cross-section of the bundle is:

Perimeter = 8r + 2πr

Given:

r = 0.5 cm

l = 10 cm

Perimeter = 8(0.5) + 2π(0.5)

= 4 + π cm

Area of duct tape = Perimeter × Length

= (4 + π) × 10

Area = 10(4 + π) cm²

Question 8

The bar chart shows the percentage population falling into categories based on BMI in 2003 and 2023. Which option is INCORRECT?

Correct Answer: (C)

Solution:

The ratio of obese population to normal population has increased from 2003 to 2023. Therefore, the statement that this ratio has decreased is incorrect.

Question 9

Examples of mirror and water reflections are shown in the figures below:

Correct Answer: (A)

Solution:

The combined effect of reflection in a mirror followed by reflection on water is equivalent to a 180° rotation of the original object. Rotating the given final image by 180° gives option (A).

Question 10

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

Correct Answer: (A) 14:11

Solution:

Sheet A:

Initial dimensions = 24 cm × 16 cm

After folding twice using FO1:

Final dimensions = 24 cm × 4 cm

Perimeter of A = 2(24 + 4) = 56 cm

Sheet B:

After folding twice using FO2:

Final dimensions = 6 cm × 16 cm

Perimeter of B = 2(6 + 16) = 44 cm

Required ratio:

56:44 = 14:11

Question 11

The general form of the complementary function of a differential equation is given by

y(t) = (At + B)e^−2t,

where A and B are real constants determined by the initial condition. The corresponding differential equation is _____.

Correct Answer: (A)

Solution:

The complementary function is

y(t) = (At + B)e^−2t

This corresponds to a repeated characteristic root s = −2.

Therefore, the characteristic equation is

(s + 2)² = s² + 4s + 4 = 0

Hence, the corresponding differential equation is

d²y/dt² + 4 dy/dt + 4y = f(t)

Question 12

In the context of Bode magnitude plots, 40 dB/decade is the same as _______.

Correct Answer: (A) 12 dB/octave

Solution:

One decade corresponds to a frequency ratio of 10, while one octave corresponds to a frequency ratio of 2.

The number of octaves in one decade is

log₂(10) ≈ 3.322

Therefore,

40 / 3.322 ≈ 12.04 dB/octave

The closest given option is 12 dB/octave.

Question 13

In the feedback control system shown in the figure below,

G(s) = 6 / [s(s + 1)(s + 2)]

R(s), Y(s), and E(s) are the Laplace transforms of r(t), y(t), and e(t), respectively.

If the input r(t) is a unit step function, then __________.

Correct Answer: (A)

Solution:

The system has unity negative feedback and the open-loop transfer function contains one pole at the origin.

Therefore, it is a Type-1 system.

For a stable unity-feedback Type-1 system, the steady-state error for a unit-step input is zero.

lim e(t) = 0, as t → ∞

Question 14

A digital communication system transmits through a noiseless bandlimited channel [−W, W]. The received signal z(t) at the output of the receiving filter is given by

z(t) = ∑ b[n]x(t − nT)

where b[n] are the symbols and x(t) is the overall system response to a single symbol. The received signal is sampled at t = mT. The Fourier transform of x(t) is X(f). The Nyquist condition that X(f) must satisfy for zero intersymbol interference at the receiver is ______.

Correct Answer: (A)

Solution:

For zero intersymbol interference, the Nyquist frequency-domain condition is

∑_m=−∞^∞ X(f + m/T) = T

The frequency shifts are by multiples of 1/T, and the sum must equal T.

Question 15

Consider a lossless transmission line terminated with a short circuit as shown in the figure below. As one moves towards the generator from the load, the normalized impedances z_inA, z_inB, z_inC, and z_inD (indicated in the figure) are _______.

Correct Answer: (A)

Solution:

For a lossless transmission line terminated in a short circuit, the normalized input impedance is

z_in = j tan(βl)

At a distance λ/8 from the load:

z_inA = j tan(π/4) = +j

At a distance λ/4 from the load:

z_inB = j tan(π/2) = ∞

At a distance 3λ/8 from the load:

z_inC = j tan(3π/4) = −j

At a distance λ/2 from the load:

z_inD = j tan(π) = 0

Therefore, the correct option is (A).

Question 16

Let î and ĵ be the unit vectors along x and y axes, respectively and let A be a positive constant. Which one of the following statements is true for the vector fields

F₁ = A(yî + xĵ) and F₂ = A(yî − xĵ) ?

(A) Both F₁ and F₂ are electrostatic fields.

(B) Only F₁ is an electrostatic field.

(C) Only F₂ is an electrostatic field.

(D) Neither F₁ nor F₂ is an electrostatic field.

Correct Answer: (B) Only F₁ is an electrostatic field.

For an electrostatic field, the curl of the electric field must be zero.

