GATE 2026 Electronics & Communication Engineering (EC) | Complete Question Paper

 

GATE Electronics & Communication Engineering (EC)

Welcome to the GATE Electronics & Communication Engineering (EC) question paper . This paper is designed to help students understand the concepts, question patterns and problem-solving approach required for the Graduate Aptitude Test in Engineering (GATE). Electronics and Communication Engineering is a broad technical field that combines Electronics, Communication Systems, Signal Processing, Networks, Control Systems, Electromagnetics and several fundamental engineering subjects.

The questions in the EC paper require more than just memorizing formulas. A strong understanding of fundamental concepts, mathematical analysis and logical reasoning is essential for solving questions accurately. Topics such as Network Theory, Signals and Systems, Analog Electronics, Digital Electronics, Control Systems, Communication Systems, Electromagnetic Theory, Electronic Devices and Engineering Mathematics form an important part of the preparation.

This question paper is useful for students who want to practice both conceptual and numerical problems. Each question can help develop the ability to identify the correct concept, apply the appropriate formula and reach the answer with minimum calculation errors. Regular practice also improves speed, accuracy, time management and confidence during the actual examination.

While solving the paper, students are encouraged to first read each question carefully and identify the given information and required quantity. For numerical problems, proper units and intermediate calculations should be checked before entering the final answer. For multiple-choice questions, understanding why the other options are incorrect can be equally valuable because it strengthens conceptual clarity.

The purpose of this practice material is not only to obtain the correct answer but also to build a systematic approach toward GATE EC preparation. Students can use these questions for self-assessment, revision and identifying topics that require additional practice. Attempting the questions without seeing the solution first can provide a more realistic assessment of preparation level.

Practice consistently • Understand concepts • Improve accuracy • Build confidence • Crack GATE EC

Question 1

Among the following options, the antonym of the word ‘nocturnal’ is ________.

Solution

Meaning of “nocturnal”:

The word nocturnal refers to something that is active or occurring during the night.

The antonym of nocturnal is diurnal, which means active or occurring during the daytime.

Therefore,

Nocturnal ↔ Diurnal

✅ Correct Answer: (B) Diurnal

Question 2

The statements (S1), (S2), (S3) pertain to the scores obtained by students in an exam. The maximum possible marks in the exam is 150.

(S1) The highest score is 100.

(S2) The fourth highest score is 76.

(S3) There are at least four students whose scores are within 25 of each other.

Which of the following options is necessarily correct?

Solution

Step-by-step Solution:

We are given:

(S1) The highest score is 100.
(S2) The fourth highest score is 76.

If the highest score is 100 and the fourth highest score is 76, then the top four scores can be represented as:

100,   x,   y,   76

Since 100 is the highest score and 76 is the fourth highest score, the other two scores must lie between 76 and 100.

Therefore, the difference between the highest and fourth highest scores is:

100 − 76 = 24

Thus, these four students have scores lying within a range of 24 marks.

Since
24 < 25

there are at least four students whose scores are within 25 of each other.

Hence,
(S1) + (S2) ⇒ (S3)

✅ Correct Answer: (A)

Question 3

The figure below has exactly three intersecting line segments with a rectangular portion WXYZ missing.

Which one of the following options P, Q, R, and S is the missing portion WXYZ?

Solution

Step-by-step Solution:

The given figure contains exactly three line segments passing through the missing rectangular portion WXYZ.

We need to match the continuation of all three line segments from outside the rectangle.

The three line segments must continue through the rectangle without changing their direction:

1. One line enters through W and continues diagonally downward towards Y.

2. Another line enters through Z and continues diagonally upward towards X.

3. The third line enters from the left side and continues towards the right side with a slight upward slope.

Among the given options, only S preserves the correct direction and continuation of all three line segments.

✅ Correct Answer: (D) S

Question 4

Real numbers y, p, n (all greater than 1) satisfy

(log_p^1/n y) (log_y^1/n p) = 16

where the logarithms are taken to the bases p^1/n and y^1/n .

The value of n is ________.

Solution

Step-by-step Solution:

Given:

(log_p^1/n y) (log_y^1/n p) = 16

Using the change of base formula,

log_p^1/n y = ln(y) / ln(p^1/n)

Since
ln(p^1/n) = (1/n) ln(p)

Therefore,
log_p^1/n y = n ln(y) / ln(p)

Similarly,
log_y^1/n p = n ln(p) / ln(y)

Multiplying both terms:

(n ln(y) / ln(p)) (n ln(p) / ln(y))

Cancelling ln(y) and ln(p) :

n² = 16

Since n > 1 ,

n = 4

✅ Correct Answer: (B) 4

Question 5

The following observation is made about the scores obtained by 100 students in an exam:

“For each student, there exists another student in the class such that their scores are at most ten marks away.”

If the above statement is false, which of the following statements is necessarily true?

Solution

Step-by-step Solution:

The given statement is:

“For each student, there exists another student whose score is at most 10 marks away.”

In logical form, this means:

For every student → there exists another student within 10 marks.

We are told that this statement is false.

Therefore, its negation must be true.

The negation of
“For every student, there exists another student within 10 marks”

is:
“There exists at least one student for whom all the other students are more than 10 marks away.”

This is exactly what option (B) states.

✅ Correct Answer: (B)

Question 6

Each one of the following clues contains a keyword that is partially filled.

Clue 1: Synonym of recognize (8 letters): _ D _ N T _ F Y

Clue 2: A story long enough to fill a book (5 letters): _ _ _ _ EL

Clue 3: Two of something (6 letters): _ _ _ PLE

Clue 4: A fraction of something, split equally into two parts (4 letters): _ _ _ F

The first letter of each of the keywords can be rearranged to form a four-letter word.

Which of the options below is a possible choice for the four-letter word?

Solution

Step-by-step Solution:

Clue 1: Synonym of recognize (8 letters)
_ D _ N T _ F Y

The word is:
IDENTIFY
First letter = I

Clue 2: A story long enough to fill a book
_ _ _ _ EL

The word is:
NOVEL
First letter = N

Clue 3: Two of something
_ _ _ PLE

The word is:
COUPLE
First letter = C

Clue 4: A fraction split equally into two parts
_ _ _ F

The word is:
HALF
First letter = H

Therefore, the four first letters are:

I, N, C, H

Rearranging these letters gives:

C H I N

Hence, the possible four-letter word is CHIN.

✅ Correct Answer: (A) CHIN

Question 7

Three children P, Q, R and two grown-ups X, Y play a badminton doubles tournament. X and Y are parents to two of the children playing. The child of X is not the same as the child of Y. Exactly one of the children does not have a parent playing in the tournament.

The following rules are followed:

(i) A parent and his/her child cannot be on the same team.
(ii) A match can feature at most one parent and his/her child, that is, a maximum of one parent-child pair can play in a match.

The following matches were played:

	TEAM 1	TEAM 2
MATCH 1	P and X	Q and R
MATCH 2	P and R	X and Y
MATCH 3	R and X	Q and Y

Which one of the following options is correct?

Solution

Step-by-step Solution:

We have three children:
P, Q, R

and two parents:
X, Y

Since X and Y are parents of two different children, each parent must have a different child among P, Q and R.

Step 1: Consider X.

In Match 3, R and X are on the same team. Therefore, R cannot be the child of X because a parent and child cannot be on the same team.

In Match 1, P and X are on the same team. Therefore, P cannot be the child of X either.

Hence, X's child must be:

Q

Step 2: Consider Y.

Since X's child is Q and X and Y have different children, Y's child must be either P or R.

In Match 3, Q and Y are on the same team. This is consistent because Q is X's child, not Y's.

In Match 2, X and Y play together.

If Y's child were P, then Match 2 would contain parent Y and child P. This would make the remaining conditions inconsistent with the given matches.

Therefore, Y's child is:

R

Thus:

X → Q
Y → R

The remaining child is:
P

Therefore, P is the only child who does not have a parent playing in the tournament.

