innddeexx · gate -2016 in set 1 : 080-617 66 222, [email protected] ©copyright reserved....

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GATE-2016

: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

IInnddeexx

11.. QQuueessttiioonn PPaappeerr AAnnaallyyssiiss

22.. QQuueessttiioonn PPaappeerr && AAnnsswweerr kkeeyyss

http://www.thegateacademy.com/

.

GATE-2016 IN-SET-1

: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com 1

NETWORK 10%

CONTROL SYSTEM 7%

SIGNAL & SYSTEM 8%

ANALOG CIRCUITS 10%

COMMUNICATION6%

Measurements 8%

Transducer 7%

Optical Instrumentation6%

GA 15%

ANALYSIS OF GATE 2016

Instrumentation Engineering


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GATE-2016 IN-SET-1


IN ANALYSIS-2016_31-JAN_Afternoon Session

SUBJECT No. OF

QUESTIONS Topics Asked in Paper

Level of Toughness

Total Marks

Networks 1 M: 4 2 M: 3

Basic Networks; Dependent Source; Sinusoidal Analysis

EASY 10

Control System 1 M: 1 2 M: 3

Nyquist Plot; Initial Value Time Response;

EASY 7

Signals & Systems

1 M: 2 2 M: 3

Convolutions ;Digital filters; Energy and Power; DFT; Periodic signal, Fourier series,

EASY 8

Digital & µP 1 M: 3 2 M: 4

Combinational Circuit ; Flip-flop; Finite State Machines* ;Memory; Converter; Counter.

MODERATE 11

Analog Circuits 1 M: 2 2 M: 4

Op-Amp; Elber-Moll Model; Current Mirror; Diode based

TOUGH 10

Com. Systems 1 M: 2 2 M: 2

SSB; F.M; Noise TOUGH 6

Transducer 1 M: 3 2 M: 2

Piezoelectric; strain gauge; Pressure measurement

MODERATE 7

Measurements 1 M: 2 2 M: 3

Flow Measurements; wheatstone bridge ; Potentiometer ; Power Measurements.

MODERATE 8

Optical Instrumentation

1 M: 2 2 M: 2

LED. MODERATE 6

Engineering Mathematics

1 M: 4 2 M: 4

Calculus ;Linear Algebra, Probability & Distributions ;Limit & Continuity ; Laplace

MODERATE 12

GA 1 M: 5 2 M: 5

Verbal Ability; Time and Work; Directions; Venn Diagrams; Mensuration and Area; Clock

EASY 15

Total 65 MODERATE 100

* Indicates Questions from New Syllabus

Faculty Feedback: Few questions came from new syllabus; General Ability was pretty easy;

question required logic for solving , qualifying is easy but scoring is tough ,

Time Management very critical. Practice previous year question papers will be

beneficial and practice online test series.


GATE-2016

Question Paper

&

Answer Keys

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GATE-2016 IN-SET-1


GATE 2016 Examination

Instrumentation Engineering

Test Date: 31/01/2016

Test Time: 2:00 PM to 5:00 PM

Subject Name: INSTRUMENTATION ENGINEERING

Section: General Aptitude

Q No. 1.

Q No. 2.


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GATE-2016 IN-SET-1


Q No. 3.

Q No. 4.

Q No. 5.


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GATE-2016 IN-SET-1


Q No. 6.


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GATE-2016 IN-SET-1


Q No. 7.

Q No. 8.


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GATE-2016 IN-SET-1


Q No. 9.

Q No. 10.


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GATE-2016 IN-SET-1


Section: Technical

Q No. 1

[Ans. *] Range: 3 to 3

Slope =y2 − y1

x2 − x1=

6 − 0

2 − 0= 3

Q No. 2

[Ans. *] Range: 0.5 to 0.5 Here ‘n’ tends to infinity,

Let us try n = 100, then √100 × 100 + 100 − √100 × 100 + 1 = 0.493

try n = 1000, then √1000 × 1000 + 1000 − √1000 × 1000 + 1 = 0.4993

try n = 10000, then √10000 × 10000 + 10000 − √10000 × 10000 + 1 = 0.49993 ∴ We observe that as n tends larger and larger the value of the limit approaches 0.5 asymptotically ∴ The value of the limit is 0.5

