Official Paper

GATE EC 2023 Official Paper (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

“I cannot support this proposal. My ________ will not permit it.”

((a))
conscious
((b))
consensus
((c))
conscience
((d))
consent

Show Answer

Answer: ((c))

conscience

The correct answer is 'conscience'.

Key Points

The correct word to fill in the blank is "conscience".
The sentence implies that the speaker has moral or ethical objections to the proposal and is unable to support it due to their conscience.
The word "conscience" refers to an individual's sense of right and wrong, and their internal moral compass.

Additional Information

Here's a brief explanation of each option:

Conscious: This word refers to being aware of one's surroundings or actions. It does not fit in the given sentence as it does not convey any sense of moral or ethical objection.
Consensus: This word refers to a general agreement or unity of opinion among a group of people. It does not fit in the given sentence as it does not convey any personal objection by the speaker.
Conscience: This word refers to an individual's sense of right and wrong, and their internal moral compass. It fits perfectly in the given sentence as it suggests that the speaker has a personal moral or ethical objection to the proposal.
Consent: This word refers to giving permission or approval for something to happen. It does not fit in the given sentence as it does not convey any sense of objection by the speaker.

Courts : _______ :: Parliament : Legislature

(By word meaning)

((a))
Judiciary
((b))
Executive
((c))
Governmental
((d))
Legal

Show Answer

Answer: ((a))

Judiciary

In India, the Government has three branches: The Executive, The legislature and the Judiciary

Parliament comes under the Legislature of the Union.

Similarly,

Courts come under the Judiciary.

Hence, the correct answer is "Option (1)".

What is the smallest number with distinct digits whose digits add up to 45?

((a))
123555789
((b))
123457869
((c))
123456789
((d))
99999

Show Answer

Answer: ((c))

123456789

Concept:

Our main goal is to find the smallest number with distinct digits with the constraint that the digits add up to 45.
First off, let's disregard the 5th option (4) 99999 because it doesn't satisfy the condition of distinct digits. It only has the digit 9 repeating.
Now, let's explore the options of distinct digits and their sums.
123555789 The sum of these digits is 1+2+3+5+5+5+7+8+9 = 45 But this option doesn't meet the criteria for having distinct digits, as '5' is repeated three times.
123457869
The sum of these digits is 1+2+3+4+5+7+8+6+9 = 45 This satisfies both conditions – the digits add up to 45 and are distinct.
123456789
The sum of these digits is 1+2+3+4+5+6+7+8+9 = 45 But this option is a higher number than option "2".
So, the correct answer is #2) 123457869 because it not only has a sum of 45, which fulfills one condition, but is also the smallest among the options provided that have distinct digits.

In a class of 100 students,

(i) there are 30 students who neither like romantic movies nor comedy movies,

(ii) the number of students who like romantic movies is twice the number of students who like comedy movies, and

(iii) the number of students who like both romantic movies and comedy movies is 20.

How many students in the class like romantic movies?

((a))
40
((b))
20
((c))
60
((d))
30

Show Answer

Answer: ((c))

Given:

There are 100 students from which so students who neither like romantic movies nor comedy movies.

Therefore,

100 -30 = 70

70 students will like either romantic or comedy movies.

The number of students who like romantic movies is twice the number of students who like comedy movies.

Let, the number of students who like comedy movie be x, then the number of students like romantic movies is 2x. The

number of students who like both romantic movies and comedy movies is 20.

The number of students in the class like romantic movies,

⇒ 2x+ x -20 = 70

<br>

⇒ 3x = 90

<br>

⇒ x =30

So,

2x = 2 x 30 = 60

Therefore, the number of students who like romantic movies are 60.

Hence, the correct option is (C).

How many rectangles are present in the given figure?

((a))
8
((b))
9
((c))
10
((d))
12

Show Answer

Answer: ((c))

The total number of rectangles will be 10, as shown below:

Hence, the correct answer is "Option (3)".

Forestland is a planet inhabited by different kinds of creatures. Among other creatures, it is populated by animals all of whom are ferocious. There are also creatures that have claws, and some that do not. All creatures that have claws are ferocious.

Based only on the information provided above, which one of the following options can be logically inferred with certainty?

((a))
All creatures with claws are animals.
((b))
Some creatures with claws are non-ferocious.
((c))
Some non-ferocious creatures have claws.
((d))
Some ferocious creatures are creatures with claws.

Show Answer

Answer: ((d))

Some ferocious creatures are creatures with claws.

The correct inference based on the information provided in the passage is: "Some ferocious creatures are creatures with claws."

Explanation:

1) "All creatures with claws are animals." — This can't be inferred with certainty because while the passage mentions creatures that are animals and creatures that have claws, it doesn't say that all clawed creatures must be animals. There may be non-animal creatures with claws.

2) "Some creatures with claws are non-ferocious." — The passage specifically says "All creatures that have claws are ferocious." Therefore, this statement is contrary to the information provided in the passage.

"Some non-ferocious creatures have claws." — This option contradicts the information provided that "All creatures that have claws are ferocious." Therefore, non-ferocious creatures cannot have claws.

4) "Some ferocious creatures are creatures with claws." — This can be supported by the text. We know that some creatures have claws, and those creatures are ferocious. Therefore, some (but not necessarily all) of the ferocious creatures on Forestland have claws. It's also possible that there are ferocious creatures without claws, but that doesn't impact the correctness of this statement.

Which one of the following options represents the given graph?

((a))
f(x) = x² 2^-|x|
((b))
f(x) = x 2-|x|
((c))
f(x) = |x| 2-x
((d))
f(x) = x 2-x

Show Answer

Answer: ((a))

f(x) = x² 2^-|x|

Concept:

1)The whole graph is above the x-axis i.e. positive for all ‘x’.

An even function is symmetrical along the vertical axis, which is usually referred to as the y-axis

These two conditions satisfy only even power of x ,f(x) = x2 2-|x| which is for any value of x the function f(x) is always positive. Also when we replace x by -x there is no change in the expression hence it is symmetrical about Y axis.

Hence, the correct option is (A)

Which one of the following options can be inferred from the given passage alone?

When I was a kid, I was partial to stories about other worlds and interplanetary travel. I used to imagine that I could just gaze off into space and be whisked to another planet.

((a))
It is a child’s description of what he or she likes.
((b))
It is an adult’s memory of what he or she liked as a child.
((c))
The child in the passage read stories about interplanetary travel only in parts.
((d))
It teaches us that stories are good for children.

Show Answer

Answer: ((b))

It is an adult’s memory of what he or she liked as a child.

The correct answer is "It is an adult’s memory of what he or she liked as a child."

Explanation:

1): "It is a child’s description of what he or she likes." — This option might be a plausible interpretation, but the passage implies an adult looking back at their childhood ("When I was a kid"), rather than a child's current perspective.

2): "It is an adult’s memory of what he or she liked as a child." — This is a valid interpretation because the passage is written from the perspective of someone reflecting on their childhood interests ("When I was a kid, I used to...") which implies that the person is now an adult.

3): "The child in the passage read stories about interplanetary travel only in parts." — This can't be inferred from the passage. While it mentions that the child was "partial to stories about other worlds and interplanetary travel," it doesn't specify that these stories were read only in parts.

4): "It teaches us that stories are good for children." — The passage does indicate that the narrator enjoyed stories as a child, but it does not make a larger point or recommendation about the benefit of stories for all children. This is a subjective experience and can't be used to create a general rule that stories are good for all children.

Out of 1000 individuals in a town, 100 unidentified individuals are covid positive. Due to lack of adequate covid-testing kits, the health authorities of the town devised a strategy to identify these covid-positive individuals. The strategy is to:

(i) Collect saliva samples from all 1000 individuals and randomly group them into sets of 5.

(ii) Mix the samples within each set and test the mixed sample for covid.

(iii) If the test done in (ii) gives a negative result, then declare all the 5 individuals to be covid negative.

(iv) If the test done in (ii) gives a positive result, all 5 individuals are separately tested for covid.

Given this strategy, no more than _______ testing kits will be required to identify all the 100 covid positive individuals irrespective of how they are grouped.

((a))
700
((b))
600
((c))
800
((d))
1000

Show Answer

Answer: ((a))

700

Explanation:

To understand why the answer is 700, we need to break down the testing process and calculate the maximum number of tests required.

Step (i): Collect saliva samples from all 1000 individuals and randomly group them into sets of 5.

To group 1000 individuals into sets of 5, we need 1000/5 = 200 sets.

Step (ii): Mix the samples within each set and test the mixed sample for covid.

We will test 200 mixed samples, each containing 5 individual samples. If all 5 individuals in a set are negative, we can declare all 5 individuals as negative without further testing. If any of the 5 individuals in a set are positive, we move to step (iv).

Step (iii): If the test done in (ii) gives a negative result, then declare all 5 individuals to be covid negative. If all 5 individuals in a set are negative, we can declare all 5 individuals as negative without further testing. Therefore, we do not need any additional tests for that set.

