GATE 2021 ECE Solutions Modified

GATE-2021 ORIGINAL PAPER – ECE/1

GENERAL APTITUDE

1. Corners are cut from an equilateral triangle to produce a regular convex hexagon as shown in the

figure above. The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle

is (A) 2 : 3 (B) 5 : 6 (C) 3 : 4 (D) 4 : 5 Sol. Let ABC be the equilateral triangle and PQRSTU be the convex hexagon, with side AB having points P,Q;

side BC having points R,S and side CA having points T,U respectively. Since PQRSTU is a regular convex hexagon and ABC is equilateral. The triangles APU, BQR and CST are also equilateral. Let the side of each of ∆APU, BQR and CST be s. Then each side of the regular polygon is also s and the side of the triangle ABC = 3s Area of the regular convex polygon : Area of ∆ ABC – 3 Area of ∆APU

√ 3 3 ∙ √ : √ 3

= 6 : 9 2: 3 Choice (A)

2. Computers are ubiquitous. They are used to improve efficiency in almost all fields from agriculture to space exploration. Artificial intelligence (AI) is currently a hot topic. AI enables computers to learn, given enough training data. For humans sitting in front of a computer for long hours can lead to health issues.

Which of the following can be deduced from the above passage?

(i) Nowadays, computers are present in almost all places.

(ii) computers cannot be used for solving problems in engineering.

(iii) For humans, there are both positive and negative effects of using computers.

(iv) Artificial intelligence can be done without data.

(A) (ii) and (iii)

(B) (i) and (iii)

(C) (ii) and (iv)

(D) (i), (iii) and (iv)

Sol. The passage states that computers are ‘ubiquitous’, hence (i) can be deduced. (ii) is not mentioned in the

passage. (iii) can be deduced from ‘long hours’. (iv) is not true, as the passage states ‘given enough data’. Choice (B)


3. p and q are positive integers and , then

(A) 9 (B) 7 (C) 11 (D) 3 Sol. Given, + = 3

Squaring on both sides ( + ( + 2 = 9

( + ( = 7 Choice (B)

4. Consider a square sheet of side 1 unit. In the first step, it is cut along the main diagonal to get two triangles. In the next step, one of the cut triangles is revolved about its short edge to from a solid cone. The volume of the resulting cone, in cubic unit, is______

(A)

(B)

(C)

(D)

Sol. In the first step a right-angled triangle with sides 1 1 √2 is formed. Let the cone formed from the second step have the radius and vertical height as and respectively. In the formation of the cone, 1 and 1 ⇒ Volume of the cone = 1 1 Choice (D)

5. Given below are two statements and two conclusions. Statements 1: All purple are green. Statements 2: All black are green. Conclusion I: Some black are purple. Conclusion II: No black is purple.

Based on the above statement and conclusion, which one of the following options if logically CORRECT? (A) Only conclusion II is correct. (B) Only conclusion I is correct. (C) Both conclusion I and II are correct. (D) Either conclusion I or II is correct. Sol. In the given statements, the middle term ‘green’ is not distributed. Hence we have two possibilities. Either

the terms ‘black’ and ‘purple’ will have some relation between them or they do not have any relation between them. Therefore either conclusion I or conclusion II is correct.

Answer: Either conclusion I or II is correct. Choice (D)

6. The current population of a city is 11,02,500. If it has been increasing at the rate of 5% per annum, what was its population 2 years ago? (A) 10,00,000 (B) 9,92,500 (B) 9,95,006 (D) 12,51,506

Sol. Let the population of the city two years ago be X

11,02,500 = X 1

X = , ,

=

= 10,00,000 Choice (A)


7. The least number of suares that must be added so that the line P–Q becomes the line of symmetry

is ____ (A) 6 (B) 7 (C) 4 (D) 3 Sol. To make PQ the line of symmetry we can draw the following diagram. P

Hence the minimum number of squares must be drawn are six. Choice (A)

8. Nostalgia is to anticipation as ______ is to _______ Which one of the following options maintains a similar logical relation in the above sentence? (A) Present, past (B) Future, present (C) Future, past (D) Past, future Sol. The first part of the analogy refers to the past and the second part to the future. Choice (D)

Q

P

Q


45

55

60

60

20

60

55

70

65

50

55

35

50

65

0 10 20 30 40 50 60 70 80

Monday

Tuesday

Wednesday

Thursday

Firday

Saturday

Sunday

Y X

9. Consider the following sentence:

(i) I woke up from sleep

(ii) I woked up from sleep

(iii) I was woken up from sleep

(iv) I was wokened up from sleep

Which of the above sentence are grammatically correct?

(A) (ii) and (iii)

(B) (i) and (iv)

(C) (i) and (iii)

(D) (i) and (ii)

Sol. ‘Woked’ and ‘wokened’ do not exist. (i) is active voice and (iii) is passive voice.

Choice (C)

10. The number of minutes spent by two students, X and Y, exercising every day in a given week are

shown in the bar chart above. The number of days in the week in which one of the students spent a minimum of 10% more than

the other student, on a given day, is (A) 4 (B) 7 (C) 5 (D) 6 Sol. Except Thursday, in the remaining days one of the students spent at least 10% more than that by the

other. Hence required answer is 6 Choice (D)



ELECTRONICS AND COMMUNICATION ENGINEERING

1. The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time t = 0.

Initially, when the switch is open, the capacitor is discharged, and the ammeter reads zero ampere.

After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is _______A.

Sol. From the given data all initial values are zero For t > 0 : - switch closed Redraw the circuit in s - domain From the given data

Vc (s) =

I(S) =

=

5 k

100PF

I = ? t = 0

10 mH

A

1A C

R

I = ?

