memory based questions of gate 2020 · 2020. 2. 10. · q.6q.6 the canadian constitution requires...

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Memory Based Questions of GATE 2020Electronics and Comm. Engg. | Afternoon Session

GENERAL APTITUDE

Q.1Q.1Q.1Q.1Q.1 Time at quarter past three angle between hour hand and minute hand is

Ans.Ans.Ans.Ans.Ans. (7.5°)(7.5°)(7.5°)(7.5°)(7.5°)

End of Solution

Q.2Q.2Q.2Q.2Q.2 Radius of circle is ‘a’ and PQR is the maximum possible area of rectangle.

P Q

S R

a

Find shaded area

(a) πa2 – 2a2 (b) 2 22a aπ −

(c) 2 23a aπ − (d) None of these

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)

End of Solution

Q.3Q.3Q.3Q.3Q.3 Implicit : Explicit : : Express : __________?(a) Compress (b) Suppress(c) Impress (d) Repress

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

End of Solution

Q.4Q.4Q.4Q.4Q.4 He was not only accused of theft _______ of conspiracy.(a) Rather than (b) Rather(c) But also (d) But even

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

End of Solution

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Q.5Q.5Q.5Q.5Q.5 The untimely loss of life is a course of serious global concern as thousands of peopleget killed ________ accidents every year while many other die _____ disease likecardiovascular disease cancer etc.(a) during, from (b) in, of(c) of, from (d) None of these


End of Solution

Q.6Q.6Q.6Q.6Q.6 The Canadian constitution requires equal importance to English and French. Last yearair Canada lost a lawsuit and had to pay a six figure fine to French speaking coupleafter they field a complaints about formal in-flight announcements in English lasting15 seconds as compared to informal announcements in French lasting only for 5 seconds.The French speaking couples were upset at(a) The English announcements being longer than French once(b) The English announcements being clearer than French(c) Equal importance given to English and French(d) The in-flight announcement being made in English


End of Solution

Q.7Q.7Q.7Q.7Q.7 The global financial crisis in 2008 as considered to be most serious worldwide FC. Whichstarted with sub-prime landing crisis (SPLC) in USA 2007? The SPLC led to bankingcrisis in 2008 with the collapse of Lehman brothers in 2008. The SPL refers to the provisionof loans to those borrowers. Who may have difficulties in repaying loans and its arisesbecause of excess liquidity following the East Asian crisis. The correct precedenceaccording to Paragraph is(a) Banking crisis — Sub-prime landing crisis — Global financial crisis — East Asian crisis(b) Sub-prime landing crisis — Global financial crisis — Banking crisis — East Asian crisis(c) Global financial crisis — East Asian crisis — Banking crisis — Sub-prime landing crisis(d) East Asian crisis — Sub-prime landing crisis — Banking crisis — Global financial crisis

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

Q.8Q.8Q.8Q.8Q.8 Following quadratic equation isax2 – bx + c = 0 and it has equal root which is β and a, b, c are the real number whichis true

(a) β3 = 22bCa (b) b ≠ 4ac

(c) β = ac (d) β3 = 2

ab


End of Solution



ELECTRONICS AND COMMUNICATIONS ENGINEERING

Q.1Q.1Q.1Q.1Q.1 Given a binary number 1100 and p, q, r represents its sign magnitude, 1’s complement,2’s complement representation respectively.

Then 6 digit 2’s complement form of p + q + r is

Ans.Ans.Ans.Ans.Ans. (110101)(110101)(110101)(110101)(110101)Given, binary number 1100

1’s complement of 1100 = –3Sign magnitude of 1100 = –42’s complement of 1100 = –4

∴ P + Q + R = –4 – 3 – 4 = –11∴ 2’s complement of (–11) = 110101

End of Solution

Q.2Q.2Q.2Q.2Q.2 Output of 4 × 1 MUX F(P, Q, R) is

R

R0

1

FMUX4 × 1

P Q

S1 S0

EN

Ans.Ans.Ans.Ans.Ans.

Output, F = PQR PQR PQ+ +00 01 11 10P

QR

0

1

1

1 1 1

00

0

0

F = QR PQ+

End of Solution

Q.3Q.3Q.3Q.3Q.3 For 10 bit D/A converter, full range of voltage 0-10, then find the output value for Vinput

13A (in hexadecimal) in volts is ______V.

Ans.Ans.Ans.Ans.Ans. (3.066)(3.066)(3.066)(3.066)(3.066)Given, n = 10

VFS = 10 VInput voltage = (13A)16 = (314)10

Output voltage = Resolution × Decimal equivalent of input

Vo = 10

10314 3.066 V

2× =

End of Solution



Q.4Q.4Q.4Q.4Q.4 Given: ( ) ( ) ( )y t W t t∞

−∞

= ⋅ φ∫

W(t) is the random variable having PSD 3W/Hz and 1 5 7

( ) .0 Otherwise

tt

≤ ≤φ =

Find the variance

of y(t)?