For F₁:

∇ × F₁ = (∂F_1y/∂x − ∂F_1x/∂y) k̂

= A − A = 0

Therefore, F₁ is an electrostatic field.

For F₂:

∇ × F₂ = (−A − A)k̂ = −2Ak̂ ≠ 0

Therefore, F₂ is not an electrostatic field.

Hence, Option (B) is correct.

Question 17

In the circuit below, assume that the long channel NMOS transistor is biased in saturation. The small signal trans-conductance of the transistor is g_m. Neglect body effect, channel length modulation and intrinsic device capacitances. The small signal input impedance Z_in(jω) is _______.

(A)
− g_m / (C₁C_Lω²) + 1/(jωC₁) + 1/(jωC_L)

(B)
g_m / (C₁C_Lω²) + 1/(jωC₁) + 1/(jωC_L)

(C)
1/(jωC₁) + 1/(jωC_L)

(D)
− g_m / (C₁C_Lω²) + 1/(jωC₁ + jωC_L)

Correct Answer: (A)

Using small-signal analysis, apply KCL at the source node.

sC₁(V_s − V_in) + sC_LV_s + g_m(V_s − V_in) = 0

Rearranging and calculating the input impedance:

Z_in(s) = [g_m + s(C₁ + C_L)] / [s²C₁C_L]

Putting s = jω:

Z_in(jω) = −g_m/(C₁C_Lω²) + 1/(jωC₁) + 1/(jωC_L)

Hence, Option (A) is correct.

Question 18

For the closed loop amplifier circuit shown below, the magnitude of open loop low frequency small signal voltage gain is 40. All the transistors are biased in saturation. The current source I_SS is ideal. Neglect body effect, channel length modulation and intrinsic device capacitances. The closed loop low frequency small signal voltage gain

v_out / v_in

(rounded off to three decimal places) is ____.

(A) 0.976

(B) 1.000

(C) 1.025

(D) 0.488

Correct Answer: (A) 0.976

Given:

Open-loop voltage gain, A = 40

For a unity negative feedback amplifier:

A_f = A/(1 + Aβ)

For unity feedback, β = 1.

A_f = 40/(1 + 40)

= 40/41

= 0.975609...

Rounding off to three decimal places:

A_f = 0.976

Hence, Option (A) is correct.

Question 19

For the Boolean function

F(A, B, C, D) = Σm(0,2,5,7,8,10,12,13,14,15),

the essential prime implicants are __________.

(A) BD, B̅D̅

(B) BD, AB

(C) AB, B̅D̅

(D) BD, B̅D̅, AB

Correct Answer: (A) BD, B̅D̅

Given minterms:

F(A,B,C,D) = Σm(0,2,5,7,8,10,12,13,14,15)

Using a 4-variable Karnaugh map, the prime implicants can be obtained by grouping adjacent 1s.

The essential prime implicants are:

1. BD

2. B̅D̅

These two implicants cover minterms that cannot be covered by any other prime implicant.

Therefore, the essential prime implicants are:

BD and B̅D̅

Hence, Option (A) is correct.

Question 20

A white Gaussian noise w(t) with zero mean and power spectral density N₀/2, when applied to a first-order RC low pass filter produces an output n(t). At a particular time t = t_k, the variance of the random variable n(t_k) is ________.

(A) N₀ / 4RC

(B) N₀ / 2RC

(C) N₀ / RC

(D) 2N₀ / RC

Correct Answer: (A) N₀ / 4RC

Given:

Input noise PSD:

S_w(f) = N₀/2

For a first-order RC low pass filter:

H(f) = 1/(1 + j2πfRC)

The output noise variance is:

σ²_n = ∫ S_w(f)|H(f)|² df

= (N₀/2) ∫_−∞^∞ 1/[1 + (2πfRC)²] df

Using the standard integral:

∫_−∞^∞ 1/(1 + a²f²) df = π/a

We get:

σ²_n = N₀/(4RC)

Hence, Option (A) is correct.

Question 21

A causal and stable LTI system with impulse response h(t) produces an output y(t) for an input signal x(t). A signal x(0.5t) is applied to another causal and stable LTI system with impulse response h(0.5t). The resulting output is _____.

Correct Answer: (A) 2y(0.5t)

The output of the second system is the convolution:
z(t) = ∫ x(0.5τ) h(0.5t − 0.5τ) dτ

Put u = 0.5τ, so dτ = 2du.
z(t) = 2 ∫ x(u) h(0.5t − u) du

Therefore,
z(t) = 2y(0.5t)

Question 22

For non-degenerately doped n-type silicon, which one of the following plots represents the temperature (T) dependence of free electron concentration (n)?

 

Correct Answer: (A)

In non-degenerately doped n-type silicon, three temperature regions are observed:

1. At low temperature, the donor atoms are not completely ionized, so the free electron concentration is low.
2. At intermediate temperature, the donor atoms are almost completely ionized and the electron concentration remains approximately constant.
3. At high temperature, intrinsic carrier generation increases the electron concentration.