SL

A

1

1

Vc(s)

5 k

100PF

Load Switch

10 mH

A

1 A Shunt Capacitor

Current Source

Ammeter


= =

= 2.5 0.1 = 0.25

% peak overshoot =

–

– 100

= –

100 = 44

– = 0.44 C(t) But C(t) = i(t) I(t)max – 1 = 0.44 I(t)max = 1.44 Amp Ans:1.4 to 1.5

2. For the circuit with an ideal OPAMP shown in the figure. VREF is fixed. If VOUT = 1 volt for VIN = 0.1 volt and VOUT = 6 volt, for Vin = 1V, where VOUT is measured across RL

connected at the output of this OPAMP. The value of RF/RIN is (A) 3.285 (B) 3.825 (C) 2.860 (D) 5.555

Sol.

=

.

=

= |– 5.55| = 5.55. Choice (D)

3. Consider a real-valued base-band signal x(t). band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) x (1 + ) is

(A) 20 kHz (B) 30 kHz (C) 60 kHz (D) 15 kHz

+

– VIN +

–

RF RIN

+12V

VOUT R1

+

–12V

R2

RL = 100 VREF –


Sol. Given, Y(f) :

Nyquist rate, = 2 x fmax = 2 x 15 = 30 kHz Choice (B)

4. The autocorrelation function RX() of a wide-sense stationary random process X(t) is shown in the figure.

The average power of X(t) is _________

Sol.

The Average power of x(t) = ? Rxx() = E[x(t) x(t + )] Rxx(0) = E[x2(t)] = Pavg = 2 watts. Ans:2 to 2

X(f)

f 10KHz –10KHz

f 10KHz –10KHz

f 5KHz –5KHz

Y(f)

f 15KHz –15KHz

RX()

–2 0 2

RX()

–2 0 2


5. Consider a super heterodyne receiver tuned to 600 kHz. If the local oscillator feeds a 1000 kHz signal to the mixer. The image frequency (in integer) is _____kHz.

Sol. fIF = fLO – fs = 1000 kHz – 600 kHz = 400 kHz fsi = fs + 2fIF = 600 kHz + 2 x 400 kHz fsi = 1400 kHz Ans: 1400 to 1400

6. For the transistor M1 in the circuit shown in the figure, μn Cox = 100 µA/V2 and = 10, where μn is

the mobility of electron, Cox is the oxide capacitance per unit area. W is the width and L is the length.

The channel length modulation coefficient is ignored. If the gate-to-source voltage VGS is 1 V to

keep the transistor at the edge of saturation. Then the threshold voltage of the transistor (rounded off to one decimal place) is _____ V.

Sol. From the given data nCox = 100A/V2 VGS = 1V

= 10

Given transistor is in saturation region VDS VGS – VT VDS = VGS – VT saturation region VDS = 1 – VT

ID = nCox [VGS – VT]2

ID = –

mA

–

mA = 100 10–6 10 [1 – VT]2

(2 + VT) = 10[1 + V – 2VT] 10V – 21VT + 8 = 0 VT = 1.6 and 0.5 But valid VT = 0.5V. Ans: 0.5 to 0.5

20 k

Vo

VG

3 V

RD = 20 k

VOUT

VDD = 3 V

VGS M1


7. Consider the circuit with an ideal OPAMP shown in the figure. Assuming | | << | | and | | << | |. The condition at which VOUT equals to zero is (A) VIN = 2 VREF

(B) VIN = 2 + VREF

(C) VIN = VREF

(D) VIN = 0.5 VREF

Sol. For ideal op-amp KCL at node V–.

+

+ = 0.

= V V

V = 0 V V = 0 V = V . Choice (C)

8. A real 2 x 2 non- singular matrix A with repeated eigenvalue is given as

A = .. .

Where x is a real positive number. The value of x (rounded off to one decimal place) is ______

Sol. Given A = 3.03.0 4.0

The characteristic equation of A is

|A – I| = 0

33 4

0

(x – ) (4 – ) + 9 = 0

2 – (x + 4) + 9 + 4x = 0 (1)

As the eigen values of A are repeated, we have from (1),

[–(x + 4)]2 – 4(1) (9 + 4x) = 0

(∵ If the roots of ax2 + bx + c = 0 are repeated then b2 – 4ac = 0)

(x2 + 8x + 16) – 36 – 16x = 0 x2 – 8x – 20 = 0 (x –10) (x + 2) = 0 x =10 (∵ x is positive) Ans.:10 to 10

– VIN +

RF R

+Vcc

– VOUT

R

+

–Vcc

VREF –

+



9. A box contains the following three coins. I. A fair coin with head on one face and tail on the other face. II. A coin with heads on both the faces. III. A coin with tails on both the faces. A coin is picked randomly from the box and tossed. Out of the remaining coins in the box. One

coin is then picked randomly and tossed. If the first toss results in a head. The probability of getting a head in the second toss is

(A)

(B)

(C)

(D)

Sol. Let A = Getting head in the first toss

And B = Getting head in the second toss

The probability of getting head in the second toss when the first toss results in a head

= P(B/A) = ∩

=

=

=

= Choice (A)

10. The propagation delays of the XOR gate, AND gate and multiplexer (MUX) in the circuit shown in the figure are 4 ns, 2ns and 1 ns respectively.

If all the inputs P, Q, R, S and T are applied simultaneously and held constant, the maximum propagation delay of the circuit is

(A) 3 ns (B) 6 ns (C) 5 ns (D) 7 ns

P

Q

0

Y MUX 1

R

S0

MUX

0

1 S

T

S0


Sol. Maximum propagation delay is 6 ns Choice (B)

11. A message signal having peak-to-peak value of 2V, root mean square value of 0.1V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is _____.