Ans.Ans.Ans.Ans.Ans. (6)(6)(6)(6)(6)

Given: φ(t) =1 5 7

0 Otherwise

t≤ ≤

Sω(f) = 3 Watts/HzRω(τ) = 3δ(τ0) = 3δ(t1 – t2)

Var[y] = E[y 2] – E[y]2

E[W(t)]2 = DC power = Area under PSD at f = 0E[W(t)]2 = 0

E[W(t)] = 0

y = ( ) ( )W t t dt∞

−∞

φ∫

E[y] = [ ( )] ( ) 0E W t t dt∞

−∞

φ =∫

y = 2( ) ( ) [ ] ( ) Energy [ (t)]W t t dt E y S f∞

ω−∞

φ → = ⋅ φ∫

= 3 × 2 = 6Var[y] = 6 – 0 = 6

Detailed explanations for :

y = ( ) ( )W t t dt∞

−∞

φ∫

E(y2) = 1 1 1 2 2 2[ ] ( ) ( ) ( ) ( )E y y E W t t dt W t t dt∞ ∞

−∞ −∞

⋅ = φ φ

∫ ∫

= 1 1 1 2 1 2( ) ( ) ( ) ( )E W t W t t t dt dt∞ ∞

−∞ −∞

φ φ ⋅

∫ ∫

= 1 2 1 2 1 2[ ( ) ( )] ( ) ( )E W t W t t t dt dt∞ ∞

−∞ −∞

⋅ φ φ∫ ∫

= 1 2 1 2 1 2( ) ( ) ( )WR t t t t dt dt∞ ∞

−∞ −∞

− φ φ∫ ∫

= 1 2 1 2 1 23 ( ) ( ) ( )t t t t dt dt∞ ∞

−∞ −∞

δ − φ φ∫ ∫



Above integration exists providedt1 = t2 = t

= 3 (0) ( ) ( )t t dt dt∞ ∞

−∞ −∞

δ φ φ∫ ∫

= 23 (0) ( )dt t dt∞ ∞

−∞ −∞

δ φ∫ ∫= 3 × 1 × Energy [φ(t)]

E[y2] = 6

End of Solution

Q.5Q.5Q.5Q.5Q.5 Given that m(t) = 4cos1000πt and c(t) = cos2πfct ; where fc = 1 MHz. Given thats(t) = m(t) cos2πfct and s(t) is demodulated using a demodulator.

s t( )

–520 Hz520

cos2 ( + 40)π f tc

y t( )

Then y(t) is


cos [2 ( + 40 ]π f tc )

S t( )–510 500

Hz

S(t) = m(t) cos 2πfctS(t) = 4 cos 1000πt cos 2π (106)t

Output of multiplier= [4 cos 1000πt cos 2π 106t] cos 2π (fct + 40)t]= [2 [cos [2π (500 + 106)t] + cos [2π (106 500)t] cos 2π(fc + 40)t= 2 cos 2π (500 + 106)t cos 2π ((fct + 40)t) + 2 cos [2π (106 – 500)t) cos 2π (fc + 40)t= cos [2π (2fc +540t) + cos(60t) + cos [2fc – 460)t + cos(540t)]Output of Low pass filter

= cos [2π (460)]t= cos 920 πt

End of Solution



Q.6Q.6Q.6Q.6Q.6 Given phase modulated and frequency modulated signalsspm = Accos(2πfct + kp m(t))

sfm =0

cos 2 ( )c c fA f t k m t dt∞

π + ⋅ ∫

where kf is given in rad/sec/voltkp is in rad/voltm(t) is given as

1 3 6 7 t

m t( )

10

Find .p

f

K

K

Ans.Ans.Ans.Ans.Ans. (2)(2)(2)(2)(2)S(t)pm = Ac cos[2π fc t + kp m(t)]

S(t)Fm =0

cos 2 ( )c c fA f t k m t dt∞

π +

∫

Instantaneous frequency are equal

fi =1

( )2

dt

dtθ

π

1 3 6 7

10

m t( )

t

⇒ fi(FM) = fi(PM)

( )2

pc

k df m tdt

+π

= ( )2

fc

kf m t+

π

max

( )2

pk dm t

dtπ=

max( )

2fk m tπ

max( )

dm t

dt= 5, max( )m t = 10

kp(5) = kf(10)

p

f

k

k= 2

End of Solution

Q.7Q.7Q.7Q.7Q.7 A digital communication system is used to transmit a block of N-bits, if the probabilityof receiving 1-bit in error is α and all bits are transmitted independently. The receivedblock is said to be erroneous if at least one bit in error. The probability of the blockto be erroneous is ______.(a) αN (b) 1 – (1 – α)N

(c) N(1 – α) (d) 1 – αN



Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Proabability of error in decoding single bit = αThen proabability of no error will be 1 – α.Total N-bits transmitted, so that probability of no error in received block

= (1 – α) (1 – α) ... N times= (1 – α)N

The probability of received block is erroneous = 1 – (1 – α)N

End of Solution

Q.8Q.8Q.8Q.8Q.8 A binary random variable takes two values as +2 and –2 with probability P(X = 2) = α;then value of α for which entropy will be maximum is ______.