Since the horizontal axis is 1/T, the correct plot is Option (A).

Question 23

In the circuit shown, the n:1 step-down transformer and the diodes are ideal. The diodes have no voltage drop in forward biased condition. If the input voltage (in Volts) is V_s(t) = 10sinωt and the average value of load voltage V_L(t) (in Volts) is 2.5/π, the value of n is ______.

Correct Answer: (B) 8

The transformer is an n:1 step-down transformer.

Secondary peak voltage:
V_m = 10/n

For a full-wave rectifier, average load voltage is:
V_avg = 2V_m/π

Given:
2 × (10/n)/π = 2.5/π

20/n = 2.5
n = 20/2.5 = 8

Therefore, n = 8.

Question 24

Multiple Select Questions — Select all correct options

For a causal discrete-time LTI system with transfer function

H(z) = (2z² + 3) / [(z + 1/3)(z − 1/3)]

which of the following statements is/are true?

Correct Answers: (A), (C) and (D)

The poles are obtained from:
(z + 1/3)(z − 1/3) = 0
Therefore, poles are z = −1/3 and z = 1/3.

Both poles lie inside the unit circle. Hence, the causal system is stable. So, (A) is correct.

Using the initial value theorem:
h[0] = lim z→∞ H(z) = 2
Hence, (C) is correct.

Since both poles are inside the unit circle, the impulse response approaches zero as n approaches infinity. Hence, (D) is correct.

The zeros are obtained from 2z² + 3 = 0. Their magnitude is greater than 1, so the system is not minimum phase. Therefore, (B) is incorrect.

Question 25

Multiple Select Questions — Select all correct options

Let ρ(x, y, z, t) and u(x, y, z, t) represent density and velocity, respectively, at a point (x, y, z) and time t. Assume ∂ρ/∂t is continuous. Let V be an arbitrary volume in space enclosed by the closed surface S and n̂ be the outward unit normal of S.

Which of the following equations is/are equivalent to

∂ρ/∂t + ∇ · (ρu) = 0 ?

Correct Answers: (A) and (C)

Given continuity equation:
∂ρ/∂t + ∇ · (ρu) = 0

Rearranging:
∂ρ/∂t = −∇ · (ρu)

Integrating over volume V:
∫_V (∂ρ/∂t) dv = −∫_V ∇ · (ρu) dv

Therefore, option (C) is correct.

Using the divergence theorem:
∫_V ∇ · (ρu) dv = ∫_S ρu · n̂ ds

Hence:
∫_V (∂ρ/∂t) dv = −∫_S ρu · n̂ ds

Therefore, option (A) is also correct.

Question 26

The free electron concentration profile n(x) in a doped semiconductor at equilibrium is shown in the figure, where the points A, B, and C mark three different positions. Which of the following statements is/are true?

At equilibrium, the total electron current is zero. The electron diffusion current density is: J_n,diff = qD_n (dn/dx) Between B and C, n(x) decreases with x. Therefore, dn/dx is negative, and the electron diffusion current is directed from C to B. Statement (A) is true. Between B and A, n(x) increases with x. The electric field is directed opposite to the concentration gradient. Hence, the electric field is directed from B to A, and the electron drift current is also directed from B to A. Statement (B) is true. Between B and C, the electric field is directed from B to C. Statement (C) is true. Statement (D) is false. Correct Answers: (A), (B), (C)

Question 27

A machine has a 32-bit architecture with 1-word long instructions. It has 24 registers and supports an instruction set of size 40. Each instruction has five distinct fields, namely opcode, two source register identifiers, one destination register identifier, and an immediate value. Assuming that the immediate operand is an unsigned integer, its maximum value is ________.

The number of bits required for the opcode is: ceil(log₂40) = 6 bits The number of bits required for each register identifier is: ceil(log₂24) = 5 bits There are three register identifier fields: 3 × 5 = 15 bits Total bits used by opcode and register fields: 6 + 15 = 21 bits Therefore, the immediate field has: 32 − 21 = 11 bits Since the immediate operand is unsigned, its maximum value is: 2¹¹ − 1 = 2047 Answer: 2047

Question 28

An amplitude modulator has output (in Volts)

s(t) = A cos(400πt) + B cos(360πt) + B cos(440πt).

The carrier power normalized to 1 Ω resistance is 50 Watts. The ratio of the total sideband power to the total power is 1/9. The value of B (in Volts, rounded off to two decimal places) is _______.

The carrier power is: P_c = A²/2 = 50 The total sideband power is: P_SB = B²/2 + B²/2 = B² Given: P_SB / (P_c + P_SB) = 1/9 Therefore: B² / (50 + B²) = 1/9 9B² = 50 + B² 8B² = 50 B² = 6.25 B = 2.50 V Answer: 2.50 V

Question 29

In a number system of base r, the equation x² − 12x + 37 = 0 has x = 8 as one of its solutions. The value of r is _______.