Sol. Vpp = VH – VL = 2V

rms value = 0.1 V

Signal power = Ps = V2rms = 0.01 wats

fmax = 5 kHz fs = 2 x 5 kHz = 10 kHz

rb = 50 kbps = nfs

50 kbps = n x 10 kHz

n = 5

SQNR of PCM =

Noise power = Pn =

√ =

= =

Pn = =

SQNR = = .

SQNR = 30.72 Ans: 30.70 to 30.74

P

Q

R

S

T = 0

0 1

0

1

2 1 MUX

(2)

(3) ns 2

4 0

P

Q

R

S

T = 1

0 1

0

1

(2)

(3)

2

(5)

6 ns


12. A 4 kHz sinusoidal message signal having amplitude 4 V is fed to a delta modulator (DM) operating at a sampling rate of 32 kHz. The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is ______ V.

Sol. To avoid slope over load error the condition is

∆

≥ (m(t))

∆

≥ 2 fm.Am.

∆ min = 2 fm. Am.Ts

= 2 × 4 × 103 ×

∆min = = 3.141V Ans:3.12 to 3.16

13. Consider a rectangular coordinate system (x, y, z) with unit vectors ax, ay and az. A plane wave traveling in the region z ≥ 0 with electric field vector E = 10 cos (2 x 108t + βz)ay is incident normally on the plane at z=0, where β is the phase constant. The region z ≥ 0 is in free space and the region z<0 is filled with a lossless medium (permittivity ε = ε0 permeability µ = 4 µ0, where ε0 = 8.85 x 10-12 F/m and µ0 = 4π x 10-7 H/m). The value of the reflection coefficient is

(A)

(B)

(C)

(D)

Sol. =

–

=

=

∵ = =

= √ – √

√ √ = √ – √

√ √ =

– = Choice (C)

14. A digital transmission system uses a (7,4) systematic linear Hamming code for transmitting data over a noisy channel. If three of the message-code word pairs in this code (mi ; ci). Where ci is the codeword corresponding to the ith message mi, are known to be (1 1 0 0 ; 0 1 0 1 1 0 0), (1 1 1 0 ; 0 0 1 1 1 1 0) and (0 1 1 0 ; 1 0 0 0 1 1 0), then which of the following is a valid codeword in this code?

(A) 1 1 0 1 0 0 1 (B) 1 0 1 1 0 1 0 (C) 0 1 1 0 1 0 0 (D) 0 0 0 1 0 1 1

Sol. C1 = 0101100 C1 ⊕ C2 = C4

C2 = 0011110 C2 ⊕ C3 = C5

C3 = 1000110 C1 ⊕ C3 = C6

C3 ⊕ C4 = C7

P1 = d1 ⊕ d2 ⊕ d4

P2 = d2 ⊕ d3 ⊕ d4

P3 = d1 ⊕ d2 ⊕ d3 Choice (A)


15. The complete Nyquist plot of the open-loop transfer function G(s) H(s) of a feedback control system is shown in the figure.

If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is

(A) 1 (B) 0 (C) 4 (D) 3

Sol. From the given data Open loop zeros in RHS(Z = 1) N = P – Z Open loop: N(0, 0) = (O.L.P – O.L.Z) 2 = OLP – 1 OLP(P) = 3 N(–1, 0) = P – Z 0 = 3 – Z Z = 3. Choice (D)

16. An asymmetrical periodic pulse train vin of 10V amplitude with on-time TON = 1 µs is applied to the circuit shown in the figure. The diode D1 is ideal.

The difference between the maximum voltage and minimum voltage of the output waveform v0 (in integer) is ______ V.

Sol. Vin = 10V; Diode is ON

Capacitor charges up to 10V. Vc = 10V Vin = 0; Diode is OFF Discharging time constant = RC = 10 m sec ≫ Vc = 10V. Vout = Vin – Vc = Vin – 10 Vin = 10V Vout = 0V ∴ Vout = 10V. Ans: 10 to 10

+

R 10V

–

10V + –

20nF

0 (–1, j0)

Re GH

G(s)H(s)-plane

j Im GH

20 nF

500 k

Vin

+

Vo

–


17. Consider the two-port network shown in the figure. The admittance parameters, in siemens are (A) y11 = 2, y12 = -4, y21 = -1, y22 = 2 (B) y11 = 2, y12 = -4, y21 = -4, y22 = 2 (C) y11 = 2, y12 = -4, y21 = -4, y22 = 3 (D) y11 = 1, y12 = -2, y21 = -1, y22 = 3 Sol. From the given data

– I1 + – 3V2 + –

= 0

I1 = V1 – 3V2 + V1 – V2 I1 = 2V1 – 4V2 (i) and

– I2 + + –

= 0

I2 = 2V2 – V1 I2 = – V1 + 2V2 (ii) From equation (i) and (ii), we can get y11 = 2Ʊ ; y12 = – 4 Ʊ

y21 = – 1Ʊ ; y22 = 2 Ʊ

[Y] = 2 – 4– 1 2

Ʊ

18. Consider the differential equation given below.

+

y = x

The integrating factor of the differential equation is (A) )-1/2

(B) )-1/4

(C) )-3/2 (D) )-3/4

Sol. Given differential equation is

= x (1)

√

√

= x

= x (2)

1

V1 I1 V2 I 2

3V2 V1 V2 1

1

I1 I 2

+

3V2 V1 V2

1

+

1 1

– –


Put = z =

= 2

(2) becomes,

2 +

z = x

+

z = (3)

(3) is a linear equation of first order where P(x) =

and Q(x) =

Integrating Factor = I. F =

=

=

=

=

I.F = 1 Choice (B)

19. In the circuit shown in the figure, the transistors M1 and M2 are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the MOSFETs M1 and M2 are gm1 and gm2, respectively, and the internal resistance of the MOSFETs M1 and M2 are r01 and r02 respectively.

