GATE 2023 EC Paper: Worked Solutions to All 65 Questions

Q.1

General Aptitude1 markMCQ

"I cannot support this proposal. My ________ will not permit it." Which word fills the blank?

Option A: conscious
Option B: consensus
Option C: conscience
Option D: consent

Full question on page 1 of the official paper (PDF)

Show answerHide answer

Key idea: The blank needs a noun for an inner moral sense, the thing that permits or forbids.

Official answer: C

A conscience is a person's sense of right and wrong, and it is what would or wouldn't permit supporting a proposal.

Conscious is an adjective, consensus is a group's agreement, not one person's, and consent is permission given, which doesn't itself permit anything. Option (C).

Q.2

General Aptitude1 markMCQ

Courts : ________ : : Parliament : Legislature (by word meaning). Which word fills the blank?

Option A: Judiciary
Option B: Executive
Option C: Governmental
Option D: Legal

Full question on page 1 of the official paper (PDF)

Show answerHide answer

Key idea: Parliament is the body that makes up the legislature; find the branch that courts make up.

Official answer: A

Parliament belongs to the legislature, the law-making branch. Courts belong to the judiciary, the branch that interprets and applies the law. Option (A).

The executive is the third branch, and governmental and legal are adjectives, not branches.

Q.3

General Aptitude1 markMCQ

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

Option A: 123555789
Option B: 123457869
Option C: 123456789
Option D: 99999

Full question on page 2 of the official paper (PDF)

Show answerHide answer

Key idea: 0 to 9 add up to 45, so the digits must be 1 to 9 (a 0 adds nothing and can't lead), placed in rising order.

Official answer: C

The digits 0 to 9 add up to 45. A number with distinct digits summing to 45 must therefore use 1 to 9 (adding 0 only makes it longer, so bigger). The smallest arrangement puts them in rising order: 123456789.

(A) repeats the 5 and (D) repeats the 9, so neither has distinct digits, and (B) uses the right digits in a larger order. Option (C).

Q.4

General Aptitude1 markMCQ

In a class of 100 students, (i) 30 students like neither romantic movies nor comedy movies, (ii) the number who like romantic movies is twice the number who like comedy movies, and (iii) 20 students like both. How many students like romantic movies?

Option A: 40
Option B: 20
Option C: 60
Option D: 30

Full question on page 2 of the official paper (PDF)

Show answerHide answer

Key idea: Students who like at least one kind: $n(R \cup C) = n(R) + n(C) - n(R \cap C)$ .

Official answer: C

Taking away the 30 who like neither leaves $100 - 30 = 70$ in the union. With $n(C) = c$ and $n(R) = 2c$ :

2c + c - 20 = 70 \quad\Rightarrow\quad c = 30,\quad n(R) = 60

\begin{aligned} 2c + c - 20 &= 70 \\ \Rightarrow\; \\ c &= 30, \\ n(R) &= 60 \end{aligned}

Option (C).

Learn the idea:

Q.5

General Aptitude1 markMCQ

How many rectangles are present in the figure? It is a square with both diagonals drawn and a tilted square joining the midpoints of its sides.

A square with both of its diagonals and a tilted square joining the midpoints of its four sides, forming a lattice of diamonds. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: 8
Option B: 9
Option C: 10
Option D: 12

Full question on page 3 of the official paper (PDF)

Show answerHide answer

Key idea: Apart from the outer square, every rectangle is tilted: its sides lie on the three lines of each diagonal direction.

Official answer: C

Inside the square there are three parallel lines in each diagonal direction: the diagonal itself and the two sides of the inner tilted square. A tilted rectangle needs two lines from each direction, and every such choice has all four corners inside the outer square:

\binom{3}{2} \times \binom{3}{2} = 9

These are the four small squares, the four rectangles made of two small squares, and the inner tilted square. With the outer square, $9 + 1 = 10$ . Option (C).

Learn the idea:

Q.6

General Aptitude2 marksMCQ

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 this information, which option can be logically inferred with certainty?

Option A: All creatures with claws are animals.
Option B: Some creatures with claws are non-ferocious.
Option C: Some non-ferocious creatures have claws.
Option D: Some ferocious creatures are creatures with claws.

Full question on page 4 of the official paper (PDF)

Show answerHide answer

Key idea: Clawed creatures exist and all of them are ferocious, so the two sets overlap.

Official answer: D

Creatures with claws exist, and every one of them is ferocious, so at least some ferocious creatures have claws: (D) follows.

(B) and (C) contradict "all creatures that have claws are ferocious", and (A) isn't stated: a clawed creature could be ferocious without being an animal. Option (D).

Q.7

General Aptitude2 marksMCQ

Which one of the following options represents the graph in the figure?

Graph of f(x): zero at x = 0, rising to a peak on each side of the origin, symmetric about the vertical axis, and decaying towards zero far from the origin on both sides. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $f(x) = x^2\, 2^{-|x|}$
Option B: $f(x) = x\, 2^{-|x|}$
Option C: $f(x) = |x|\, 2^{-x}$
Option D: $f(x) = x\, 2^{-x}$

Full question on page 5 of the official paper (PDF)

Show answerHide answer

Key idea: The graph is symmetric about the y-axis and never negative, so $f$ must be even and non-negative.

Official answer: A

Test each option for an even, non-negative function:

(A) $x^2 2^{-|x|}$ : even, zero at 0, positive elsewhere, and decays as $|x|$ grows. It matches.
(B) $x\, 2^{-|x|}$ is odd: negative for $x < 0$ .
(C) $|x|\, 2^{-x}$ grows without bound as $x \to -\infty$ .
(D) $x\, 2^{-x}$ is negative for $x < 0$ and grows in size as $x \to -\infty$ .

Option (A). Its peaks are where $2x - x^2 \ln 2 = 0$ , at $x = \pm 2/\ln 2 \approx \pm 2.885$ .

Learn the idea:

Q.8

General Aptitude2 marksMCQ

Which one of the following options can be inferred from the 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." [Excerpt from The Truth about Stories by T. King]

Option A: It is a child's description of what he or she likes.
Option B: It is an adult's memory of what he or she liked as a child.
Option C: The child in the passage read stories about interplanetary travel only in parts.
Option D: It teaches us that stories are good for children.

Full question on page 6 of the official paper (PDF)

Show answerHide answer

Key idea: "When I was a kid" and the past tense mean the speaker is no longer a child.

Official answer: B

The narrator looks back ("When I was a kid", "I used to imagine"), so this is an adult remembering a childhood liking. Option (B).

(A) needs the speaker to be a child now. Partial to means fond of, not reading in parts, so (C) misreads it, and the passage draws no lesson, so (D) goes beyond it.

Q.9

General Aptitude2 marksMCQ

Out of 1000 individuals in a town, 100 unidentified individuals are covid positive. To identify them with few testing kits, the authorities (i) collect saliva samples from all 1000 and randomly group them into sets of 5, (ii) mix the samples in each set and test the mixed sample, (iii) if that test is negative, declare all 5 negative, and (iv) if it is positive, test all 5 separately. With this strategy, no more than ____ testing kits will be required to identify all 100 positives, however they are grouped.

Option A: 700
Option B: 600
Option C: 800
Option D: 1000

Full question on page 7 of the official paper (PDF)

Show answerHide answer

Key idea: The worst case puts every positive in a different group, so as many groups as possible need five more tests.

Official answer: A

There are $1000 / 5 = 200$ groups, so 200 kits test the mixed samples. Each positive group then needs 5 more kits. The most positive groups there can be is 100, one positive in each:

200 + 100 \times 5 = 700

Option (A).

Q.10

General Aptitude2 marksMCQ

A 100 cm × 32 cm rectangular sheet is folded 5 times. Each time, the long edge aligns with its opposite side. In the end, the folded sheet is a rectangle of 100 cm × 1 cm. How many creases are visible when the sheet is unfolded?

Option A: 32
Option B: 5
Option C: 31
Option D: 63

Full question on page 8 of the official paper (PDF)

Show answerHide answer

Key idea: Each fold halves the 32 cm side, and in the end the creases split it into 1 cm strips.

Official answer: C

Five halvings take the 32 cm side to $32 / 2^5 = 1$ cm. Unfolded, the sheet is divided into 32 strips of 1 cm, and 32 strips have 31 creases between them. Fold by fold, the creases add up as $1 + 2 + 4 + 8 + 16 = 31$ . Option (C).

Q.11

Engineering Mathematics1 markMCQ

Let $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$ . What value of the coefficient $\alpha$ in $\mathbf{v}_1 = \alpha \mathbf{v}_2 + \mathbf{e}$ minimises the length of the error vector $\mathbf{e}$ ?

Option A: $\dfrac{7}{2}$
Option B: $-\dfrac{2}{7}$
Option C: $\dfrac{2}{7}$
Option D: $-\dfrac{7}{2}$

Full question on page 9 of the official paper (PDF)

Show answerHide answer

Key idea: The error is shortest when it is perpendicular to $\mathbf{v}_2$ , which makes $\alpha \mathbf{v}_2$ the projection of $\mathbf{v}_1$ .

Official answer: C

Setting $\mathbf{e} \cdot \mathbf{v}_2 = 0$ gives the projection coefficient:

\alpha = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} = \frac{2 + 2 + 0}{4 + 1 + 9} = \frac{4}{14} = \frac{2}{7}

\begin{aligned} \alpha &= \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \\ &= \frac{2 + 2 + 0}{4 + 1 + 9} \\ &= \frac{4}{14} \\ &= \frac{2}{7} \end{aligned}

Option (C).

Learn the idea:

Q.12

Engineering Mathematics1 markMCQ

What is the rate of increase of the scalar field $f(x, y, z) = xyz$ in the direction $\mathbf{v} = (2, 1, 2)$ at the point $(0, 2, 1)$ ?

Option A: $\dfrac{2}{3}$
Option B: $\dfrac{4}{3}$
Option C: 2
Option D: 4

Full question on page 9 of the official paper (PDF)

Show answerHide answer

Key idea: The rate of increase along a direction is the gradient dotted with the unit vector in that direction.

Official answer: B

\nabla f = (yz,\ xz,\ xy) = (2,\ 0,\ 0) \ \text{at}\ (0, 2, 1), \qquad \hat{\mathbf{v}} = \frac{(2, 1, 2)}{3}

\begin{aligned} \nabla f &= (yz,\ xz,\ xy) \\ &= (2,\ 0,\ 0) \ \text{at}\ (0, 2, 1), \\ \hat{\mathbf{v}} &= \frac{(2, 1, 2)}{3} \end{aligned}

\nabla f \cdot \hat{\mathbf{v}} = \frac{2 \times 2}{3} = \frac{4}{3}

Option (B). Forgetting to normalise $\mathbf{v}$ gives 4, option (D).

Learn the idea:

Q.13

Engineering Mathematics1 markMCQ

Let $w^4 = 16j$ . Which of the following cannot be a value of $w$ ?