Ans.Ans.Ans.Ans.Ans. (1/2)(1/2)(1/2)(1/2)(1/2)Given that P(X = 2) = αEntropy will be maximum; provided probabilities are equal.

i.e. P(X = 2) = 1( 2)

2P X = − = α =

α = 12

End of Solution

Q.9Q.9Q.9Q.9Q.9 Given that X(ω) is Fourier transform of x(t) then find 2( )X d∞

−∞

ω ω∫

3

3

1

–1 0 1 2

x( )t

t


2( )X d∞

−∞

ω ω∫ = 2 22 ( ) 2 ( )t dt y t dt∞ ∞

−∞ −∞

π = π∫ ∫x

3

3

1

–1 0 1 2

x( )t

t

=0

2

2

2 2 ( )y t dt−

× π ∫

=1 0

2 2

2 1

2 2 ( 2) (2 3)t dt t dt−

− −

× π + + +

∫ ∫

=03 3

1

( 2) (2 3)43 3 2

t t

−

+ + + ×



=30 3 1 1 264 4

3 6 3 6t − − π + = π +

= 1 26 1 134 43 6 3 3

π × + = π × +

=14 56

43 3

ππ × =

End of Solution

Q.10Q.10Q.10Q.10Q.10 Given that 3

0

( ) ( 1) ( ).k

K

ky n n k

=

== − −∑ x Then which one of the following is the correct pole

zero plot of above given system?

(a)

Z-plane

4 orderth

1(b)

Z-plane

3 orderrd

1

(c)

Z-plane

3 orderrd

1(d)

Z-plane

3 orderrd

1


y(n) =3

0

( 1) ( )K

K

n K=

− −∑ x

= x(n) – x(n – 1) + x(n – 2) – x(n – 3)⇒ Y(z) = X(z ) – z –1 X(z) + z–2 X(z) – z –3 X(z)

⇒ H(z) =1 2 3( ) 1

( )Y z z z zX z

− − −= − + −

=3 2 2

3 3

1 ( 1) ( 1)z z z z zz z

− + − − +=

Pole zero plot:Z-plane

3 orderrd

1

End of Solution

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Q.11Q.11Q.11Q.11Q.11 Discrete time signals output is y(n) = max[X(k)], –∞ ≤ K ≤ n(a) unit impulse (b) unit step function(c) a constant function (d) zero


End of Solution

Q.12Q.12Q.12Q.12Q.12 A finite duration discrete time signals x(n) is obtained by sampling x(t) = cos[200πt]

of sample instant = ,400n

n = 0, 1, 2, ... 7. The 8-point DFT of x(n) is defined as

74

0

[ ] ( ) , 0,1, ... 7k

n

n

X k n e kπ−

== ⋅ =∑

ix

(a) X (3) and X(5) are non-zero (b) X(4) is non-zero(c) X(2) and X(6) are non-zero (d) All X (k) are non-zero

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)x(t) = cos200πt

t =400n

x(n) = cos 200 cos ; 0,1,...,7400 2n n nπ π = =

=3 5 7cos0,cos ,cos ,cos ,cos2 ,cos ,cos3 ,cos

2 2 2 2π π π π π π π

= 1, 0, –1, 0, 1, 0, –1, 0 DFT→← X(k)

Suppose, y(n) = 1, –1, 1, –1 DFT→← Y(k)

[Y(k)] = [ ]4, 0, 4, 0

1 1 1 1 1

1 1 1[ ] [ ( )] 0

1 1 1 1 1

1 1 1

N N

j jW y n

j j

=

− − − = = − − − − −

Now, as we know,

If for a, b, c, d DFT→← A, B, C, D

Then for a, 0, b, 0, c, 0, d, 0 DFT→← A, B, C, D, A, B, C, D

Similarly, for y(n) = 1, –1, 1, –1 DFT→← Y(k) = 0, 0, 4, 0

Here, for x(n) = 1, 0, –1, 0, 1, 0, –1, 0X(k) = 0, 0, 4, 0, 0, 0, 4, 0

End of Solution



Q.13Q.13Q.13Q.13Q.13 The transfer function of a stable discrete time LTI system is ( )( )( 0.5)K zH zz

− α=+

when α

and K are constants and 1.α > If the magnitude response is constant for all frequencies

then the value of α is ____.