Ignoring the body effect, the ac small signal voltage gain (dVout/ dVin) of the circuit is (A) -gm2 (r01 II r02)

(B) -gm2 ( II r02)

(C) -gm1 ( II r01 II r02)

(D) -gm2 ( II r01 II r02)

Sol. MOSFET M2 outs as an common source amplifier.

Vcc

Vin

c

VDD

M2

Vout

M1

Vin


Drain to gate connect MOSFET M1 acts as load. Reg = ∥

=

r02 ∥

|AV| = –g r ∥ ∥ Choice (D)

20. A silicon P-N junction is shown in the figure. The doping in the P region is 5 x 1016 cm-3 and doping in the N region is 10 x 10–16 cm–3. The parameters given are

Built-in voltage (bi) = 0.8 V Electro charge (q) = 1.6 x 10-19 C Vacuum permittivity of silicon (εsi) = 12

The magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is ________ V.

Sol.

NA = 5 1016 cm–3 and ND = 10 1016 cm–3 Vbi = 0.8V. εsi = 11.7 VR = ?

W =

= . . –

. –

[0.8 + VR]

W = 12.46 10 10– 0.2 0.1 0.8

(Wn + Wp) = 3.738 10– 0.8 ND.Wn = NA.Wp

Wp = .Wn

= 0.4 m W = Wn + Wp = 0.6m 0.6 10–4 = 3.738 0.8 10–

VR = 8.23V Ans: 8.1 to 8.4

P N

1.2 m 0.2 m

P N

1.2 m 0.2 m



21. The impedance matching Network shown in figure is to match a lossless line having characteristic impedance = 50 Ω with a load impedance . A quarter – Wave line having a characteristic impedance = 75 Ω is connected to . Two stubs having characteristic impedance of 75 Ω each are connected to this quarter – wave line. One is a short – circuited (S.C) stub of length 0.25 connected across PQ and the other one in an open – Circuted (O.C) stub of length 0.5 connected across RS.

The impedance matching is achieved when the real part of .is (A) 33.3 Ω (B) 75.0 Ω (C) 50.0 Ω (D) 112.5 Ω

Sol. Sc of the length /4 ⟹OC OC of length /2 ⟹ OC at distance of /4 ⟹

New impedance = =

This is in parallel with OC [Sc modified]

∴ // OC =

This is matched with = 50 Ω

= 50 ⟹ =

= 112.5 Ω Choice (D)

22. The switch in the circuit in the figure is in position P for a long time and then moved to position Q at time t = 0

The value of at t = is

(A) 3 V/s (B) –5V/s (C) 0 V/s (D) –3V/s

R 0.25 P

S

S.C

Stu

b

Z1 =

75

Ω

Z1 = 75 Ω

O.C

Stu

b Z

1 =

75

Ω

0. 5

Z0 = 50 Ω 0.25

Q

ZL

5 k

1 mF

t = 0

V(t) +

20V –

5 k

20 k

P

10 k Q

0.1 mH

+ –


Sol.

at t = 0+

For t < 0 :- switch connected at position P At t = 0 : Circuit is in steady state

iL (0+) =

= 1 mA

Vc (0) = 10K 10 mA = 10 V

For t > 0 : - Switch connected at position Q At t = 0+

ic (0+) + 2 mA + 1 mA = 0 ic (0+) = – 3 mA

ic = C

=

at t = 0+

= –

V/sec

= – 3V/sec Choice (D)

23. The refractive indices of the core and cladding of an optical fiber are 1.50 and 1.48 respectively. The critical propagation angle. Which is defined as the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection, (rounded off to two decimal places) is ______ degree.

Sol. sinA = NA = – = √1. 5 – 1.48 = √0.0596 = 0.24413 A = 14.13°

5 k

1 mF

t = 0

Vc(t) +

20V –

5 k

20 k

P 10 k

Q 0.1 mH

5 k

iL (0-) Vc(0-) 20V

5 k

20 k ±

10 k

10 k

1mA

Q

10V

5 k

ic(0+)


24. Consider a carrier signal which is amplitude modulated by a single – tone sinusoidal message signal with a modulation index of 50%. If the carrier and one of the side bands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is ______

Sol. % power saved = =

for SSB - SC

=

% power saved =

Given µ = 0.5

% power saving = .

. × 100 = 94.44. Ans: 94.2 to 94.6

25. For a vector filed D = p + ∅ in a cylindrical coordinate system (p, ,z) with unit vectors , and , the net flux of D leaving the closed surface of the cylinder (p = 3, 0 Z 2) rounded off to two decimals places ) is _______

Sol. φ = ∯ .

φ = 2ʃ0 2

ʃ0

3 ∅d∅dz

= 2ʃ0 2

ʃ0

9 ∅

d∅dz

= 2ʃ0 ∅

∅dz

2ʃ0(2 – 0) dz

= × 2 = 9 × ×2 = 18 = 56.548 Ans: 56.5 to 56.6

26. An antenna with a directive gain of 6dB is radiating a total power of 16 kw. The amplitude of the electric field in free space at a distance of 8km from the antenna in the direction of 6 dB gain(rounded off to three decimal places is ________ V/m.