Option A: $2e^{j2\pi/8}$
Option B: $2e^{j\pi/8}$
Option C: $2e^{j5\pi/8}$
Option D: $2e^{j9\pi/8}$

Full question on page 10 of the official paper (PDF)

Show answerHide answer

Key idea: Write $16j$ in polar form; its four fourth roots have modulus 2 and angles a quarter turn apart.

Official answer: A

16j = 16 e^{j(\pi/2 + 2\pi k)} \quad\Rightarrow\quad w = 2 e^{j(\pi/8 + k\pi/2)},\ k = 0, 1, 2, 3

\begin{aligned} 16j &= 16 e^{j(\pi/2 + 2\pi k)} \\ \Rightarrow\; \\ w &= 2 e^{j(\pi/8 + k\pi/2)},\ k \\ &= 0, 1, 2, 3 \end{aligned}

The angles are $\pi/8$ , $5\pi/8$ , $9\pi/8$ and $13\pi/8$ . $2\pi/8$ is not among them: $(2e^{j\pi/4})^4 = 16e^{j\pi} = -16$ . Option (A).

Learn the idea:

Q.14

Engineering Mathematics1 markMCQ

What is the value of the contour integral $\oint_C \dfrac{z + 2}{z^2 + 2z + 2}\, dz$ , where $C$ is the circle $\{z : |z + 1 - \tfrac{3}{2} j| = 1\}$ , taken counter-clockwise?

Option A: $-\pi(1 + j)$
Option B: $\pi(1 + j)$
Option C: $\pi(1 - j)$
Option D: $-\pi(1 - j)$

Full question on page 10 of the official paper (PDF)

Show answerHide answer

Key idea: Find which poles lie inside the circle, then use the residue theorem.

Official answer: B

The poles are the roots of $z^2 + 2z + 2$ , $z = -1 \pm j$ . The circle has centre $-1 + 1.5j$ and radius 1, so $-1 + j$ (distance 0.5) is inside and $-1 - j$ (distance 2.5) is not.

\text{Res}_{z = -1 + j} = \frac{z + 2}{z - (-1 - j)}\bigg|_{z = -1 + j} = \frac{1 + j}{2j}

\begin{aligned} \text{Res}_{z = -1 + j} &= \frac{z + 2}{z - (-1 - j)}\bigg|_{z = -1 + j} \\ &= \frac{1 + j}{2j} \end{aligned}

\oint_C = 2\pi j \times \frac{1 + j}{2j} = \pi(1 + j)

\begin{aligned} \oint_C &= 2\pi j \times \frac{1 + j}{2j} \\ &= \pi(1 + j) \end{aligned}

Option (B).

Learn the idea:

Q.15

Engineering Mathematics1 markMCQ

The eigenvalues and eigenvectors of a matrix $B$ are $\{\lambda_k \mid 1 \le k \le n\}$ and $\{\mathbf{v}_k \mid 1 \le k \le n\}$ . For any invertible matrix $P$ , what are the eigenvalues and eigenvectors of the matrix $A$ , where $B = P^{-1} A P$ ?

Option A: $\{\lambda_k \det(A)\}$ and $\{P\mathbf{v}_k\}$
Option B: $\{\lambda_k\}$ and $\{\mathbf{v}_k\}$
Option C: $\{\lambda_k\}$ and $\{P\mathbf{v}_k\}$
Option D: $\{\lambda_k\}$ and $\{P^{-1}\mathbf{v}_k\}$

Full question on page 11 of the official paper (PDF)

Show answerHide answer

Key idea: Similar matrices share eigenvalues; multiply $B\mathbf{v}_k = \lambda_k \mathbf{v}_k$ by $P$ to see the eigenvectors.

Official answer: C

P^{-1} A P \mathbf{v}_k = \lambda_k \mathbf{v}_k \quad\Rightarrow\quad A (P\mathbf{v}_k) = \lambda_k (P\mathbf{v}_k)

\begin{aligned} P^{-1} A P \mathbf{v}_k &= \lambda_k \mathbf{v}_k \\ \Rightarrow\; \\ A (P\mathbf{v}_k) &= \lambda_k (P\mathbf{v}_k) \end{aligned}

So $A$ has the same eigenvalues $\lambda_k$ , with eigenvectors $P\mathbf{v}_k$ . Option (C).

Learn the idea:

Q.16

Electronic Devices1 markMCQ

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

Option A: degenerate n-type.
Option B: degenerate p-type.
Option C: non-degenerate n-type.
Option D: non-degenerate p-type.

Full question on page 11 of the official paper (PDF)

Show answerHide answer

Key idea: A Fermi level near the conduction band means n-type; one inside a band means degenerate doping.

Official answer: A

Doping with donors pushes $E_F$ up towards $E_C$ . Very heavy donor doping pushes it past $E_C$ into the conduction band, where Boltzmann statistics no longer hold: the semiconductor is degenerate. Option (A).

A degenerate p-type semiconductor has $E_F$ inside the valence band instead.

Learn the idea:

Q.17

Electronic Devices1 markMCQ

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

Option A: All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.
Option B: All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons.
Option C: All energy states in the valence and conduction band are filled with holes.
Option D: All energy states in the valence and conduction band are filled with electrons.

Full question on page 12 of the official paper (PDF)

Show answerHide answer

Key idea: At 0 K no electron has the thermal energy to cross the band gap.

Official answer: A

At 0 K the Fermi function is a step at $E_F$ , which lies in the gap: every state below it is full and every state above it is empty. So the valence band is full and the conduction band is empty, which is why an intrinsic semiconductor is an insulator at 0 K. Option (A).

Q.18

Networks, Signals and Systems1 markMCQ

A series $RLC$ circuit has a quality factor $Q$ of 1000 at a centre frequency of $10^6$ rad/s. What are possible values of $R$ , $L$ and $C$ ?

Option A: $R = 1\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option B: $R = 0.1\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option C: $R = 0.01\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option D: $R = 0.001\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F

Full question on page 12 of the official paper (PDF)

Show answerHide answer

Key idea: For a series RLC, $\omega_0 = 1/\sqrt{LC}$ and $Q = \omega_0 L / R$ .

Official answer: D

Every option has $L = 1\ \mu$ H and $C = 1\ \mu$ F, so $\omega_0 = 1/\sqrt{10^{-12}} = 10^6$ rad/s, as required. Then

R = \frac{\omega_0 L}{Q} = \frac{10^6 \times 10^{-6}}{1000} = 0.001\ \Omega

\begin{aligned} R &= \frac{\omega_0 L}{Q} \\ &= \frac{10^6 \times 10^{-6}}{1000} \\ &= 0.001\ \Omega \end{aligned}

Option (D).

Learn the idea:

Q.19

Electronic Devices1 markMCQ

For a MOS capacitor, $V_{fb}$ and $V_t$ are the flat-band and threshold voltages. Which graph in the figure best represents the depletion width $W_{dep}$ as the gate voltage $V_g$ varies?

Four sketches of depletion width against gate voltage across accumulation, depletion and inversion. A: zero until V_fb, rising steeply and flattening at a value reached before V_t. B: zero until V_fb, rising through depletion and levelling off at V_t, flat in inversion. C: constant in accumulation and depletion, falling to zero at V_t. D: rising from V_fb to a peak and falling back to zero at V_t. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: Graph (A)
Option B: Graph (B)
Option C: Graph (C)
Option D: Graph (D)

Full question on page 13 of the official paper (PDF)

Show answerHide answer

Key idea: There is no depletion layer in accumulation; it grows through depletion and stops growing once the inversion layer forms.

Official answer: B

Below $V_{fb}$ (accumulation) majority carriers pile up at the surface and $W_{dep} = 0$ . Between $V_{fb}$ and $V_t$ the depletion layer widens as $V_g$ rises. At $V_t$ the inversion layer forms and screens any further charge, so $W_{dep}$ stays at its maximum, $W_{max}$ , through inversion.

Graph (B) shows exactly this: zero, a smooth rise through depletion, and a flat line beyond $V_t$ . (A) saturates before $V_t$ , and (C) and (D) fall to zero at $V_t$ . Option (B).

Q.20

Electromagnetics1 markMCQ

A narrow-band signal propagates in a lossless dielectric medium ( $\varepsilon_r = 4$ , $\mu_r = 1$ ) with phase velocity $v_p$ and group velocity $v_g$ . Which statement is true? ( $c$ is the velocity of light in vacuum.)

Option A: $v_p > c$ , $v_g > c$
Option B: $v_p < c$ , $v_g > c$
Option C: $v_p > c$ , $v_g < c$
Option D: $v_p < c$ , $v_g < c$

Full question on page 14 of the official paper (PDF)

Show answerHide answer

Key idea: A lossless dielectric with constant $\varepsilon_r$ is not dispersive, so $v_g = v_p$ .

Official answer: D

v_p = \frac{\omega}{\beta} = \frac{c}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{2}

$\beta = \omega\sqrt{\mu\varepsilon}$ is proportional to $\omega$ , so $v_g = d\omega/d\beta$ is also $c/2$ . Both are less than $c$ . Option (D).

Q.21

Analog Circuits1 markMCQ

In the circuit in the figure, $V_1$ and $V_2$ are bias voltages. Based on its input and output impedances, the circuit behaves as a

A PMOS transistor with gate at bias V1 and source at VDD, its drain joined at the output vOUT to the drain of an NMOS transistor with gate at bias V2. The NMOS source connects through resistor R1 to the input source vIN, which goes to ground. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: voltage controlled voltage source.
Option B: voltage controlled current source.
Option C: current controlled voltage source.
Option D: current controlled current source.

Full question on page 15 of the official paper (PDF)

Show answerHide answer

Key idea: Low input impedance means the input is a current; high output impedance means the output is a current.

Official answer: D

The NMOS is a common-gate stage: its gate is held at $V_2$ and the signal enters at its source. Looking into a source the impedance is about $1/g_m$ , which is low, so the stage takes a current as its input.

The output is taken at the drains, where the NMOS drain and the PMOS current source (gate at $V_1$ ) both look like high impedances, about $r_o$ each, so the output behaves as a current. Low input impedance and high output impedance make a current controlled current source, a current buffer. Option (D).

Q.22

Analog Circuits1 markMCQ

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

Option A: the closed loop gain is less than 1 and the phase shift is less than 180°.
Option B: the closed loop gain is greater than 1 and the phase shift is less than 180°.
Option C: the closed loop gain is less than 1 and the phase shift is greater than 180°.
Option D: the closed loop gain is greater than 1 and the phase shift is greater than 180°.

Full question on page 15 of the official paper (PDF)

Show answerHide answer

Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The key awards marks to all for this question and doesn't say why, so we don't pick an option.

The idea behind it: a feedback amplifier oscillates when the loop gain reaches 1 in magnitude at the frequency where the loop's phase shift makes the feedback positive (the Barkhausen condition). One likely reason for the ruling is that the options state the condition in terms of the closed-loop gain and a strict inequality on the phase, so none of them states the condition exactly.