Q No. 3

[Ans. *] Range: 2.5 to 2.5 V1 = 100 ± 1.5 V V2 = 150 ± 2 V V3 = V1 + V2 Standard deviation in V3 will be

σV3= √(

∂V3

∂V1)2

σV12 + (

∂V3

∂V2)2

∂V3

∂V1=

∂

∂V2 (V1 + V2) = 1

∂V3

∂V1=

∂

∂V1

(V1 + V2) = 1

σV3 = √12 × (1.5)2 + 12 × (2)2

= √6.25 = 2.5 V


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GATE-2016 IN-SET-1


Q No. 4

[Ans. D]

We know that if a⃗ and b⃗ are perpendicular

Then a⃗ ⋅ b⃗ = 0 Options (A), (B), (C) are perpendicular. Option (D) is not perpendicular.

Q No. 5

[Ans. A] f(z) is of the from 1 + x + x2 + ⋯ = (1 + x)−1 Please note f(z) four is expanded in neighborhood of z = 1, so the series will converge

f(z) =1

1 − (1 − z)=

1

z

Q No. 6

[Ans. *] Range: 5 to 5 By KCL clearly i1 = i2 + i3 put here i1, i2, i3 have different phase so, the phases diagram will look like

∴ I3 = √32 + 42 = 5A

i2 = 3 sinωt −i2 = −3 sinωt

i1 = 4 cosωt i3


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GATE-2016 IN-SET-1


Q No. 7


Q =ωL

R= 5 at 100 kHz (given)

⇒L

R=

5

ω=

5

2π × 100 kHz……… .①

At 20 kHzQ = ω ×L

R= 2π × 20 kHz ×

5

2π × 100 kHz [Using

L

R from ①]

∴ Q = 1

Q No. 8


V =1

c∫ i. dt = V =

1

1 ∫8dt

2

1

= 8[t]12

= 8 Volts

Q No. 9

8A

i(t) =

1s 2s

x

t


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GATE-2016 IN-SET-1


[Ans. *] Range: 2 to 2 1 1 −1 0 1 1 1 −1 0 1 1 ① −1 0

−1 −1 −1 +1 0 0 0 0 0 0

∴ y(n) ={1,2, −1,−2,1,0,0}

↑

∴ y(−1) = 2

Q No. 10

[Ans. *] Range: 0.28 to 0.283

V1 = 1 sin (2π 10000 t) V V2 = 1 sin(2π 30000 t) V ∴ f1 = 10000 Hz f2 = 30000 Hz = 3 f1 For LC – 2

ω0 =1

√LC⇒ f0 =

1

2π√LC

So, for f1 = f0|LC=2 ⇒ Resonance It act as open so I = 0 For LC − 1 to be open circuit i. e. , for resonance

(3 f1)30,000 =1

2π√LC

(3f1)30,000 =1

(4π)2 × 9 × 108 × 100 × 10−6= 0.28μF

~

C 2.53 μF

LC − 1 LC − 2

R

100 μH 100 μH

V1 + V2

I


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GATE-2016 IN-SET-1


Q No. 11


limt→∞

f(t) = lims→0

sF(s) =S × 20(S + 2)

(S + 1)(S + 3)2= 0

Q No. 12


G(s) =s − 1

s + 1 G(jω) =

jω − 1

jω + 1

= 1∠180° − 2 tan−1 ω On varying ω = 0+to + ∞ Now on varying ω = 0−to − ∞ G(jω)|ω=0 = 1∠180° G(jω)|ω=0− = 1∠180° G(jω)|ω=1 = 1∠90° G(jω)|ω=−1 = 1∠270° G(jω)|ω=∞ = 1∠0° G(jω)|ω=−∞ = 1∠360° On plotting

∴ No. of times the Nyquist plot encircle the origin clockwise = 1

Q No. 13

[Ans. D]

Routh array

s3 1 2 s2 3 90

s1 6 − 0

3 0

s0 90 0 6 − 90

3> 0

⇒ 6 > 90

ω = +∞

ω = −∞

ω = 0+

ω = 0−

Im [G(jω)]

Re [G(jω)]


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GATE-2016 IN-SET-1


Q No. 14

[Ans. *] Range: 0.14 to 0.15 Here x(t) is a sinusoid (Eigen function of a system) ∴ |y(t)| = |G(jω|. |x(t)| ω = 3r/s