Sol. GT = 4[∵ 6dB]

=

E2 =

= = 0.06

E = √0.06 = 0.2449 ≅ 0.25V/m Ans: 0.224 to 0.264

27. The vector function F(r) = - xi + yj is defined over a circular are C shown in the figure. The line integral of ʃ F(r). dr is

(A)

(B)

(C)

(D)

j

i

45°

1


Sol. Given F – i + yj We have to find F . dr

F . dr

= x dx y dy

(1)

Put x = cos and y = sin where 0

dx = –sin d and dy = cos d (1) becomes,

F . dr

= cos sin d sin cos d

= cos. sin sin . cos d

= sin2 d

=

= cos cos 0

= Choice (D)

28. In a high school having equal number of boy students and girl students, 75% of the students study Science and remaining 25% students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is _______ bits.

Sol. From the given data

P(B) = and P(G) =

P(s) = and P(c) =

P = 2p (B/s)

P(B) = P(B/C) P(c) + P(B/S) P(S)

= P(B/C) + P(B/S)

2 = P (B/C) + 3 × P(B/C)

2.5 P(B/C) = 2

P(B/C) = .

=

P(C/B) = /

P(C/B) =

=

P(C) = P P(B) + P P (G)

= + P ×

P =

I =

I = – P

I = – 10

I = 3.3219 bits Ans: 3.32 to 3.35


29. A unity feedback system that uses proportional – integral (PI) control is shown in the figure. The stability of the overall system is controlled by tuning the PI control parameters and The

maximum value of that can be chosen so as to keep the overall system stable or, in the worst case, marginally – stable (rounded off to three decimal places) is ______

Sol. Redrawing the given figure, From figure

G(s) =

.

=

H(S) = 1 Characteristic equation 1 + G(S) H(S) = 0

1 +

= 0

Characteristic equation S4 + 4S3 + 5S2 + [2KP + 2]S + 2KI = 0 Applying Routh Table

S4 1 5 2KI S3 4 2KP + 2

S2 2 18

4 2Ki

S1 32 2 18 2 2

2 18

S0 2KI Necessary Condition From Hurwitz Polynomial Condition all coefficients of characteristic equation needs to be of same sign

(Here > 0) Comparing characteristic Equation 2Ki > 0 Ki > 0 2KP + 2 > 0 KP > –1

Y(s) 2

4 5 2

+ –

X(s)

Y(s)

–

+ X(s) Kp +


Sufficient Condition From Routh table all elements of First column to be positive.

From Row S2

> 0

18 2 > 0 18 2

or KP < 9

KP Range is –1 < KP < 9 From Row S

> 0

32 2 18 2 2 > 0 32 36 32 4

Ki >

Range of Ki

0 < Ki <

If KP = –1 KI = 0 If KP = 9 KI = 0 To find Max value of KI on the equation he find

0

32 – 8 = 0 KP = 4

When KP = 4, KI =

= 3.125

= 3.125 Ans: 3.125 to 3.125

30. A speech signal, band limited to 4 kHz, is sampled at 1.25 times the Nyquist rate. The speech samples, assumed to be statistically independent and uniformly distributed in the range – 5V to + 5V, are subsequently quantized in an 8 – bit uniform quantizer and then transmitted over a voice – grade AWGN telephone channel. If the ratio of transmitted signal power to channel noise power is 26 dB, the minimum channel bandwidth required to ensure reliable transmission of the signal with arbitrarily small probability of transmission error (rounded off to two decimal places) is ______kHz.

Sol. From the given data fs = 1.25. Nyquist rate = 1.25 × 2fm = 1.25 × 4 kHz fs = 10 kHz

N = 8 and = 26 dB

For small probability of error : C Rb.

∴ C = B.log2 1

P(x)

0 -5 5 x


∴ B

B

B =

.

B =

.

B = 9.25 kHz. Ans:9.24 to 9.28

31. A circuit with an ideal OPAMP is shown in the figure. A plus VIN of 20 ms duration is applied to the input. The capacitors are initially uncharged.

The output voltage VOUT of this circuit at = 0+ (in integer) is _________ V. Sol: At t = 0+, capacitor is short circuit V– = Vin = 5v V+ = 0V V– > V+

Vout = –Vsaf = – 12V. Ans: –12 to –12

32. A bar of silicon is doped with boron concentration of 1016 cm–3 and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of 1020 cm–2S–1. If the recombination lifetime is 100 µs, intrinsic carrier concentration of silicon is 1010 cm–3 and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

(A) 1020 cm–6 (B) 2 × 1032 cm–6 (C) 1032 cm–6 (D) 2 × 1020 cm–6

5V –

+ –12V

+12V

R

Vout

5V

0 V

t = 0 t = 20 ms

1 F

1 F

VIN

–12 V

+12 V

VOUT

10 Ωk

+

–


Sol. From the given data NA 1016 cm–3 G = 1020/cm3 – sec Recombination life time () = 100S i = 1010 cm–3

We know = p. and p Pp =

= = = 104 cm–3

= p = Gp.p

= 1020 100 10–6 p = 1016 cm–3

Product of steady state e– and hole concentration due to the right = ? ( + ) ( + p) = ? = (104 + 1016)(1016 + 1016) 1016 1016 2 Approximate Value 2 1032 cm–6. Choice (B)

33. Two continuous random variables X and Y are related as Y = 2x + 3 Let and denote the variances of X and Y, respectively. The variances are related as (A) = 5 (B) = 25 (C) = 4 (D) = 2

Sol. Sol. y = 2x + 3 var [x] = E[x2] – E[x]2 var [y] = E[y2] – E[y]2 = E[(2x + 3)2] – (E(2x + 3))2 = E[4x2 + 12x + 9] – E2 [2x] + E2 [3] – 2E[2x] E[3] = 4E[x2] + 12 E[x] + 9 –4E2 [x] –9 – 12E[x]] var [y] = 4[E[x2] – E2[x]] var [y] = 4 var[x] Choice (C)

34. A standard air-filled rectangular waveguide with dimensions a = 8 cm, b = 4 cm, operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is p. The value

(rounded off to two decimal places) of p/c, where c denotes the velocity of light, is ________.