Ans.Ans.Ans.Ans.Ans. (–2)(–2)(–2)(–2)(–2)System is all-pass filter.For digital all-pass filter, condition is

Zero =1

Pole∗ ...(i)

By given transfer function,Zero = αPole = –0.5

Using condition (i),

α =1

20.5

= −−

End of Solution

Q.14Q.14Q.14Q.14Q.14 For the impedance Z = jX, having X values from –∞ to ∞, which of the following is correctin the Smith chart?(a) A circle of radius 1, with centre at (0, 0)(b) A point at the centre of the Smith chart(c) A circle of radius 0.5 with centre at (0.5, 0)(d) A line passing through the centre of the Smith chart

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)For given impedance Normalized impedance is

0

ZZ

=0

jXZ

Z = jX⇒ Z = 0 + jXNormalized resistance = 0 ⇒ r = 0X = –∞ to ∞r = 0 and X from –∞ to ∞ is a unit circle (radius 1) and centre (0, 0) on a complexreflection coefficient plane:

Γr

r = 0Γi

End of Solution



Q.15Q.15Q.15Q.15Q.15 Which one of the following is correct?

(a) 2 0A∇ = then ‘A’ is irrotational

(b) ( ) ( )( )2·A A A∇ ∇ = ∇ × ∇ × − ∇

(c) · 0A∇ = then vector is A is solenoidal

(d) A∇ × is also a vector quantity


End of Solution

Q.16Q.16Q.16Q.16Q.16 Given the magnetic field intensity of a uniform plane wave in vacuum is

ˆ ˆ ˆ( , , , ) ( 2 )cos( 3 )y zH y z t a a ba t y z= + + ω + − −xx x

then find value of b _____.

Ans.Ans.Ans.Ans.Ans. (1)(1)(1)(1)(1)For uniform plane wave

ˆ ˆHa aρ⋅ = 0

ˆHa is unit vector in magnetic field direction aρ is unit vector in power flow direction

ˆHa =2 2 2

ˆ ˆ ˆ1 2

1 2y za a ba

b

+ +

+ +x

aρ =2 2 2

ˆ ˆ ˆ3

3 1 1y za a a− + +

+ +x

ˆ ˆHa aρ⋅ = 0

ˆ ˆ ˆ ˆ ˆ ˆ( 2 ) ( 3 )y z y za a ba a a a+ + ⋅ − + +x x = 0

–3 + 2 + b = 0b = 1

End of Solution

Q.17Q.17Q.17Q.17Q.17 For a given two port ideal lossless transformer, the parameter S21 for a referenceimpedance of 1 Ω is

2 : 1



Ans.Ans.Ans.Ans.Ans. (0.8)(0.8)(0.8)(0.8)(0.8)For ideal transformer of n : 1, the scattering matrix is

11 12

21 22

S SS S

=

2

2 2

2

2 2

1 21 1

2 11 1

n nn n

n nn n

−

+ + − + +

S21 = 2 2

2 2(2) 40.8

51 2 1

nn

= = =+ +

End of Solution

Q.18Q.18Q.18Q.18Q.18 For an infinitely small dipole in free space, the electric field Eo in the far field is proportional

to sin ,jkrer

− θ

where 2

.Kπ=

λ A vertical infinitesimally small electric dipole (δI << λ)

is placed at a distance h(h > 0) above an infinite ideal conduction plane as shown infigure. The minimum value of ‘h’, for which one of maxima in the far field radiation patternoccurs at θ = 60° is _____.

θ

I δl

h

z

Infinite conduction plate

y

Ans.Ans.Ans.Ans.Ans. (((((λλλλλ/4)/4)/4)/4)/4)

End of Solution

Q.19Q.19Q.19Q.19Q.19 For a given transistor T1, base region has uniform doping of 1017/cm3 and transistorT2 has base region with doping varying linearly with x. If all other parameters of thetransistor are same, then find the ratio of common emitter current gain of transistor T1

to transistor T2?

10 /cm17 3

1014/cm3

W0

(a) 0.3 times that of T2 (b) 0.7 times that of T2

(c) 2 times that of T2 (d) 2.5 times that of T2



Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

1

2

ββ

=2

1

0

0

( )

( )

W

A

W

A

N d

N d

∫

∫

x x

x x

=( )

17

17 14

102

110 10

2

W

W

×

× × −

End of Solution

Q.20Q.20Q.20Q.20Q.20 For a solar cell, Pin = 100 mW/cm2, surface area = 1 cm2, Voc = 0.7V, fill factor = 0.8,efficiency = 15%. If thickness of the cell is 200 µm, then the light generation ratewill be


η =( ) OC SC

in

FF VP

I

0.15 =0.8 0.7

100 mWSC× × I

⇒ ISC =15

mA0.56

GL = Area × thicknessSC

q ×I

=3

19 4

15 10

0.56 1.6 10 1 200 10

−

− −×

× × × × ×

= 19 19 31510 0.837 10 / cm /second

0.56 32× = ×

×

End of Solution

Q.21Q.21Q.21Q.21Q.21 Given that a semiconductor has density of state function in conduction band NC is halfof density of state function in valence band NV, then the shift in the intrinsic fermi levelfrom the centre of forbidden gap is ____ meV.