✅ Correct Answer: (A) P does not have any parent playing

Question 8

Let P_k represent the perimeter of a square with sides of length k .

The value of the expression P₁ + P₂ + P₃ + ⋯ + P₁₀ is ________.

Solution

Step-by-step Solution:

The perimeter of a square is:

P_k = 4k

Therefore,
P₁ = 4
P₂ = 8
P₃ = 12
and so on.

Hence,
P₁ + P₂ + ⋯ + P₁₀

= 4(1 + 2 + 3 + ⋯ + 10)

Using
1 + 2 + ⋯ + 10 = 10 × 11 / 2 = 55

Therefore,
4 × 55 = 220

✅ Correct Answer: (C) 220

Question 9

The city of Atlantis was crafted by the God of the seas, Poseidon. It was made of alternating concentric circular rings of land (shaded) and water (not shaded) as represented in the figure (not to scale).

The radius of Inner Island was 2.5 stades (a unit of length used in ancient Greece).

The water surrounding Inner Island was one stade wide (length AB ).

This was surrounded by two pairs of alternating rings of land and water. The first pair of land and water was two stades wide each (lengths BC and CD ), and the outer pair is three stades wide each (lengths DE and EF ).

The ratio of the surface area of the land to that of the water in the city of Atlantis is ________ (round off to two decimal places).

Solution

Step-by-step Solution:

Radius of Inner Island:
r₀ = 2.5

The successive radii are:

r₁ = 2.5 + 1 = 3.5
r₂ = 3.5 + 2 = 5.5
r₃ = 5.5 + 2 = 7.5
r₄ = 7.5 + 3 = 10.5
r₅ = 10.5 + 3 = 13.5

Land Area:

A_L = π[2.5² + (5.5² − 3.5²) + (10.5² − 7.5²)]

A_L = π[6.25 + 18 + 54]

A_L = 78.25π

Water Area:

A_W = π[(3.5² − 2.5²) + (7.5² − 5.5²) + (13.5² − 10.5²)]

A_W = π[6 + 26 + 72]

A_W = 104π

Therefore,

A_L / A_W = 78.25π / 104π = 0.7524

Rounding off to two decimal places:

0.75

✅ Correct Answer: (C) 0.75

Question 10

If

Ref Fig

12  34

→

12  34

Ref Fig

Then

GATE - 2026

Aptitude Test

→

?

Solution:

Reference figure में transformation को ध्यान से देखें। इसमें दोनों lines की position interchange होती है और प्रत्येक अक्षर/character का vertical reflection होता है।

इसलिए:

• GATE - 2026 नीचे चला जाएगा।
• Aptitude Test ऊपर चला जाएगा।
• दोनों lines के letters vertically reflected होंगे।
• लेकिन words का original order नहीं बदलेगा।

यह pattern Option R में मिलता है।

Correct Answer: (C) R

Question 11

Consider the differential equation dw⃗/dt = Aw⃗ , with

w⃗(t = 0) = 1
1

If

w⃗(t) = e^t u⃗_x + e^−2t u⃗_y

be the solution to the equation where u⃗_x and u⃗_y are unit vectors along the positive x and y axes respectively, then which of the following options is the correct matrix representing A?

Solution:

Given

w⃗(t) = e^t u⃗_x + e^−2t u⃗_y

Therefore, in matrix form:

w⃗(t) = e^t
e^−2t

Differentiating with respect to t:

dw⃗/dt = e^t
−2e^−2t

Since dw⃗/dt = Aw⃗ , the matrix A must multiply the first component by 1 and the second component by −2.

A = 1   0
0  −2

Correct Answer: (A)

Question 12

A surface is given by z² = 2x² − y² and n⃗ and −n⃗ are unit normal vectors to the surface at the point

P⃗ = î + √2 k̂

Which of the following vectors can be n⃗, where î, ĵ and k̂ are the unit vectors along x, y and z axes, respectively?

Solution:

Given surface:

z² = 2x² − y²

Rearranging:

F(x,y,z) = z² − 2x² + y² = 0

A normal vector to the surface is given by ∇F.

∇F = −4x î + 2y ĵ + 2z k̂

At the point P = (1, 0, √2):

∇F = −4î + 2√2 k̂

Its magnitude is:

|∇F| = √(16 + 8) = 2√6

Therefore, a unit normal vector can be written as

n⃗ = −4î + 2√2 k̂ 2√6

Simplifying:

n⃗ = − ^√2⁄_√3 î + ¹⁄_√3 k̂

The opposite unit normal is also valid:

n⃗ = ^{√2 î − k̂}⁄_√3

Hence the correct option is:

Correct Answer: (D)

Question 13

The Laplace Transform of the signal

x(t) = u(t − 2) * (t u(t))

is given by which of the following expressions?

[“*” represents convolution operator]

Solution:

Given:

x(t) = u(t − 2) * (t u(t))

Using the convolution property of Laplace Transform:

L{f(t) * g(t)} = F(s)G(s)

We know:

L{u(t − 2)} = e^−2s/s
L{t u(t)} = 1/s²

Therefore,

X(s) = e^−2s s³

Correct Answer: (D)

Question 14

Consider carrier transport in a Zener diode in the breakdown region.

Which is the dominant transport mechanism for current flow in this case?

Solution

Step-by-step Solution:

Step 1: A Zener diode is a heavily doped p-n junction diode designed to operate in the reverse-bias breakdown region. Due to heavy doping, its depletion region is extremely thin.

Step 2: When a high reverse-bias voltage is applied, a very strong electric field is established across this thin depletion region.

Step 3: This intense electric field enables electrons to undergo quantum mechanical tunneling directly from the valence band of the p-side to the conduction band of the n-side.

Step 4: Consequently, quantum mechanical tunneling becomes the dominant mechanism governing current transport in a Zener diode under breakdown.

✅ Correct Answer: (C) Tunneling

Question 15

Two analog signals x₁(t) and x₂(t) (t in second), are sampled at a rate

F_s = 40 Hz,

where

x₁(t) = cos(20πt),   t ≥ 0
x₂(t) = cos(100πt),   t ≥ 0

The first ten samples (starting from t = 0) are considered for the analysis.

Which of the following statements is TRUE?

Solution:

Sampling frequency is

F_s = 40 Hz

Therefore, the sampling time is

T_s = 1/F_s = 1/40

For the n-th sample, put t = nT_s = n/40.

For the first signal:

x₁[n] = cos(20πn/40) = cos(πn/2)

For the second signal:

x₂[n] = cos(100πn/40) = cos(5πn/2)

Since

5πn/2 = πn/2 + 2πn

and cosine is periodic with period 2π,

x₂[n] = cos(πn/2 + 2πn) = cos(πn/2) = x₁[n]

Hence, all the samples of the two signals are equal. Therefore, the fourth to seventh samples are also equal.

Correct Answer: (D)

Question 16

The response of a discrete time system y[n] obeys the following relation:

y[n] = ⁵⁄₆ y[n − 1] − ¹⁄₆ y[n − 2] + x[n]

The input to the system is

x[n] = δ[n] − ¹⁄₃ δ[n − 1]

Which of the following options is TRUE for y[n]?

Solution:

Given difference equation:

y[n] = ⁵⁄₆y[n−1] − ¹⁄₆y[n−2] + x[n]

For the corresponding homogeneous equation:

y[n] − ⁵⁄₆y[n−1] + ¹⁄₆y[n−2] = 0

The characteristic equation is:

r² − ⁵⁄₆r + ¹⁄₆ = 0

Multiplying by 6:

6r² − 5r + 1 = 0

Factoring:

(3r − 1)(2r − 1) = 0

Therefore, the poles are:

r = ¹⁄₃,    r = ¹⁄₂

Both poles lie inside the unit circle:

|1/3| < 1   and   |1/2| < 1

Hence the system is stable.

Since the difference equation depends only on present and past input/output samples, the corresponding causal response is obtained from the causal system.

Correct Answer: (A) Stable and causal response

Question 17

The ideal OP-AMP circuit shown in the Figure produces output voltage V_o = x when the Switch, S, is open.