Ruling in the key: The key gives marks to all (MTA) for this question: every candidate received its 1 mark.

Q.23

Digital Circuits1 markMCQ

In the circuit in the figure, P and Q are the inputs of a 2 × 1 multiplexer: $I_0$ is grounded, $I_1 = P$ and the select line is Q. What logical function does it realise?

A 2 by 1 multiplexer with input I0 connected to ground, input I1 connected to P, select line connected to Q, and output Y. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $Y = PQ$
Option B: $Y = P + Q$
Option C: $Y = \overline{PQ}$
Option D: $Y = \overline{P + Q}$

Full question on page 16 of the official paper (PDF)

Show answerHide answer

Key idea: A 2 × 1 multiplexer gives $Y = \bar{S} I_0 + S I_1$ .

Official answer: A

Y = \bar{Q} \cdot 0 + Q \cdot P = PQ

Y is 1 only when Q selects $I_1$ and P is 1: an AND gate. Option (A).

Learn the idea:

Q.24

Digital Circuits1 markMCQ

The synchronous sequential circuit in the figure, three D flip-flops in a chain sharing one clock, works at a clock frequency of 1 GHz. What are its throughput, in Mbits/s, and its latency, in ns?

Three D flip-flops in series: Input to D0, Q0 to D1, Q1 to D2, and Q2 to Output, all clocked by CLK = 1 GHz. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: 1000, 3
Option B: 333.33, 1
Option C: 2000, 3
Option D: 333.33, 3

Full question on page 17 of the official paper (PDF)

Show answerHide answer

Key idea: A new bit enters on every clock edge, and each bit needs one clock per flip-flop to reach the output.

Official answer: A

Like a pipeline, the chain accepts one new bit every clock period of 1 ns, so its throughput is one bit per ns, 1000 Mbit/s.

Each bit moves one flip-flop per clock, so it appears at $Q_2$ three clocks after it enters: a latency of 3 ns. Option (A).

Learn the idea:

Q.25

Control Systems1 markMCQ

The open-loop transfer function of a unity negative feedback system is $G(s) = \dfrac{k}{s(1 + sT_1)(1 + sT_2)}$ , where $k$ , $T_1$ and $T_2$ are positive constants. What is the phase crossover frequency, in rad/s?

Option A: $\dfrac{1}{\sqrt{T_1 T_2}}$
Option B: $\dfrac{1}{T_1 T_2}$
Option C: $\dfrac{1}{T_1\sqrt{T_2}}$
Option D: $\dfrac{1}{T_2\sqrt{T_1}}$

Full question on page 18 of the official paper (PDF)

Show answerHide answer

Key idea: At the phase crossover frequency the phase of $G(j\omega)$ is $-180°$ .

Official answer: A

-90^\circ - \tan^{-1}(\omega T_1) - \tan^{-1}(\omega T_2) = -180^\circ

\begin{aligned} &-90^\circ \\ &\qquad - \tan^{-1}(\omega T_1) \\ &\qquad - \tan^{-1}(\omega T_2) \\ &\quad = -180^\circ \end{aligned}

So $\tan^{-1}(\omega T_1) + \tan^{-1}(\omega T_2) = 90°$ , which needs $\omega T_1 \cdot \omega T_2 = 1$ :

\omega_{pc} = \frac{1}{\sqrt{T_1 T_2}}

Option (A).

Q.26

Networks, Signals and Systems1 markMCQ

A system has input $x(t)$ and output $y(t) = x(e^t)$ . The system is

Option A: Causal and time invariant.
Option B: Non-causal and time varying.
Option C: Causal and time varying.
Option D: Non-causal and time invariant.

Full question on page 18 of the official paper (PDF)

Show answerHide answer

Key idea: Causal means $y(t)$ uses no input later than $t$ ; time invariant means a delayed input gives the same output, delayed.

Official answer: B

Causality: $e^t > t$ for every $t$ , so $y(t)$ always uses the input at a later time. At $t = 0$ , $y(0) = x(1)$ . The system is non-causal.

Time invariance: delaying the input by $t_0$ gives $x(e^t - t_0)$ , but delaying the output gives $y(t - t_0) = x(e^{t - t_0})$ . These differ, so the system is time varying. Option (B).

Q.27

Communications1 markMCQ

Let $m(t)$ be a strictly band-limited signal with bandwidth $B$ and energy $E$ . Assuming $\omega_0 = 10B$ , what is the energy in the signal $m(t)\cos\omega_0 t$ ?

Option A: $\dfrac{E}{4}$
Option B: $\dfrac{E}{2}$
Option C: $E$
Option D: $2E$

Full question on page 19 of the official paper (PDF)

Show answerHide answer

Key idea: Modulation splits the spectrum into two half-height copies; with $\omega_0 > B$ they don't overlap.

Official answer: B

The Fourier transform of $m(t)\cos\omega_0 t$ is

\tfrac{1}{2}\left[M(\omega - \omega_0) + M(\omega + \omega_0)\right]

Each copy has a quarter of the energy density, $|M|^2/4$ , and since $\omega_0 = 10B$ the copies don't overlap, so the energies add: $E/4 + E/4 = E/2$ . Option (B). In the time domain, $\cos^2\omega_0 t = \tfrac{1}{2}(1 + \cos 2\omega_0 t)$ , and the $\cos 2\omega_0 t$ part integrates to zero.

Learn the idea:

Q.28

Networks, Signals and Systems1 markMCQ

What is the Fourier transform $X(\omega)$ of $x(t) = e^{-t^2}$ ? (Note: $\int_{-\infty}^{\infty} e^{-y^2}\, dy = \sqrt{\pi}$ .)

Option A: $\sqrt{\pi}\, e^{\omega^2/2}$
Option B: $\dfrac{e^{-\omega^2/4}}{2\sqrt{\pi}}$
Option C: $\sqrt{\pi}\, e^{-\omega^2/4}$
Option D: $\sqrt{\pi}\, e^{-\omega^2/2}$

Full question on page 19 of the official paper (PDF)

Show answerHide answer

Key idea: Complete the square in the exponent, then use the given integral.

Official answer: C

X(\omega) = \int_{-\infty}^{\infty} e^{-t^2 - j\omega t}\, dt = e^{-\omega^2/4} \int_{-\infty}^{\infty} e^{-(t + j\omega/2)^2}\, dt = \sqrt{\pi}\, e^{-\omega^2/4}

\begin{aligned} X(\omega) &= \int_{-\infty}^{\infty} e^{-t^2 - j\omega t}\, dt \\ &= e^{-\omega^2/4} \int_{-\infty}^{\infty} e^{-(t + j\omega/2)^2}\, dt \\ &= \sqrt{\pi}\, e^{-\omega^2/4} \end{aligned}

A quick check: $X(0)$ must equal $\int x(t)\, dt = \sqrt{\pi}$ , which rules out (B), and (A) grows with $\omega$ . Option (C).

Q.29

Networks, Signals and Systems1 markMCQ

Match each signal type with its spectral characteristics. Signal types: (i) continuous, aperiodic; (ii) continuous, periodic; (iii) discrete, aperiodic; (iv) discrete, periodic. Spectral characteristics: (a) continuous, aperiodic; (b) continuous, periodic; (c) discrete, aperiodic; (d) discrete, periodic.

Option A: (i) → (a), (ii) → (b), (iii) → (c), (iv) → (d)
Option B: (i) → (a), (ii) → (c), (iii) → (b), (iv) → (d)
Option C: (i) → (d), (ii) → (b), (iii) → (c), (iv) → (a)
Option D: (i) → (a), (ii) → (c), (iii) → (d), (iv) → (b)

Full question on page 20 of the official paper (PDF)

Show answerHide answer

Key idea: Periodic in one domain means discrete in the other, and discrete in one domain means periodic in the other.

Official answer: B

(i) Continuous, aperiodic: the Fourier transform, continuous and aperiodic, (a).
(ii) Continuous, periodic: the Fourier series, discrete lines, aperiodic, (c).
(iii) Discrete, aperiodic: the DTFT, continuous and periodic in $2\pi$ , (b).
(iv) Discrete, periodic: the DFS, discrete and periodic, (d).

Option (B).

Learn the idea:

Q.30

Communications1 markMSQ

For a real signal, which of the following is/are valid power spectral density/densities? Options (C) and (D) are the plots in the figure.

Two sketches of S_X(omega). (C): equal to 1 over a band of positive omega and equal to -1 over the matching band of negative omega. (D): equal to 1 over a band of positive omega and zero for all negative omega. — Figure from the official GATE 2023 EC paper, IIT Kanpur

One or more options are correct.

Option A: $S_X(\omega) = \dfrac{2}{9 + \omega^2}$
Option B: $S_X(\omega) = e^{-\omega^2}\cos^2\omega$
Option C: The plot labelled (C) in the figure
Option D: The plot labelled (D) in the figure

Full question on page 21 of the official paper (PDF)

Show answerHide answer

Key idea: The PSD of a real signal is real, never negative, and even in $\omega$ .

Official answer: A and B

(A) $2/(9 + \omega^2)$ is positive and even: valid. It is the PSD of $e^{-3|\tau|}/3$ .
(B) $e^{-\omega^2}\cos^2\omega$ is never negative and is even: valid.
(C) is negative for $\omega < 0$ : not a PSD.
(D) is not even: for a real signal $S_X(-\omega) = S_X(\omega)$ , so a one-sided band isn't valid.

Options (A) and (B).

Q.31

Digital Circuits1 markNAT

The signal-to-noise ratio (SNR) of an ADC with a full-scale sinusoidal input is 61.96 dB. What is the resolution of the ADC, in bits?

Answer format: bits, rounded off to the nearest integer

Full question on page 21 of the official paper (PDF)

Show answerHide answer

Key idea: For an ideal $N$ -bit ADC and a full-scale sine, $\text{SNR} = 6.02N + 1.76$ dB.

Official answer: 10

N = \frac{61.96 - 1.76}{6.02} = \frac{60.2}{6.02} = 10

\begin{aligned} N &= \frac{61.96 - 1.76}{6.02} \\ &= \frac{60.2}{6.02} \\ &= 10 \end{aligned}

The formula comes from a full-scale sine's power, $(2^N \Delta)^2/8$ , over the quantisation noise power, $\Delta^2/12$ : their ratio is $1.5 \times 4^N$ , which is $1.76 + 6.02N$ dB. Answer: 10 bits.

Learn the idea:

Q.32

Networks, Signals and Systems1 markNAT

In the circuit in the figure, what is the current $i$ flowing through the 200 Ω resistor, in mA?

Node A at the top left connects to ground through a 2 V source (positive terminal up) in series with 2 kilohms, through a separate 2 kilohm resistor, and through a 1 mA current source pointing up into node A. A 1 kilohm resistor joins node A to node B, with a 1 mA current source in parallel pointing from A to B. Node B connects to ground through 200 ohms, carrying the current i downwards. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: mA, rounded off to two decimal places

Full question on page 22 of the official paper (PDF)

Show answerHide answer

Key idea: Two unknown node voltages: write KCL at each, working in volts, mA and kΩ.