Ans.Ans.Ans.Ans.Ans. (9.01)(9.01)(9.01)(9.01)(9.01)

2C V

FE E

E+

− i = ln2

C

V

kT NN

2

VC

NN = ∵

=0.026

ln0.5 9.01 meV2

= −

End of Solution

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Q.22Q.22Q.22Q.22Q.22 In a MOS capacitor, threshold voltage is –0.16V, metal work function is 3.7V, bandgap

of semiconductor is 1 eV whose band diagram is shown below. C ′x = 100 pF/cm2. Oxide

is free from non-idealities. When the voltage across the capacitor is equal to thresholdvoltage, what will be the depletion charge present in the semiconductor?

4 eV

Ec

E0

Ei

0.5 eV

0.2 eVEF

Ans.Ans.Ans.Ans.Ans.MOS capacitance

φm = 3.7, φs = 4.8, φms = –1.1

VT = 2ox dms Fp

ox ox

Q QC C

′ ′φ − − + φ

φFp = Ei – EF = 0.5 – 0.2 = 0.3

–0.16 = 1.1 0 2 0.3d

ox

QC

′− − − + ×

d

ox

QC

′= 0.6 + 0.16 – 1.1 = –0.34

Qb = –0.34 × Cox

= –0.34 × 100 × 10–9

= –34 nC/cm2

Q.23Q.23Q.23Q.23Q.23 Consider the following MOSFET, find the equivalent Norton’s resistance

R

g rm ds,




vπ = –Vx

Vx = (Ix – gm Vx) rds + IxR

g vm π rds

R

+–

+

–

vπ

Ix

DG

SVx(1 + gm rds) = (rds + R)Ix

RN =1

ds

m ds

V R rg r+=

+x

xI

End of Solution

Q.24Q.24Q.24Q.24Q.24 A sinusoidal voltage is applied to the given circuit, then the steady state output voltage Vo

is _______ V.

CVrms = 230 V

CVo


Voltage doubles, Vo = 2 Vm = 2 230 2 650.4 V× ≅

End of Solution

Q.25Q.25Q.25Q.25Q.25 Consider the op-amp shown below:

–

+1 kΩ

+5

–5VVi = 2V

1 kΩ

I0

1 kΩ

The current “I0” is ________ mA.


Vo

1 kΩIo2 V

1 kΩ

1 kΩ



Vo = (1 + 1) × 2 = 4 V

2 4 0 41 1o− −+ +I = 0

–2 + Io – 4 = 0⇒ Io = 6 mA

End of Solution

Q.26Q.26Q.26Q.26Q.26

+

–1 Ω

+5V

–5V

1 Ω

I0

Vi

Vi

V0

t

1V

–1V

The output V0 is(a) square wave with an amplitude of 5V(b) sinusoidal of amplitude 10V(c) inverted sinusoidal of input(d) constant signal with either +5V and –5V

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)The given circuit is a Schmitt trigger, which produces a square wave at the output.

End of Solution

Q.27Q.27Q.27Q.27Q.27 Find the state matrix (consider v and i as state variables).

2 Ω

0.5 H

1 ΩI1

i+

–v 0.25 F I2

Ans.Ans.Ans.Ans.Ans.From source transformation,

0.5 H

2I1

i+

–v 0.25 F I2

2 Ω 1 Ω

(1)

KVL in loop 1,

2I1 = 2 0.5d

vdt

+ +ii



ddt

i= –2V + 4i + 4I1 (i)

KCL at node,

i = 20.251

Vdvdt

−+ I

dvdt

= –4V + 4i + 4I2 (ii)

( )

( )

dv tdt

d tdt

i=

1

2

4 4 ( ) 0 4

2 4 ( ) 4 0

v tt

− + − −

IIi

End of Solution

Q.28Q.28Q.28Q.28Q.28 When base current of given BJT is negligible. Given that Vt = 26 mV and VBE = 0.7V.