Which of the options represents the output voltage when S is closed?

Solution:

The non-inverting terminal is connected to V_i = 1 V.

For an ideal non-inverting amplifier:

V_o = V_i ←   1 + ^R_f⁄_R1

When the switch S is open, the feedback resistance is:

R_f = 2 kΩ + 1 kΩ = 3 kΩ

Therefore,

x = 1 × ← 1 + ³⁄₁ → = 4 V

When the switch S is closed, the 1 kΩ resistor in the feedback path is short-circuited. Hence,

R_f = 2 kΩ

Thus the new output voltage is:

V_o = 1 × ← 1 + ²⁄₁ → = 3 V

Since x = 4 V,

V_o = ³⁄₄x

Correct Answer: (C) 3/4 x

Question 18

Consider the circuit shown in the Figure with V_i = 3 V and V_CC = 12 V.

Assume V_BE = 0.7 V and β_dc = 99 for the BJT.

Which of the following options is the correct value of the current I_o?

Solution:

Since the op-amp is ideal and negative feedback is present, the voltage at the inverting terminal is equal to the voltage at the non-inverting terminal.

V_E = V_i = 3 V

The emitter resistor is 1 kΩ. Therefore, emitter current is:

I_E = ^V_E⁄_{1 kΩ} = ^{3 V}⁄_{1 kΩ} = 3 mA

For a BJT:

I_E = (β + 1) I_B

Given β = 99:

3 mA = (99 + 1) I_B

3 mA = 100 I_B

Hence,

I_o = I_B = 30 μA

Correct Answer: (D) 30 μA

Question 19

A control system is shown in the Figure.

Which option represents the correct transfer function of the system?

Solution:

Let the output of the first summing junction be E(s).

The forward transfer function is

G(s) = 1 s + 4

Therefore,

C(s) = G(s)E(s)

The lower summing junction receives:

• Positive input: C(s)
• Negative input: G(s)E(s)

Hence its output is

C(s) − G(s)E(s) = 0

Thus the feedback signal to the first summing junction is zero. Therefore,

E(s) = R(s)

Consequently,

C(s) R(s) = 1 s + 4

Correct Answer: (B)

Question 20

In the circuit shown in the Figure, A and B are logic inputs and Y is the logic output.

Which of the following logic operations is realized by the circuit?

Solution:

The circuit is a CMOS logic circuit.

The PMOS transistors form the pull-up network and the NMOS transistors form the pull-down network.

The output Y becomes HIGH only when both inputs A = 0 and B = 0.

Y = A̅ · B̅

Using De Morgan's theorem:

Y = A̅ · B̅ = (A + B)̅

Therefore, the circuit implements a NOR operation.

Correct Answer: (A) NOR

Question 21

Consider the Friis’ transmission equation

P_R = P_TG_TG_Rλ² (4πD)²

where P_R and P_T are the received and the transmitted powers, respectively.

G_T and G_R are the gain of transmitting and receiving antennas, respectively, D is the distance between the transmitting and receiving antennas, and λ is the wavelength in free space.

Given: G_T = G_R = 1.0, λ = 0.30 m and P_T = +10 dBm.

Choose the distance D, in km, from the following options at which the received power, P_R = −90 dBm.

Solution:

Given:

G_T = G_R = 1

λ = 0.30 m

P_T = +10 dBm

The received power is P_R = −90 dBm.

Therefore, the power ratio in dB is:

P_R − P_T = −90 − 10 = −100 dB

Hence,

10^−100/10 = 10⁻¹⁰

From the Friis equation:

10⁻¹⁰ = λ² (4πD)²

Taking square root:

10⁻⁵ = 0.30 4πD

Therefore,

D = 0.30 × 10⁵ 4π m

Converting into km:

D = 15 4π km

Correct Answer: (A) 15/(4π) km

Question 22

Consider a discrete memoryless source with an alphabet of four source symbols.

s(t) is a multi-level (−1, 0, +1, +2) signal representing a long sequence of random symbols from the above source which is generating 10⁴ symbols per second.

Which of the following options is the correct value of equivalent Nyquist bandwidth of s(t)?

Solution:

Number of source symbols:

M = 4

Since there are 4 possible symbols, the number of bits represented by each symbol is:

log₂(4) = 2 bits/symbol

Symbol rate is:

R_s = 10⁴ symbols/s

Therefore, the bit rate is:

R_b = R_s × log₂(M)

= 10⁴ × 2 = 2 × 10⁴ bits/s

For the Nyquist criterion, the minimum required bandwidth is:

B = R_b/2

B = 20,000/2 = 10,000 Hz = 10 kHz

Correct Answer: (A) 10 kHz

Question 23

The relation between the input current (I) and the output voltage (V) of a circuit is governed by the equation:

C dV dt = I(t) − m(t)

The circuit is excited by I(t) = qδ(t), where q is a real valued constant. V at t = 0⁻ is V₀.

Which of the following is an equivalent representation of the above case?

Solution:

Given:

C dV/dt = I(t) − m(t)

and

I(t) = qδ(t)

Integrating the equation around t = 0:

C[V(0⁺) − V(0⁻)] = q

Since V(0⁻) = V₀, we get:

V(0⁺) = V₀ + q/C

For t > 0, the impulse disappears, so I(t) = 0. Therefore:

C dV/dt = −m(t)

Thus the impulse can be represented by an updated initial voltage of V₀ + q/C.

Note: The printed options use V(t = 0⁻), although the mathematically updated value is at t = 0⁺. Option (A) is the intended answer.

Correct Answer: (A)

Question 24

The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogeneous medium is given by

Ẽ(z,t) = E₀ [ x̂ cos(ωt − kz) + ŷ sin(ωt − kz) ]

in a right-handed orthogonal co-ordinate system.

Which of the following is the correct polarization of the electromagnetic wave?

(A) Right-handed circularly polarized

(B) Left-handed circularly polarized

(C) Linearly polarized

(D) Linearly polarized with −45° angle to x̂

Solution:

The electric field is

E = E₀ [x̂ cos(ωt − kz) + ŷ sin(ωt − kz)]

The x and y components have equal amplitudes and are 90° out of phase. Therefore, the wave is circularly polarized.

Since the wave is travelling in the +z direction and the electric-field vector rotates from x̂ towards ŷ with time, the polarization is

Left-handed circularly polarized

Therefore, the correct answer is:
(B) Left-handed circularly polarized

Question 25

Consider a p-n junction diode when it is forward biased with 2 V.

Which of the following is/are the correct magnitude(s) of the energy difference between quasi Fermi-levels E_fn in the n-side and E_fp in the p-side?

(A) 2 eV

(B) 1 eV

(C) 2 V

(D) 1 V

Solution:

For a forward-biased p-n junction, the separation between the quasi Fermi levels is related to the applied voltage by

E_fn − E_fp = qV

Given,
V = 2 V

Therefore,
|E_fn − E_fp| = q(2 V) = 2 eV

Hence, the correct answer is:
(A) 2 eV

Question 26

Consider the matrix

M = ⎡ 2   1   1 ⎤
⎢ 1   3   0 ⎥
⎣ −1   a   b ⎦

Which of the following options is/are TRUE if det(M) ≠ 0?

(A) a = −1/2 and b = −1/2

(B) a = 1/2 and b = 1/2

(C) a = −3 and b = 0

(D) a = 1/2 and b = −3

Solution:

Given,

det(M) = a + 5b + 3

Since det(M) ≠ 0,
a + 5b + 3 ≠ 0

(A)
a = −1/2, b = −1/2
a + 5b + 3 = −1/2 − 5/2 + 3 = 0
✗ Not true

(B)
a = 1/2, b = 1/2
a + 5b + 3 = 1/2 + 5/2 + 3 = 6 ≠ 0
✓ True

(C)
a = −3, b = 0
a + 5b + 3 = −3 + 0 + 3 = 0
✗ Not true

(D)
a = 1/2, b = −3
a + 5b + 3 = 1/2 − 15 + 3 = −23/2 ≠ 0
✓ True

Correct Answers: (B) and (D)

Question 27

A binary ripple counter is designed to count (0)₁₀ to (64)₁₀.