=8

(√ω2 + 102)2 × 2 =

16

109= 0.1467

Q No. 15


Let the diode is ON The diode is replaced by short circuit KCL at node B: VB − 3

110+

VB

5− 1 × 10−3 = 0

⇒ VB =10

3= 3.33 V

⇒ ID =3 − 3.33

1 × 103= −ve not possible

⇒ Diode is off So the current passing through diode = 0 ⇒ Vab = V1K = 0 Volts

1 k b VB

1 mA 5 k 3 V

a ID


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GATE-2016 IN-SET-1


Q No. 16

[Ans. *] Range: −𝟏 𝐭𝐨 − 𝟏 Thevenin’s circuit of input is [This is the important point here]

∴ The overall circuit looks like

V0 = −1V

Q No. 17

20 k

10 k Vo

10 k

−

+ 10 k

1V

10 k


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GATE-2016 IN-SET-1


[Ans. *] Range: 1.55 to 1.65 During positive half cycle When vi > 0.6 V ⇒ D1 − ON and D2 − OFF Then equivalent circuit

V0 = −vi [16

10] = −1.6 vi

For negative half cycle when vi < −0.6 V ⇒ D1 − OFF and D2 − ON

Then magnitude of the negative peak value of the output Vo is 1.6 V

Q No. 18

[Ans. C] XY + (X̅ + Y̅)Z = XY + XY̅̅̅̅ Z

= (XY + XY̅̅̅̅ )(XY + Z) = XY + Z = (X + Z) (Y + Z)

Vo = 0

−

+ 10 kΩ

16 k

0.6 V

Vi

Vo

−

+ 10 kΩ

16 k

0.6 V

Vi


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GATE-2016 IN-SET-1


Q No. 19

[Ans. B]

f = (x. y.̅̅ ̅̅ ̅ ). (y̅. z) = (x. y̿̿ ̿̿ ) + (y̅z̿̿̿)

By Demorgan’s law f = xy + y̅z

Q No. 20


Accuracy = ±1% of Range =±1

100× 200 = ±2 mA

∴Reading can be 100 mA ± 2 mA

x

y

𝑧

(x. y̅̅ ̅̅ ). (y̅. z̅̅ ̅̅ )̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅

y̅

x. y̅̅ ̅̅

y̅. z̅̅ ̅̅


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GATE-2016 IN-SET-1


Q No. 21

[Ans. *] Range: 1.1234 to 1.1234 The unknown voltage Vx will be Vx = IP [11 resistors of 10Ω + Slide wire resistance]

= 10 × 10−3 (11 × 10 +234

1000× 10)

= 1.1234 V

Q No. 22

[Ans. *] Range: 20 to 20 R = RF ∥ Ri to cancel Bias current ⇒ R = 60 k ∥ 30 k = 20 kΩ


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GATE-2016 IN-SET-1


Q No. 23

[Ans. *] Range: 0 to 0 0 Volts, PIEZO CANNOT SENSE STATIC LOAD

Q No. 24

[Ans. D] SSB signal is given by = ACm(t)cos(ωct) ± ACm̂ sin(ωct) In phase component = ACm(t) Quadrature phase component = ACm̂(t) Now, the output of envelop detector can be given by

|S(t)| = AC√m2(t) + m̂2(t)

= √cos2 θ + sin2 θ = 1

Q No. 25

[Ans. B]

Q No. 26

[Ans. *] Range: −𝟏𝟑 𝒕𝒐 − 𝟏𝟑 f ′(x) = 6x2 − 4x3 = 0 formin or max point

= x2(6 − 4x) = 0

⇒ x = 0 or 6 − 4x = 0 ⇒ x =3

2not possible


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GATE-2016 IN-SET-1


x is in (−1 , 1)as given f ′′(x) = 12x − 12x2

= 0@x = 0 f(x) = 0 & f ′′(x) = 0 @ x = 0 ∴ x is not amin point f(x)@ − 1 = −2 − 1 − 10 = −13 f(x)@1 = 2 − 1 − 10 = −9 f(x)@0 = 0 − 0 − 10 = −10 ∴ f(x)has amin value @ x = −1; f(x) = −13

Q No. 27

[Ans. A] P(red ball) = P(green in 1 trial followed by red) or P(red in 1 trail followed by red) Here, we get to pick 12 balls in TRIAL 1 and 13 balls in TRIAL 2