Step (iv): If the test done in (ii) gives a positive result, all 5 individuals are separately tested for covid.

If any of the 5 individuals in a set are positive, we must test each separately.

Therefore, we need 5 additional tests for that set. So, the maximum number of tests required to identify all 100 covid positive individuals is: 200 (mixed sample tests) + 5 x 100 (individual tests for positive sets) = 700 tests

Therefore, no more than 700 testing kits will be required to identify all the 100 covid positive individuals irrespective of how they are grouped."

A 100 cm × 32 cm rectangular sheet is folded 5 times. Each time the sheet is folded, the long edge aligns with its opposite side. Eventually, the folded sheet is a rectangle of dimensions 100 cm × 1 cm.

The total number of creases visible when the sheet is unfolded is _______.

((a))
32
((b))
5
((c))
31
((d))
63

Show Answer

Answer: ((c))

Each time the sheet is folded, the long edge aligns with its opposite side It means that the paper is folded along one side.

Total folds done = 5

therefore, the number of creases visible when the sheet is unfolded = (2⁵ - 1)

= 32 - 1

= 31

Hence, the correct answer is "Option 3".

Electronics and Communication Engineering (55 questions)

Let $\rm v_1=\begin{bmatrix}1\ 2\\ 0\end{bmatrix}$ and $\rm v_2=\begin{bmatrix}2\ 1\\ 3\end{bmatrix}$ be two vectors. The value of the coefficient 𝛼 in the expression v₁ = αv₂ + e, which minimizes the length of the error vector 𝒆, is

((a))
$\frac{7}{2}$
((b))
$\frac{-2}{7}$
((c))
$\frac{2}{7}$
((d))
$\frac{-7}{2}$

Show Answer

Answer: ((c))

$\frac{2}{7}$

Given: e = V₁ - αV₂

e = (i + 2k + 0k) - α(2i + j + 3k)

ê = (1 - 2α)î + (2 - α)ĵ + (0 - 3α)k̂

$|\hat{e}|=\sqrt{(1-2 \alpha)^2+(2-\alpha)^2+(-3 \alpha)^2}$

|ê|² = 5 + 14α² - 8α to be minimum at $\frac{\partial e^2}{\partial \alpha}=28 \alpha-8=0$

∴ $\alpha = \frac{2}{7}$ stationary point

The rate of increase, of a scalar field 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧, in the direction 𝒗 = (2, 1, 2) at a point (0, 2, 1) is

((a))
$\frac{2}{3}$
((b))
$\frac{4}{3}$
((c))
2
((d))
4

Show Answer

Answer: ((b))

$\frac{4}{3}$

Concept:

The gradient of a function f(x, y, z) is given by:

$\rm ∇ f = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j + \frac{\partial f}{\partial z} k$

where i, j, and k are the unit vectors along the x, y, and z-axis respectively.

The gradient of any function f(x, y, z) represents the direction of the greatest rate of increase of any scalar field function.

Calculation:

Given, f(x, y, z) = xyz

$∇ f = \frac{\partial (xyz)}{\partial x} ̂ i + \frac{\partial (xyz)}{\partial y} ̂ j + \frac{\partial (xyz)}{\partial z} ̂ k$

∇ f = (yz)î + (xz)ĵ + (xy)k̂

Gradient of a function f(x, y, z) at P = (0, 2, 1) is:

∇ f = 2î

The rate of increase, of a field in the direction of 𝒗 = $∇ f.{\hat v}$

= $\frac{ 2î .( 2î +{\hat j} +2{\hat k})}{\sqrt{2^2+1^2+2^2}}$

= $\frac{4}{3}$

Let 𝑤⁴ = 16𝑗. Which of the following cannot be a value of 𝑤?

((a))
$\rm 2e^{\frac{j2\pi}{8}}$
((b))
$\rm 2e^{\frac{j\pi}{8}}$
((c))
$\rm 2e^{\frac{j5\pi}{8}}$
((d))
$\rm 2e^{\frac{j9\pi}{8}}$

Show Answer

Answer: ((a))

$\rm 2e^{\frac{j2\pi}{8}}$

The correct option is 1

Given: w⁴ = 16j

w = (2)j^1/4

w = 2(0 + j)^1/4

$w=2\left[e^{j(2 n+1) π / 2}\right]^{1 / 4}=2\left[e^{\frac{j(2 n+1) π}{8}}\right]$

For n = 0, w = 2e^jπ/8

For n = 2, w = 2e⁵π/8

For n = 4, w = 2e9π/8

So, only option (A) cannot be the value of w.

Hence, the correct option is (A).

The value of the contour integral, $\rm \oint_c\left(\frac{z+2}{z^2+2z+2}\right)dz$ , where the contour C is $\rm \left{z:\left|z+1-\frac{3}{2}j\right|=1\right}$ taken in the counter clockwise direction, is

((a))
−𝜋(1 + 𝑗)
((b))
𝜋(1 + 𝑗)
((c))
𝜋(1 − 𝑗)
((d))
−𝜋(1 − 𝑗)

Show Answer

Answer: ((b))

𝜋(1 + 𝑗)

Concept:

$I=\oint_c \frac{z+2}{z^2+2 z+2} d z ; c=\left|z+1-\frac{3}{2} i\right|=1$

Poles are given (z + 1)² + 1 = 0

z + 1 = ±√-1

z = -1 + j, -1 - j

where -1 - i lies outside 'c'

z = (-1, 1) lies inside 'c'.

Using Cauchy’s Theorem:

$\int f\left( z \right)dz=2\pi \left[ sumofresidue \right]$

$\oint_c f(z) d z=2 π i$ Res (f(z), z = -1 + j)

$=2 π i\left(\frac{z+2}{2(z+1)}\right)_{z=-1+i}$

$=2 π i\left(\frac{-1+j+2}{2(-1+j+1)}\right)$

= π(1 + j)

Let the sets of eigenvalues and eigenvectors of a matrix 𝐵 be {𝜆_𝑘 |1 ≤ 𝑘 ≤ 𝑛} and {𝒗_𝑘 |1 ≤ 𝑘 ≤ 𝑛}, respectively. For any invertible matrix 𝑃, the sets of eigenvalues and eigenvectors of the matrix 𝐴, where 𝐵 = 𝑃⁻¹𝐴𝑃, respectively, are

((a))
{𝜆_𝑘 det(𝐴) |1 ≤ 𝑘 ≤ 𝑛} and {𝑃𝒗_𝑘 |1 ≤ 𝑘 ≤ 𝑛}
((b))
{𝜆_𝑘 |1 ≤ 𝑘 ≤ 𝑛} and {𝒗_𝑘 |1 ≤ 𝑘 ≤ 𝑛}
((c))
{𝜆_𝑘 |1 ≤ 𝑘 ≤ 𝑛} and {𝑃𝒗𝑘 |1 ≤ 𝑘 ≤ 𝑛}
((d))
{𝜆_𝑘 |1 ≤ 𝑘 ≤ 𝑛} and {𝑃⁻¹𝒗_𝑘 | 1 ≤ 𝑘 ≤ 𝑛}

Show Answer

Answer: ((c))

{𝜆_𝑘 |1 ≤ 𝑘 ≤ 𝑛} and {𝑃𝒗𝑘 |1 ≤ 𝑘 ≤ 𝑛}

The correct option is 2

Concept:

Matrix B has Eigen value 𝜆𝑘 and Eigenvector V_k .

$\therefore$ BV_K = 𝜆𝑘V_K ............(1)

Also given 𝐵 = 𝑃−1𝐴𝑃

Putting the value of B in equation (1) We get,

(𝑃−1𝐴𝑃)VK = 𝜆𝑘VK

Multiplying with P on both sides, we get

𝐴(𝑃VK) = 𝜆𝑘(𝑃VK)

$\therefore$ Eigenvalue of A =𝜆𝑘

Eigen vector of matrix A = P𝜆𝑘

In a semiconductor, if the Fermi energy level lies in the conduction band, then the semiconductor is known as

((a))
degenerate n-type.
((b))
degenerate p-type.
((c))
non-degenerate n-type.
((d))
non-degenerate p-type.

Show Answer

Answer: ((a))

degenerate n-type.

Concept:

Degenerate semiconductors are those materials that have a very high doping density of the order of 1020.

In n-type semiconductors as the doping density increases, the Fermi level which was just below the conduction band now penetrates the conduction band that is fermi energy level lies inside the conduction band

Similarly, in p-type semiconductors, the doping density increases. The Fermi level penetrates the valence band.

The figure shows a band diagram of a degenerate semiconductor

For an intrinsic semiconductor at temperature 𝑇 = 0 𝐾, which of the following statement is true?

((a))
All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.
((b))
All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons.
((c))
All energy states in the valence and conduction band are filled with holes.
((d))
All energy states in the valence and conduction band are filled with electrons.

Show Answer

Answer: ((a))

All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.

The correct option is 1.