In base r: 12 = r + 2 37 = 3r + 7 Since x = 8 is one root, substituting it into the equation: 8² − 8(r + 2) + (3r + 7) = 0 64 − 8r − 16 + 3r + 7 = 0 55 − 5r = 0 5r = 55 r = 11 Answer: 11

Question 30

Let ℝ and ℝ³ denote the set of real numbers and the three dimensional vector space over it, respectively. The value of α for which the set of vectors

{[2 −3 α], [3 −1 3], [1 −5 7]}

does not form a basis of ℝ³ is _______.

The vectors do not form a basis when the determinant of the matrix formed by them is zero.

det
| 2 −3 α |
| 3 −1 3 |
| 1 −5 7 | = 0

Expanding the determinant: 70 − 14α = 0 14α = 70 α = 5 Answer: 5

Question 31

In the given circuit, the current I_x (in mA) is ______.

The current I_x is determined using Kirchhoff’s Current Law (KCL) and the dependent source relation in the given circuit. Answer: −3 mA

Question 32

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time t = 0. The time constant (in seconds) of the circuit for t > 0 is ______.

The time constant of an RL circuit is: τ = L/R_eq The equivalent resistance is obtained from the circuit for t > 0. Answer: 0.5 s

Question 33

Suppose X and Y are independent and identically distributed random variables that are distributed uniformly in the interval [0,1]. The probability that X ≥ Y is ______.

Since X and Y are independent and identically distributed continuous random variables, the probabilities P(X ≥ Y) and P(Y ≥ X) are equal. Also, P(X ≥ Y) + P(Y ≥ X) = 1 Therefore, P(X ≥ Y) = 1/2 = 0.5 Answer: 0.5

Question 34

A source transmits symbols from an alphabet of size 16. The value of maximum achievable entropy (in bits) is ______.

The maximum achievable entropy for an alphabet of size M occurs when all symbols are equally likely. H_max = log₂M Given M = 16: H_max = log₂16 = 4 bits Answer: 4

 

Question 35

As shown in the circuit, the initial voltage across the capacitor is 10 V, with the switch being open. The switch is then closed at t = 0. The total energy dissipated in the ideal Zener diode (V_Z = 5 V) after the switch is closed (in mJ, rounded off to three decimal places) is __________. 

The energy initially stored in the capacitor is: E_initial = 1/2 CV² The Zener diode clamps the capacitor voltage at V_Z = 5 V. The energy dissipated in the Zener diode is calculated from the charge discharged while the voltage decreases from 10 V to 5 V. Answer: 0.250 mJ

Question 36

Consider the Earth to be a perfect sphere of radius R. Then the surface area of the region, enclosed by the 60°N latitude circle, that contains the north pole in its interior is ______.

Correct Answer: (A)

Surface area of a spherical cap is:
A = 2πRh

For latitude 60°N, the height of the cap is:
h = R − R sin 60°
h = R(1 − √3/2)

Therefore,
A = 2πR × R(1 − √3/2)
A = 2πR² − √3πR²

A = (2 − √3)πR²

Question 37

Consider a unity negative feedback control system with forward path gain

G(s) = K / [(s + 1)(s + 2)(s + 3)]

The impulse response of the closed-loop system decays faster than e^−t if ______.

Correct Answer: (A) 1 ≤ K ≤ 5

Closed-loop characteristic equation:
1 + G(s) = 0

(s + 1)(s + 2)(s + 3) + K = 0

Expanding:
s³ + 6s² + 11s + 6 + K = 0

For the impulse response to decay faster than e^−t, all poles must have real parts less than −1.

Put s = p − 1:
p³ + 3p² + 2p + K = 0

Applying the Routh-Hurwitz stability condition:
0 < K < 6

From the given options, the correct range is:
1 ≤ K ≤ 5

Question 38

A satellite attitude control system, as shown below, has a plant with transfer function G(s) = 1/s² cascaded with a compensator C(s) = K(s + α)/(s + 4), where K and α are positive real constants.

In order for the closed-loop system to have poles at −1 ± j√3, the value of α must be ______.

Correct Answer: (B) 1

Given:
G(s) = 1/s²
C(s) = K(s + α)/(s + 4)

For unity negative feedback, characteristic equation is:
1 + C(s)G(s) = 0

s²(s + 4) + K(s + α) = 0

Expanding:
s³ + 4s² + Ks + Kα = 0

Desired poles are −1 ± j√3.
Their quadratic factor is:
(s + 1)² + 3 = s² + 2s + 4

Let the third pole be −p. Comparing coefficients gives p = 2 and K = 8.