V0

10 kΩ

Vi

10V

–10V

10 kΩ

20 kΩ

Voltage gain at low frequencies is

Ans.Ans.Ans.Ans.Ans. (–89.423)(–89.423)(–89.423)(–89.423)(–89.423)

IEQ =10 0.7

0.465 mA20− =

gm =0.465

A/V26

EQ

TV=

I

out

in

VV

=0.465

( ) 5000 89.42326m e Lg R R||− = × = −

End of Solution



Q.29Q.29Q.29Q.29Q.29 In the given Nyquist contour the pole zero locations are indicated as shown. If the givencontour is transformed from s-plane to G(s) H(s) plane then which one of the followingis true

(a) encircles the origin in clockwise(b) encircles the –1 + j0 in clockwise(c) encircles the –1 + j0 in anticlockwise(d) encircles the origin in anticlockwise


End of Solution

Q.30Q.30Q.30Q.30Q.30 Given op-amp circuit

–

+

R

R

2R

C

V0

2C

Vi

R = 10 kΩ, C = 0.1 µF, then the 3dB cut-off frequency of op-amp circuit is ____ Hz.

Ans.Ans.Ans.Ans.Ans. (159.2)(159.2)(159.2)(159.2)(159.2)Op-amp active filter (LPF) inverting type 3 dB cut-off frequency,

fc =4 7

1 1 10002 22 10 10RC −= =

π ππ × ×

= 159.2 Hz

End of Solution



Q.31Q.31Q.31Q.31Q.31 In the following circuit, assume that IB = 0, VBE = 0.7V for both the transistors. If Vin = 15V(unregulated) and V0 = 9V (regulated), then the resistance R is equal to _______ Ω.

R

1 k Ω

V0 = 9VVin = +15V

+3.3V


91kR

R×

+ Ω = 4

9R = 4R + 4 kΩ5R = 4K

R =4000

8005

= Ω

End of Solution

Q.32Q.32Q.32Q.32Q.32 Consider the below figure.

G s( )x( ) = 5cos3t t 110

y t( ) = cos(3 – 1.8 )t

G(s) =1

( 1) ( )s s+ + αFind the value of α.


Given that, G(jω) = 1(1 ) ( )j j+ ω α + ω

; ( )G jω = 2 2 2

1

( 1) ( )ω + ω + αAccording to question,

3( )G j ω =ω =

15 10

⇒1 2 2

1

( 1) ( )ω + ω + α=

15 10

⇒2

1

10( )a a+=

15 10

α2 + 9 = 25α2 = 16α = 4

End of Solution



Q.33Q.33Q.33Q.33Q.33 Given: ( ) ( ) ( )( 1) 1

,3 1

K sC s G s

s s s+= =

+ +. If steady state error for ramp input is 0.1, then

find the value of k or negative feedback unity gain system

C s( ) G s( )R s( ) Y s( )+–


Open loop transfer function for the system = C(s) × G(s) = ( 1) 1

( 3) ( 1)K ss s s

+ ×+ +

Since the system is type-1 so far a given unit ramp input steady state

ess = 1

VK

where, KV = lim( 3) 3s

K Kss s→ ∞

× =+

So, ess =1 3/ 3K K

=

Given that, ess = 0.1

So, 0.1 =3

30KK

⇒ =

End of Solution

Q.34Q.34Q.34Q.34Q.34 If the value of current i = 10 cos 54

t π − , the value of inductance L is

R

L+

V t t( ) = 200 cos 5–


Z =VI

= 200 0

20 4510 45

∠ = ∠∠ −

Z = 10 2 10 2j+

XL = 10 2

ωL = 10 2

L =10 2

5 = 2.828 H

End of Solution



Q.35Q.35Q.35Q.35Q.35 For given open loop transfer function ( )

( )( )11

( ) ( ) .2 8

k sG s H s

s s s+

=+ +

Find value of k for which

system is marginally stable.

Ans.Ans.Ans.Ans.Ans. (160)(160)(160)(160)(160)Characteristic equation q(s) for the given open loop system will be

q(s) = s3 + 10s2 + 16s + Ks + 11K = 0Using R-H criteria,

3

2

1

0

1 16

10 10

10(16 ) 110

10

11 0

s K

s KK K

s

s K

+

+ −

For system to be marginally stabled

10(16 ) 1110K K+ −

= 0

160 + 10K – 11K = 0K = 160

End of Solution

Q.36Q.36Q.36Q.36Q.36 A two port network has an impedance [Z] = 40 60

60 120

, find [ZL] for maximum power

transfer. Find ZL = ?