Intrinsic Semiconductors:

A pure semiconductor is called an intrinsic semiconductor. It has thermally generated current carriers.
They have four electrons in the outermost orbit of the atom and atoms are held together by a covalent bond.
Because of fewer charge carriers at room temperature, intrinsic semiconductors have low conductivity so they have no practical use.

Explanation:

An ideal, perfectly pure semiconductor (with no impurities) is called an intrinsic semiconductor.
At absolute zero temperature valence band is full of electrons and the conduction band is empty, hence there are no free electrons in the conduction band and holes in the valence band.
The charge carrier concentration is zero. Hence intrinsic semiconductor behaves like an insulator

A series 𝑅𝐿𝐶 circuit has a quality factor 𝑄 of 1000 at a center frequency of 10⁶ rad/s. The possible values of 𝑅, 𝐿 and C are

((a))
𝑅 = 1 Ω, 𝐿 = 1 𝜇𝐻 and 𝐶 = 1 𝜇F
((b))
𝑅 = 0.1 Ω, 𝐿 = 1 𝜇𝐻 and 𝐶 = 1 𝜇F
((c))
𝑅 = 0.01 Ω, 𝐿 = 1 𝜇𝐻 and 𝐶 = 1 𝜇F
((d))
𝑅 = 0.001 Ω, 𝐿 = 1 𝜇𝐻 and 𝐶 = 1 𝜇F

Show Answer

Answer: ((d))

𝑅 = 0.001 Ω, 𝐿 = 1 𝜇𝐻 and 𝐶 = 1 𝜇F

The correct option is 4

Concept:

RLC CIRCUIT:

An RLC circuit is an electrical circuit consisting of an inductor (L), Capacitor (C), and Resistor (R) it can be connected either parallel or in series.
When the LCR circuit is set to resonate (XL = XC), the resonant frequency is expressed as

$\Rightarrow f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}}}$ ⇒ ${\omega _O} = {1 \over {\sqrt {LC}}}$

Quality factor is

$\Rightarrow Q=\frac{{{\omega }_{0}}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$

Where, XL & XC = Impedance of inductor and capacitor, L, R & C = Inductance, resistance, and capacitance, f = frequency and, ω0 = angular resonance frequency

Calculation:

Given: Q=1000 & ${\omega _O}$ =106 rad/s.

Only option 4 satisfies both these equations.

Hence the correct option is 4

For a MOS capacitor, V_fb and V_t are the flat-band voltage and the threshold voltage, respectively. The variation of the depletion width (W_dep) for varying gate voltage (V_g) is best represented by

((a))
((b))
((c))
((d))

Show Answer

Answer: ((b))

∵ We know V_G < V_FB then accumulation mode.

∴ In accumulation mode W_d = 0 because there is no depletion charge.

Now, V_FB < V_G < V_T⇒ then depletion and inversion mode.

∴ Depletion width is available.

∴ V_G > V_T ⇒ Strong inversion.

∴ Depletion width W_d ⇒ Constant.

And $W_d=\sqrt{\frac{2 \in\left|\phi_S\right|}{q N_S}}$ and $\left|\phi_S\right| \propto V_G$

But after strong inversion, W_d remains constant.

∴ Correction option is (b).

Consider a narrow band signal, propagating in a lossless dielectric medium (𝜀_𝑟 = 4, 𝜇_𝑟 = 1), with phase velocity 𝑣_𝑝 and group velocity 𝑣_𝑔. Which of the following statement is true? (𝑐 is the velocity of light in vacuum.)

((a))
v_p > c, v_g > c
((b))
vp < c, vg > c
((c))
vp > c, vg < c
((d))
vp < c, vg < c

Show Answer

Answer: ((d))

vp < c, vg < c

The correct option is 4

Concept:

In a dielectric medium, the phase velocity (V_P) is given by:

${{\rm{V}}_{\rm{p}}} = \frac{c}{{\sqrt {{\rm{\mu_r }}\epsilon_r} }}$

where, c is the speed of light in a vacuum, ε_r and μ_r are the relative permittivity and permeability of the medium, respectively.

Given: ε_r= 4 and μ_r = 1, we have:

${{\rm{V}}_{\rm{p}}} = \frac{c}{{\sqrt {{\rm{4\times }}1} }}$ ⇒ V_p = $\frac{c}{2}$

Therefore, the phase velocity is less than the speed of light in a vacuum. ( V_P< c)

Now, for a lossless medium, the phase velocity (V_P) is always equal to the group velocity (V_g). So Vg is also less than c.

Hence, the correct answer is:

vp < c, vg < c

In the circuit shown below, V₁ and V₂ are bias voltages. Based on input and output impedances, the circuit behaves as a

((a))
voltage controlled voltage source.
((b))
voltage controlled current source.
((c))
current controlled voltage source.
((d))
current controlled current source.

Show Answer

Answer: ((d))

current controlled current source.

The correct option is 4

Concept:

From the given figure we may conclude that the upper transistor will act as an active load while the lower one will act as a Common gate Amplifier.

Common gate amplifier has low input impedance and high output impedance.

Properties of MOSFET amplifier

Amplifier	Input Resistance	Output Resistance	Voltage gain	Current gain
Common source	High	High	High	High
Common drain	Low	High	High	1
Common gate	High	Low	<1	High

The given circuit acts as a current-controlled current source.

A cascade of common-source amplifiers in a unity gain feedback configuration oscillates when

((a))
the loop gain is less than 1 and the phase shift is less than 180°.
((b))
the loop gain is greater than 1 and the phase shift is less than 180°
((c))
the loop gain is less than 1 and the phase shift is greater than 180°.
((d))
the loop gain is greater than 1 and the phase shift is greater than 180°.

Show Answer

Answer: ((d))

the loop gain is greater than 1 and the phase shift is greater than 180°.

This question has been modified. In the official GATE question, the condition was mentioned on the basics of a closed loop system which is not possible.

Hence MTA was given. Here We have modified our question and the solution is done on the basics of a modified solution.

Concept:

An amplifier is said to oscillate when there is a feedback loop that maintains the ability of the system to continue to generate waveforms. This occurs under a very specific condition that is given by the Barkhausen Stability Criterion.

The Barkhausen Stability Criterion is a mathematical condition to determine when a linear electronic circuit will oscillate. It states that a circuit will oscillate if it meets these two conditions at the same frequency:

The total phase shift around a loop is 0 degrees or an integral multiple of 360 degrees
The total gain around the loop is equal to or greater than 1.

<br>

When we apply the Barkhausen Stability Criterion to the options given, the answer becomes clear.

Option 1: The closed loop gain is less than 1 and the phase shift is less than 180°. This condition will not cause oscillation because the gain is less than 1 and the phase shift condition for oscillation isn't met.

Option 2: The closed loop gain is greater than 1 and the phase shift is less than 180° This condition will not cause oscillation because despite the gain being greater than 1, the phase shift condition for oscillation isn't met.

Option 3: The closed loop gain is less than 1 and the phase shift is greater than 180°. This condition will not cause oscillation because despite the phase shift being greater than 180°, the gain condition for oscillation isn't met.

Option 4: The closed loop gain is greater than 1 and the phase shift is greater than 180°. This condition can cause oscillation. Although having the phase shift exactly 180° is the ideal case for sustained oscillation, a phase shift greater than 180° might also lead to instability which may result in oscillations.

So, option 4 is the closest to fulfilling the conditions of the Barkhausen Stability Criterion for a system to oscillate. But remember, ideal sustained oscillations precisely occur when the product of gain and the phase shift is 1 and 360° (or 0°), respectively. As said before, actual systems may oscillate even when these exact conditions are not met, due to various other complexities.

In the circuit shown below, P and Q are the inputs. The logical function realized by the circuit shown below is

((a))
Y = PQ
((b))
Y = P + Q
((c))
Y = $\rm \overline{PQ}$
((d))
Y = $\rm \overline{P+Q}$

Show Answer

Answer: ((a))

Y = PQ

Concept:

For a 2 × 1 MUX as shown above, the output function F is expressed as:

F = S̅1 I0 + S1I1

i.e. when S1 = 0, I0 is transmitted to the output.

And when S1 = 1, I1 is transmitted to the output.

Calculation:

For the given figure

$Y=\bar Q.0+Q.P$

Y = PQ

The synchronous sequential circuit shown below works at a clock frequency of 1 GHz. The throughput, in Mbits/s, and the latency, in ns, respectively, are

((a))
1000, 3
((b))
333.33, 1
((c))
2000, 3
((d))
333.33, 3

Show Answer

Answer: ((a))

1000, 3

The given circuit is a type of SISO.