Sol. Vp =

=

fC =

= ∵ TE10 mode.

f = 3.4 × 109

fC =

– = × 108 = 1.875 × 109

=

..

= .

= 1.198 1.2 Ans: 1.15 to 1.25

S.C Boron IIIrd Group

P-type



35. Consider the circuit shown in the figure. The current I flowing through the 7Ω resistor between P and Q (rounded off to one decimal place)

is _______ A. Sol.

Redraw the given circuit and simplicity I V transformation

I = – –

= 0.5 A Ans: 0.5 to 0.5

3

7

6 2

2

5 A

Q

P I

6 3

7

2 2

5 A

I P

Q

2

7

5 V

I P Q

∓

1

5 A

2 Ω 6 Ω

7 Ω

P Q

I


36. Consider the signals x[n] = 2n–1 u[–n + 2] and y[n] = 2–n+2 u[n + 1], where u[n] is the unit step sequence. Let X(ej) and Y(ej) be the discrete-time Fourier transform of x[n] and y [n], respectively. The value of the integral

–

(round off to one decimal place) is _____

Sol. Given, x[n] = 2n-1 u[-n+2] y[n] = 2-n+2 u[n+1] y[n] Y(ejw) y[-n] Y(e-jw)

∴ y[-n] = 2n+2u[-n+1]

y )dw = x n ∗ y n |

∴ x[n] * y[-n] = ∑ 2 2 2 1 = ∑ 2 1

x n ∗ y n | = ∑ 2 1 = ∑ 2 2 2 2 2 = 8 Ans: 7.9 to 8.1

37. Consider the circuit shown in the figure.

The value of o (round off to one decimal place) is _______ Sol.

Vo = ? From the above data ix = io + 2mA io =

= vo mA

ix = (Vo + 2) mA Applying KVL in loop 1 – ix 1K + 4 – Vo = 0 – (Vo + 2) + 4 – Vo = 0 2Vo = 2 Vo = 1V Ans:1 to 1

DTFT

DTFT

1 k

6 mA

1 k 2 mA

1 k

4V

i1

Vo

ix

+

– –

+

6 mA 1 kΩ

1 kΩ 2 mA 4 V

1 kΩ

+

– V0

+ –


38. For an n-channel silicon MOSFET with 10nm gate oxide thickness, the substrate sensitivity ( / | |) is found to be 50 mV/V at a substrate voltage | | = 2V, where VT is the threshold voltage of the MOSFET. Assume that, | | >> 2ϕB, where qϕB is the separation between the Femi energy level EF and the intrinsic level Ei in the bulk. Parameters given are

Electron charge (q) = 1.6 × 10–9C Vacuum permittivity (εo) = 8.85 × 10–12F/m Relative permittivity of silicon (εSi) = 12 Relative permittivity of oxide (εox) = 4 The doping concentration of the substrate is (A) 2.37 × 1015 cm–3 (B) 4.37 × 1015 cm–3

(C) 9.37 × 1015 cm–3

(D) 7.37 × 1015 cm–3 Sol. From the given data q = 1.6 x 10–19 C 0 = 8.85 x 10–12 F/m si = 12 0x = 4 tox = 10nm

| |

= 50 mv/v at VBS = 2V

|VBS| >> 2B,

The doping concentration of the substrate = ? Vt = Vto + |2Φ | |2Φ |

| |

=

| | = 50 mv/v

= 0.1 x √2

= = 0.1 x √2

=

=

= 3.54 x 10–3 F/m2 . . = x 0.01 1.6 × 10–19 × NA × 12 × 8.85 × 10–12 = 10–12 × (3.54 × 10–13)2

NA = .

.

NA = 7.37 × 1021 /m3 NA = 7.37 × 1015 cm–3. Choice (D)

39. Consider the vector field F = ax(4y – c1z) + ay (4x + 2z) + az(2y + z) in a rectangular coordinate system (x, y, z) with unit vectors ax, ay and az. If the field F is irrotational (Conservative), then the constant c1 (in integer) is ______

Sol. Given = (4y + c1z) ai + (4x + 2z) aj + (2y + z) is irrotational Curl = 0

i j k∂

∂x

∂

∂y

∂

∂z

4y c1z 4x 2z 2y z

= 0

i (2 – 2) – j (0 – c1) + k (4 – 4) = 0 c1 = 0 Ans: 0 to 0


40. Addressing of a 32K × 16 memory is realized using a single decoder. The minimum number of AND gates required for the decoder is

(A) 219 (B) 28

(C) 232

(D) 215

Sol. Address of a 32 × 16 memory is realized using a single decoder. The minimum number of AND gates required for the decoder is 2 . Choice (D)

41. Consider the integral ∮

Where C is a counter-clockwise oriented circle defined as |x – i| = 2. (A) – sin(2i)

(B) sin(2i)

(C) sin(2i)

(D) – sin(2i)

Sol.