=7

12×

5

13+

5

12×

6

13=

35 + 30

156=

65

156

Q No. 28

[Ans. *] Range: −𝟔 𝒕𝒐 − 𝟔 If E value of ‘A’ is λ1, λ2, λ3

Then by property E value of Ak = λ1k, λ2

k, λ3k

E Value of μA = μλ1, μλ2, μλ3 (μ is ‘n’ constant) ∴ E Value of A3 = 13, 23, 33 = 1, 8, 27 E Value of A2 = 12, 22, 32 = 1, 4, 9 E Value of 3A2 = 3, 12, 27

TRACE OF A MATRIX = Σ (E Value) ∴ TRACE OF A3 = 36 TRACE OF 3A2 = 42 ∴ TRACE OF A3 − 3A2 = −6


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GATE-2016 IN-SET-1


Q No. 29

[Ans. A] Taking L.T on both sides, (M2s + Bs + K)X(s) = F(s)

∴X(s)

F(s)=

1

M2s + Bs + K=

1

0.1s2 + 2s + 10=

10

s2 + 20s + 100

Q No. 30

[Ans. *] Range: 1 to 1 1

2πj ∫

z2 + 1

(z2 − 1)c

dz =1

2πj ∫

z2 + 1

(z − 1)(z + 1)c

Given circle is |z − 1| = 1 pole z = 1 lies inside C pole z = −1 lies outside C

Res f(z)at z = 1 is = limz→1

(z − 1)z2 + 1

(z − 1)(z + 1)=

2

2= 1

Res f(z)at z = – 1 is = 0 By Cauchy’s residue theorem 1

2πj ∫

z2 + 1

z2 − 1 C

dz =1

2πj× 2πj(1 + 0) = 1


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GATE-2016 IN-SET-1


Q No. 31

[Ans. *] Range: 10 to 10 By nodal analysis at ‘x’ we get Vx − 1

100+

Vx

100− 10 mA = 0

∴2Vx

100= 10 mA + 10 mA

∴ Vx = 50(20 mA) = 1V

∴ Ix =1V

100 Ω= 10 mA

Q No. 32

[Ans. *] Range: 100 to 100 According to statement given circuit is at resonance

Y =1

(R + jXL)+

1

−jXC

Y =R − jXL

R2 + XL2 +

j

XC

Imaginary part is zero Is = Vs ⋅ Yreal part

= 101 ⋅ [10

102 + 1002]

=101 × 10

10100= 0.1A

= 100 mA


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GATE-2016 IN-SET-1


Q No. 33

[Ans. *] Range: 3463 to 3465

Q No. 34

[Ans. *] Range:2.9 to 3.1

f =1

2πRC= 150 Hz (given)

∴ C =1

2πRF

Thevenin’s equation of bridge. Here, we need RTH only of bridge to find C RTH of bridge Seen pay cap is

R1 R2

R4 R3

V

R

C


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GATE-2016 IN-SET-1


R1 ∥ R4 + R2 ∥ R3

=R(1 − X)

2+

R(1 + X)

2

= R = 350 Ω

∴ C =1

2πRf= 3.03 μF

Q No. 35

[Ans. *] Range: 1.62 to 1.64 For t < 0; ckt is in steady state with V(t) = 3V ∴ iL(θ) = 1A

After t = 0

Here the final inductor current would be 2A

∴ iL(t) = ifinal + (iinitial − ifinal)e−t τ⁄

∴ iL(t) = 2 + (1 − 2)e−t τ⁄ = 2 − e−t τ⁄

τ =L

Req where Req is Rthevein seen by τ

+ 1 Ω

1 Ω

1 Ω 6V −

2 A 2 A

4 A i(∞)

+ 1 Ω

1 Ω

1 Ω 6V −

3V

(L is like short)

1 Ω

1 Ω

1 Ω

2A

iL(0−)

1 A 1 A


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GATE-2016 IN-SET-1


Here Req is

∴ τ = 1s ∴ iL(t) = 2 − e−t ∴ iL(1s) = 2 − e−1 = 1.632 A

Q No. 36

[Ans. A] Given signal satisfies odd and half wave symmetry so it contains only odd harmonics. ∴ Power in 10th harmonic will be zero.