Which of the following is/are the number of flip-flops required to design the counter?

(A) 6

(B) 7

(C) 4

(D) 5

Solution:

The counter has to count from

0 to 64

Therefore, total number of states required is
64 − 0 + 1 = 65 states

For n flip-flops, the number of available states is
2ⁿ

We need:
2ⁿ ≥ 65

Since,
2⁶ = 64 < 65
and
2⁷ = 128 ≥ 65

Therefore, the required number of flip-flops is
n = 7

Correct Answer: (B) 7

Question 28

Which option(s) represents/represent the dielectric loss tangent of a substrate?

(A) Ratio of the real to imaginary parts of the total displacement current

(B) (ωε″ + σ)/(ωε′)

(C) Ratio of the electric susceptibility to permittivity

(D) Ratio of the polarization vector P⃗ to the displacement vector D⃗

Solution:

For a lossy dielectric, the complex permittivity can be written as

ε = ε′ − jε″

The total current density is
J = (σ + ωε″ + jωε′)E

Hence, the dielectric loss tangent is
tan δ = (σ + ωε″)/(ωε′)

Therefore, option (B) is directly correct.

Also, the loss tangent is the ratio of the real to imaginary parts of the total current (displacement/conduction current representation), so option (A) is also correct.

Correct Answers: (A) and (B)

Question 29

Figure shows the output characteristics of two different Bipolar Junction Transistors (BJT), BJT 1 with magnitude of Early voltage |V_A1|, and BJT 2 with magnitude of Early voltage |V_A2|.

Which of the following options is/are correct regarding the Early voltages?

(A) |V_A1| > |V_A2|

(B) |V_A1| is infinitely large

(C) |V_A1| < |V_A2|

(D) |V_A2| is finite

Solution:

Due to the Early effect, the output characteristics of a BJT have a finite slope.

The magnitude of Early voltage is inversely related to the slope of the output characteristic:

|V_A| ∝ 1 / (slope)

From the figure, the characteristic of BJT 1 has a larger slope than that of BJT 2.

Therefore,
|V_A1| < |V_A2|

Also, BJT 2 has a non-zero slope, hence its Early voltage is finite.

Therefore, the correct answers are:

(C) |V_A1| < |V_A2|
(D) |V_A2| is finite

Question 30

The output voltage V_o (in Volt) for the network given in the Figure is __.

(rounded off to two decimal places)

Enter your answer in Volts:

Solution:

Let the top node voltage be V_o. Applying KCL and using the 6 V source constraint gives

V_o = 1 V

Therefore, rounded off to two decimal places:

V_o = 1.00 V

Question 31

Consider the circuit shown in the Figure, where the input v_i(t) is in Volt.

The average power (in mW) dissipated in the load resistance of 1 kΩ at the resonant frequency is __.

(rounded off to two decimal places)

Enter your answer in mW:

Solution:

Given,

v_i(t) = 12 sin(ωt)

At parallel resonance, the capacitor and inductor currents cancel each other. Therefore, the 1 kΩ load resistance has the full source voltage across it.

RMS voltage:
V_rms = 12 / √2

Average power:
P = V_rms² / R

= (12/√2)² / 1000

= 72 × 10⁻³ W

= 72 mW

Therefore, rounded off to two decimal places:

Answer = 72.00 mW

Question 32

A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel and a maximum-likelihood receiver. Consider the information bit rate from source to be 4 × 10⁶ bits per second.

The minimum transmission bandwidth (in MHz) of the modulated signal necessary for optimum recovery of information at the receiver is __.

(rounded off to two decimal places)

Enter your answer in MHz:

Solution:

For 16-QAM,

Bits per symbol = log₂(16) = 4

Symbol rate:
R_s = (4 × 10⁶)/4 = 1 × 10⁶ symbols/s

The minimum Nyquist transmission bandwidth is
B = R_s/2

Therefore,
B = 0.5 MHz

Answer = 0.50 MHz

Question 33

The cutoff frequency (in GHz) for the dominant TE₁₀ mode of an air-filled rectangular waveguide of inner dimension 0.28 inch × 0.14 inch is __.

(rounded off to two decimal places)

Enter your answer in GHz:

Solution:

For the dominant TE₁₀ mode of an air-filled rectangular waveguide,

f_c10 = c / (2a)

Given,
a = 0.28 inch

Since,
1 inch = 0.0254 m
a = 0.28 × 0.0254 = 0.007112 m

Therefore,
f_c10 = (3 × 10⁸) / (2 × 0.007112)

f_c10 ≈ 21.09 GHz

Answer = 21.09 GHz

Question 34

For a lossless passive two-port network, |S₁₁| and |S₂₁| intersect at −3 dB.

For a lossy passive two-port network, |S₁₁| and |S₂₁| intersect at −4 dB.

The percentage of power dissipated in the lossy network at the intersection frequency is __.

(rounded off to two decimal places)

Enter your answer in %:

Solution:

At the intersection frequency of the lossy network,

|S₁₁| = |S₂₁| = −4 dB

Converting dB to power ratio:
|S|² = 10^−4/10

|S|² ≈ 0.3981

Therefore, the total reflected + transmitted power is
0.3981 + 0.3981 = 0.7962

Hence, the power dissipated is
P_d = 1 − 0.7962 = 0.2038

Percentage dissipated:
0.2038 × 100 ≈ 20.38%

Answer = 20.38%

Question 35

The negative edge triggered JK flip-flop in the Figure has J and K inputs tied to Logic High and a square wave of 10 cycles/second is applied to its clock (C) input.

The frequency of the output Q (in cycles/second) is __.

(rounded off to two decimal places)

Enter your answer (cycles/second):

Solution:

Since,
J = K = 1

A JK flip-flop with J = K = 1 operates in toggle mode. Therefore, the output Q changes state at every active clock edge.

The clock frequency is
f_CLK = 10 cycles/second

Since the flip-flop toggles once at every negative edge, one complete output cycle requires two clock cycles.

Therefore,
f_Q = f_CLK/2 = 10/2 = 5 cycles/second

Answer = 5.00 cycles/second

Question 36

Consider the two series, S_A and S_B, where

S_A = Σ_n=1^∞ n²/2ⁿ

S_B = 1 + 1/2 + 1/8 + 1/16 + 1/64 + 1/128 + 1/512 + …

Which of the following statements is correct for the two given series?

(A)  Both S_A and S_B converge.

(B)  Neither S_A nor S_B converges.

(C)  S_A converges but S_B does not converge.

(D)  S_B converges but S_A does not converge.

Solution:

For S_A = Σ n²/2ⁿ , the exponential term 2ⁿ grows faster than the polynomial term n².

Using the ratio test:

a_n = n²/2ⁿ

a_n+1/a_n = (n+1)²/(2n²) → 1/2

Since 1/2 < 1,
S_A converges.

For S_B:

S_B = 1 + 1/2 + 1/8 + 1/16 + 1/64 + 1/128 + …

The terms can be grouped as
(1 + 1/2) + (1/8 + 1/16) + (1/64 + 1/128) + …

Each group forms a geometric sequence with ratio 1/8. Therefore S_B also converges.

Hence,
Both S_A and S_B converge.

Correct Answer: (A)

Question 37

The continuous time signal x(t) is real, periodic with period T and satisfies the Dirichlet conditions.

The Fourier series representation of x(t) is

x(t) = Σ_n=-∞^∞ a_n e^j(2πnt/T)

and x(t) satisfies the following:

x(t − T/2) = −x(t)

For any integer m, which of the following options is correct?