∴ Latency = n × T_clk .... n = number of flip flops

= 3 × 1 ..... T_clk = $\frac{1}{f_{clk}}$ = 1 ns

= 3 ns

Now, Throughput = Number of bits/sec

∵ 1 bit = 1 n-sec

∴ Throughput = 10⁹ bits/sec

= 1000 Mbps

Additional Information

SISO:

In an n-bit shift register, to enter n bits of data n clock pulses are required
However, to output data serially (n-1) clock pulses are required

SIPO:

For n bit serial input data, n clock pulses are required
To output data, no clock pulse is required

PISO:

To store n bit data – 1 clock pulse required
To output data (n-1) clock pulses are required

PIPO:

1 clock pulse is required no input data
No clock pulse is required to output data

The open loop transfer function of a unity negative feedback system is $\rm G(s)=\frac{k}{s(1+sT_1)(1+sT_2)}$ , where 𝑘, 𝑇₁ and 𝑇₂ are positive constants. The phase crossover frequency, in rad/s, is

((a))
$\rm \frac{1}{\sqrt{T_1T_2}}$
((b))
$\rm \frac{1}{{T_1T_2}}$
((c))
$\rm \frac{1}{T_1\sqrt{T_2}}$
((d))
$\rm \frac{1}{T_2\sqrt{T_1}}$

Show Answer

Answer: ((a))

$\rm \frac{1}{\sqrt{T_1T_2}}$

Concept:

Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.

It is determined from the gain at the phase cross-over frequency.

$GM = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}$

Phase crossover frequency (ωpc): It is the frequency at which the phase angle of G(s) H(s) is -180°.

$\angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{pc}}}} = - 180^\circ$

Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.

$PM = 180^\circ + \angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{gc}}}} = - 180^\circ$

It is determined from the phase at the gain cross-over frequency.

Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.

Calculation:

Given: $\rm G(s)=\frac{k}{s(1+sT_1)(1+sT_2)}$

$\rm G( j ω )=\frac{k}{j ω (1+ j ω T_1)(1+ j ω T_2)}$

At ${\rm{\omega }} = {{\rm{\omega }}_{{\rm{pc}}}} \Rightarrow \angle {\rm{G}}\left( {{\rm{j\omega }}} \right){\rm{H}}\left( {{\rm{j\omega }}} \right) = - 180^\circ$

$\begin{array}{l} - 90^\circ - {\tan ^{ - 1}}\left( {\rm{\omega }{T_1}} \right) - {\tan ^{ - 1}}\left( {\rm{\omega }{{T_2}}} \right) = - 180^\circ \ {\tan ^{ - 1}}\left( {\rm{\omega }{T_1}} \right) + {\tan ^{ - 1}}\left( {{\rm{\omega }}{T_2}} \right) = 90^\circ \ {\tan ^{ - 1}}\left( {\frac{{{\rm{\omega }{T_1}} + {\rm{\omega }{T_2}}}}{{1 - {{\rm{\omega }}^2}{T_1}{T_2}}}} \right) = 90^\circ \ \therefore 1 - {{\rm{\omega }}^2{T_1}{T_2}} = 0 \Rightarrow {\rm{\omega }} = {{\rm{\omega }}_{{\rm{pc}}}} = \frac{1}{\sqrt {{T_1}{T_2}}} {{{\rm{rad}}}}/{{{\rm{sec}}}} \end{array}$

Consider a system with input 𝑥(𝑡) and output 𝑦(𝑡) = 𝑥(𝑒^𝑡). The system is

((a))
Causal and time invariant.
((b))
Non-causal and time varying.
((c))
Causal and time varying.
((d))
Non-causal and time invariant.

Show Answer

Answer: ((b))

Non-causal and time varying.

Given: 𝑦(𝑡) = 𝑥(𝑒𝑡).

Causality: A system is causal, if the output of the system does not depend on future inputs, but only on past input.

Time-Invariance: If the input to a time-invariant system is shifted in time, its output remains the same signal, but is shifted equally in time.

If the output for an input x(t) is y(t), then for a time shift of t0 in the input gives the t0 shift in the output.

x(t) → y(t), then x(t – t0) → y(t – t0)

First Condition

Putting t= 0 we have

y(0) =𝑥(𝑒0)=𝑥(1)

Here we can see that the present output depends on the future input.

Hence the system is non-causal

Second Condition

Delaying the input,

y(t) = 𝑥(e^t - t_o) ........................(1)

Again delaying the output

y(t -t₀) = 𝑥(et-t₀) ...........................(2)

Here we may conclude that both equations are not equal.

Hence it is a time-variant system.

Let 𝑚(𝑡) be a strictly band-limited signal with bandwidth 𝐵 and energy 𝐸. Assuming 𝜔₀ = 10𝐵, the energy in the signal 𝑚(𝑡) cos 𝜔₀𝑡 is

((a))
$\frac{E}{4}$
((b))
$\frac{E}{2}$
((c))
E
((d))
2E

Show Answer

Answer: ((b))

$\frac{E}{2}$

$Energy: (E) =\frac{1}{2 \pi} \int_{-B}^B(1)^2 \cdot d ω$

$E =\frac{B}{\pi}$

Now, let y(t) = m(t) cos ω₀t

Y(ω) = $\frac{1}{2}$ [M(ω - ω₀) + M(ω + ω₀)]

Here; ω₀ = 10B

Now,

$\text { Energy }\left(E^{\prime}\right)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|Y(\omega)|^2 \cdot d \omega \$

$=\frac{1}{2 \pi}\left[\int_{-11 B}^{-9 B}\left(\frac{1}{2}\right)^2 \cdot d \omega+\int_{9 B}^{11 B}\left(\frac{1}{2}\right)^2 \cdot d \omega\right]$

$=\frac{1}{2 \pi}\left[\frac{1}{4} \times 2 B+\frac{1}{4} \times 2 B\right]$

$E^{\prime} =\frac{B}{2 \pi}=\frac{1}{2}\left(\frac{B}{\pi}\right)=\frac{E}{2}$

The Fourier transform 𝑋(𝜔) of $\rm x(t)=e^{-t^2}$ is

Note: $\rm \int_{-\infty}^{\infty}e^{-y^2}dy=\sqrt{\pi}$

((a))
$\rm \sqrt{\pi}e^{\frac{\omega^2}{2}}$
((b))
$\rm \frac{e^{-\frac{\omega^2}{4}}}{2\sqrt{\pi}}$
((c))
$\rm \sqrt{\pi}e^{-\frac{\omega^2}{4}}$
((d))
$\rm \sqrt{\pi}e^{-\frac{\omega^2}{2}}$

Show Answer

Answer: ((c))

$\rm \sqrt{\pi}e^{-\frac{\omega^2}{4}}$

Calculation:

$e^{-\pi t^2}\leftrightarrow e^{-\pi f^2}$

By the time scaling property of Fourier transform

$e^{-\pi t^2}\leftrightarrow e^{-\pi f^2}$

$As \ x(t)=e^{-at^2}$

$e^{-\pi(\sqrt{\frac{a}{\pi}}t)^2}\leftrightarrow\frac{\sqrt{\pi}}{\sqrt{a}}e^{\frac{-\pi^2f^2}{a}}$

$e^{-\pi(\sqrt{\frac{a}{\pi}}t)^2}\leftrightarrow\sqrt{\frac{\pi}{a}}e^{-\omega^2/4a}$

$e^{-at^2}\leftrightarrow\sqrt{\frac{\pi}{a}}e^{-\omega^2/4a}$ ............(1)

Given: $\rm x(t)=e^{-t^2}$

Substituting the value of a = 1 in the equation 1, we get

$e^{-t^2}\leftrightarrow\sqrt{\frac{\pi}{1}}e^{-\omega^2/4}$

Hence the correct option is 3

In the table shown below, match the signal type with its spectral characteristics.

Signal type	Spectral characteristics
I.	Continuous, aperiodic	a.	Continuous, aperiodic
II.	Continuous, periodic	b.	Continuous, periodic
III.	Discrete, aperiodic	c.	Discrete, aperiodic
IV.	Discrete, periodic	d.	Discrete, periodic

((a))
(𝑖) → (𝑎), (𝑖𝑖) → (𝑏), (𝑖𝑖𝑖) → (𝑐), (𝑖𝑣) → (𝑑)
((b))
(𝑖) → (𝑎), (𝑖𝑖) → (𝑐), (𝑖𝑖𝑖) → (𝑏), (𝑖𝑣) → (𝑑)
((c))
(𝑖) → (𝑑), (𝑖𝑖) → (𝑏), (𝑖𝑖𝑖) → (𝑐), (𝑖𝑣) → (𝑎)
((d))
(𝑖) → (𝑎), (𝑖𝑖) → (𝑐), (𝑖𝑖𝑖) → (𝑑), (𝑖𝑣) → (𝑏)

Show Answer

Answer: ((b))

(𝑖) → (𝑎), (𝑖𝑖) → (𝑐), (𝑖𝑖𝑖) → (𝑏), (𝑖𝑣) → (𝑑)

The correct option is 2

Concept:

A continuous and aperiodic signal has a continuous and aperiodic spectrum.

(2) A continuous and periodic signal has a discrete and aperiodic spectrum.

(3) A discrete and aperiodic signal has a continuous and periodic spectrum.