Given ∮

; c:| | 2

x = 0, ± 2i are the singularities of the integrand. of these x = 0 and x = +2i lie inside C and x = 2i lies outside C

Let f(x) =

∮

= ∮

2 ------(1)

lim→

(∵ x=0 is a pole of order 2)

= lim→

=lim→

→(2)

2→

(∵x = 2i is a simple pole)

= lim→

=

Im

3i

C 2i

i

–i

0

–2i

Re


→(3)

Substituting (2) and (3) in (1), we get

∮

2

Choice (No option correct)

42. For a unit step input u[n], a discreate-time LTI system produces an output signal (

– ). Let y[n] be the output of the system for an input . The value of y[0] is

_______

Sol. Step response, = S(n) Impulse Response, [n] = S(n) – S(n – 1) [n] = 2 [n + 1] + [n] + [n – 1] – 2[n] – [n – 1] – [n – 2] [n] = 2 [n + 1] – [n] – [n – 2]

For input x[n] = u[n] :

Output, y[n] = 2 [n + 1] u[n] – [n] u[n] – [n – 2] u[n]

y[n] = u(n + 1) – u(n) – u(n – 2)

y(0) = 1 – 1 = 0 Ans: 0 to 0

43. An 8-bit unipolar (all analog output values are positive) digital-to-analog converter (DAC) has a full-scale voltage range from 0 V to 7.68 V. If the digital input code is 10010110 (the leftmost bit is MSB). Then the analog output voltage of the DAC (rounded off to one decimal place) is _______ V.

Sol. VFS = 7.68 V

Step size = .

– = 0.03

VPAC = 0.3 [150] = 4.5 V Ans: 4.5 to 4.5

44. If (1235)x = (3033)y, where x and y indicate the corresponding numbers, then (A) x = 7 and y = 5 (B) x = 8 and y = 6

(C) x = 9 and y = 7

(D) x = 6 and y = 4

Sol. x3 + 2x2 + 3x + 5 = 3y3 + 3y + 3 Option (B) will satisfy the equation Choice (B)

45. Consider a polar non-return to Zero (NRZ) waveform, using +2 V and –2V for representing binary ‘1’ and ‘0’ respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance 0.4 V2. If the a priori probability of transmission of a binary ‘1’ is 0.4, the optimum threshold voltage for a maximum a posteriori (MAP) receiver (rounded off to two decimal places) is _______ V.

Sol. From the given data Binary ‘1’ ⇒ +2 Binary ‘0’ ⇒ –2 = 0.4 V2

P( x = 0 ) = 0.6 and P(x = 1) = 0.4.

=

+

. ℓn

= E [x + N] = E (x) + E[N] =E [2] + 0 = 2 = E [x] + E [N] = E [-2] = – 2.

Vth = + .

ℓn

.

. = 0.1 ℓn

.

.

= 0.0405V


46. Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while Z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k] = y[k] is/are

(A) k = 0 and k = 15 (B) k = 15

(C) k = 0

(D) k = 0, 1, 2, . . . , 15 Sol. x[n] X(k) : Length =16 h[n] H(k) : Length =16 Length of y[n] = 16 + 16 – 1 = 31 y[n] = , , … . . z[n] = IDFT z[n] = x[n] h[n] : circular convolution Length of z[n] =16 Z[n] = , , … . . Z0 = y0 + y16 Z1 = y1 + y17 . . Z14 = y14 +y30 Z15 = y15 + 0 = y15

∴ k =15. Choice (B)

47. The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e., the Fermi energy level EF is constant) is shown in the figure. The valance band EV is sloped since doping is non-uniform along the bar. The different between the energy levels of the valence band at the two edges of the bar is Δ.

If the charge of an electron is q, then the magnitude of the electric field developed inside the

semiconductor bar is

(A) ∆

(B) ∆

(C) ∆

(D) ∆

DTF

DTF

p-type

z=0 z=L

Δ

EV

EF


Sol.

E = E = .

E = .

E = Choice (C)

48. A sinusoidal message signal having root mean square value of 4 V and frequency of 1kHz is fed to a phase modulator with phase deviation constant 2 rad/volt. If the carrier signal is

c(t) = 2cos (2π106t), the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is ________ Hz.

Sol. Vrms = √

= 4

Am = 4 x √2 volts fm = 1KHz = 2 rad/volt = Kpm(t) fc = 106 Ac = 2 fi = Ac cos [(t)] = Ac cos [2 fct + Kp m(t)]

fi = [(f)]

(fi)max = fc +

= 2 × 103 × 4√2

(fi)max = 1000 × 103 + 2 4√2 10

= 106 + 8√2 × 103 = (1000 + 11.3137) KHz = 1011313.7 Hz Ans: 1011313.5 to 1011313.9

49. In the circuit shown in the figure, the switch is closed at time t = 0, while the capacitor is initially charges to –5V (i.e., ( C(0) = – 5V).

The time after which the voltage across the capacitor becomes zero (rounded off to three decimal

places) is _______ ms.

P-type

EFi

L

Ev

250 500⁄ Vc (t) 5 V

t = 0

+

–

0.6F

VR – +

250

+ –


Sol. From the given data Vc (0) = – 5V Switch closed at t = 0 Then if Vc (t) = 0 t = ? For t > 0 : - switch closed At t = 0+ Vc (0+) = Vc (0) = – 5V As t ∞ ; circuit is in steady state

–

+ + = 0

2Vc – 10 + VR + 2Vc = 0 But 5 – VR – Vc = 0 VR = 5 – Vc 4Vc – 10 + 5 – Vc = 0

Vc = Vc (∞) = V

Vc (t) = Vc (∞) + Vc (0+) – Vc (∞) e–t/

= Req.Ceq

Req or Rth : - Redraw the equivalent circuit and find Rth value

Rth =

But VR = – Vt

– It + + + = 0

It = – +

It = Vt

∓ 250 500 Vc 5 V

t = 0

+

– 0.6F

VR – +

250

∓ 250 500 5 V

Vc

+

–

VR – +

250

Vc()

± 250 500

It

VR – +

250

Rth


= Req = Rth =

= 0.6 10–6 = 0.1m sec

Vc (t) = + – 5 – – /

Vc (t) = 0 an t = ?