(A)  a_2m = 0

(B)  a_2m = 1

(C)  a_2m = a_2m+1

(D)  a_2m = −1

Solution:

Given,
x(t − T/2) = −x(t)

The Fourier series is
x(t) = Σ a_n e^j2πnt/T

Replacing t by t − T/2:
x(t − T/2) = Σ a_n e^{j2πn(t−T/2)/T}

Therefore,
x(t − T/2) = Σ a_n (−1)ⁿ e^j2πnt/T

But,
x(t − T/2) = −x(t)

Hence,
(−1)ⁿ a_n = −a_n

For even n = 2m,
(−1)^2m = 1
so,
a_2m = −a_2m

Therefore,
a_2m = 0

Correct Answer: (A)

Question 38

Let X, N, Y and Z be random variables. The variables X and N are independent of each other. X is uniformly distributed between −1 and 1; N follows Normal distribution with zero mean and unity variance.

Y and Z are defined as,

Y = X + N      and      Z = X² + N

Which of the following pairs represents the values of correlation between X and Y and that between X and Z?

(A)  1/3 and 0

(B)  1/3 and 1/9

(C)  1/3 and 1/3

(D)  1 and 0

Solution:

Since X is uniformly distributed over [−1,1],
E[X] = 0 and E[X²] = 1/3 .

Also, X and N are independent and
E[N] = 0.

For Y = X + N:

E[XY] = E[X(X+N)] = E[X²] + E[XN]

Due to independence, E[XN] = E[X]E[N] = 0. Therefore,
E[XY] = 1/3.

For Z = X² + N:

E[XZ] = E[X(X²+N)] = E[X³] + E[XN].

The distribution of X is symmetric about zero, hence
E[X³] = 0 and, again, E[XN] = 0.

Thus,
E[XZ] = 0.

Hence the required pair is
1/3 and 0

Correct Answer: (A)

Question 39

In the given circuit, L = 1 μH and C = 1 μF. The phasor diagram for I_C and I_L is also shown. Assume that the phase (θ₁ + θ₂) is 90° at a frequency of 159.15 kHz.

Among the following options, what is the nearest integer value of R_C × R_L?

(rounded off to the nearest integer)

Enter your answer:

Solution:

For the RC series branch,

tan θ₁ = 1 / (ωR_CC)

For the RL series branch,

tan θ₂ = ωL / R_L

Given: θ₁ + θ₂ = 90°

Therefore,

tan θ₁ × tan θ₂ = 1

[1/(ωR_CC)] × [ωL/R_L] = 1

L/(R_CR_LC) = 1

R_CR_L = L/C

Since L = 1 μH and C = 1 μF,

R_CR_L = 1

Correct Answer = 1

Question 40

For the control system shown in the Figure, the transfer function of a plant, G(s) = 1/[(s+1)(s+2)] is connected in cascade with a compensator C(s) = K(s + α), where K and α are positive real valued constants.

Which of the following pairs (K, α) represent the correct values for the closed loop system to have poles at (−3 ± j√5)?

(A)   2, 3

(B)   3, 4

(C)   2, 4

(D)   3, 3

Solution:

Open-loop transfer function:
C(s)G(s) = K(s+α) / [(s+1)(s+2)]

For unity negative feedback, the characteristic equation is
(s+1)(s+2) + K(s+α) = 0

Given closed-loop poles:
s = −3 ± j√5 Therefore, the desired characteristic polynomial is
(s+3)² + 5 = s² + 6s + 14

Now,
(s+1)(s+2) + K(s+α) = s² + (3+K)s + (2+Kα) Comparing with
s² + 6s + 14 we get:

3 + K = 6
⇒ K = 3

2 + Kα = 14
⇒ 3α = 12
⇒ α = 4

Therefore,
(K, α) = (3, 4)

Hence, the correct answer is (B).

Question 41

The state and output equations for a control system are:

ẋ = −4    −1.5 4      0 x + 2 0 u

y = [ 1.5    0.625 ] x

Which of the following expressions correctly represents the transfer function Y(s)/U(s) of the system with zero initial conditions?

(A)   3s s² + 4s + 6

(B)   3s + 5 s² + 4s − 6

(C)   3s + 5 s² + 4s + 6

(D)   3s s² + 4s + 6

Solution:

For the state-space model,

G(s) = C(sI − A)⁻¹B

Here,
A = [ −4   −1.5 ; 4   0 ]
B = [ 2 ; 0 ]
C = [ 1.5   0.625 ]

Therefore,
sI − A = [ s+4   1.5 ; −4   s ]

Its determinant is
(s+4)s + 6 = s² + 4s + 6

Now,
(sI−A)⁻¹B = 1/(s²+4s+6) [ 2s ; 8 ]

Hence,
Y(s)/U(s) = [1.5   0.625] [2s ; 8] /(s²+4s+6)

= (3s + 5)/(s² + 4s + 6)

Therefore, the correct answer is:

(C)

Question 42

The address of the first location of a 256 kilo byte (KB) memory is (2500)_H.

Choose the correct address of the last location of the memory.

(A)   (2FFF)_H

(B)   (124FF)_H

(C)   (424FF)_H

(D)   (324FF)_H

Solution:

Memory size = 256 KB

Since
1 KB = 1024 bytes = 2¹⁰ bytes

Therefore,
256 KB = 256 × 1024 = 2¹⁸ = 0x40000

First address = (2500)_H

The last location is:
Last address = First address + Memory size − 1

= 2500_H + 40000_H − 1

= 2500_H + 3FFFF_H

= 424FF_H

Therefore, the correct answer is:

(C) (424FF)_H

Question 43

Consider a real signal x(t), −∞ < t < ∞, such that:

x(t) = 0 for t < 0, x(t) = 2 for 0 ≤ t < 1, and x(t) = 0 for t ≥ 1.

Let

E[x(t)] = ∫_−∞^∞ |x(t)|² dt.

Which of the following options correctly represents the ratio

E[x(t)] / E[3x(−3t + 5)]?

Enter Answer:

Solution

Given:
x(t) = 2 for 0 ≤ t < 1

Therefore,

E[x(t)] = ∫₀¹ 2² dt = 4

For the signal 3x(−3t + 5), the amplitude becomes 6 and the time duration becomes 1/3.

E[3x(−3t+5)] = 6² × 1/3 = 12

Ratio = 4/12 = 1/3

Correct Answer: (C) 1/3

Question 44

A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with an E_b/N₀ = 8.4 dB at the input of a coherent QPSK demodulator. A maximum-likelihood reception method is used in the demodulator.

Assume the complementary error function

erfc(u) ≅ [1/(u√π)] exp(−u²)

Which is the nearest bit error rate (BER) at the output of the demodulator?

(A)   10⁻³

(B)   10⁻⁴

(C)   10⁻⁵

(D)   10⁻⁶

Solution:

For coherent QPSK with maximum-likelihood detection,

BER = Q(√(2E_b/N₀))

Also,
Q(x) = 1/2 erfc(x/√2)

Therefore,
BER = 1/2 erfc(√(E_b/N₀))

Given,
E_b/N₀ = 8.4 dB

Convert dB into linear scale:
E_b/N₀ = 10^8.4/10 ≈ 6.92

Hence,
u = √6.92 ≈ 2.63

Using the given approximation,
erfc(u) ≅ [1/(u√π)] exp(−u²)

erfc(2.63) ≈ [1/(2.63√π)] exp(−6.92) ≈ 2.1 × 10⁻⁴

Therefore,
BER ≈ 1/2 × 2.1 × 10⁻⁴

BER ≈ 1.05 × 10⁻⁴

The nearest option is:

(B) 10⁻⁴

Question 45

What is the 10's complement of (47)₁₀?

(A)   52

(B)   53

(C)   54

(D)   55

Solution:

Given number:

47

For a two-digit decimal number, the 10's complement is:

10² − Number

Therefore,
10² − 47 = 100 − 47 = 53

Hence, the 10's complement of 47 is:

(B) 53

Question 46

Consider a real baseband signal x(t) = e^−2t, for t (in seconds) ≥ 0.

If 99% of energy of x(t) lies within B Hz, then which of the following options is TRUE for the value of B?