(4) A discrete and periodic signal has a discrete and periodic spectrum.

For a real signal, which of the following is/are valid power spectral density/densities?

((a))
$\rm S_X(\omega)=\frac{2}{9+\omega^2}$
((b))
$\rm S_X(\omega)=e^{-\omega^2}\cos^2\omega$
((c))
((d))

Show Answer

Answer: ((a))

$\rm S_X(\omega)=\frac{2}{9+\omega^2}$

The correct option is 1 & 2

Concept:

The power spectral density (PSD) is a metric used to quantify the magnitude of energy of a signal for each frequency component present within it. A few properties should hold true for a function to qualify as a valid power spectral density (PSD), for a real signal:

1) The PSD should be real and non-negative for all frequencies, i.e., S_x(ω) ≥ 0 for all ω because power/energy cannot be negative.

2) The PSD should be an even function, i.e., S_X(-ω) = S_X(ω) for all ω.

Clearly, we can see that option 1 & option 2 only satisfy these two conditions

The signal-to-noise ratio (SNR) of an ADC with a full-scale sinusoidal input is given to be 61.96 dB. The resolution of the ADC is ________ bits (rounded off to the nearest integer).

In the circuit shown below, the current i flowing through 200 Ω resistor is ______ mA (rounded off to two decimal places).

For the two port network shown below, the [Y]-parameters is given as

$\rm [Y]=\frac{1}{100}\begin{bmatrix}2&-1\\ -1&4/3\end{bmatrix}S$

The value of load impedance Z_L, in Ω, for maximum power transfer will be _______ (rounded off to the nearest integer).

For the circuit shown below, the propagation delay of each NAND gate is 1 ns. The critical path delay, in ns, is __________ (rounded off to the nearest integer).

In the circuit shown below, switch S was closed for a long time. If the switch is opened at 𝑡 = 0, the maximum magnitude of the voltage V_R, in volts, is _________ (rounded off to the nearest integer).

A random variable X, distributed normally as 𝑁(0, 1), undergoes the transformation Y = h(X), given in the figure. The form of the probability density function of 𝑌 is

(In the options given below, 𝑎, 𝑏, 𝑐 are non-zero constants and 𝑔(𝑦) is piece-wise continuous function)

((a))
𝑎𝛿(𝑦 − 1) + 𝑏𝛿(𝑦 + 1) + 𝑔(𝑦)
((b))
𝑎𝛿(𝑦 + 1) + 𝑏𝛿(𝑦) + 𝑐𝛿(𝑦 − 1) + 𝑔(𝑦)
((c))
𝑎𝛿(𝑦 + 2) + 𝑏𝛿(𝑦) + 𝑐𝛿(𝑦 − 2) + 𝑔(𝑦)
((d))
𝑎𝛿(𝑦 + 2) + 𝑏𝛿(𝑦 − 2) + 𝑔(𝑦)

Show Answer

Answer: ((b))

𝑎𝛿(𝑦 + 1) + 𝑏𝛿(𝑦) + 𝑐𝛿(𝑦 − 1) + 𝑔(𝑦)

X = N(0, 1)

$f_x(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}$

Y = -1; x ≤ -2

0; -1 ≤ x ≤ 1

1; x ≥ 2

x + 1; -2 ≤ x ≤ -1

x - 1; 1 ≤ x ≤ 2

Given that the random variable X, transforms Y = h(X)

From the given figure, it can be concluded that Y takes the discrete set of values {– 1, 0, 1}.

So, the probability density function of Y will consist of impulses at y =-1, y = 0 and y =1.

Y is taking a discrete set of values and a continuous range of values, so it is a mixed random variable.

From the given options, the density function of 'Y' will be.

f_Y(y) = aδ(y + 1) + bδ(y) + cδ(y - 1) + g(y)

The value of the line integral $\rm \int_P^Q(z^2dx+3y^2dy+2xzdz)$ along the straight line joining the points 𝑃 (1, 1, 2) and 𝑄 (2, 3, 1) is

((a))
20
((b))
24
((c))
29
((d))
-5

Show Answer

Answer: ((b))

Given points are P(1, 1, 2) and Q(2, 3, 1)

The line equation is

$\frac{{{\rm{x}} - 1}}{{2 - 1}} = \frac{{{\rm{y}} - 1}}{{3 - 1}} = \frac{{{\rm{z}} - 2}}{{1 - 2}} = {\rm{t}}$

x = t+1, y = 2t+1 and z = 2-t

dx = dt, dy = 2dt and dz = -dt

Here t will vary from 0 to 1

$\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \mathop \smallint \limits_0^1 (z^2dx+3y^2dy+2xzdz)$

$\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \mathop \smallint \limits_0^1 {[(2-t)^2+3(2t+1)^2-2(t+1)(2-t)}]{dt}$

$\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \mathop \smallint \limits_0^1 27{{\rm{t}}^2}{\rm{dt}} + 18{\rm{tdt}}+6{dt}$

$\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \frac{{27}}{3}\left[ {{{\rm{t}}^3}} \right]_0^1 + \frac{18}{2}\left[ {{{\rm{t}}^2}} \right]_0^1+6$

= 9 + 9+ 6

= 24

Let 𝒙 be an 𝑛 × 1 real column vector with length $\rm l=\sqrt{x^Tx}$ . The trace of the matrix 𝑃 = 𝒙𝒙^𝑇 is

((a))
l²
((b))
$\frac{l^2}{4}$
((c))
l
((d))
$\frac{l^2}{2}$

Show Answer

Answer: ((a))

l²

Given,

$l =\sqrt{x^T x}, P=\left(x x^T\right)_{n \times n}$

$(x)_{n \times 1} =\left[\begin{array}{l} x_1 \ x_2 \ x_3 \ \cdot \ \cdot \ x_n \end{array}\right]$

$l =\sqrt{x^{\top} x}=\sqrt{x_1^2+x_2^2+x_3^2+\ldots x_n^2}$

P = xx^T

$=\left[\begin{array}{l} x_1 \ x_2 \ x_3 \ \cdot \ \cdot \ x_n \end{array}\right]\left[\begin{array}{llll} x_1 x_2 & x_3 \ldots x_n \end{array}\right]$

$P=\left[\begin{array}{lllll} x_1^2 & & & & \ & x_1^2 & & & \ & & - & & \ & & & - & \ & & & & x_n^2 \end{array}\right]$

Trace of P = $x_1^2+x_2^2+\ldots . .+x_n^2=l^2$

The $\rm \frac{V_{OUT}}{V_{IN}}$ of the circuit shown below is

((a))
$\rm -\frac{R_4}{R_3}$
((b))
$\rm \frac{R_4}{R_3}$
((c))
$\rm 1+\frac{R_4}{R_3}$
((d))
$\rm 1-\frac{R_4}{R_3}$

Show Answer

Answer: ((a))

$\rm -\frac{R_4}{R_3}$

Here, A₁ is an inverting amplifier and A₂ is a non-inverting amplifier.

$V_{01}=\frac{-R_2}{R_1} V_{i n}$

$V_{02}=\left(1+\frac{R_2}{R_1}\right) V_{i n}$

Also, A₃ is an inverting summing amplifier,

$V_{\text {out }} =\frac{-R_4}{R_3} V_{01}-\frac{R_4}{R_3} V_{02}$ = $\frac{-R_4}{R_3}\left[\frac{R_2}{R_1} V_{\text {in }}+\left(1+\frac{R_2}{R_1}\right) V_{\text {in }}\right]$

$V_{\text {out }} =\frac{-R_4}{R_3} V_{\text {in }}$

$\frac{V_{\text {out }}}{V_{\text {in }}} =\frac{-R_4}{R_3}$

In the circuit shown below, D₁ and D₂ are silicon diodes with cut-in voltage of 0.7 V. V_IN and V_OUT are input and output voltages in volts. The transfer characteristic is

((a))
((b))
((c))
((d))

Show Answer

Answer: ((a))

Case I:

V_γ= 0.7 V

For the +ve half cycle if input V_in,

D₁ → ON and D₂ → OFF

For diode D₁: V_in - 1V > 0.7

V_in > 1.7V

V₀ = V_in - 0.7

Case II:

For the +ve half cycle if input Vin,

D1 → OFF and D2 → ON

For diode D2: 1 - Vin > 0.7

Vin < 0.3V

V0 = Vin + 0.7

Case III:

0.3V < V_in < 1.7V

D₁ → OFF and D₂ → OFF

V₀ = 1V

Transfer characteristics,

A closed loop system is shown in the figure where k > 0 and α > 0. The steady state error due to a ramp input (R(s) = α/s²) is given by

((a))
$\rm \frac{2\alpha}{k}$
((b))
$\rm \frac{\alpha}{k}$
((c))
$\rm \frac{\alpha}{2k}$
((d))
$\rm \frac{\alpha}{4k}$

Show Answer

Answer: ((a))

$\rm \frac{2\alpha}{k}$

Concept:

The steady-state error for a system is defined as:

${e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}$

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Calculation:

Given r(t) = α tu(t)

Applying Laplace Transform

R(s) = α /s²

From the figure G(s)H(s) = $\frac{k}{S(S+2)}$

Steady-state error for a unit ramp input

${e_{ss}} = \frac{\alpha}{{{K_V}}}$

Where

KV = Velocity constant

${K_V} = \mathop {\lim }\limits_{s \to 0} s;G\left( s \right)H\left( s \right)$

${K_V} = \mathop {\lim }\limits_{s \to 0} s\frac{{k}}{{s(s + 2})}$ = $\frac{k}{2}$

Putting the value of K_V we get

${e_{ss}} = \frac{2\alpha}{{{K}}}$

Hence the correct answer is 1

In the following block diagram, R(s) and D(s) are two inputs. The output Y(s) is expressed as Y(s) = G₁(s)R(s) + G₂(s)D(s).