10 Ω

120 V

I1I2

Z ZL

+

–

V1

+

–VL

Ans.Ans.Ans.Ans.Ans. (48)(48)(48)(48)(48)From maximum power transfer theorem

ZL = Zth

Zth = 12 2122

5 11

Z ZZ

R Z−

−+

For given data

Zth =60 60

120 4810 40

×− = Ω+

ZL = 48Ω

End of Solution



Q.37Q.37Q.37Q.37Q.37 Find value of Vth

1 A

2 V 2 Ω

1 Ω 2 A

4 Ω

Vth

+

–

2 Ω

Ans.Ans.Ans.Ans.Ans. (3.6)(3.6)(3.6)(3.6)(3.6)By applying source transformation

1 A

– +2 Ω 4 Ω2 V

2 A 2 Ω+–

1 Ω

1 A 3 Ω

4 Ω

2 A 2 Ω

3.6 V

4 Ω

+–

Vth

Vth = 3.6 V

Ω65

Vth = 3.6 V

End of Solution

Q.38Q.38Q.38Q.38Q.38 Find the value of Io

Z j = 80 – 35

Z

Z

Io

120 –90°∠+

–

–

+120 –30°∠




+

+

–

–

Z

Z

Z j = 80 – 35I0

120 –90°∠

120 –30°∠

I1

I2

I0 = [I1 + I2]

I0 =120 90 120 30

80 35 80 35j j ∠ − ° ∠ − °+ − −

I0 = 2.38 ∠143.7°

End of Solution

Q.39Q.39Q.39Q.39Q.39 In a transmission line given Z0 = 50, ZL = 400, 34λ=l , find Zin

Ans.Ans.Ans.Ans.Ans. (6.25)(6.25)(6.25)(6.25)(6.25)l = 3 /4λ

Zo = 50 Ω

Zin

ZL = 400 Ω

Zin for (l = λ/4) =2 20 50 25 6.25

400 4L

ZZ

= = = Ω

End of Solution

Q.40Q.40Q.40Q.40Q.40 For the given characteristic equationQ(s) = s3 + 3s2 + (K + 2)s + 3K = 0The root locus plot will have breakaway or break-in point in the region.(a) (0, –1) (b) (–1, –3)(c) (–∞, –3) (d) None of these

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)Q(s) = 1 + G(s) H(s) = 0

s3 + 3s2 + 2s + ks + 3k = 0

–k =3 23 2

3s s s

s+ +

+

dkds

− =2 3 2

2( 3)(3 6 2) ( 3 2 )

0( 3)

s s s s s s

s

+ + + − + + =+



3s3 + 6s2 + 2s + 9s2 + 18s + 6 – s3 – 3s2 – 2s = 02s3 + 12s2 + 18s + 6 = 0

s = –0.46, –3.87, –1.65

–3 –2 –1 0

–0.46

Im

Re

End of Solution

Q.41Q.41Q.41Q.41Q.412

26 9 0

d y dydd

− + =xx

. Find the solution of differential equation

(a) (C1 + C2 x)e3x (b) (C1 + C2 x)e–3x

(c) C1 e3x (d) None of these


End of Solution

Q.42Q.42Q.42Q.42Q.42 In sequential circuit, the maximum clock frequency at which given circuit can operatereliably?

tpd = 3 nstsetup = 5 nsthold = 1 nsCLK

FF1

tpd = 8 nstsetup = 4 nsthold = 3 ns

FF2IN

tpd = 2 nstpd = 2 ns

Ans.Ans.Ans.Ans.Ans. (76.92)(76.92)(76.92)(76.92)(76.92)Total propagation delay = (tpd + tset-up)max = 8ns = 5 ns = 13 ns

∴ Frequency of operations =1000

MHz = 76.92 MHz13

End of Solution



Q.43Q.43Q.43Q.43Q.43 f (x, y, z) = e(1 – x cosy) 21/1 yze− ++ x partial differentiation of f (x, y, z) with respect to x

at point (1, 0, e) will be


End of Solution

Q.44Q.44Q.44Q.44Q.44 f (x2 + x2) ≥ f(x1) + f(x2)

(a)1x (b) square root of x

(c) e x (d) e –x

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Verify with options

Option (a):Option (a):Option (a):Option (a):Option (a):1 2

1+x x

≥1 2

1 1+x x

12 3+

≥1 12 3

+

15

≥56

0.2 ≥ 0.8 which is wrong.

Option (b):Option (b):Option (b):Option (b):Option (b): 1 2+x x ≥ 1 2+x x

2 3+ ≥ 2 3+

5 ≥ 1.414 + 1.732

2.23 ≥ 3.146 which is wrong.

Option (c):Option (c):Option (c):Option (c):Option (c): 1 2e +x x ≥ 1 2e e+x x

e1 + 2 ≥ e1 + e2

20.085 ≥ 2.718 + 7.389 Satisfying.