(A)   B > 1 kHz

(B)   63/π Hz < B < 64/π Hz

(C)   126/π Hz < B < 128/π Hz

(D)   B < 1 Hz

Solution:

Given:

x(t) = e^−2t,   t ≥ 0

Its Fourier transform is:
X(f) = 1/(2 + j2πf)

Therefore, the energy spectral density is:
|X(f)|² = 1/(4 + 4π²f²)

Total energy:
E = ∫_−∞^∞ |X(f)|² df = 1/4

The fraction of energy contained within −B ≤ f ≤ B is:
E_B/E = (2/π) tan⁻¹(πB)

Since 99% of energy lies within B Hz:
(2/π) tan⁻¹(πB) = 0.99

Therefore,
tan⁻¹(πB) = 0.495π

Hence,
πB = tan(0.495π)

Since
tan(0.495π) ≈ 63.66

Therefore,
B ≈ 63.66/π ≈ 20.26 Hz

Now,
63/π ≈ 20.05 Hz
64/π ≈ 20.37 Hz

Thus,
63/π < B < 64/π

Therefore, the correct answer is:

(B) 63/π Hz < B < 64/π Hz

Question 47

Consider the discrete time system (S) with input x[n] and output y[n] as shown in the figure. The two sub-systems represented by their impulse responses h₁[n] and h₂[n] are linear and time invariant.

Which of the following statements is necessarily TRUE?

(A)   S is causal.

(B)   S is linear and time invariant.

(C)   S is linear and time varying.

(D)   S is non-linear.

Solution:

The two sub-systems having impulse responses h₁[n] and h₂[n] are given to be linear and time invariant (LTI).

The overall system consists of:

• LTI systems h₁[n] and h₂[n]
• Adders
• Delay elements z⁻¹

An adder is a linear and time-invariant operation.

A delay element
y[n] = x[n − 1]
is also linear and time invariant.

Since the overall system is formed by interconnection of linear time-invariant blocks, the complete system S is also linear and time invariant.

Therefore, the statement that is necessarily TRUE is:

(B) S is linear and time invariant.

Question 48

A Boolean function, f(x,y,z) with x as MSB and z as LSB is realized by a 4:1 multiplexer (MUX) with select lines, S₁ and S₀ (S₁ is MSB, S₀ is LSB) and inputs I₀, I₁, I₂, I₃ as shown in the figure.

Which of the following options is the correct expression of f(x,y,z)?

(A)   xz + y

(B)   xy + z

(C)   xy + z

(D)   xy + z

Solution:

For a 4:1 MUX, the output is:

f = I₀x'z' + I₁x'z + I₂xz' + I₃xz

From the given MUX connections:

I₀ = 1
I₁ = 0
I₂ = 1
I₃ = y

Therefore,
f = x'z' + 0(x'z) + xz' + xyz

f = x'z' + xz' + xyz

Taking z' common from the first two terms:
f = z'(x' + x) + xyz

Since,
x' + x = 1

Hence,
f = z' + xyz

Using the Boolean identity A + A'B = A + B,
z' + xyz = z' + xy

Therefore,

f(x,y,z) = xy + z'

Hence, the correct answer is:

(C) xy + z'

Question 49

A shift-left Shift Register (SR) and a D flip-flop are connected to a synchronized clock as shown in the figure. Assume that the SR and D flip-flops are initially cleared and the XOR gate has no propagation delay.

Which of the following options gives the correct binary representation (b₇b₆b₅b₄ b₃b₂b₁b₀) of the content of the shift register immediately after the 5^th clock transition (positive edge)?

(A)   00011111

(B)   10111111

(C)   00111111

(D)   11000011

Solution:

Initially, both the shift register and the D flip-flop are cleared.

Therefore,
b₇ = b₆ = ... = b₀ = 0

The D flip-flop receives the feedback from b₇. The XOR gate has one input equal to 1, so:

D = b₇ ⊕ 1 = b₇'

During the first five clock transitions, b₇ remains 0 because the first inserted bit has not yet reached b₇. Therefore,
Q = 0 ⊕ 1 = 1

So a 1 is inserted into the shift register at each of the first five clock transitions.

Starting from:
00000000

After 1st clock:
00000001

After 2nd clock:
00000011

After 3rd clock:
00000111

After 4th clock:
00001111

After 5th clock:
00011111

Hence, the required content is:

(A) 00011111

Question 50

A small signal source, V_i(t) = A cos(10⁵t) + B sin(10⁷t) is applied to a BJT circuit as shown in the figure.

Assume zero source resistance, V_BE = 0.7 V, β_dc = 99, Early voltage = 100 V and thermal voltage = 25 mV. Effect of internal parasitic capacitances of the BJT may be neglected.

Which expression is the best approximation of the output voltage V_o(t)?

(A)   −9.1[A cos(10⁵t) + B sin(10⁷t)]

(B)   9.1[A cos(10⁵t) − B sin(10⁷t)]

(C)   −190.4[A cos(10⁵t) + B sin(10⁷t)]

(D)   190.4[A cos(10⁵t) − B sin(10⁷t)]

Solution:

For DC analysis, the base bias voltage is:

V_B = 12 × 25/(100 + 25) = 2.4 V

Thevenin resistance of the bias network:
R_th = 100 kΩ || 25 kΩ = 20 kΩ

Since β = 99,
R_E = 500 Ω + 1 kΩ = 1.5 kΩ

Base current:
I_B = (2.4 − 0.7)/(20 kΩ + 100 × 1.5 kΩ)
I_B ≈ 15.88 μA

Therefore,
I_C = 99I_B ≈ 1.572 mA

Transconductance:
g_m = I_C/V_T = 1.572 mA/25 mV ≈ 0.0629 S

At both given frequencies, the coupling and bypass capacitors can be treated approximately as AC shorts. The 1 kΩ emitter resistor is bypassed, while the 500 Ω resistor remains in the emitter path.

Hence, the voltage gain is approximately:
A_v = −g_mR_C/ (1 + g_mR_E})

With R_C = 5 kΩ and R_E = 500 Ω:
A_v = −(0.0629 × 5000)/ (1 + 0.0629 × 500)

A_v ≈ −9.1

A common-emitter amplifier produces a 180° phase inversion, so both frequency components receive the same negative gain.

Therefore,
V_o(t) ≈ −9.1[A cos(10⁵t) + B sin(10⁷t)]

Hence, the correct answer is:

(A) −9.1[A cos(10⁵t) + B sin(10⁷t)]

Question 51

Consider the two-port network as shown in the figure.

Which of the following options provides the correct set of values of A, B, C and D parameters?

(A)   A = 1/2, B = 3 × 10³ Ω, C = 10⁻³ Ω⁻¹, D = 1/2

(B)   A = 2, B = 3 Ω, C = 1 Ω⁻¹, D = 2

(C)   A = 2, B = 6 × 10³ Ω, C = 2 × 10⁻³ Ω⁻¹, D = 2

(D)   A = 2, B = 3 × 10³ Ω, C = 10⁻³ Ω⁻¹, D = 2

Solution:

The network consists of:

• Left series resistance: Z₁ = 1 kΩ
• Shunt resistance: R = 1 kΩ
• Right series resistance: Z₂ = 1 kΩ

For the ABCD parameters, the transmission matrices are:

Series impedance Z:   [1  Z; 0  1]

Shunt admittance Y:   [1  0; Y  1]

Here,
Z = 1 kΩ = 1000 Ω
Y = 1/R = 1/1000 = 10⁻³ S

Therefore,
ABCD = [1  1000; 0  1] [1  0; 10⁻³  1] [1  1000; 0  1]

Multiplying the matrices gives:

A = 1 + ZY
= 1 + (1000)(10⁻³)
= 2

B = Z₁ + Z₂ + Z₁Z₂Y
= 1000 + 1000 + (1000)(1000)(10⁻³)
= 1000 + 1000 + 1000
= 3000 Ω = 3 × 10³ Ω

C = Y = 10⁻³ Ω⁻¹

D = 1 + ZY = 2

Hence,
A = 2, B = 3 × 10³ Ω, C = 10⁻³ Ω⁻¹, D = 2

Therefore, the correct answer is:

(D) A = 2, B = 3 × 10³ Ω, C = 10⁻³ Ω⁻¹, D = 2

Question 52

A complex load (in Ω) is represented as Γ_L = 0.5∠30° on the Smith chart.