G₁(s) and G₂(s) are given by

((a))
$\rm G_1(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$ and $\rm G_2(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$
((b))
$\rm G_1(s)=\frac{G(s)}{1+G(s)+H(s)}$ and $\rm G_2(s)=\frac{G(s)}{1+G(s)+H(s)}$
((c))
$\rm G_1(s)=\frac{G(s)}{1+G(s)+H(s)}$ and $\rm G_2(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$
((d))
$\rm G_1(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$ and $\rm G_2(s)=\frac{G(s)}{1+G(s)+H(s)}$

Show Answer

Answer: ((a))

$\rm G_1(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$ and $\rm G_2(s)=\frac{G(s)}{1+G(s)+G(s)H(s)}$

$Y(s)=\underbrace{G_1(s) R(s)}{Y_1(s)}+\underbrace{G_2(s) D(s)}{Y_2(s)}$

Considering first R(s) only, then Y(s) is Y₁(s)

$\frac{Y_1(s)}{R(s)} =\frac{\frac{G(s)}{1+G(s) H(s)}}{1+\frac{G(s)}{1+G(s) H(s)}}$

$\frac{Y_1(s)}{R(s)} =\frac{G(s)}{1+G(s) H(s)+G(s)}$

$Y_1(s) =\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] R(s)$

$G_1(s) =\frac{G(s)}{1+G(s)+G(s) H(s)}$

Now considering D(s) only, then Y(s) is Y₂(s)

$\frac{Y_2(s)}{D(s)} =\frac{G(s)}{1+G(s)[1+H(s)]}$

$Y_2(s) =\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] D(s)$

∴ $G_2(s) =\frac{G(s)}{1+G(s)+G(s) H(s)}$

Hence G₁(s) and G₂(s) both are equal.

The state equation of a second order system is

𝒙̇ (𝑡) = A𝒙(𝑡), 𝒙(0) is the initial condition.

Suppose λ₁ and λ₂ are two distinct eigenvalues of A and 𝒗₁ and 𝒗₂ are the corresponding eigenvectors. For constants 𝛼₁ and 𝛼₂, the solution, 𝒙(𝑡), of the state equation is

((a))
$\rm\displaystyle\Sigma^2_{i=1}\alpha_ie^{\lambda_1t}v_i$
((b))
$\rm\displaystyle\Sigma^2_{i=1}\alpha_ie^{2\lambda_1t}v_i$
((c))
$\rm\displaystyle\Sigma^2_{i=1}\alpha_ie^{3\lambda_1t}v_i$
((d))
$\rm \displaystyle\Sigma^2_{i=1}\alpha_ie^{4\lambda_1t}v_i$

Show Answer

Answer: ((a))

$\rm\displaystyle\Sigma^2_{i=1}\alpha_ie^{\lambda_1t}v_i$

The correct option is 1

Concept:

Given: 𝒙̇ (𝑡) = A𝒙(𝑡)

If λ is the eigenvalue of matrix A then 𝒙̇ (𝑡) = λ𝒙(𝑡).

As there are 2 eigenvalues λ1 and λ2 of matrix A, the solution of the state equation will be, $\rm\displaystyle\Sigma^2_{i=1}\alpha_ie^{\lambda_1t}v_i$

Hence, the correct option is (A).

The switch S₁ was closed and S₂ was open for a long time. At 𝑡 = 0, switch S₁ is opened and S₂ is closed, simultaneously. The value of i_c(0⁺), in amperes, is

((a))
1
((b))
-1
((c))
0.2
((d))
0.8

Show Answer

Answer: ((b))

-1

At t = 0^-; S₁ → closed, S₂ → opened

$i_L\left(0^{-}\right)=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A}$

$v_c\left(0^{-}\right)=\frac{1}{5} \times 100=20 \mathrm{~V}$

At t = 0+; S1 → opened, S2 → closed

Applying KCL we get:

$i_x=\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A}$

By KCL; -i_c = i_x + 0.2 = 0.8 + 0.2

⇒ i_c = -1 A

Let a frequency modulated (FM) signal

$\rm x(t)=A\cos(\omega_ct+k_f\int_{-\infty}^{t}m(\lambda)d\lambda)$ , where 𝑚(𝑡) is a message signal of bandwidth W. It is passed through a non-linear system with output 𝑦(𝑡) = 2𝑥(𝑡) + 5(𝑥(𝑡))² . Let 𝐵_𝑇 denote the FM bandwidth. The minimum value of 𝜔_𝑐 required to recover 𝑥(𝑡) from 𝑦(𝑡) is

((a))
B_T + W
((b))
$\rm \frac{3}{2}B_T$
((c))
2B_T + W
((d))
$\rm \frac{5}{2}B_T$

Show Answer

Answer: ((b))

$\rm \frac{3}{2}B_T$

$x(t)=A \cos \left[\omega_c t+K_f \int_{-\infty}^t m(\lambda) d \lambda\right]$

$B.W. [x(t)] → B T=2[\Delta f+\omega]$

$\left.x^2(t) → \begin{array}{c} \Delta f^{\prime}=2 \Delta f \ \omega_C^{\prime}=2 \omega_C \end{array}\right}$

$\mathrm{BW}\left[x^2(t)\right] =2\left[\Delta f^{\prime}+\omega\right]$

$=2[2 \Delta f+\omega]=B T+2 \Delta f$

y(t) = 2x(t) + 5x²(t)

To recover x(t) → $2 \omega_C-\frac{B_T}{2}-\Delta f>\omega_C+\frac{B_T}{2}$

$\omega_C>\Delta f+B T$

$\omega_C>\Delta f+2 \Delta f+2 \omega$

$\omega_C>3 \Delta f+2 \omega$

$\omega_C>\frac{3}{2}{2[\Delta f+\omega]}-\omega$

$\omega_C>\frac{3}{2} B_T-\omega$

Compared to FM BW, message BW is very small. So, that it can be ignored.

$\omega_C >\frac{3}{2} B_T$

${\left[\omega_C\right]_{\min } } =\frac{3}{2} B_T$

The h-parameters of a two port network are shown below. The condition for the maximum small signal voltage gain $\rm \frac{V_{out}}{V_s}$ is

((a))
h₁₁ = 0, h₁₂ = 0, h₂₁ = very high and h₂₂ = 0
((b))
h₁₁ = very high, h₁₂ = 0, h₂₁ = very high and h₂₂ = 0
((c))
h₁₁ = 0, h₁₂ = very high, h₂₁ = very high and h₂₂ = 0
((d))
h₁₁ = 0, h₁₂ = 0, h₂₁ = very high and h₂₂ = very high

Show Answer

Answer: ((a))

h₁₁ = 0, h₁₂ = 0, h₂₁ = very high and h₂₂ = 0

In this question marks to all were given by GATE. The dependent current source notation was wrongly given. It should have been h₂₁I₁ instead of h₂₁V₁

Concept:

The dependent current source should have h₂₁ I₁ instead of h₂₁ V₁ according to the h-parameter.

$A_v=\frac{V_{\text {out }}}{V_s}=\frac{-h_{21} I_1 \times\left(\frac{1}{h_{22}} | R_L\right)}{h_{11} I_1+h_{12} V_2}$

To achieve maximum $\frac{V_{\text {out }}}{V_s}$

h₁₁ = 0, h₁₂ = 0

h₂₁ = Very high, h₂₂ = 0

Hence, the answer should be (a) according to h₂₁I₁.