Official answer: 1.30 to 1.40

Call the top-left node A and the node above the 200 Ω resistor B. In mA and kΩ, KCL at A (the 1 mA into A from below and the 1 mA out of A to B cancel):

\frac{V_A - 2}{2} + \frac{V_A}{2} + \frac{V_A - V_B}{1} = 1 - 1 = 0 \quad\Rightarrow\quad 2V_A - V_B = 1

\begin{aligned} &\frac{V_A - 2}{2} \\ &\qquad + \frac{V_A}{2} \\ &\qquad + \frac{V_A - V_B}{1} \\ &\quad = 1 - 1 \\ &\quad = 0 \\ \Rightarrow\; \\ 2V_A - V_B &= 1 \end{aligned}

KCL at B, with the 1 mA source feeding it:

\frac{V_B - V_A}{1} + \frac{V_B}{0.2} = 1 \quad\Rightarrow\quad -V_A + 6V_B = 1

\begin{aligned} &\frac{V_B - V_A}{1} + \frac{V_B}{0.2} \\ &\quad = 1 \\ \Rightarrow\; \\ -V_A + 6V_B &= 1 \end{aligned}

So $V_A = (1 + V_B)/2$ , and $5.5\,V_B = 1.5$ gives $V_B = 0.2727$ V. Then $i = V_B / 0.2\ \text{k}\Omega = 1.36$ mA.

Learn the idea:

Q.33

Networks, Signals and Systems1 markNAT

For the two-port network in the figure, $[Y] = \dfrac{1}{100}\begin{bmatrix} 2 & -1 \\ -1 & 4/3 \end{bmatrix}$ S. It is driven by a 120 V source through 10 Ω at port 1. What load impedance $Z_L$ , in Ω, takes the maximum power?

A 120 V source in series with 10 ohms drives port 1 of a two-port network labelled [Y]; port 2 is loaded by Z_L. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: Ohms, rounded off to the nearest integer

Full question on page 22 of the official paper (PDF)

Show answerHide answer

Key idea: Maximum power needs $Z_L$ equal to the Thevenin impedance seen at port 2, with the source shorted.

Official answer: 80

With the source shorted, port 1 sees $I_1 = -V_1/10$ , so $-0.1V_1 = 0.02V_1 - 0.01V_2$ , which gives $V_1 = V_2/12$ . Then

I_2 = -0.01 V_1 + \tfrac{4}{300} V_2 = \left(-\tfrac{0.01}{12} + 0.01333\right) V_2 = 0.0125\, V_2

\begin{aligned} I_2 &= -0.01 V_1 + \tfrac{4}{300} V_2 \\ &= \left(-\tfrac{0.01}{12} + 0.01333\right) V_2 \\ &= 0.0125\, V_2 \end{aligned}

So $Z_{th} = 1/0.0125 = 80$ Ω, and $Z_L = 80$ Ω. Equivalently, $Y_{out} = Y_{22} - Y_{12}Y_{21}/(Y_{11} + 0.1)$ . Answer: 80.

Learn the idea:

Q.34

Digital Circuits1 markNAT

For the circuit in the figure, two cross-coupled NAND gates with inputs A and B and outputs Q and $\overline{Q}$ , the propagation delay of each NAND gate is 1 ns. What is the critical path delay, in ns?

Two NAND gates, cross-coupled: input A and the second gate's output feed the first gate, whose output is Q; input B and Q feed the second gate, whose output is Q bar. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: ns, rounded off to the nearest integer

Full question on page 23 of the official paper (PDF)

Show answerHide answer

Key idea: A change only settles once it has gone through one gate and been fed back through the other.

Official answer: 2

Pull A low to set the latch: after 1 ns the first gate's output Q goes high. Only then does the second gate see the new Q, and after another 1 ns $\overline{Q}$ goes low and holds Q. The outputs are settled after two gate delays, so the critical path is 2 ns.

Learn the idea:

Q.35

Networks, Signals and Systems1 markNAT

In the circuit in the figure, a 2 V source drives, through switch S and 1 Ω, a 2 Ω resistor in parallel with a 1 H inductor. S was closed for a long time and is opened at $t = 0$ . What is the maximum magnitude of the voltage $V_R$ across the 2 Ω resistor, in volts?

A source Vs = 2 V in series with switch S and a 1 ohm resistor, feeding a 2 ohm resistor (voltage V_R across it) in parallel with a 1 H inductor, back to the source. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: volts, rounded off to the nearest integer

Full question on page 23 of the official paper (PDF)

Show answerHide answer

Key idea: The inductor current can't change instantly, and once S opens it has nowhere to go but the 2 Ω resistor.

Official answer: 4

Before $t = 0$ the inductor is a short, so $V_R = 0$ and the whole current, $2/1 = 2$ A, flows through the inductor.

After S opens, the inductor and the 2 Ω resistor form a loop, and the 2 A must flow through the resistor: $|V_R(0^+)| = 2 \times 2 = 4$ V. It then decays as $4e^{-2t}$ , so the maximum magnitude is 4 V.

Q.36

Engineering Mathematics2 marksMCQ

A random variable $X$ , distributed normally as $N(0, 1)$ , undergoes the transformation $Y = h(X)$ in the figure. What is the form of the probability density function of $Y$ ? (In the options, $a$ , $b$ , $c$ are non-zero constants and $g(y)$ is a piecewise continuous function.)

Graph of y = h(x): equal to -1 for x below -2, rising linearly to 0 between x = -2 and x = -1, equal to 0 between -1 and 1, rising linearly to 1 between 1 and 2, and equal to 1 for x above 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $a\delta(y - 1) + b\delta(y + 1) + g(y)$
Option B: $a\delta(y + 1) + b\delta(y) + c\delta(y - 1) + g(y)$
Option C: $a\delta(y + 2) + b\delta(y) + c\delta(y - 2) + g(y)$
Option D: $a\delta(y + 2) + b\delta(y - 2) + g(y)$

Full question on page 24 of the official paper (PDF)

Show answerHide answer

Key idea: Wherever $h$ is flat over an interval of $x$ , all that probability lands on one value of $y$ : an impulse in the PDF.

Official answer: B

$h$ is flat at three levels: $y = -1$ for $x < -2$ , $y = 0$ for $-1 < x < 1$ and $y = 1$ for $x > 2$ . Each of these ranges has non-zero probability under $N(0, 1)$ , so the PDF has impulses at $y = -1$ , $0$ and $1$ :

P(Y = 0) = P(|X| < 1) \approx 0.683, \qquad P(Y = \pm 1) = P(X > 2) \approx 0.0228

\begin{aligned} P(Y = 0) &= P(|X| < 1) \\ &\approx 0.683, \\ P(Y = \pm 1) &= P(X > 2) \\ &\approx 0.0228 \end{aligned}

The sloped parts map $1 < |x| < 2$ continuously onto $0 < |y| < 1$ , which gives $g(y)$ . Option (B).

Learn the idea:

Q.37

Engineering Mathematics2 marksMCQ

What is the value of the line integral $\int_P^Q (z^2\, dx + 3y^2\, dy + 2xz\, dz)$ along the straight line joining $P(1, 1, 2)$ and $Q(2, 3, 1)$ ?

Option A: 20
Option B: 24
Option C: 29
Option D: -5

Full question on page 25 of the official paper (PDF)

Show answerHide answer

Key idea: The integrand is an exact differential, so the integral depends only on the end points.

Official answer: B

z^2\, dx + 2xz\, dz = d(xz^2), \qquad 3y^2\, dy = d(y^3)

\begin{aligned} &z^2\, dx + 2xz\, dz \\ &\quad = d(xz^2), \\ 3y^2\, dy &= d(y^3) \end{aligned}

\int_P^Q d(xz^2 + y^3) = (2 \cdot 1 + 27) - (1 \cdot 4 + 1) = 29 - 5 = 24

\begin{aligned} &\int_P^Q d(xz^2 + y^3) \\ &\quad = (2 \cdot 1 + 27) \\ &\qquad - (1 \cdot 4 + 1) \\ &\quad = 29 - 5 \\ &\quad = 24 \end{aligned}

Option (B). Parametrising the line as $(1 + t, 1 + 2t, 2 - t)$ gives the same 24.

Learn the idea:

Q.38

Engineering Mathematics2 marksMCQ

Let $\mathbf{x}$ be an $n \times 1$ real column vector with length $l = \sqrt{\mathbf{x}^T\mathbf{x}}$ . What is the trace of the matrix $P = \mathbf{x}\mathbf{x}^T$ ?

Option A: $l^2$
Option B: $\dfrac{l^2}{4}$
Option C: $l$
Option D: $\dfrac{l^2}{2}$

Full question on page 25 of the official paper (PDF)

Show answerHide answer

Key idea: The diagonal of $\mathbf{x}\mathbf{x}^T$ holds the squares $x_i^2$ .

Official answer: A

\operatorname{tr}(\mathbf{x}\mathbf{x}^T) = \sum_i x_i^2 = \mathbf{x}^T\mathbf{x} = l^2

\begin{aligned} \operatorname{tr}(\mathbf{x}\mathbf{x}^T) &= \sum_i x_i^2 \\ &= \mathbf{x}^T\mathbf{x} \\ &= l^2 \end{aligned}

Option (A).

Learn the idea:

Q.39

Analog Circuits2 marksMCQ

In the circuit in the figure, $v_{IN}$ drives an inverting amplifier ( $R_1$ , $R_2$ ) and a non-inverting amplifier ( $R_1$ , $R_2$ ); their outputs feed, each through $R_3$ , the inverting input of a third op-amp with feedback $R_4$ and its non-inverting input grounded. What is $V_{OUT}/V_{IN}$ ?

Op-amp circuit. The top op-amp is an inverting amplifier: v_IN through R1 to its inverting input, R2 feedback, non-inverting input grounded. The bottom op-amp is a non-inverting amplifier: v_IN to its non-inverting input, R1 from its inverting input to ground and R2 feedback. Both outputs go through resistors R3 to the inverting input of a third op-amp with feedback R4 and non-inverting input grounded; its output is v_OUT. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $-\dfrac{R_4}{R_3}$
Option B: $\dfrac{R_4}{R_3}$
Option C: $1 + \dfrac{R_4}{R_3}$
Option D: $1 - \dfrac{R_4}{R_3}$

Full question on page 26 of the official paper (PDF)

Show answerHide answer

Key idea: The third op-amp is an inverting summer, and the first two outputs add to exactly $v_{IN}$ .