Q No. 37


x[n] = sin [301

4 πn]

N0 =2π

ω0=

2π

301π/4× m

=8

301× 301 = 8

1 Ω

1 Ω

1 Ω = 1.5 Ω

Req


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GATE-2016 IN-SET-1


Q No. 38

[Ans. B] Since the initial slope is 0 dB/decade, thus it a type zero system ∴ 20 log K = 0 K = 1 now at ω1 = 10° Slope goes to 40 dB/decade ∴ Two zeros added at ω1 = 10° and at ω2 = 102 Slope goes to 40 dB/dec ∴ Two poles added at ω2 = 102

Thus T. F =K (

sω1

+ 1)2

(s

ω2+ 1)

2 =K(

s1

+ 1)2

(s

100+ 1)

2

= 104(s + 1)2

(s + 100)2


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GATE-2016 IN-SET-1


Q No. 39

[Ans. C]

For C(s) = 3 +9

2

Characteristic equation 1 + C(s) G(s) = 0

1 + (3 +9

2) (

1

(s + 1)2) = 0

1 + (3s + 9

s) (

1

s2 + 2s + 1) = 0

s3 + 2s2 + 4s + 9 = 0 Using RH criteria

s3 1 4 s2 2 9

s1 5 − 9

2 0

s0 9 0 ∵ 1st column of RH table does not contains positive sign only ∴ System become unstable.


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GATE-2016 IN-SET-1


Q No. 40

[Ans. B]

Q No. 39

[Ans. C]

For C(s) = 3 +9

2

Characteristic equation 1 + C(s) G(s) = 0


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GATE-2016 IN-SET-1


1 + (3 +9

2) (

1

(s + 1)2) = 0

1 + (3s + 9

s) (

1

s2 + 2s + 1) = 0

s3 + 2s2 + 4s + 9 = 0 Using RH criteria

s3 1 4 s2 2 9

s1 5 − 9

2 0

s0 9 0 ∵ 1st column of RH table does not contains positive sign only ∴ System become unstable.

Q No. 40

[Ans. B]

P(s) =−1

s + 1

Response C(s) = P(s) ⋅ u(s) =−1

s + 1⋅1

s

c(∞) = lims→0

s. c(s) = lim s→0

s [−1

s + 1⋅1

s] = −1

∴ P(s) ⇒ 1 Match

Q(s) =2(s − 1)

(s + 10)(s + 2)

C(s) = Q(s). u(s) =2(s − 1)

(s + 10)(s + 2)⋅1

s


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GATE-2016 IN-SET-1


c(∞) = lims→0

SC(s) = lims→0

s [2(s − 1)

(s + 10)(s + 2)⋅1

s] =

−1

2= −0.1

Q(s) ⇒ 2 Match

R(s) =1

(s + 1)2; C(s) =

1

(S + 1)2⋅1

s

C(∞) = lims→0

s C(s) = lims→0

1

(s + 1)2⋅1

s= 1

R(s) ⇒ 3 Match; P(s) → 1, Q(s) → 2, R(s)

Q No. 41

[Ans. *] Range: 40 to 41 Q No. 42

[Ans. *] Range: 100 to 100


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GATE-2016 IN-SET-1


Apply KCL at node A 0 − Va

1k=

Va − V0

10k

⇒ Va =V0

11

Using virtual ground concept

VA = VB = Vy =V0

11

Apply KCL at Node B Vy − Vx

1K+

Vy

1/CS+

Vy − V0

1/CNS= 0

Vx = Vy [Given]

Vy(CS) + (Vy − V0)CN. S = 0 V0

11(CS) + (

−10

11)V0CNS = 0

CN = 0.1C = 0.1 nF = 100 pF

Q No. 43

+

−

A B

C

1 kΩ

1 kΩ

10 kΩ

CN

Vo

Vx


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GATE-2016 IN-SET-1


[Ans. *] Range: 100 to 100

Apply KCL at Node A 0 − Va

100 k=

Va − V0

100 k

⇒ Va =V0

2

Using virtual ground concept

Va = VB =V0

2

Apply KCL at Node B VB − 1

10k+

VB − V0

10K+ IL = 0

⇒ IL = 0.1 mA = 100μA

Q No. 44

[Ans. A] F(x, y, z) = y z(x) + yz(0) + y z(x) + yz(1)

= xy z + xy z + yz Σm(2, 3, 4, 7)