A co-axial cable with a characteristic impedance of 50 Ω is connected to the load. The new input impedance of the load now moves to a diametrically opposite point on the same Γ circle on the Smith chart.

Which option is the nearest input impedance of the cable connected load (in Ω)?

(A)   20.7 − j5.1

(B)   17.7 − j11.8

(C)   97.5 − j65.0

(D)   97.5 + j65.0

Solution:

Given:
Γ_L = 0.5∠30°

The diametrically opposite point on the same Γ circle is obtained by changing the phase by 180°.

Therefore,
Γ_in = 0.5∠(30° + 180°) = 0.5∠210°

In rectangular form:
Γ_in = −0.433 − j0.25

The normalized impedance is related to reflection coefficient by:

z = (1 + Γ)/(1 − Γ)

Substituting Γ = −0.433 − j0.25:
z ≈ 0.354 − j0.236

Given characteristic impedance:
Z₀ = 50 Ω

Hence,
Z_in = Z₀z
= 50(0.354 − j0.236)

Z_in ≈ 17.7 − j11.8 Ω

Therefore, the correct answer is:

(B) 17.7 − j11.8 Ω

Question 53

A circuit using an ideal OP-AMP is shown in the figure.

Which of the following options gives the correct value of the current I_x?

(A)   2.0 mA

(B)   1.5 mA

(C)   3.0 mA

(D)   0 mA

Solution:

Since the OP-AMP is ideal and the non-inverting terminal is grounded,

V₋ = V₊ = 0 V

Therefore, the inverting terminal is at virtual ground.

The input current through the 1 kΩ resistor is:

I = (2 − 0)/1 kΩ = 2 mA

Since the input current of an ideal OP-AMP is zero, the entire 2 mA flows through the feedback 1 kΩ resistor.

Let the right-side node voltage be V_A.

Then:
V_A/1 kΩ = 2 mA

Hence,
V_A = 2 V

The right-side 2 kΩ resistor is connected from this node to ground. Therefore, current through it is:

I_load = 2 V / 2 kΩ = 1 mA

At the same node, 2 mA comes through the feedback resistor and 1 mA flows through the load resistor.

Therefore, the remaining current must flow through the 2 kΩ resistor connected to the OP-AMP output:

I_x = 2 mA + 1 mA

I_x = 3 mA

Hence, the correct answer is:

(C) 3.0 mA

Question 54

Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV.

Given: Planck's constant, h = 6.63 × 10⁻³⁴ J s and speed of light in free space is 3 × 10⁸ m s⁻¹.

In which of the following wavelength ranges the LED will NOT emit?

(A)   1410 ± 20 nm

(B)   1090 ± 20 nm

(C)   950 ± 20 nm

(D)   510 ± 20 nm

Solution:

For an LED based on a direct bandgap semiconductor,

E = E_g = 1.3 eV

We know:
E = hc/λ

Therefore,
λ = hc/E

Convert the bandgap energy into joules:
E = 1.3 × 1.6 × 10⁻¹⁹ J

Hence,
λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.3 × 1.6 × 10⁻¹⁹)

λ ≈ 9.56 × 10⁻⁷ m

Therefore,
λ ≈ 956 nm

So the LED emission wavelength is around 950–960 nm.

A wavelength of 510 ± 20 nm corresponds to approximately 490–530 nm, which is much shorter than the required wavelength.

Therefore, the LED will NOT emit in this wavelength range.

Hence, the correct answer is:

(D) 510 ± 20 nm

Question 55

Let the relevant bandwidth (B) of a digital communication system be 1 MHz and kT = −174 dBm/Hz, where k is Boltzmann's constant and T is equivalent noise temperature of the receiver. The power (S) of signal received through an additive Gaussian channel is −80 dBm.

Which of the following options is/are TRUE about Shannon capacity (C) of the channel?

(A)   C = B

(B)   C = 2B

(C)   C > 3B

(D)   C < B

Solution:

Shannon capacity is given by:

C = B log₂(1 + S/N)

Given:
B = 1 MHz
S = −80 dBm
Noise spectral density:
N₀ = −174 dBm/Hz

Total noise power over 1 MHz bandwidth:

N = −174 + 10 log₁₀(10⁶)

N = −174 + 60 = −114 dBm

Therefore, SNR in dB is:
S/N = −80 − (−114) = 34 dB

Thus,
S/N ≈ 10^3.4 ≈ 2512

Hence,
C/B = log₂(1 + 2512)

C/B ≈ log₂(2513) ≈ 11.3

Therefore,
C ≈ 11.3B

Clearly,
C > 3B

Hence, the correct answer is:

(C) C > 3B

Question 56

Consider the four-variable Boolean function,

f(w,x,y,z) = Σm(0, 2, 5, 7, 8, 10, 13, 14, 15)

with w as MSB and z as LSB.

Which of the following expressions is/are the valid form(s) of f(w,x,y,z)?

(A)   x̄z̄ + xz + wxy

(B)   xz + wxy + wx̄z̄ + w̄xyz̄

(C)   x̄z̄ + wxy + wx̄z̄ + w̄xyz̄

(D)   x̄z̄ + xz + wyz̄

Solution:

Given:

f(w,x,y,z) = Σm(0,2,5,7,8,10,13,14,15)

We need to check which expression produces exactly the same minterms.

Option (A):

x̄z̄ + xz + wxy

The expression gives the minterms:

{0, 2, 5, 7, 8, 10, 13, 14, 15}

This is exactly the given set of minterms.

Therefore,
Option (A) is correct.

Option (D):

x̄z̄ + xz + wyz̄

This expression also produces the minterms:

{0, 2, 5, 7, 8, 10, 13, 14, 15}

Hence, it is also equivalent to the given Boolean function.

Therefore,
Option (D) is correct.

Options (B) and (C) do not generate the complete required minterm set.

Final Answer: (A) and (D)

Question 57

Consider a real, narrowband signal x(t) = A(t) cos[2πf_ct + θ(t)] where the maximum frequency components of A(t) and θ(t) are f_m and f_c = 1000f_m, respectively.

Which of the following statements is/are correct for −∞ < t < ∞?

(A)   x(t) represents a PSK modulated signal for suitable choices of A(t) and θ(t).

(B)   x(t) represents an amplitude modulated signal for suitable choices of A(t) and θ(t).

(C)   x(t) represents a band-limited Gaussian noise process.

(D)   x(t) never represents a narrowband FM signal.

Solution:

Given general narrowband signal:

x(t) = A(t) cos[2πf_ct + θ(t)]

This is a general amplitude-phase representation of a narrowband signal.

Option (A): PSK

In PSK, the amplitude remains constant while the phase changes according to the information signal.
Therefore, for suitable choices such as

A(t) = constant
and θ(t) carrying the phase information,

x(t) = A cos[2πf_ct + θ(t)]

can represent a PSK signal.

Hence, (A) is correct.


Option (B): AM

In amplitude modulation, the amplitude varies according to the message while the phase remains fixed.

For example,
θ(t) = constant
and A(t) contains the message information.

Therefore the given representation can represent an AM signal.

Hence, (B) is correct.


Option (C): Band-limited Gaussian noise

A band-limited Gaussian noise process can also be represented in narrowband amplitude-phase form as

x(t) = A(t) cos[2πf_ct + θ(t)]

where the envelope and phase are random processes.

Thus, the given representation is capable of representing a band-limited Gaussian noise process.

Hence, (C) is correct.


Option (D): Narrowband FM

In FM, the amplitude can remain constant while the phase varies according to the integral of the message signal.

Therefore, by choosing
A(t) = constant
and a suitable θ(t), the given expression can represent a narrowband FM signal.

Hence the statement that it never represents FM is false.

Therefore, (D) is incorrect.