Official answer: A

v_1 = -\frac{R_2}{R_1} v_{IN}, \qquad v_2 = \left(1 + \frac{R_2}{R_1}\right) v_{IN}, \qquad v_1 + v_2 = v_{IN}

\begin{aligned} v_1 &= -\frac{R_2}{R_1} v_{IN}, \\ v_2 &= \left(1 + \frac{R_2}{R_1}\right) v_{IN}, \\ v_1 + v_2 &= v_{IN} \end{aligned}

The summer gives $v_{OUT} = -\dfrac{R_4}{R_3}(v_1 + v_2) = -\dfrac{R_4}{R_3} v_{IN}$ . Option (A).

Learn the idea:

Q.40

Analog Circuits2 marksMCQ

In the circuit in the figure, $D_1$ and $D_2$ are silicon diodes with a cut-in voltage of 0.7 V, connected in anti-parallel between $V_{IN}$ and $V_{OUT}$ ; $V_{OUT}$ connects through 1 kΩ to a 1 V source. Which transfer characteristic is correct?

Top: the circuit, V_IN connected to V_OUT through two diodes D1 and D2 in anti-parallel, and V_OUT connected through 1 kilohm to a 1 V source to ground. Below: four transfer characteristics A to D, as described in the options. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $V_{OUT} = 1$ for $0.3 \le V_{IN} \le 1.7$ , and slope 1 outside that range
Option B: $V_{OUT} = 0$ for $-0.7 \le V_{IN} \le 0.7$ , and slope 1 outside that range
Option C: $V_{OUT} = 1$ for $-1.7 \le V_{IN} \le 1.7$ , and slope 1 outside that range
Option D: $V_{OUT} = 0$ for $0.3 \le V_{IN} \le 1.7$ , and slope 1 outside that range

Full question on page 27 of the official paper (PDF)

Show answerHide answer

Key idea: With no diode on, no current flows in the 1 kΩ, so $V_{OUT}$ sits at 1 V.

Official answer: A

While $|V_{IN} - 1| < 0.7$ , neither diode conducts, no current flows and $V_{OUT} = 1$ V: that is $0.3 \le V_{IN} \le 1.7$ .

For $V_{IN} > 1.7$ , $D_1$ conducts and $V_{OUT} = V_{IN} - 0.7$ . For $V_{IN} < 0.3$ , $D_2$ conducts and $V_{OUT} = V_{IN} + 0.7$ . Both have slope 1, joining the flat part at 1 V. Option (A).

Learn the idea:

Q.41

Control Systems2 marksMCQ

The closed-loop system in the figure has unity negative feedback round $G(s) = \dfrac{k}{s(s + 2)}$ , where $k > 0$ and $\alpha > 0$ . What is the steady-state error due to a ramp input, $R(s) = \alpha/s^2$ ?

Block diagram: R(s) = alpha/s^2 into a summing junction (plus), then E(s) into a block k/(s(s+2)) giving Y(s), which is fed back to the junction with a minus sign. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $\dfrac{2\alpha}{k}$
Option B: $\dfrac{\alpha}{k}$
Option C: $\dfrac{\alpha}{2k}$
Option D: $\dfrac{\alpha}{4k}$

Full question on page 28 of the official paper (PDF)

Show answerHide answer

Key idea: A type-1 loop follows a ramp with a constant error, $\alpha / K_v$ , where $K_v = \lim_{s \to 0} sG(s)$ .

Official answer: A

K_v = \lim_{s \to 0} s \cdot \frac{k}{s(s + 2)} = \frac{k}{2}, \qquad e_{ss} = \frac{\alpha}{K_v} = \frac{2\alpha}{k}

\begin{aligned} K_v &= \lim_{s \to 0} s \cdot \frac{k}{s(s + 2)} \\ &= \frac{k}{2}, \\ e_{ss} &= \frac{\alpha}{K_v} \\ &= \frac{2\alpha}{k} \end{aligned}

This holds when the loop is stable, which it is for every $k > 0$ : the characteristic equation $s^2 + 2s + k = 0$ has both roots in the left half-plane. Option (A).

Q.42

Control Systems2 marksMCQ

In the block diagram in the figure, $R(s)$ and $D(s)$ are two inputs, and the output is $Y(s) = G_1(s)R(s) + G_2(s)D(s)$ . What are $G_1(s)$ and $G_2(s)$ ?

Block diagram: R(s) enters a first summing junction (plus), whose output enters a second junction (plus) together with D(s) (plus). The second junction feeds G(s), whose output is Y(s). Y(s) is fed back through H(s) to the second junction with a minus sign, and directly to the first junction with a minus sign. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $G_1 = \dfrac{G}{1 + G + GH}$ and $G_2 = \dfrac{G}{1 + G + GH}$
Option B: $G_1 = \dfrac{G}{1 + G + H}$ and $G_2 = \dfrac{G}{1 + G + H}$
Option C: $G_1 = \dfrac{G}{1 + G + H}$ and $G_2 = \dfrac{G}{1 + G + GH}$
Option D: $G_1 = \dfrac{G}{1 + G + GH}$ and $G_2 = \dfrac{G}{1 + G + H}$

Full question on page 29 of the official paper (PDF)

Show answerHide answer

Key idea: Write the signal into $G$ and solve once for $Y$ ; both inputs enter at the same point, so they share a denominator.

Official answer: A

The second junction adds $D$ to the first junction's output, $R - Y$ , and subtracts $HY$ :

Y = G\left[(R - Y) + D - HY\right]

Collecting the $Y$ terms:

Y(1 + G + GH) = G R + G D

So $G_1 = G_2 = G/(1 + G + GH)$ . Option (A).

Q.43

Control Systems2 marksMCQ

The state equation of a second-order system is $\dot{\mathbf{x}}(t) = A\mathbf{x}(t)$ , with initial condition $\mathbf{x}(0)$ . $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $A$ , and $\mathbf{v}_1$ and $\mathbf{v}_2$ the corresponding eigenvectors. For constants $\alpha_1$ and $\alpha_2$ , what is the solution $\mathbf{x}(t)$ ?

Option A: $\sum_{i=1}^{2} \alpha_i e^{\lambda_i t} \mathbf{v}_i$
Option B: $\sum_{i=1}^{2} \alpha_i e^{2\lambda_i t} \mathbf{v}_i$
Option C: $\sum_{i=1}^{2} \alpha_i e^{3\lambda_i t} \mathbf{v}_i$
Option D: $\sum_{i=1}^{2} \alpha_i e^{4\lambda_i t} \mathbf{v}_i$

Full question on page 30 of the official paper (PDF)

Show answerHide answer

Key idea: Try $\mathbf{x} = e^{\lambda t}\mathbf{v}$ : it solves the equation exactly when $A\mathbf{v} = \lambda\mathbf{v}$ .

Official answer: A

\frac{d}{dt}\left(e^{\lambda_i t}\mathbf{v}_i\right) = \lambda_i e^{\lambda_i t}\mathbf{v}_i = A\left(e^{\lambda_i t}\mathbf{v}_i\right)

\begin{aligned} &\frac{d}{dt}\left(e^{\lambda_i t}\mathbf{v}_i\right) \\ &\quad = \lambda_i e^{\lambda_i t}\mathbf{v}_i \\ &\quad = A\left(e^{\lambda_i t}\mathbf{v}_i\right) \end{aligned}

Distinct eigenvalues give independent eigenvectors, so every solution is $\alpha_1 e^{\lambda_1 t}\mathbf{v}_1 + \alpha_2 e^{\lambda_2 t}\mathbf{v}_2$ , with $\alpha_1, \alpha_2$ fixed by $\mathbf{x}(0)$ . In (B) to (D), $\dot{\mathbf{x}}$ would be $k\lambda_i$ times each term, not $A\mathbf{x}$ . Option (A).

Learn the idea:

Q.44

Networks, Signals and Systems2 marksMCQ

In the circuit in the figure, switch $S_1$ was closed and $S_2$ was open for a long time. At $t = 0$ , $S_1$ is opened and $S_2$ is closed, simultaneously. $S_2$ shorts the 100 Ω resistor that is in series with the 1 H inductor. What is $i_C(0^+)$ , in amperes?

A 1 A current source connects through switch S1 to a top node. From the top node to the bottom rail: a 0.01 F capacitor carrying i_C downwards; a 100 ohm resistor in series with a 1 H inductor, with switch S2 across the 100 ohm resistor; and a 25 ohm resistor. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: 1
Option B: −1
Option C: 0.2
Option D: 0.8

Full question on page 31 of the official paper (PDF)

Show answerHide answer

Key idea: Find the capacitor voltage and inductor current before switching; both carry over to $t = 0^+$ .

Official answer: B

Before $t = 0$ the capacitor is open and the inductor a short, so the 1 A splits between the 100 Ω branch and the 25 Ω resistor:

i_L(0^-) = 1 \times \frac{25}{125} = 0.2\ \text{A}, \qquad v_C(0^-) = 25 \times 0.8 = 20\ \text{V}

\begin{aligned} i_L(0^-) &= 1 \times \frac{25}{125} \\ &= 0.2\ \text{A}, \\ v_C(0^-) &= 25 \times 0.8 \\ &= 20\ \text{V} \end{aligned}

At $t = 0^+$ the source is gone, $v_C = 20$ V and $i_L = 0.2$ A. With the 100 Ω shorted, the inductor still carries 0.2 A, and the 25 Ω carries $20/25 = 0.8$ A. The capacitor supplies both, so the current down into it is $i_C(0^+) = -(0.2 + 0.8) = -1$ A. Option (B).

Learn the idea:

Q.45

Communications2 marksMCQ

An FM signal $x(t) = A\cos\left(\omega_c t + k_f \int_{-\infty}^{t} m(\lambda)\, d\lambda\right)$ , where $m(t)$ has bandwidth $W$ , is passed through a non-linear system with output $y(t) = 2x(t) + 5x^2(t)$ . Let $B_T$ denote the FM bandwidth. What is the minimum value of $\omega_c$ required to recover $x(t)$ from $y(t)$ ?

Option A: $B_T + W$
Option B: $\dfrac{3}{2}B_T$
Option C: $2B_T + W$
Option D: $\dfrac{5}{2}B_T$

Full question on page 32 of the official paper (PDF)

Show answerHide answer

Key idea: Squaring an FM wave gives a DC term and an FM wave at $2\omega_c$ with twice the deviation, so twice the bandwidth.

Official answer: B

x^2(t) = \frac{A^2}{2} + \frac{A^2}{2}\cos\left(2\omega_c t + 2k_f \int m\right)

\begin{aligned} x^2(t) &= \frac{A^2}{2} \\ &\quad + \frac{A^2}{2}\cos\left(2\omega_c t + 2k_f \int m\right) \end{aligned}

The term at $2\omega_c$ occupies $2\omega_c \pm B_T$ (twice the deviation, so twice the bandwidth), while $x(t)$ occupies $\omega_c \pm B_T/2$ , and the DC term sits at 0. A bandpass filter can keep $x(t)$ alone if the bands don't overlap:

\omega_c + \frac{B_T}{2} \le 2\omega_c - B_T \quad\Rightarrow\quad \omega_c \ge \frac{3}{2} B_T

\begin{aligned} \omega_c + \frac{B_T}{2} &\le 2\omega_c - B_T \\ \Rightarrow\; \\ \omega_c &\ge \frac{3}{2} B_T \end{aligned}

The DC term needs only $\omega_c > B_T/2$ , which this already meets. Option (B).