Consider a discrete-time periodic signal with period 𝑁 = 5. Let the discrete-time Fourier series (DTFS) representation be $\rm x[n]=\Sigma_{k=0}^4a_ke^{\frac{jk2\pi n}{s}}$ , where 𝑎₀ = 1, 𝑎₁ = 3𝑗, 𝑎₂ = 2𝑗, 𝑎₃ = −2𝑗 and 𝑎₄ = −3𝑗. The value of the sum $\rm \Sigma_{n=0}^4x[n]\sin\frac{4\pi n}{5}$ is

((a))
-10
((b))
10
((c))
-2
((d))
2

Show Answer

Answer: ((a))

-10

Let $I=\sum_{n=0}^4 x(n) \sin \frac{4 \pi n}{5}$

$=\frac{1}{2 j} \sum_{n=0}^4 x(n) \cdot\left[e^{j \frac{4 \pi n}{5}}-e^{-j \frac{4 \pi n}{5}}\right]$

$=\frac{1}{2 j}\left[\sum_{n=0}^4 x(n) e^{j \frac{4 \pi n}{5}}-\sum_{n=0}^4 x(n) \cdot e^{-j \frac{4 \pi n}{5}}\right]$ ....(1)

As we know, $a_k =\frac{1}{N} \sum_{n=0}^4 x(n) \cdot e^{-j k \cdot \frac{2 \pi}{N} n}$

$=\frac{1}{5} \sum_{n=0}^4 x(n) \cdot e^{-j \frac{2 \pi}{N} k n}$

Put K = 2; $a_2 =\frac{1}{5} \sum_{n=0}^4 x(n) \cdot e^{-j \frac{4 \pi n}{5}}$

Put K = -2; $a_{-2} =\frac{1}{5} \sum_{n=0}^4 x(n) \cdot e^{j \frac{4 \pi n}{5}}$

From equation (i), $I =\frac{1}{2 j}\left[5 a_{-2}-5 a_2\right]$

$=\frac{5}{2 j}\left[a_3-a_2\right]$ $\left[\begin{array}{rl} a_{-2} & =a_{-2+N} \ & =a_{-2+5} \ & =a_3 \end{array}\right]$

⇒ $I =\frac{5}{2 j}[-2 j-2 j]=-10$

Let an input 𝑥[𝑛] having discrete time Fourier transform

𝑋(𝑒^𝑗𝛺) = 1 − 𝑒^−𝑗𝛺 + 2𝑒^−3𝑗𝛺 be passed through an LTI system. The frequency response of the LTI system is $\rm H(e^{j\Omega)}=1-\frac{1}{2}e^{-j2\Omega}$ . The output 𝑦[𝑛] of the system is

((a))
$\rm \delta[n]+\delta[n-1]-\frac{1}{2}\delta[n-2]-\frac{5}{2}\delta[n-3]+\delta[n-5]$
((b))
$\rm \delta[n]-\delta[n-1]-\frac{1}{2}\delta[n-2]-\frac{5}{2}\delta[n-3]+\delta[n-5]$
((c))
$\rm \delta[n]-\delta[n-1]-\frac{1}{2}\delta[n-2]+\frac{5}{2}\delta[n-3]-\delta[n-5]$
((d))
$\rm \delta[n]+\delta[n-1]+\frac{1}{2}\delta[n-2]+\frac{5}{2}\delta[n-3]+\delta[n-5]$

Show Answer

Answer: ((c))

$\rm \delta[n]-\delta[n-1]-\frac{1}{2}\delta[n-2]+\frac{5}{2}\delta[n-3]-\delta[n-5]$

The correct option is 3

Given: 𝑦[𝑛] = x[n] ∗ h[n]

We know that the convolution in one domain results in convolution in another domain.

Y(𝑒𝑗𝛺) =𝑋(𝑒𝑗𝛺) H(𝑒𝑗𝛺)

= (1 − 𝑒−𝑗𝛺 + 2𝑒−3𝑗𝛺)( $1-\frac{1}{2}e^{-j2\Omega}$ )

=1 - 𝑒−𝑗𝛺 + 2.5𝑒−3𝑗𝛺 - 0.5𝑒−2𝑗𝛺 - 𝑒−5𝑗𝛺

Taking IDFT we have:

y[n] = $\rm \delta[n]-\delta[n-1]-\frac{1}{2}\delta[n-2]+\frac{5}{2}\delta[n-3]-\delta[n-5]$

Let 𝑥(𝑡) = 10 cos(10.5 𝑊𝑡) be passed through an LTI system having impulse response $\rm h(t)=\pi \left(\frac{\sin Wt}{\pi t}\right)^2\cos 10Wt$ . The output of the system is

((a))
$\rm \left(\frac{15W}{4}\right)\cos(10.5Wt)$
((b))
$\rm \left(\frac{15W}{2}\right)\cos(10.5Wt)$
((c))
$\rm \left(\frac{15W}{8}\right)\cos(10.5Wt)$
((d))
(15𝑊) cos(10.5𝑊𝑡)

Show Answer

Answer: ((a))

$\rm \left(\frac{15W}{4}\right)\cos(10.5Wt)$

Given h(t) is Real and Even. When sinusoidal input applied to LTI system having even impulse response, then output will also be sinusoidal.

here, $y(t) =\left.H(W)\right|_{W=10.5 W} \cdot 10 \cos (10.5 W t)$

let, h(t) = f(t) cos 10 Wt

where, $f(t) =\pi\left(\frac{\sin W t}{\pi t}\right)^2$

Now, $H(W)=\frac{1}{2}[F(W+10 W)+F(W-10 W)]$

∴ $\left.H(W)\right|_{W=10.5 W} =\frac{3}{8} W$

Hence, $y(t) =\left(\frac{3}{8} W\right)(10 \cos 10.5 W t)$

= $=\frac{15}{4} W \cos 10.5 W t$

Let x₁(t) and x₂(t) be two band-limited signals having bandwidth 𝐵 = 4𝜋 × 10³ rad/s each. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample y(t), is

((a))
20𝜋 × 10³
((b))
40𝜋 × 10³
((c))
8𝜋 × 10³
((d))
32𝜋 × 10³

Show Answer

Answer: ((d))

32𝜋 × 10³

Given that, x₁(t) and x₂(t) are two bandlimited signals having bandwidth B = 4π × 10³ rad/sec

and y(t) = x₂(t)cos(12π × 10³t) + x₁(t) cos(4π × 10³t)

So, Nyquist rate = 2ω_max

= 2[16π × 10³]

= 32π × 10³ rad/sec

The S-parameters of a two port network is given as

$\rm [S]=\begin{bmatrix}S_{11}&S_{12}\\ S_{21}&S_{22}\end{bmatrix}$

with reference to 𝑍₀. Two lossless transmission line sections of electrical lengths 𝜃₁ = 𝛽𝑙₁ and 𝜃₂ = 𝛽𝑙₂ are added to the input and output ports for measurement purposes, respectively. The S-parameters [𝑆′] of the resultant two port network is

((a))
$\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}$
((b))
$\rm \begin{bmatrix}S_{11}e^{j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{j2\theta_2}\end{bmatrix}$
((c))
$\rm \begin{bmatrix}S_{11}e^{j2\theta_1}&S_{12}e^{j(\theta_1+\theta_2)}\\ S_{21}e^{j(\theta_1+\theta_2)}&S_{22}e^{j2\theta_2}\end{bmatrix}$
((d))
$\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{j(\theta_1+\theta_2)}\\ S_{21}e^{j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}$

Show Answer

Answer: ((a))

$\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}$

$\left[\begin{array}{l} V_1^{-} \ V_2^{-} \end{array}\right]=\left[\begin{array}{ll} S_{11} & S_{12} \ S_{21} & S_{22} \end{array}\right]\left[\begin{array}{l} V_1^{+} \ V_2^{+} \end{array}\right]$ ....(1)

To be calculated,

$\begin{aligned} & {\left[\begin{array}{l} V_{1 N}^{-} \ V_{2 N}^{-} \end{array}\right]=\left[\begin{array}{ll} S_{11 N} & S_{12 N} \ S_{21 N} & S_{22 N} \end{array}\right]\left[\begin{array}{l} V_{1 N}^{+} \ V_{2 N}^{+} \end{array}\right]} \ \end{aligned}$

$V_1^{+}=V_{1 N}^{+} e^{-j \theta_1}$

$V_{1 N}^{-}=V_1^{-} e^{-j \theta_1}=V_1^{-}=V_{1 N}^{-} e^{j \theta_1}$

$V_2^{+}=V_{2 N}^{+} e^{-j \theta_2}$

$V_{2 N}^{-}=V_2^{-} e^{-j \theta_2}=V_2^{-}=V_{2 N}^{-} e^{j \theta_2}$

Feeding the data in equation (i),

$\left[\begin{array}{c} V_{1 N}^{-} \ V_{2 N}^{-} \end{array}\right]=\left[\begin{array}{cc} S_{11 N} e^{-j\left(2 \theta_1\right)} & S_{12 N} e^{-j\left(\theta_1+\theta_2\right)} \ S_{21 N} e^{-j\left(\theta_1+\theta_2\right)} & S_{22 N} e^{-j 2\left(\theta_2\right)} \end{array}\right]$

Hence, the correct option is (A).