Final Answer: (A), (B) and (C)

Question 58

Let x₁(t) = cos(2πnt) and x₂(t) = 2sin(4πnt) represent two sinusoids for a positive integer n and −∞ < t < ∞.

Which of the following statements about x₁(t) and x₂(t) is/are valid?

(A)   x₁(t) and x₂(t) are orthogonal to each other over 0 ≤ t < 1/n.

(B)   x₁(t) and x₂(t) are orthonormal to each other over 0 ≤ t < 1/n.

(C)   x₂(t) is a harmonic of x₁(t).

(D)   x₁(t) and x₂(t) are non-orthogonal to each other over 0 ≤ t < 1/(2n).

Solution:

Given:

x₁(t) = cos(2πnt)
x₂(t) = 2sin(4πnt)

### Orthogonality

Two signals are orthogonal over an interval if:

∫ x₁(t)x₂(t)dt = 0

Over 0 ≤ t < 1/n:

∫ cos(2πnt) · 2sin(4πnt) dt = 0

Therefore, the two signals are orthogonal over 0 ≤ t < 1/n.

Hence (A) is correct.

### Orthonormality

Orthogonal signals are not necessarily normalized. For x₁(t):

∫₀^1/n |x₁(t)|²dt = 1/(2n)

For x₂(t):

∫₀^1/n |x₂(t)|²dt = 1/n

Their energies are not equal to 1, so they are not orthonormal.

Therefore, (B) is incorrect.

### Harmonic Relationship

Frequency of x₁(t):
f₁ = n

Frequency of x₂(t):
f₂ = 2n

Since:
f₂ = 2f₁

x₂(t) is the second harmonic of x₁(t).

Hence (C) is correct.

Therefore, the correct options are:

(A) and (C)

Question 59

Consider the unity negative feedback control system shown in the figure. The value of gain K (> 0) at which the given system will remain marginally stable is ____.

Answer in integer.

Enter your answer:

Solution:

Open-loop transfer function:

G(s) = K / [s(s + 7)(s + 11)]

For unity negative feedback, the characteristic equation is:

1 + G(s) = 0

Therefore,

s(s + 7)(s + 11) + K = 0

Expanding:
s³ + 18s² + 77s + K = 0

Routh array:

s³    1    77
s²    18    K
s¹    (1386 − K)/18    0
s⁰    K

For marginal stability:

1386 − K = 0

Hence,
K = 1386

Therefore, the correct answer is:

1386

Question 60

An n-channel MOSFET is connected as shown in the figure. Assume V_TH = 1 V, V_DD = 5 V, and μC_ox(W/L) = 2 mA V⁻² and neglect channel length modulation effects.

The gate voltage (V_G) of the n-channel MOSFET (in Volt) is ____.

Round off to two decimal places.

Enter your answer:

Solution:

Gate is connected to drain, so the MOSFET operates in saturation region.

Drain current:

I_D = 1/2 μC_ox(W/L) (V_G − V_TH)²

Given:
μC_ox(W/L) = 2 mA/V²
V_TH = 1 V

Therefore,
I_D = (V_G − 1)² mA

Current through the 1 kΩ resistor:

I_D = (5 − V_G) / 1kΩ

In mA:
I_D = 5 − V_G

Equating both currents:

(V_G − 1)² = 5 − V_G

Let:
x = V_G − 1

Then:
x² = 4 − x

Therefore:
x² + x − 4 = 0

Positive root:
x = (−1 + √17)/2

Hence:
V_G = 1 + (−1 + √17)/2

V_G ≈ 2.56 V

Therefore, the correct answer is:

2.56 V

Question 61

Consider the square region R in the X-Y plane as shown with the dark shading in the figure.

The value of ∬_R (x² + y² − 1) dxdy is _____.

(Rounded off to two decimal places)

Solution:

The square has vertices:

(0,0), (1,1), (2,0), (1,-1)

Shift the origin by putting:

u = x − 1,   v = y

Then the region becomes:
|u| + |v| ≤ 1

Also,
x² + y² − 1 = u² + v² + 2u

Because the region is symmetric about the y-axis,
∬_R 2u dA = 0

Therefore,
∬_R(u² + v²)dA

By symmetry:

∬u²dA = ∬v²dA = 1/3

Hence,
Integral = 1/3 + 1/3 = 2/3

2/3 = 0.6667 ≈ 0.67

Correct Answer: 0.67

Question 62

Consider an ideal OP-AMP circuit. The resistances R₁ = R₂ = R₃ = R₄ = 50 kΩ.

The magnitude of the closed loop gain is _____.

(Rounded off to two decimal places)

Solution:

Since the OP-AMP is ideal and the non-inverting terminal is grounded:

V₋ = 0

The feedback network consists of:

R₂ = 50 kΩ in series with
R₃ || R₄

Since R₃ = R₄ = 50 kΩ,

R₃ || R₄ = (50 × 50)/(50 + 50) = 25 kΩ

Therefore, equivalent feedback resistance is:

R_f = 50 + 25 = 75 kΩ

The magnitude of the inverting closed-loop gain is:

|A_v| = R_f/R₁

|A_v| = 75/50 = 1.5

Therefore,

Correct Answer: 1.50

Question 63

The average bit error rate at the input of a (7, 4, 1) Hamming decoder is 0.10.

The probability that the decoder will fail to decode a received word correctly is _____.

(Rounded off to two decimal places)

Solution:

A (7,4,1) Hamming code can correct a maximum of 1 bit error.

Therefore, decoder failure occurs when there are 2 or more errors in the 7-bit received word.

Given:
p = 0.10,   q = 1-p = 0.90

Probability of 0 errors:
P(0) = (0.9)⁷

Probability of exactly 1 error:
P(1) = 7(0.1)(0.9)⁶

Hence probability of successful decoding:
P(success) = P(0)+P(1)

Therefore,
P(failure) = 1 − [(0.9)⁷ + 7(0.1)(0.9)⁶]

P(failure) ≈ 0.1497

Thus,
Correct Answer: 0.15

Question 64

Consider that the concentration of electrons in a semiconductor bar varies linearly from 2 × 10¹⁷ cm⁻³ at x = 1 μm to 1 × 10¹⁶ cm⁻³ at x = 4 μm along the x-direction.

Assume that the concentration of electrons is not varying along any other directions.

Given the mobility of electron is 1400 cm²V⁻¹s⁻¹, thermal voltage is 25 mV and electronic charge is 1.6 × 10⁻¹⁹ C.

The density of electron diffusion current (in A/mm²) is _____.

(Rounded off to two decimal places)

Solution:

Einstein relation:

D_n = μ_n V_T

D_n = 1400 × 0.025 = 35 cm²/s

Electron concentration gradient:

dn/dx = (1 × 10¹⁶ − 2 × 10¹⁷) / (4 − 1) μm

Since 3 μm = 3 × 10⁻⁴ cm

dn/dx = −6.333 × 10²⁰ cm⁻⁴

Electron diffusion current density magnitude:

|J_n| = qD_n|dn/dx|

= (1.6 × 10⁻¹⁹) (35) (6.333 × 10²⁰)

= 3546.67 A/cm²

Conversion:
1 A/cm² = 0.01 A/mm²

Therefore,
|J_n| = 35.47 A/mm²

Correct Answer: 35.47

Question 65

Consider the ideal diodes D1 and D2 with cut-in voltage V_γ = 0 V and v_i(t) is in Volt.

The input voltage is:

v_i(t) = 6 sin(ωt)

The maximum voltage (Volt) of the output v_o(t) is _____.

(Rounded off to two decimal places)

Solution:

The diodes are ideal, therefore:

V_γ = 0 V

During the positive half-cycle, D2 limits the output voltage to the positive reference level of 4 V.

Since the diode is ideal, the limiting voltage is exactly:

v_o,max = 4 V

The input has a peak value of 6 V, so without clipping the output could rise above 4 V. D2 therefore conducts and clamps the positive peak.

Hence,

Correct Answer: 4.00 V