Q.46

Networks, Signals and Systems2 marksMCQ

The h-parameter model of a two-port network is shown in the figure, driven by $V_s$ and loaded by $R_L$ . What is the condition for the maximum small-signal voltage gain $V_{out}/V_s$ ?

h-parameter two-port: source Vs drives the input through h11 into a voltage source h12 v2; at the output, a dependent current source labelled h21 v1 in parallel with h22 and the load R_L, across which Vout is taken. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $h_{11} = 0$ , $h_{12} = 0$ , $h_{21}$ = very high and $h_{22} = 0$
Option B: $h_{11}$ = very high, $h_{12} = 0$ , $h_{21}$ = very high and $h_{22} = 0$
Option C: $h_{11} = 0$ , $h_{12}$ = very high, $h_{21}$ = very high and $h_{22} = 0$
Option D: $h_{11} = 0$ , $h_{12} = 0$ , $h_{21}$ = very high and $h_{22}$ = very high

Full question on page 33 of the official paper (PDF)

Show answerHide answer

Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The key awards marks to all for this question and doesn't say why, so we don't pick an option.

The idea behind it: with $V_s$ driving port 1 directly, the gain of an h-parameter two-port is $-h_{21}R_L / \left(h_{11} + \Delta h\, R_L\right)$ with $\Delta h = h_{11}h_{22} - h_{12}h_{21}$ , so a large $h_{21}$ and small $h_{11}$ , $h_{12}$ and $h_{22}$ raise it. One likely reason for the ruling is the figure: it labels the output source $h_{21}v_1$ where the h-parameter model has $h_{21}i_1$ , and with $h_{11} = 0$ and $h_{12} = 0$ the input current isn't set by $V_s$ at all.

Ruling in the key: The key gives marks to all (MTA) for this question: every candidate received its 2 marks.

Q.47

Networks, Signals and Systems2 marksMCQ

A discrete-time periodic signal has period $N = 5$ and DTFS $x[n] = \sum_{k=0}^{4} a_k e^{jk2\pi n/5}$ , with $a_0 = 1$ , $a_1 = 3j$ , $a_2 = 2j$ , $a_3 = -2j$ and $a_4 = -3j$ . What is $\sum_{n=0}^{4} x[n]\sin\dfrac{4\pi n}{5}$ ?

Option A: −10
Option B: 10
Option C: −2
Option D: 2

Full question on page 34 of the official paper (PDF)

Show answerHide answer

Key idea: $\sum_n x[n] e^{-jk2\pi n/N} = N a_k$ , so write the sine as two exponentials.

Official answer: A

\sin\frac{4\pi n}{5} = \frac{e^{j4\pi n/5} - e^{-j4\pi n/5}}{2j}

$\sum_n x[n] e^{-j2(2\pi n/5)} = 5a_2 = 10j$ , and $\sum_n x[n] e^{+j2(2\pi n/5)} = 5a_{-2} = 5a_3 = -10j$ . So

\sum_{n=0}^{4} x[n]\sin\frac{4\pi n}{5} = \frac{-10j - 10j}{2j} = -10

\begin{aligned} &\sum_{n=0}^{4} x[n]\sin\frac{4\pi n}{5} \\ &\quad = \frac{-10j - 10j}{2j} \\ &\quad = -10 \end{aligned}

Option (A).

Learn the idea:

Q.48

Networks, Signals and Systems2 marksMCQ

An input $x[n]$ with DTFT $X(e^{j\Omega}) = 1 - e^{-j\Omega} + 2e^{-j3\Omega}$ is passed through an LTI system with frequency response $H(e^{j\Omega}) = 1 - \tfrac{1}{2}e^{-j2\Omega}$ . What is the output $y[n]$ ?

Option A: $\delta[n] + \delta[n-1] - \tfrac{1}{2}\delta[n-2] - \tfrac{5}{2}\delta[n-3] + \delta[n-5]$
Option B: $\delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] - \tfrac{5}{2}\delta[n-3] + \delta[n-5]$
Option C: $\delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]$
Option D: $\delta[n] + \delta[n-1] + \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] + \delta[n-5]$

Full question on page 34 of the official paper (PDF)

Show answerHide answer

Key idea: Multiply the two polynomials in $e^{-j\Omega}$ ; each power $e^{-jk\Omega}$ is a $\delta[n-k]$ .

Official answer: C

Y = \left(1 - e^{-j\Omega} + 2e^{-j3\Omega}\right)\left(1 - \tfrac{1}{2}e^{-j2\Omega}\right) = 1 - e^{-j\Omega} - \tfrac{1}{2}e^{-j2\Omega} + \tfrac{5}{2}e^{-j3\Omega} - e^{-j5\Omega}

\begin{aligned} Y &= \left(1 - e^{-j\Omega} + 2e^{-j3\Omega}\right)\left(1 - \tfrac{1}{2}e^{-j2\Omega}\right) \\ &= 1 \\ &\quad - e^{-j\Omega} \\ &\quad - \tfrac{1}{2}e^{-j2\Omega} \\ &\quad + \tfrac{5}{2}e^{-j3\Omega} \\ &\quad - e^{-j5\Omega} \end{aligned}

The $e^{-j3\Omega}$ term collects $2$ from $X$ and $+\tfrac{1}{2}$ from $(-e^{-j\Omega})(-\tfrac{1}{2}e^{-j2\Omega})$ . So $y[n] = \delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]$ . Option (C).

Learn the idea:

Q.49

Networks, Signals and Systems2 marksMCQ

$x(t) = 10\cos(10.5Wt)$ is passed through an LTI system with impulse response $h(t) = \pi\left(\dfrac{\sin Wt}{\pi t}\right)^2 \cos 10Wt$ . What is the output?

Option A: $\left(\dfrac{15W}{4}\right)\cos(10.5Wt)$
Option B: $\left(\dfrac{15W}{2}\right)\cos(10.5Wt)$
Option C: $\left(\dfrac{15W}{8}\right)\cos(10.5Wt)$
Option D: $(15W)\cos(10.5Wt)$

Full question on page 35 of the official paper (PDF)

Show answerHide answer

Key idea: $\sin Wt/(\pi t)$ is an ideal low-pass of height 1 and cut-off $W$ ; squaring it makes a triangle in frequency.

Official answer: A

$\dfrac{\sin Wt}{\pi t}$ transforms to a rectangle of height 1 on $|\omega| < W$ . Its square transforms to $\tfrac{1}{2\pi}$ times the rectangle convolved with itself: a triangle of peak $\tfrac{2W}{2\pi} = \tfrac{W}{\pi}$ , reaching zero at $|\omega| = 2W$ . Times $\pi$ , the peak is $W$ :

T(\omega) = W\left(1 - \frac{|\omega|}{2W}\right), \quad |\omega| < 2W

\begin{aligned} T(\omega) &= W\left(1 - \frac{|\omega|}{2W}\right), \\ |\omega| &< 2W \end{aligned}

Multiplying by $\cos 10Wt$ shifts half of it to each of $\pm 10W$ : $H(\omega) = \tfrac{1}{2}[T(\omega - 10W) + T(\omega + 10W)]$ . At $\omega = 10.5W$ , 0.5W from the centre:

H(10.5W) = \frac{1}{2} \cdot W\left(1 - \frac{0.5W}{2W}\right) = \frac{3W}{8}

\begin{aligned} H(10.5W) &= \frac{1}{2} \cdot W\left(1 - \frac{0.5W}{2W}\right) \\ &= \frac{3W}{8} \end{aligned}

$H$ is real and positive, so there is no phase shift: $y(t) = 10 \cdot \tfrac{3W}{8}\cos(10.5Wt) = \tfrac{15W}{4}\cos(10.5Wt)$ . Option (A).

Q.50

Networks, Signals and Systems2 marksMCQ

$x_1(t)$ and $x_2(t)$ are band-limited signals, each with bandwidth $B = 4\pi \times 10^3$ rad/s. In the figure, $y(t) = x_1(t)\cos(4\pi \times 10^3 t) + x_2(t)\cos(12\pi \times 10^3 t)$ . What Nyquist sampling frequency, in rad/s, is required to sample $y(t)$ ?

x1(t) is multiplied by cos(4 pi 10^3 t) and x2(t) by cos(12 pi 10^3 t); the two products are added to give y(t). — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $20\pi \times 10^3$
Option B: $40\pi \times 10^3$
Option C: $8\pi \times 10^3$
Option D: $32\pi \times 10^3$

Full question on page 36 of the official paper (PDF)

Show answerHide answer

Key idea: Find the highest frequency in $y(t)$ and double it.

Official answer: D

Multiplying by a cosine shifts the spectrum to the carrier. $x_1$ reaches $4\pi + 4\pi = 8\pi \times 10^3$ rad/s, and $x_2$ reaches $12\pi + 4\pi = 16\pi \times 10^3$ rad/s. The highest frequency is $16\pi \times 10^3$ , so

\omega_s = 2 \times 16\pi \times 10^3 = 32\pi \times 10^3\ \text{rad/s}

\begin{aligned} \omega_s &= 2 \times 16\pi \times 10^3 \\ &= 32\pi \times 10^3\ \text{rad/s} \end{aligned}

Option (D).

Learn the idea:

Q.51

Electromagnetics2 marksMCQ

The S-parameters of a two-port network are $[S] = \begin{bmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{bmatrix}$ , with reference to $Z_0$ . Two lossless transmission line sections of electrical lengths $\theta_1 = \beta l_1$ and $\theta_2 = \beta l_2$ are added to the input and output ports, respectively. What are the S-parameters $[S']$ of the resulting two-port?

Option A: $\begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{-j(\theta_1 + \theta_2)} \\ S_{21}e^{-j(\theta_1 + \theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}$
Option B: $\begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{-j(\theta_1 + \theta_2)} \\ S_{21}e^{-j(\theta_1 + \theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix}$
Option C: $\begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{j(\theta_1 + \theta_2)} \\ S_{21}e^{j(\theta_1 + \theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix}$
Option D: $\begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{j(\theta_1 + \theta_2)} \\ S_{21}e^{j(\theta_1 + \theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}$

Full question on page 37 of the official paper (PDF)

Show answerHide answer

Key idea: Each pass along a lossless line delays a wave by its electrical length, a factor $e^{-j\theta}$ .

Official answer: A

A wave reflected at port 1 crosses the input line twice, in and back out: $S'_{11} = S_{11}e^{-j2\theta_1}$ . Likewise $S'_{22} = S_{22}e^{-j2\theta_2}$ .

A wave passing through crosses both lines once each: $S'_{21} = S_{21}e^{-j(\theta_1 + \theta_2)}$ , and the same for $S'_{12}$ . Every factor is a delay, so every exponent is negative. Option (A).