+

−

10 k

100 k 100 k

10 k

RL IL

VB

VA

1 V

Vo

A


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GATE-2016 IN-SET-1


Q No. 45

[Ans. B] Note: OBSERVE QA is toggling on every clock ∴ JA = 1 KA = 1 JB KB can be both connected QA This matches with the option given

Q No. 46

[Ans. D] Y5 is connect to CS of memory will be active only when A11 is ‘1’ & A10 = ′0′ & A14, A13, A12 = 101


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GATE-2016 IN-SET-1


A14 A13 A12

1 0 1|A11 A10 A9 …… . . A0

1 0 |

μP has 16 address Bus and A15 is not used for decoding when A15 is “0” as well as A15 = 1 with above condition met CS of memory will be selected

∴ Address Dec is 5800

(A15 = 0)−5BFF &

D800(A15 = 1)

+ DBFF

Q No. 47

[Ans. *] Range: 10 to 10 C1 = 110 pF, C2 = pF

n =f2f1

=2f

f= 2

Cd =C1 − n2C2

n2 − 1=

110 − 4 × 20

3= 10 pF

Q No. 48


−

+

−

+

Low-Pass

Filter Vo

VXOR

V01

V02

V2

V1


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GATE-2016 IN-SET-1


Output of low pass filter is the average value of output of XOR GATE

Vdc =1

2π[∫ 5dt

π/5

0

+ ∫ 5dt6π/5

0

] =1

2π[5 (

π

5) + 5 (

6π

5− π)]

=1

2π[π + π] = 1 Volt

Q No. 49

[Ans. *] Range: 199 to 201

Q No. 50


Output Waveform

π 2π 3π 4π 5π V01

VXOR π/5 6π/5 11π/5 16π/5 21π/5 26π/5

π/5 6π/5 11π/5 16π/5 21π/5 26π/5 π 2π 3π 4π 5π

V02


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GATE-2016 IN-SET-1


𝑅𝜃 = 1000[1 + 0.004 𝜃] For 1° C rise in temperature 𝑅𝜃 = 1000[1 + 0.004]

= 1004 𝛺 Redraw the CKT Voltage at Node A by potential divider

𝑉𝐴 =(−5)10

10 + 10= −2.5 𝑉

Using virtual ground 𝑉𝐴 = 𝑉𝐵 = −2.5 𝑉 Apply KCL at Node B 𝐼1 = 𝐼2 𝑉𝐵 − (−5)

1000=

𝑉0 − 𝑉𝐵

1004

𝑉0 = 10 𝑚𝑉 Sensitivity = 10 mV/°C

Q No. 51

[Ans. *] Range: 2.0 to 2.1 OUTPUT current through photo diode (V0) = Intensity × Sensitivity × Area

V0 =4W

m2× 0.5

A

W× 10 × 10−6 m2 = 20 μA

Total current flow through photo diode = IP + ID (Dark current) OUTPUT voltage i to v converter

V0 = (920 + 1)μA ×100 mV

μA= 2100 mV = 2.1 V

−

+

+ −

5 V 10 kΩ

1 kΩ

10 kΩ

I1

I2 1004 Ω

Vo


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GATE-2016 IN-SET-1


Q No. 52

[Ans. *] Range: 4.9 to 5.1

V = √2gΔP ; ΔP = [Sm

Sw− 1] × h

V = √2 × 9.81 × 1.176 ; ΔP = [13.5 − 1] ×94.1

1000

≃ 5 m/s

Q No. 53

[Ans. *] Range: 1.9 to 2.1

Eg =hc

λ

=4.13567 × 10−15 × 2.998 × 108

620 × 10−9= 2 eV

Q No. 54

[Ans. *] Range: 60.0 to 60.2


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GATE-2016 IN-SET-1


m(t) =sin(100πt)

100 πt= sin c(100t)

Kf = 30 kHz Volt⁄

β =KfAm

fm=

30 × 1

50

β =3

5K = 600

BW = (β + 1)2f1 = (600 + 1) × 100 BW = 60.1 kHz

Q No. 55

[Ans. *] Range: 15 to 16 Signal power = 3 W

Resolution (Δ) =3.5 − (−3.5)

23=

7

8

Quantization noise power

N =Δ2

12= 0.0638

Signal to noise ratio S

N=

3

0.0638

(S

N)dB

= 10 log10(47.02)

= 16.72 dB

f

M(f)

−50 0 50


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