The standing wave ratio on a 50 Ω lossless transmission line terminated in an unknown load impedance is found to be 2.0. The distance between successive voltage minima is 30 cm and the first minimum is located at 10 cm from the load. 𝑍_𝐿 can be replaced by an equivalent length 𝑙_𝑚 and terminating resistance 𝑅_𝑚 of the same line. The value of 𝑅_𝑚 and 𝑙_𝑚, respectively, are

((a))
𝑅_𝑚 =100 Ω, 𝑙_𝑚 = 20 cm
((b))
𝑅_𝑚 = 25 Ω, 𝑙_𝑚 = 20 cm
((c))
𝑅_𝑚 =100 Ω, 𝑙_𝑚 = 5 cm
((d))
𝑅_𝑚 = 25 Ω, 𝑙_𝑚 = 5 cm

Show Answer

Answer: ((a))

𝑅_𝑚 =100 Ω, 𝑙_𝑚 = 20 cm

Given S = 2, Z_min = 10 cm, Z₀ = 50 Ω

As we know, $|\Gamma|=\frac{S-1}{S+1}=\frac{2-1}{2+1}=\frac{1}{3}$

Now, the distance between successive voltage minima = 30 cm

⇒ $\frac{\lambda}{2} =30 \mathrm{~cm}$

⇒ $\lambda =60 \mathrm{~cm}$

Also, for minima,

$2 \beta Z_{\min }=(2 n+1) \pi+\theta_{\Gamma}$

At n = 0, 1st minima Z_min = 10 cm

$\frac{4 \pi}{\lambda} Z_{\min } =\pi+\theta_{\Gamma}$

⇒ $\frac{4 \pi}{60} * 10 =\pi+\theta_{\Gamma}$

⇒ $\frac{2 \pi}{3}-\pi =\theta_{\Gamma}$

⇒ $\theta_{\Gamma} =\frac{-\pi}{3} \quad \therefore \Gamma=\frac{1}{3} \angle-60^{\circ}$

Now, $\Gamma =\frac{Z_L-Z_0}{Z_L+Z_0}$

The impedance at a distance of L is

$Z_L =Z_0[\frac{1+\Gamma}{1-\Gamma}]$

$Z_L = 50[\frac{1+0.33 e^{-jπ/3}}{1-0.33 e^{-jπ/3}}]$

⇒Z_L = $67.97 \angle-32.67^{\circ}$

Also ${{Z}{in}}={{Z}{0}}\left( \frac{{{Z}{L}}+j{{Z}{0}}\tan \beta l}{{{Z}{0}}+j{{Z}{L}}\tan \beta l} \right)$

Substituting Z_L =R_m & Z₀ = 50 Ω

${{Z}{in}}={{Z}{0}}\left( \frac{{R_m}+j50\tan \beta l}{{50}+j{R_m}\tan \beta l} \right)$

Substituting the values of Rm & lm, we get

Both option B & C satisfies this option

Hence the correct option is B, C

The electric field of a plane electromagnetic wave is

𝑬 = 𝒂_𝒙𝐶_1𝑥 cos(𝜔𝑡 − 𝛽𝑧) + 𝒂_𝒚𝐶_1𝑦 cos(𝜔𝑡 − 𝛽𝑧 + 𝜃) V/m.

Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?

((a))
𝐶_1𝑥 = 1, 𝐶_1𝑦 = 1, 𝜃 = 𝜋/4
((b))
𝐶1𝑥 = 2, 𝐶1𝑦 = 1, 𝜃 = 𝜋/2
((c))
𝐶1𝑥 = 1, 𝐶1𝑦 = 2, 𝜃 = 3𝜋/2
((d))
𝐶1𝑥 = 2, 𝐶1𝑦 = 1, 𝜃 = 3𝜋/4

Show Answer

Answer: ((a))

𝐶_1𝑥 = 1, 𝐶_1𝑦 = 1, 𝜃 = 𝜋/4

Given, $ \vec{E}=\hat{a}x C{1 x} \cos (ω t-\beta z)+\hat{a}y C{1 y} \cos (ω t-\beta z+\theta) $

at z = 0

$\vec{E}=C_{1 x} \cos ω t \hat{a}x+C{1 y} \cos (ω t+\theta) \hat{a}_y $

Going by options,

Option (a) $\vec{E}=\cos ω t \hat{a}_x+\cos (ω t+π / 4) \hat{a}_y$

at t = 0, ωt = 0, $\vec{E}=\hat{a}_x+\frac{1}{\sqrt{2}} \hat{a}_y$

at t = T/4, ωt = π/2, $\vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_y$

Hence, it is LHEP.

option (b) $\vec{E}=2 \cos \omega t \hat{a}_x+\cos (\omega t+\pi / 2) \hat{a}_y$

at t = 0, ωt = 0, $\vec{E}=2\hat{a}_x$

at t = T/4, ωt = π/2, $\vec{E}=-1\hat{a}_x$

⇒ Hence it is LHEP.

option (c) $\vec{E}=\cos \omega t \hat{a}_x+2 \cos (\omega t+3 \pi / 2) \hat{a}_y$

at t = 0, ωt = 0, $\vec{E}=\hat{a}_x$

at t = T/4, ωt = π/2, $\vec{E}=2\hat{a}_y$

⇒ Hence, it is RHEP.

Option (d) $\vec{E}=2 \cos \omega t \hat{a}_x+\cos (\omega t+3 \pi / 4) \hat{a}_y$

at t = 0, ωt = 0, $\vec{E} =2 \hat{a}_x-\frac{1}{\sqrt{2}} \hat{a}_y$

at t = T/4, ωt = π/2, $\vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_y=\frac{-1}{\sqrt{2}} \hat{a}_y$

⇒ Hence, it is LHEP.

∴ Option (a), (b) and (d) are correct.

The following circuit(s) representing a lumped element equivalent of an infinitesimal section of a transmission line is/are

((a))
((b))
((c))
((d))

Show Answer

Answer: ((a))

Option b,c,d

Any transmission line can be represented in terms of primary constants R, L, G and C as shown below

<br>

It can be seen from the above representation that R and L are in series, G and C are in parallel.

With this logic, options (B), (C) and (D) will be the appropriate answer.

Hence, the correct options are (B), (C) & (D).

The value of the integral $\rm \iint_R xydxdy$ over the region R, given in the figure, is _______ (rounded off to the nearest integer).

In an extrinsic semiconductor, the hole concentration is given to be 1.5𝑛_𝑖 where 𝑛_𝑖 is the intrinsic carrier concentration of 1 × 10¹⁰ 𝑐𝑚⁻³ . The ratio of electron to hole mobility for equal hole and electron drift current is given as __________ (rounded off to two decimal places).

The asymptotic magnitude Bode plot of a minimum phase system is shown in the figure. The transfer function of the system is $\rm (s)=\frac{k(s+z)^a}{s^b(s+p)^c}$ , where 𝑘, 𝑧, 𝑝, 𝑎, 𝑏 and 𝑐 are positive constants. The value of (𝑎 + 𝑏 + 𝑐) is __________(rounded off to the nearest integer).

Let x₁(t) = u(t + 1.5) − u(t − 1.5) and x₂(t) is shown in the figure below. For y(t) = x₁(t) ∗ x₂(t), the $\rm \int_{-\infty}^{\infty}y(t)dt$ is ________ (rounded off to the nearest integer).

Let 𝑋(𝑡) be a white Gaussian noise with power spectral density $\frac{1}{2}$ W/Hz. If 𝑋(𝑡) is input to an LTI system with impulse response 𝑒^−𝑡𝑢(𝑡). The average power of the system output is ___________ W (rounded off to two decimal places).

A transparent dielectric coating is applied to glass (𝜀_𝑟 = 4, 𝜇_𝑟 = 1) to eliminate the reflection of red light (𝜆₀ = 0.75 μm). The minimum thickness of the dielectric coating, in μm, that can be used is _________ (rounded off to two decimal places).

In a semiconductor device, the Fermi-energy level is 0.35 eV above the valence band energy. The effective density of states in the valence band at T = 300 K is 1 × 10¹⁹ cm⁻³ . The thermal equilibrium hole concentration in silicon at 400 K is ____________ × 10¹³ cm⁻³ (rounded off to two decimal places).

Given kT at 300 K is 0.026 eV

A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of 1 μs. The time for the sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of 1 V with 12-bit accuracy is _______ μs (rounded off to two decimal places).

In a given sequential circuit, initial states are Q₁ = 1 and Q₂ = 0. For a clock frequency of 1 MHz, the frequency of signal Q₂ in kHz, is ___________ (rounded off to the nearest integer).

In the circuit below, the voltage V_L is ____________ V (rounded off to two decimal places).

The frequency of occurrence of 8 symbols (a-h) is shown in the table below. A symbol is chosen and it is determined by asking a series of “yes/no” questions which are assumed to be truthfully answered. The average number of questions when asked in the most efficient sequence, to determine the chosen symbol, is _______ (rounded off to two decimal places).

Symbol	a	b	c	d	e	f	g	h
Frequency of occurrence	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{16}$	$\frac{1}{32}$	$\frac{1}{64}$	$\frac{1}{128}$	$\frac{1}{128}$

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