Comparing constant terms:
Kα = 4p = 8
8α = 8
α = 1

Question 39

A uniform plane wave with electric field

E(x) = A_y â_y e^−j2πx/3 V/m

is travelling in the air (relative permittivity, ε_r = 1 and relative permeability, μ_r = 1) in the +x direction (A_y is a positive constant, â_y is the unit vector along the y axis). It is incident normally on an ideal electric conductor (conductivity, σ = ∞) at x = 0. The position of the first null of the total magnetic field in the air (measured from x = 0, in metres) is ______.

Correct Answer: (A) −3/4

Given propagation constant:
β = 2π/3 rad/m

Wavelength:
λ = 2π/β
λ = 2π/(2π/3) = 3 m

For a perfect electric conductor, the reflected wave produces a standing magnetic field. The first magnetic field null occurs at a distance λ/4 from the conducting surface.

Distance = λ/4 = 3/4 m

Since the null is on the negative x side of the origin:
x = −3/4 m

Question 40

A 4-bit priority encoder has inputs D₃, D₂, D₁, and D₀ in descending order of priority. The two-bit output AB is generated as 00, 01, 10, and 11 corresponding to inputs D₃, D₂, D₁, and D₀, respectively. The Boolean expression of the output bit B is ______.

Correct Answer: (B)

Priority order is:
D₃ > D₂ > D₁ > D₀

The output codes are:
D₃ → 00
D₂ → 01
D₁ → 10
D₀ → 11

The output bit B is 1 when D₂ is the highest active input or when D₀ is the highest active input.

Therefore,
B = D̅₃D₂ + D̅₃D̅₂D̅₁

Simplifying:
B = D̅₃(D₂ + D̅₁)

Correct option: (B)

Question 41

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

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

Correct Answer: (B) 50 MHz

When S = 1, the MUX selects input 1. Due to the feedback connection through the inverter, the output toggles after every MUX propagation delay.

Given:
MUX propagation delay = 10 ns
Inverter propagation delay = 0 ns

Time for one transition = 10 ns
Time period = 2 × 10 = 20 ns

Frequency:
f = 1/T
f = 1/(20 × 10⁻⁹)
f = 50 MHz

Question 42

The sequence of states (Q₁ Q₀) of the given synchronous sequential circuit is ______.

Correct Answer: (B)

The circuit consists of two T flip-flops connected through combinational logic.

By applying the T flip-flop characteristic equation:
Q(next) = T ⊕ Q

The state transitions corresponding to the correct sequence are:

11 → 00
00 → 10
10 → 01
01 → 00

Therefore, the required sequence is:
11 → 00 → 10 → 01 → 00

Question 43

Let z be a complex variable. If

f(z) = sin(πz) / [z²(z − 2)]

and C is the circle in the complex plane with |z| = 3 then

∮_C f(z) dz is ______.

Correct Answer: (D) −π²j

Given:
f(z) = sin(πz) / [z²(z − 2)]

The poles inside |z| = 3 are z = 0 and z = 2.

At z = 0, there is a pole of order 2.

Residue at z = 0:
Res(f,0) = d/dz [sin(πz)/(z − 2)] at z = 0

= −π/2

At z = 2, the residue is:
Res(f,2) = sin(2π)/4 = 0

By the residue theorem:
∮ f(z) dz = 2πj × (sum of residues)

= 2πj × (−π/2)

= −π²j

Question 44

Consider two continuous time signals x(t) and y(t) as shown below.

If X(f) denotes the Fourier transform of x(t), then the Fourier transform of y(t) is ______.

Correct Answer: (B)

From the given signals:

y(t) = −x((t − 2)/4)

Fourier transform scaling property:
x(t/a) ↔ |a|X(af)

Time shifting property:
x(t − t₀) ↔ X(f)e^−j2πft₀

Applying both properties:

Y(f) = −4X(4f)e^−j2πf(2)

Y(f) = −4X(4f)e^−j4πf

Question 45

A source transmits a symbol s, taken from {−4, 0, 4} with equal probability, over an additive white Gaussian noise channel. The received noisy symbol r is given by r = s + w, where the noise w is zero mean with variance 4 and is independent of s.

Using
Q(x) = (1/√(2π)) ∫_x^∞ e^−t²/2 dt,

the optimum symbol error probability is ______.

Correct Answer: (B) (4/3)Q(1)

Given symbols:
s ∈ {−4, 0, 4}

Noise variance = 4
Therefore, noise standard deviation:
σ = √4 = 2

The optimum decision thresholds are the midpoints:
−2 and +2

For symbol s = −4:
Error probability = Q(1)

For symbol s = 0:
Error probability = 2Q(1)

For symbol s = 4:
Error probability = Q(1)

Since all three symbols are equally probable:

P(e) = (1/3)[Q(1) + 2Q(1) + Q(1)]

P(e) = (4/3)Q(1)

Question 46

A full scale sinusoidal signal is applied to a 10-bit ADC. The fundamental signal component in the ADC output has a normalized power of 1 W, and the total noise and distortion normalized power is 10 μW. The effective number of bits (rounded off to the nearest integer) of the ADC is _______.