Learn the idea:

Q.52

Electromagnetics2 marksMSQ

The standing wave ratio on a 50 Ω lossless line terminated in an unknown load is 2.0. Successive voltage minima are 30 cm apart, and the first minimum is 10 cm from the load. $Z_L$ can be replaced by an equivalent length $l_m$ of the same line terminated in a resistance $R_m$ . What are $R_m$ and $l_m$ ?

One or more options are correct.

Option A: $R_m = 100\ \Omega$ , $l_m = 20$ cm
Option B: $R_m = 25\ \Omega$ , $l_m = 20$ cm
Option C: $R_m = 100\ \Omega$ , $l_m = 5$ cm
Option D: $R_m = 25\ \Omega$ , $l_m = 5$ cm

Full question on page 38 of the official paper (PDF)

Show answerHide answer

Key idea: The impedance is purely resistive at every voltage minimum ( $Z_0/S$ ) and maximum ( $Z_0 S$ ), and these repeat every $\lambda/2$ , so the line can be extended past the load to one of them.

Official answer: B and C

Minima are $\lambda/2$ apart, so $\lambda = 60$ cm. At a voltage minimum the line looks like $Z_0/S = 25$ Ω, and at a maximum like $Z_0 S = 100$ Ω. The load is then the input impedance of a length $l_m$ of line ending in that resistance, measured from the load away from the generator.

Minima sit at 10 cm, 40 cm, ... towards the generator, so also at $10 - 30 = -20$ cm, 20 cm beyond the load: $R_m = 25$ Ω with $l_m = 20$ cm.
Maxima sit $\lambda/4 = 15$ cm from each minimum, at 25 cm and at $-5$ cm, 5 cm beyond the load: $R_m = 100$ Ω with $l_m = 5$ cm.

Options (B) and (C).

Learn the idea:

Q.53

Electromagnetics2 marksMSQ

The electric field of a plane wave is $\mathbf{E} = \mathbf{a}_x C_{1x}\cos(\omega t - \beta z) + \mathbf{a}_y C_{1y}\cos(\omega t - \beta z + \theta)$ V/m. Which combination(s) give a left-handed elliptically polarised (LHEP) wave?

One or more options are correct.

Option A: $C_{1x} = 1$ , $C_{1y} = 1$ , $\theta = \pi/4$
Option B: $C_{1x} = 2$ , $C_{1y} = 1$ , $\theta = \pi/2$
Option C: $C_{1x} = 1$ , $C_{1y} = 2$ , $\theta = 3\pi/2$
Option D: $C_{1x} = 2$ , $C_{1y} = 1$ , $\theta = 3\pi/4$

Full question on page 39 of the official paper (PDF)

Show answerHide answer

Key idea: The wave travels along $+z$ ; when $E_y$ leads $E_x$ ( $0 < \theta < \pi$ ), the field turns from $+x$ towards $-y$ , left-handed about $+z$ .

Official answer: A, B and D

At $z = 0$ , take $\theta = \pi/2$ : $E_x = C_{1x}\cos\omega t$ and $E_y = -C_{1y}\sin\omega t$ . As $t$ grows from 0, $\mathbf{E}$ moves from $+x$ towards $-y$ : clockwise as seen looking along $+z$ , opposite to the right-hand rule about the direction of travel. That is left-handed, and the same holds for any $0 < \theta < \pi$ , since the sign of $\sin\theta$ sets the sense. $\pi < \theta < 2\pi$ gives right-handed.

(A) $\theta = \pi/4$ : left-handed, and elliptical (equal amplitudes, but $\theta \neq \pi/2$ ). Valid.
(B) $\theta = \pi/2$ with unequal amplitudes: left-handed elliptical. Valid.
(C) $\theta = 3\pi/2$ : right-handed. Not valid.
(D) $\theta = 3\pi/4$ : left-handed elliptical. Valid.

Options (A), (B) and (D). This uses the IEEE convention, in which handedness is judged looking along the direction of travel.

Q.54

Electromagnetics2 marksMSQ

Which of the circuits in the figure represent a lumped-element equivalent of an infinitesimal section $\Delta z$ of a transmission line?

Four lumped circuits, as described in the options: (A) an L section with C delta z over 2 at the input and G delta z over 2 at the output; (B) an L section with G delta z and C delta z at the input; (C) a pi section with G delta z over 2 and C delta z over 2 at each end; (D) a T section with R delta z over 2 and L delta z over 2 on each side of a shunt G delta z and C delta z. — Figure from the official GATE 2023 EC paper, IIT Kanpur

One or more options are correct.

Option A: Circuit (A): series $L\Delta z$ and $R\Delta z$ , with $C\Delta z/2$ across the input and $G\Delta z/2$ across the output
Option B: Circuit (B): $G\Delta z$ and $C\Delta z$ across the input, then series $L\Delta z$ and $R\Delta z$
Option C: Circuit (C): series $L\Delta z$ and $R\Delta z$ , with $G\Delta z/2$ and $C\Delta z/2$ across each end
Option D: Circuit (D): series $R\Delta z/2$ and $L\Delta z/2$ on each side of a shunt $G\Delta z$ and $C\Delta z$

Full question on page 40 of the official paper (PDF)

Show answerHide answer

Key idea: A section of length $\Delta z$ must contain the full series $R\Delta z$ , $L\Delta z$ and the full shunt $G\Delta z$ , $C\Delta z$ , however they are split.

Official answer: B, C and D

As $\Delta z \to 0$ , it doesn't matter where the shunt elements sit (an L, $\pi$ or T section all reduce to the telegrapher's equations), but the totals must be $R\Delta z$ , $L\Delta z$ , $G\Delta z$ and $C\Delta z$ .

(A) has only $C\Delta z/2$ and $G\Delta z/2$ in total: half the shunt elements. Not valid.
(B) is an L section with the full shunt elements. Valid.
(C) is a $\pi$ section: two halves of each shunt element. Valid.
(D) is a T section: two halves of each series element. Valid.

Options (B), (C) and (D).

Q.55

Engineering Mathematics2 marksNAT

What is the value of $\iint_R xy\, dx\, dy$ over the region $R$ in the figure, the square bounded by $y = x$ , $y = -x$ , $y = x + 2$ and $y = -x + 2$ ?

The region R: a square with corners at the origin, (-1, 1), (0, 2) and (1, 1), bounded by the lines y = x, y = -x, y = x + 2 and y = -x + 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

Full question on page 41 of the official paper (PDF)

Show answerHide answer

Key idea: The region is symmetric about the y-axis and $xy$ is odd in $x$ .

Official answer: 0

Replacing $x$ by $-x$ maps $R$ onto itself and changes the sign of $xy$ , so the two halves cancel and the integral is 0.

Directly, for $0 \le y \le 1$ the region spans $-y \le x \le y$ , and for $1 \le y \le 2$ it spans $-(2 - y) \le x \le 2 - y$ ; each inner integral $\int x\, dx$ over a symmetric range is 0. Answer: 0.

Learn the idea:

Q.56

Electronic Devices2 marksNAT

In an extrinsic semiconductor, the hole concentration is $1.5n_i$ , where $n_i = 1 \times 10^{10}$ cm⁻³ is the intrinsic carrier concentration. What ratio of electron to hole mobility makes the hole and electron drift currents equal?

Answer format: rounded off to two decimal places

Full question on page 41 of the official paper (PDF)

Show answerHide answer

Key idea: Equal drift currents in the same field need $n\mu_n = p\mu_p$ , and $np = n_i^2$ .

Official answer: 2.20 to 2.30

n = \frac{n_i^2}{p} = \frac{n_i}{1.5}, \qquad \frac{\mu_n}{\mu_p} = \frac{p}{n} = \frac{1.5 n_i}{n_i/1.5} = 2.25

\begin{aligned} n &= \frac{n_i^2}{p} \\ &= \frac{n_i}{1.5}, \\ \frac{\mu_n}{\mu_p} &= \frac{p}{n} \\ &= \frac{1.5 n_i}{n_i/1.5} \\ &= 2.25 \end{aligned}

The value of $n_i$ cancels. Answer: 2.25.

Learn the idea:

Q.57

Control Systems2 marksNAT

The figure shows the asymptotic magnitude Bode plot of a minimum-phase system: $-20$ dB/decade up to $\omega_1$ , flat from $\omega_1$ to $\omega_2$ , and $-40$ dB/decade beyond $\omega_2$ . Its transfer function is $G(s) = \dfrac{k(s + z)^a}{s^b(s + p)^c}$ , where $k$ , $z$ , $p$ , $a$ , $b$ and $c$ are positive constants. What is $a + b + c$ ?

Asymptotic Bode magnitude plot: a -20 dB/decade line falling to omega_1, a flat line from omega_1 to omega_2, and a -40 dB/decade line beyond omega_2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

Full question on page 41 of the official paper (PDF)

Show answerHide answer

Key idea: Each pole at the origin sets $-20$ dB/decade from the start; a zero adds $+20$ and a pole $-20$ dB/decade at its corner.

Official answer: 4

$-20$ dB/decade from the start: one pole at the origin, $b = 1$ .
At $\omega_1$ the slope rises by $20$ to 0: a single zero, $a = 1$ , $z = \omega_1$ .
At $\omega_2$ the slope falls by $40$ : a double pole, $c = 2$ , $p = \omega_2$ .

So $a + b + c = 1 + 1 + 2 = 4$ .

Learn the idea:

Q.58

Networks, Signals and Systems2 marksNAT

Let $x_1(t) = u(t + 1.5) - u(t - 1.5)$ , and let $x_2(t)$ be as in the figure. For $y(t) = x_1(t) * x_2(t)$ , what is $\int_{-\infty}^{\infty} y(t)\, dt$ ?

x2(t): an impulse of strength 1 at t = -3, a rectangular pulse of height 1 from t = -1 to t = 1, and an impulse of strength 2 at t = 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

Full question on page 42 of the official paper (PDF)

Show answerHide answer

Key idea: The area of a convolution is the product of the two areas.

Official answer: 15

\int y\, dt = \left(\int x_1\, dt\right)\left(\int x_2\, dt\right)

$x_1$ is a pulse of height 1 and width 3, area 3. $x_2$ has area $1 + (1 \times 2) + 2 = 5$ : the two impulses and the pulse. So $\int y\, dt = 3 \times 5 = 15$ .

Q.59

Communications2 marksNAT

$X(t)$ is white Gaussian noise with power spectral density $\tfrac{1}{2}$ W/Hz. It is input to an LTI system with impulse response $e^{-t}u(t)$ . What is the average power of the system output, in W?

Answer format: W, rounded off to two decimal places

Full question on page 42 of the official paper (PDF)

Show answerHide answer

Key idea: For white noise of PSD $S_0$ , the output power is $S_0 \int |h(t)|^2 dt$ , by Parseval.