0 = – e–t/

= e–t/

ln 0.25 = – t/ – 1.386 =

–

t = 1.386 = 1.386 0.1m sec t = 0.1386 m sec Ans: 0.132 to 0.146

50. The content of the registers are R1 = 25H, R2 = 30H and R3 = 40H. The following machine instructions are executed.

PUSH R1 PUSH R2 PUSH R3 POP R1 POP R2 POP R3

After execution, the content of registers R1, R2, R3 are (A) R1 = 40H, R2 = 30H, R3 = 25H (B) R1 = 40H, R2 = 25H, R3 = 30H

(C) R1 = 25H, R2 = 30H, R3 = 40H

(D) R1 = 30H, R2 = 40H, R3 = 25H

Sol. Given register R1 = 25H, R2 = 30H, R3 = 40H pushing R1, R2 and R3

On popping out the values from the stack, which gives values in First in Last out fashion or Last in First out Fashion. So, it pops 40H, 30H and 25H values consecutively and it stores into R1, R2 and R3 correspondingly. The resultant values of R1, R2 and R3 are 40H, 30H, 25H respectively. Choice (A)

51. The propagation delay of the exclusive-OR (XOR) gate in the circuit in the figure is 3 ns. The propagation delay of all the flip-flops is assumed to be Zero. The clock (Clk) frequency provided to the circuit is 500MHz.

Starting from the initial value of the flip-flop outputs Q2Q1Q0 = 1 1 1 with D2 = 1, the minimum number of triggering clock edges after which the flip-flop outputs Q2Q1Q0 becomes 1 0 0 (in integer) is _______

40H

30H

25H

Q0 Q1 Q2

D2 D1 D0

Clk


Sol. Initially

Q2 1 1 1 0 0 1

Q1 1 1 1 1 0 0

Q0 1 1 1 1 1 0

D = 1 Clk Clk Clk Clk Clk

1 2 3 4 5

Total 5 clocks are required. Ans: 5

52. The block diagram of a feedback control system is shown in the figure.

The transfer function of the system is

(A)

(B)

(C)

(D)

Sol. Converting to signal flow graph

Forward path-1 Gain G1 P1 Forward path-2 Gain G2 P2 1 = 1 2 = 1 = 1 – –G1H = 1 + G1H By Mason Gain formula Transfer function =

∑ Δ

K = 2

T.F =

[G1 + G2] = Choice (B)

X(s) > > > > 1 2 1 3

1 1 Y(s)

G2

G1

–H1

G2

H

G1

–

+ X(s)

+ +

Y(s)


53. The exponential Fourier series representation of a continuous-time periodic signal X(t) is defined as x(t) = ∑ –

Where ω0 is the fundamental angular frequency of x(t) and the coefficients of the series are ak.

The following information is given about x(t) and ak. I. x(t) is real and even, having a fundamental period of 6 II. The average value of x(t) is 2

III. ak = ,

,

The average power of the signal x(t) (rounded off one decimal place) is _______

Sol. x(t) is Real + Even. So, ak is also Real + Even i.e ak = a–k Average of x(t) is ‘2’ a0 = 2 K, 1 K 3 ak = 0, K > 3 a1 = 1 a2 = 2 & a3 = 3 a–1 = 1 a–2 = 2 & a–3 = 3 Using Parseval’s theorem, Power of x(t) = ∑ | | – = ∑ | | – = + + + + + + = 32 + 22 + 12 + 22 + 12 + 22 + 32 = 32 Ans: 31.9 to 32.1

54. If the vectors (1.0, –1.0, 2.0), (7.0, 3.0, x) and (2.0, 3.0, 1.0) in R3 are linearly dependent, the value of x is ______

Sol. Given that the vectors (1.0, -1.0, 2), (7, 3, x) and (2, 3, 1) are linearly dependent The determinant of the matrix with the three vectors as columns is zero

1 7 21 3 3

2 10

1 (3 – 3x) –7(–1 – 6) +2 (–x –6)= 0 3 – 3x + 49 –2x –12 =0 –5x + 40 = 0 x = 8 Ans: 8 to 8

55. The electrical system shown in the figure converts input source current is(t) to output voltage O (t).

Current iL(t) in the inductor and voltage C(t) across the capacitor ate taken as the state variables, both assumed to be initially equal to Zero, i.e., iL(0) = 0 and c (0) = 0. The system is

(A) neither state controllable nor observable (B) completely state controllable but not observable

(C) completely observable but not state controllable

(D) completely state controllable as well as completely observable

1H

1F

1Ω

vo(t)

vc(t) ij (t) 1Ω

iL(t)

+

–


Sol.

VL = L and IC = C.

– is + ic + = 0

is = ic + Vo

ic = -Vo – is = -VC – is

C = = V i -------(i)

i = i + i

i = i +

i = i + L.

L. = i – i

I = = i + is---------------(ii) From equations (i) and (ii)

= 0

0 + is

= 1 00 1

+ 11

is ---------(ii)

Vo = Vc -----------(iv) Form the above state equations

A = 1 00 1

and B = 11

C = 0 1 and D = 0 Controllability Qc = B AB If | | 0 ⇒ system controllable otherwise uncontrollable observability Q0 = For observability | | 0 Qc =

AB = 1 00 1

11

↓

= 11

Qc =1 11 1

| Qc | = 0 Qo =

= 1 00 1

01

= 01

Qo = 0 01 1

| Qc | = 0 ∴ The system is neither controllable nor observable Choice (A)

1H

(t)

1 Ω +

1F VC(t) –

1 Ω

iL(t)

is(t)

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GATE 2021 ECE Solutions Modified