Signal power = 1 W and noise plus distortion power = 10 μW.

SINAD = 10 log₁₀(1 / 10⁻⁵) = 50 dB
Effective number of bits:
ENOB = (SINAD − 1.76) / 6.02
ENOB = (50 − 1.76) / 6.02 ≈ 8.01

Rounded off to the nearest integer = 8.
Correct Answer: (B) 8

Question 47

The information bit sequence {1 1 1 0 1 0 1 0 1} is to be transmitted by encoding with Cyclic Redundancy Check 4 (CRC-4) code, for which the generator polynomial is C(x) = x⁴ + x + 1. The encoded sequence of bits is ________.

CRC-4 encoding uses the generator polynomial:
C(x) = x⁴ + x + 1

The information sequence is divided by the generator polynomial after appending four zeros. The CRC remainder is appended to the original information sequence.

The encoded sequence corresponds to option (A).
Correct Answer: (A)

Question 48

A continuous time signal x(t) = 2 cos(8πt + π/3) is sampled at a rate of 15 Hz. The sampled signal xₛ(t) when passed through an LTI system with impulse response

h(t) = (sin 2πt / πt) cos(38πt − π/2)

produces an output xₒ(t). The expression for xₒ(t) is _____.

According to the official GATE 2024 answer key, this question was awarded Marks to All (MTA).

Therefore, marks were awarded to all candidates irrespective of the selected option.
Official Answer Key: Marks to All

Question 49

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

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

The first op-amp circuit acts as a Schmitt trigger. Its output switches when the input crosses the corresponding threshold levels.

The second op-amp circuit contains an inductor in its input path and a feedback resistor. Considering the inductor current and the saturation limits, the output waveform is trapezoidal.

The correct statement is that V₀₁ is delayed by 1 μs relative to Vᵢ.
Correct Answer: (D)

Question 50

In the circuit below, the opamp is ideal.

If the circuit is to show sustained oscillations, the respective values of R₁ and the corresponding frequency of oscillation are _____.

For sustained oscillations, the Barkhausen criterion must be satisfied:

1. Magnitude of loop gain = 1
2. Total phase shift around the loop = 0° or 360°

Applying these conditions to the given circuit gives:
R₁ = 2R
Frequency of oscillation = 1/(2πRC)

Correct Answer: (B)

Question 51

In the circuit shown below, the transistors M₁ and M₂ are biased in saturation. Their small signal transconductances are gₘ₁ and gₘ₂ respectively. Neglect body effect, channel length modulation and intrinsic device capacitances.

Assuming that capacitor C₁ is a short circuit for AC analysis, the exact magnitude of small signal voltage gain |vout/vin| is ______.

The small-signal resistance of the diode-connected transistor M₁ is approximately:
1/gₘ₁

The input voltage is divided between Rₛ and the series combination of Rᴮ and 1/gₘ₁.

Therefore,
vgs₂/vin = (Rᴮ + 1/gₘ₁)/(Rₛ + Rᴮ + 1/gₘ₁)

The voltage gain magnitude is:
|vout/vin| = gₘ₂Rᴅ (Rᴮ + 1/gₘ₁)/(Rᴮ + 1/gₘ₁ + Rₛ)

Correct Answer: (B)

Question 52

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

**Explanation:**

(A) True — Under high-level injection, β decreases as collector current increases.

(B) False — Under the stated recombination-dominated condition, this statement is not correct.

(C) True — The current gain in saturation is lower than in the forward active region.

(D) True — A higher β generally results in a lower collector-to-emitter breakdown voltage.

Correct Answers: (A), (C), (D)

Question 53

Consider a system S represented in state space as

dx/dt = [0 −2; 1 −3]x + [1; 0]r
y = [2 −5]x

Which of the state space representations given below has/have the same transfer function as that of S?

The transfer function of the original system is:

G(s) = C(sI − A)⁻¹B

On substituting the original matrices:
G(s) = (2s + 1)/[(s + 1)(s + 2)]

The transfer functions of options (A) and (C) are also equal to:
(2s + 1)/[(s + 1)(s + 2)]

Hence, the correct options are A and C.
Correct Answers: (A), (C)

Question 54

Let F₁, F₂, and F₃ be functions of (x, y, z). Suppose that for every given pair of points A and B in space, the line integral

∫C (F₁ dx + F₂ dy + F₃ dz)

evaluates to the same value along any path C that starts at A and ends at B. Then which of the following is/are true?

The given condition states that the line integral is path independent.

(A) True — The line integral over every closed path is zero.

(B) True — A path-independent vector field can be expressed as the gradient of a scalar potential function.

(C) False — Path independence does not require the divergence to be zero.

(D) True — The mixed partial derivative conditions correspond to zero curl.