End of Solution

Q.45Q.45Q.45Q.45Q.45s

d dydz∫ ∫ ∫ x x

x

y

z

10

3

1



Ans.Ans.Ans.Ans.Ans. (9/4)(9/4)(9/4)(9/4)(9/4)x : 0 to 3y : 0 to 1z : 0 to 1 – y

=11 3

0 0 0

y

y z

d dy dz−

= = =∫ ∫ ∫

x

x x

=

311 2

0 0 02

y

ydz dy

−

=

∫ ∫x

= ( )1

10

0

92

yz dy−∫

=11 2

0 0

9 9(1 )

2 2 2y

y dy y

− = − ∫

=9 1 912 2 4

− =

End of Solution

Q.46Q.46Q.46Q.46Q.46 If x is having uniform distribution between [– 2, 10] and y = 2x – 6 then 75

yP < > x

Ans.Ans.Ans.Ans.Ans. (0.3)(0.3)(0.3)(0.3)(0.3)x follows uniform distribution over [–2, 10]

∴ f(x) = 1 1 110 ( 2) 12b a

= =− − −

Given: y = 2x – 6

⇒ x =6

2y +

For y = 7

x =7 6 13

6.52 2+ = =

75

yP

< > x =

6.55

P < <

xx

= [ 5 and 6.5]

[ 5]P

P> <

>x x

x

=

6.5 6.5

5 510 10

5 5

1( )

12[5 6.5][ 5] 1

( )12

f d dP

Pf d d

< < = =>

∫ ∫

∫ ∫

x x xx

xx x x

=6.55105

( ) 1.50.3

5( )= =x

x

End of Solution



Q.47Q.47Q.47Q.47Q.47 ( )1dy

yd

= − xx solution of equation satisfies.

(a) ln 1y − = 0.5x2 + c and y = – 1 (b) ln 1y − = 2x2 + c and y = 1

(c) ln 1y − = 0.5x2 + c and y = 1 (d) ln 1y − = 2x2 + c and y = –1


Given differential equation: ( 1)dy

yd

= − xx

By variable separable method:

1dy

y −∫ = d∫ x x

ln 1y − =2

2c+x

(where y ≠ 1)

Thus, solution is valid for y ≠ 1.

End of Solution

Q.48Q.48Q.48Q.48Q.48 Two sides of a fair coin are labelled as 0 and 1. The coin is tossed two times independently.Let M and N denote lables corresponding to the outcomes of those process for randomvariable X, where X = min(M, N), E(X) = _______.

Ans.Ans.Ans.Ans.Ans. (0.25)(0.25)(0.25)(0.25)(0.25)s = ( H, H), (H, T), (T, H), (T, T)

M =1 1 0 0

H H T T

of first toss

N =H T H T

1 0 1 0

of second toss

Now, X = Min M, N∴ X = Min H, H = Min(1, 1) = 1

X = Min H, T = Min1, 0 = 0X = MinT, H = Min0, 1 = 0X = MinT, T = Min0, 0 = 0

∴ P(X = 1) =1

,4

P(X = 0) = 34

We know that, E(X) = ( )X P∑ i ii

x

=1 3 1

1 0 0.254 4 4

× + × = =

End of Solution



Q.49Q.49Q.49Q.49Q.49 In a digital communication system, 4 symbols are transmitted (s1 = –3, s2 = –1, s3 = 1,s4 = 2) through an AWGN channel. The variable at the input of decision device is si + W,where W is a Gaussian random variable with mean zero and unit variance. ML decodingis used. The conditional error probability when symbol si is transmitted is Pi. The valueof “i ” that result in maximum Pi is _____.

Ans.Ans.Ans.Ans.Ans. (3)(3)(3)(3)(3)Since the noise variable is Gaussian with zero mean and ML decoding is used, thedecision boundary between two adjacent signal points will be their arithmetic mean.

In the following graphs, the shaded area indicates the conditional probability of decodinga symbol correctly when it is transmitted.

+1 +2–2

–1–3

+2–2

–1–3 +10

+1+2 0–1 +11.5

+1+2 +1–1 1.52

P1 = 1 – (shaded area)




By comparing the above graphs, we can conclude that P3 is larger among the four.

End of Solution



Q.50Q.50Q.50Q.50Q.50 V1, V2, V3, V4, V5, V6 six vector in R4 which of the following statements is false?(a) Any 4 base for R4

(b) If V1, V3, V5, V7 span R4 then it forms a basis of R4

(c) These vectors are not linearly independent(d) It is not necessary that vector span R4


End of Solution

Q.51Q.51Q.51Q.51Q.51 In 8085 microprocessor to access a memory of 16 kB number of address lines requiredis ____.

Ans.Ans.Ans.Ans.Ans. (14)(14)(14)(14)(14)2n = Nn → Number of address linesN → Number of Memory locations

∴ 2n = 16 kB= 24 (210) [∵ 1 kB = 210]= 214

n = 14

End of Solution

memory based questions of gate 2020 · 2020. 2. 10. · q.6q.6 the canadian constitution requires...

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