Official answer: 0.24 to 0.26

P = \frac{1}{2} \int_0^\infty e^{-2t}\, dt = \frac{1}{2} \times \frac{1}{2} = 0.25\ \text{W}

\begin{aligned} P &= \frac{1}{2} \int_0^\infty e^{-2t}\, dt \\ &= \frac{1}{2} \times \frac{1}{2} \\ &= 0.25\ \text{W} \end{aligned}

In the frequency domain, $|H(f)|^2 = 1/(1 + 4\pi^2 f^2)$ , whose integral over all $f$ is also $\tfrac{1}{2}$ . Answer: 0.25.

Q.60

Electromagnetics2 marksNAT

A transparent dielectric coating is applied to glass ( $\varepsilon_r = 4$ , $\mu_r = 1$ ) to eliminate the reflection of red light ( $\lambda_0 = 0.75$ µm). What is the minimum thickness of the coating, in µm?

Answer format: µm, rounded off to two decimal places

Full question on page 42 of the official paper (PDF)

Show answerHide answer

Key idea: A quarter-wave layer cancels reflection when its index is the geometric mean of the media on either side.

Official answer: 0.12 to 0.14

Glass has $n = \sqrt{4} = 2$ and air $n = 1$ , so the coating needs $n_c = \sqrt{1 \times 2} = \sqrt{2}$ and a thickness of a quarter wavelength in the coating:

d = \frac{\lambda_0}{4 n_c} = \frac{0.75}{4\sqrt{2}} = 0.1326\ \mu\text{m}

\begin{aligned} d &= \frac{\lambda_0}{4 n_c} \\ &= \frac{0.75}{4\sqrt{2}} \\ &= 0.1326\ \mu\text{m} \end{aligned}

Answer: 0.13 µm.

Q.61

Electronic Devices2 marksNAT

In a semiconductor device, the Fermi level is 0.35 eV above the valence band edge. The effective density of states in the valence band at $T = 300$ K is $1 \times 10^{19}$ cm⁻³. What is the thermal equilibrium hole concentration in silicon at 400 K, in units of $10^{13}$ cm⁻³? Given $kT$ at 300 K is 0.026 eV.

Answer format: × 10¹³ cm⁻³, rounded off to two decimal places

Full question on page 42 of the official paper (PDF)

Show answerHide answer

Key idea: $p = N_V e^{-(E_F - E_V)/kT}$ , and both $N_V$ (as $T^{3/2}$ ) and $kT$ change with temperature.

Official answer: 60.00 to 70.00

Scale both to 400 K, keeping $E_F - E_V = 0.35$ eV as the question gives it:

N_V(400) = 10^{19}\left(\frac{400}{300}\right)^{3/2} = 1.540 \times 10^{19}\ \text{cm}^{-3}, \qquad kT = 0.026 \times \frac{400}{300} = 0.03467\ \text{eV}

\begin{aligned} N_V(400) &= 10^{19}\left(\frac{400}{300}\right)^{3/2} \\ &= 1.540 \\ &\quad \times 10^{19}\ \text{cm}^{-3}, \\ kT &= 0.026 \times \frac{400}{300} \\ &= 0.03467\ \text{eV} \end{aligned}

p = 1.540 \times 10^{19}\, e^{-0.35/0.03467} = 1.540 \times 10^{19} \times 4.12 \times 10^{-5} = 6.34 \times 10^{14}\ \text{cm}^{-3}

\begin{aligned} p &= 1.540 \\ &\quad \times 10^{19}\, e^{-0.35/0.03467} \\ &= 1.540 \\ &\quad \times 10^{19} \\ &\quad \times 4.12 \\ &\quad \times 10^{-5} \\ &= 6.34 \times 10^{14}\ \text{cm}^{-3} \end{aligned}

That is $63.4 \times 10^{13}$ cm⁻³. Answer: about 63.4. Leaving out the $T^{3/2}$ factor gives 41.2, outside the key's range.

Learn the idea:

Q.62

Digital Circuits2 marksNAT

A sample and hold circuit uses a resistive switch and a capacitor with a time constant of 1 µs. How long must the sampling switch stay closed to charge the capacitor adequately to a full-scale voltage of 1 V with 12-bit accuracy, in µs?

Answer format: µs, rounded off to two decimal places

Full question on page 42 of the official paper (PDF)

Show answerHide answer

Key idea: 12-bit accuracy means the remaining error, $e^{-t/\tau}$ of full scale, must fall to $2^{-12}$ .

Official answer: 8.30 to 8.34

e^{-t/\tau} = 2^{-12} \quad\Rightarrow\quad t = 12\,\tau \ln 2 = 12 \times 1\ \mu\text{s} \times 0.6931 = 8.32\ \mu\text{s}

\begin{aligned} e^{-t/\tau} &= 2^{-12} \\ \Rightarrow\; \\ t &= 12\,\tau \ln 2 \\ &= 12 \times 1\ \mu\text{s} \times 0.6931 \\ &= 8.32\ \mu\text{s} \end{aligned}

Answer: 8.32 µs. (Asking for half an LSB, $2^{-13}$ , would give 9.01 µs, which is outside the key's range.)

Learn the idea:

Q.63

Digital Circuits2 marksNAT

In the sequential circuit in the figure, two D flip-flops share a 1 MHz clock, with $D_1 = Q_2$ and $D_2 = \overline{Q_1}$ . The initial states are $Q_1 = 1$ and $Q_2 = 0$ . What is the frequency of the signal $Q_2$ , in kHz?

Two D flip-flops clocked by CLK = 1 MHz. Q2 is fed back to D1 along the top, and the Q1 bar output is connected to D2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: kHz, rounded off to the nearest integer

Full question on page 43 of the official paper (PDF)

Show answerHide answer

Key idea: Step the state $(Q_1, Q_2)$ clock by clock until it repeats; the cycle length divides the clock frequency.

Official answer: 250, or 500

On each edge, $Q_1 \leftarrow Q_2$ and $Q_2 \leftarrow \overline{Q_1}$ :

(1, 0) \to (0, 0) \to (0, 1) \to (1, 1) \to (1, 0)

\begin{aligned} (1, 0) &\to (0, 0) \\ &\to (0, 1) \\ &\to (1, 1) \\ &\to (1, 0) \end{aligned}

The state repeats every 4 clocks, and $Q_2$ is high for two of them and low for two, so it is a square wave at $1\ \text{MHz}/4 = 250$ kHz.

The key accepts 250 or 500. 500 kHz is what you get by reading $D_2$ as connected to $Q_1$ instead: then the state alternates $(1, 0) \to (0, 1) \to (1, 0)$ , a cycle of 2 clocks. The line in the figure is drawn from the $\overline{Q_1}$ output.

Learn the idea:

Q.64

Analog Circuits2 marksNAT

In the circuit in the figure, a 1 mA reference flows through a diode-connected PMOS (W/L = 1). It is mirrored by PMOS devices with W/L = 10 and W/L = 7. The W/L = 10 PMOS feeds a diode-connected NMOS (W/L = 10), which is mirrored by an NMOS with W/L = 5. The output node, where the W/L = 7 PMOS and W/L = 5 NMOS drains meet, drives a 1 kΩ load to ground. What is the voltage $V_L$ , in V?

Current-mirror circuit: a 1 mA source pulls current from a diode-connected PMOS of W/L = 1 at VDD, whose gate drives two PMOS mirrors of W/L = 10 and W/L = 7. The W/L = 10 PMOS feeds a diode-connected NMOS of W/L = 10, whose gate drives an NMOS of W/L = 5. The W/L = 7 PMOS and W/L = 5 NMOS drains meet at the output V_L, which has a 1 kilohm resistor to ground. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: V, rounded off to two decimal places

Full question on page 43 of the official paper (PDF)

Show answerHide answer

Key idea: In saturation, a mirror copies the reference current scaled by the ratio of W/L.

Official answer: 2

PMOS W/L = 10: $10 \times 1 = 10$ mA, into the diode-connected NMOS of W/L = 10.
NMOS W/L = 5: $\tfrac{5}{10} \times 10 = 5$ mA, drawn from the output node.
PMOS W/L = 7: $7 \times 1 = 7$ mA, pushed into the output node.

The difference, $7 - 5 = 2$ mA, flows through the 1 kΩ, so $V_L = 2 \times 1 = 2.00$ V, assuming every transistor stays in saturation and channel-length modulation is ignored.

Q.65

Communications2 marksNAT

Eight symbols a to h occur with frequencies $\tfrac{1}{2}$ , $\tfrac{1}{4}$ , $\tfrac{1}{8}$ , $\tfrac{1}{16}$ , $\tfrac{1}{32}$ , $\tfrac{1}{64}$ , $\tfrac{1}{128}$ and $\tfrac{1}{128}$ . A symbol is chosen, and it is identified by a series of truthfully answered yes/no questions. What is the average number of questions when they are asked in the most efficient sequence?

Answer format: rounded off to two decimal places

Full question on page 44 of the official paper (PDF)

Show answerHide answer

Key idea: The best question sequence is a Huffman code: each question is one bit.

Official answer: 1.97 to 1.99

Ask "is it a?", then "is it b?", and so on: a takes 1 question, b 2, ..., f 6, and g and h 7 each. This is the Huffman code for these probabilities, so no sequence does better on average:

\bar{L} = \frac{1}{2}(1) + \frac{1}{4}(2) + \frac{1}{8}(3) + \frac{1}{16}(4) + \frac{1}{32}(5) + \frac{1}{64}(6) + 2 \cdot \frac{1}{128}(7) = \frac{127}{64} = 1.984

\begin{aligned} \bar{L} &= \frac{1}{2}(1) \\ &\quad + \frac{1}{4}(2) \\ &\quad + \frac{1}{8}(3) \\ &\quad + \frac{1}{16}(4) \\ &\quad + \frac{1}{32}(5) \\ &\quad + \frac{1}{64}(6) \\ &\quad + 2 \cdot \frac{1}{128}(7) \\ &= \frac{127}{64} \\ &= 1.984 \end{aligned}

Every probability is a power of $\tfrac{1}{2}$ , so this equals the entropy exactly. Answer: 1.98.

GATE 2023 EC Paper: Worked Solutions to All 65 Questions

Q.1

Q.2

Q.3

Q.4

Q.5

Q.6

Q.7

Q.8

Q.9

Q.10

Q.11

Q.12

Q.13

Q.14

Q.15

Q.16

Q.17

Q.18

Q.19

Q.20

Q.21

Q.22

Q.23

Q.24

Q.25

Q.26

Q.27

Q.28

Q.29

Q.30

Q.31

Q.32

Q.33

Q.34

Q.35

Q.36

Q.37

Q.38

Q.39

Q.40

Q.41

Q.42

Q.43

Q.44

Q.45

Q.46

Q.47

Q.48

Q.49

Q.50

Q.51

Q.52

Q.53

Q.54

Q.55

Q.56

Q.57

Q.58

Q.59

Q.60

Q.61

Q.62

Q.63

Q.64

Q.65

Sources

More GATE ECE