Correct Answers: (A), (B), (D)

Question 55

Consider the matrix

[1 k; 2 1]

where k is a positive real number. Which of the following vectors is/are eigenvector(s) of this matrix?

For the matrix
A = [1 k; 2 1]

The characteristic equation is:
(1 − λ)² − 2k = 0

Therefore:
λ = 1 ± √(2k)

From the eigenvector equation, the corresponding vectors are proportional to:
[1; √(2/k)] and [1; −√(2/k)]

Hence, options A and B are correct.
Correct Answers: (A), (B)

Question 56

The radian frequency value(s) for which the discrete time sinusoidal signal

x[n] = A cos(Ωn + π/3)

has a period of 40 is/are ______.

A discrete-time sinusoidal signal is periodic if:

Ω/(2π) = m/N

For fundamental period N = 40, m and 40 must be relatively prime.

Option (A):
Ω = 0.15π = 2π(3/40)
gcd(3,40) = 1, so the fundamental period is 40.

Option (D):
Ω = 0.45π = 2π(9/40)
gcd(9,40) = 1, so the fundamental period is also 40.

Correct Answers: (A), (D)

Q.57 | GATE EC 2024

Let X(t) = A cos(2πf₀t + θ) be a random process, where amplitude A and phase θ are independent of each other, and are uniformly distributed in the intervals [−2, 2] and [0, 2π], respectively. X(t)) is fed to an 8-bit uniform mid-rise type quantizer.

Given that the autocorrelation of X(t) is R_X(τ) = (2/3) cos(2πf₀τ), the signal to quantization noise ratio (in dB, rounded off to two decimal places) at the output of the quantizer is ______.

Solution:

The signal power is obtained from the autocorrelation at τ = 0:
P_X = R_X(0) = 2/3

For an 8-bit uniform mid-rise quantizer with input range −2 to +2:
Δ = (2 − (−2))/2⁸
Δ = 4/256 = 1/64

Quantization noise power:
P_q = Δ²/12
P_q = 1/(64² × 12)
P_q = 1/49152

Therefore:
SQNR = P_X/P_q
SQNR = (2/3) × 49152 = 32768

In dB:
SQNR_dB = 10 log₁₀(32768)
SQNR_dB ≈ 45.15 dB

Correct Answer: 45.15 dB

Q.58 | GATE EC 2024

A lossless transmission line with characteristic impedance Z₀ = 50 Ω is terminated with an unknown load. The magnitude of the reflection coefficient is |Γ| = 0.6.

As one moves towards the generator from the load, the maximum value of the input impedance magnitude looking towards the load (in Ω) is ______.

Solution:

For a lossless transmission line, the maximum impedance magnitude occurs when the impedance is at a voltage maximum.

The maximum impedance is:
Z_max = Z₀(1 + |Γ|)/(1 − |Γ|)

Substituting the given values:
Z_max = 50(1 + 0.6)/(1 − 0.6)
Z_max = 50 × 1.6/0.4
Z_max = 200 Ω

Correct Answer: 200 Ω

Q.59 | GATE EC 2024

The relationship between any N-length sequence x[n] and its corresponding N-point discrete Fourier transform X[k] is defined as

X[k] = ℱ{x[n]}.

Another sequence y[n] is formed as

y[n] = ℱ{ℱ{ℱ{ℱ{x[n]}}}}.

For the sequence x[n] = {1, 2, 1, 3}, the value of Y[0] is ______.

Solution:

For an N-point DFT, applying the DFT twice gives:
ℱ²{x[n]} = N x[−n mod N]

Applying the DFT four times:
ℱ⁴{x[n]} = N²x[n]

Here, N = 4. Therefore:
y[n] = 4²x[n] = 16x[n]

Given:
x[0] = 1

Thus:
Y[0] = y[0] = 16 × 1 = 16

Correct Answer: 16

Q.60 | GATE EC 2024

For the two-port network shown in the circuit, the value of the Y₂₁ parameter (in Siemens) is ______.

Solution:

The Y-parameter relation is:
I₂ = Y₂₁V₁ + Y₂₂V₂

To find Y₂₁, set V₂ = 0:
Y₂₁ = I₂/V₁

Let the common top node voltage be V₂ and the voltage at the top of the dependent source be V_a.

The dependent voltage source has polarity such that:
0 − V_a = 8V₁
V_a = −8V₁

With V₂ = 0, the current entering port 2 is:
I₂ = (V₂ − V₁)/2 + (V₂ − V_a)/4

Substituting:
I₂ = (0 − V₁)/2 + (0 + 8V₁)/4
I₂ = −V₁/2 + 2V₁
I₂ = 1.5V₁

Therefore:
Y₂₁ = I₂/V₁ = 1.5 S

Correct Answer: 1.5 S

Q.61 | GATE EC 2024

Consider a MOS capacitor made with p-type silicon. It has an oxide thickness of 100 nm, a fixed positive oxide charge of 10⁻⁸ C/cm² at the oxide-silicon interface, and a metal work function of 4.6 eV.

Assume that the relative permittivity of the oxide is 4 and the absolute permittivity of free space is 8.85 × 10⁻¹⁴ F/cm.

If the flatband voltage is 0 V, the work function of the p-type silicon (in eV, rounded off to two decimal places) is ______.

Solution:

Flatband voltage is:
V_FB = φ_m − φ_s − Q_ox/C_ox

Given:
V_FB = 0 V
φ_m = 4.6 eV
Q_ox = 10⁻⁸ C/cm²

Oxide capacitance per unit area:
C_ox = ε_ox/t_ox
ε_ox = 4 × 8.85 × 10⁻¹⁴
t_ox = 100 nm = 10⁻⁵ cm

C_ox = 3.54 × 10⁻⁸ F/cm²

Therefore:
φ_s = 4.6 − (10⁻⁸/3.54 × 10⁻⁸)
φ_s ≈ 4.3175 eV

Correct Answer: 4.32 eV

Q.62 | GATE EC 2024

In the network shown below, maximum power is to be transferred to the load R_L.

The value of R_L (in Ω) is ______.

Solution:

Maximum power transfer के लिए:
R_L = R_th

Independent 50 V source को deactivate करके short circuit करते हैं। अब load terminals पर test voltage V और test current I लगाकर equivalent resistance निकालते हैं।

Dependent voltage source की relation:
V_x = V − V_c

Circuit equations solve करने पर:
V = 2.5 I

इसलिए:
R_th = V/I = 2.5 Ω

Maximum power transfer condition से:
R_L = R_th

Correct Answer: 2.50 Ω

Q.63 | GATE EC 2024

A non-degenerate n-type semiconductor has 5% neutral dopant atoms. Its Fermi level is located at 0.25 eV below the conduction band (E_C) and the donor energy level (E_D) has a degeneracy of 2.

Assuming the thermal voltage to be 20 mV, the difference between E_C and E_D (in eV, rounded off to two decimal places) is ______.

Solution:

Neutral donor fraction:
N_D⁰/N_D = 0.05

Ionized donor fraction:
N_D⁺/N_D = 0.95

Therefore:
N_D⁰/N_D⁺ = 0.05/0.95 = 1/19

Donor occupancy relation:
N_D⁰/N_D⁺ = g_D exp[(E_F − E_D)/V_T]

Given g_D = 2 and V_T = 0.02 eV:
1/19 = 2 exp[(E_F − E_D)/0.02]

Taking natural logarithm:
E_F − E_D = 0.02 ln(1/38)
E_F − E_D ≈ −0.0726 eV

Given:
E_C − E_F = 0.25 eV

Hence:
E_C − E_D = 0.25 − 0.0726
≈ 0.1774 eV

Correct Answer: 0.18 eV

Q.64 | GATE EC 2024

An NMOS transistor operating in the linear region has I_DS of 5 μA at V_DS of 0.1 V. Keeping V_GS constant, the V_DS is increased to 1.5 V.

Given that μ_nC_oxW/L = 50 μA/V², the transconductance at the new operating point (in μA/V, rounded off to two decimal places) is ______.

Solution:

MOSFET linear-region current:
I_D = K[(V_GS − V_T)V_DS − V_DS²/2]

Let V_OV = V_GS − V_T. Given:
K = 50 μA/V²
I_D = 5 μA
V_DS = 0.1 V

5 = 50[0.1V_OV − (0.1)²/2]
0.1 = 0.1V_OV − 0.005
V_OV = 1.05 V

At the new V_DS = 1.5 V:
V_DS > V_OV
Therefore, the transistor is in saturation.

Saturation transconductance:
g_m = K V_OV
g_m = 50 × 1.05
g_m = 52.5 μA/V

Correct Answer: 52.50 μA/V

Q.65 | GATE EC 2024

The photocurrent of a PN junction diode solar cell is 1 mA. The voltage corresponding to its maximum power point is 0.3 V. If the thermal voltage is 30 mV, the reverse saturation current of the diode (in nA, rounded off to two decimal places) is _____.

Solution:

Solar cell current equation:
I = I_ph − I₀(e^V/V_T − 1)

At maximum power point:
d(VI)/dV = 0
I + V(dI/dV) = 0

Since:
dI/dV = −(I₀/V_T)e^V/V_T

Substitute into the maximum power condition:
I_ph − I₀(e^V/V_T − 1) − (V/V_T)I₀e^V/V_T = 0

Given:
I_ph = 1 mA
V = 0.3 V
V_T = 0.03 V
V/V_T = 10

Therefore:
I₀ = I_ph / [11e¹⁰ − 1]

I₀ ≈ 4.127 × 10⁻⁹ A

Correct Answer: 4.13 nA