GATE 2018 · 2018-02-19 · TM ® Since 2004 GATE 2018 Instrumentation Engineering (Afternoon Session - 04.02.2018) • GATE ACADEMY has taken due care in making questions and solutions

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GATE 2018 [Afternoon Session]Instrumentation Engineering

Since 2004

PAGE

1

.General Aptitude.

Question 1

Arrange the following three-dimensional objects in the descending order of their volumes :

(i) a cuboid with dimensions 10 cm, 8 cm and 6 cm.

(ii) a cube of side 8 cm

(iii) a cylinder with base radius 7 cm and height 7 cm

(iv) a sphere of radius 7 cm

(A) (i), (ii), (iii), (iv) (B) (ii), (i), (iv), (iii) (C) (iii), (ii), (i), (iv) (D) (iv), (iii), (ii), (i) Ans. (D)

Sol. (i) A cuboid with dimensions 10 cm, 8 cm and 6 cm.

The volume of cuboid is given by,

10 8 6 480cmcuboidV lbh …(i)

(ii) A cube of side 8 cm.

The volume of cube is given by,

3cubeV a where a is side length

3 38 512cmcubeV …(ii)

(iii) Cylinder with base radius 7 cm and height 7 cm.

The volume of cylinder is given by,

2cylinderV r h

27 7 1077.566cylinderV …(iii)

(iv) Sphere of radius 7 cm.

The volume of rapture in given by,

34

3sphereV r

where, r = Radius

347 1436.753

3sphereV

…(iv)

Form equation (i), (ii), (iii) and (iv),

Descending order of the volume is given by,

(iv) > (iii) > (ii) > (i)

Question 2

“When she fell down the ________, she received many _______ but little help.” The words that best fill the blanks in the above sentence are

(A) stairs, stares (B) stairs, stairs (C) stares, stairs (D) stares, stares Ans. (A)

Sol. Stares means the act of looking directly at someone or something for a long time the act of staring.

Stairs means a series of steps that go from one level or floor to another.


Since 2004

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2

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Question 3

The area of an equilateral triangle is 3 . What is the perimeter of the triangle?

(A) 2 (B) 4 (C) 6 (D) 8

Ans. (C)

Sol. Given : The area of an equilateral triangle 3

The area of an equivalent triangle ABC of side length a is given by,

Area, 23( )

4A a

233

4a

2a

Since length never negative. So, 2a

The perimeter of the triangle is given by,

Perimeter 3 3 2 6a

Question 4

"In spite of being warned repeatedly, he failed to correct his ______ behaviour."

The word that best fills the blank in the above sentence is

(A) rational (B) reasonable (C) errant (D) good

Ans. (C)

Sol. In spite of being warned repeatedly, he failed to correct his errant behaviour.

Question 5

For 0 2x , sin and cosx x are both decreasing functions in the interval ____________.

(A) 0,2

(B) ,2

(C) 3

,2

(D) 3

, 22

Ans. (B)

Sol. The waveform sin x and cos x

Over the interval 0 2X is given by,

2

3

2

20

sin x

2

3

2

20

cos x

A

BC a

a a



Since 2004

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3

S. No. Function Status Interval

1. sin x Increasing

Increasing

Decreasing

02

x

32

2x

3

2 2x

2. cos x Decreasing

Increasing

0 x

2x

,2

function sin x and cos x both are decreasing function.

Question 6

An automobile travels from city A to city B and returns to city A by the same route. The speed of the vehicle during the onward and return journeys were constant at 60 km/h and 90 km/h respectively. What is the average speed in km/h for the entire journey?

(A) 72 (B) 73 (C) 74 (D) 75

Ans. (A)

Sol. Given : Speed of onward 60 km/hr

Speed of return 90 km/hr

Let, total distance = x km

Average speed = 72 km/hr

60 90

x xx x

Question 7

If 2 1 0x x what is the value of 44

1?x

x

(A) 1 (B) 5 (C) 7 (D) 9

Ans. (C)

Sol. Given : 2 1 0x x

2 1x x

Dividing x on both side,

1

1xx

1

1xx

…(i)


Since 2004

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4



2

4 24 2

1 12x x

x x

224

4

1 12 2x x

x x

44

1x

x 22

1 2 2 9 2 7

Question 8

A set of 4 parallel lines intersect with another set of 5 parallel lines. How many parallelograms are formed?

(A) 20 (B) 48 (C) 60 (D) 72

Ans. (C)

Sol. Given : A set of 4 parallel lines intersects with another set of 5 lines

Number of parallelograms formed is given by,

2 25 4 60C Cn .

Question 9

In a detailed study of annual crow births in India, it was found that there was relatively no growth during the period 2002 to 2004 and a sudden spike from 2004 to 2005. In another unrelated study, it was found that the revenue from cracker sales in India which remained fairly flat from 2002 to 2004, saw a sudden spike in 2005 before declining again in 2006. The solid line in the graph below refers to annual sale of crackers and the dashed line refers to the annual crow births in India. Choose the most appropriate inference from the above data.

(A) There is a strong correlation between crow birth and cracker sales.

(B) Cracker usage increases crow birth rate.

(C) If cracker sale declines, crow birth will decline.

(D) Increased birth rate of crows will cause an increase in the sale of crackers

Ans. (A)

2001 2003 2005 2007

An

nu

alcro

wbirth

sin

India

An

nu

alsa

leo

fcr

ack

ers

inIn

dia



Since 2004

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5

Sol. From the above graph there is strong correlation between crow birth and cracker sales. Thus from above option, option (A) is the correct answer.

Question 10

To pass a test, a candidate needs to answer at least 2 out of 3 questions correctly. A total of 6,30,000 candidates appeared for the test. Question A was correctly answered by 3,30,000 candidates. Question B was answered correctly by 2,50,000 candidates, Question C was answered correctly by 2,60,000 candidates. Both questions A and B were answered correctly by 1,00,000 candidates. Both questions B and C were answered correctly by 90,000 candidates. Both questions A and C were answered correctly by 80,000 candidates. If the number of students answering all questions correctly is the same as the number answering none, how many candidates failed to clear the test?

(A) 30,000 (B) 2,70,000 (C) 3,90,000 (D) 4,20,000

Ans. (D)

Sol. Given :

(i) Total number of candidates appeared = 6,30,000

(ii) Question A was correctly answered by 3,30,000 candidates.

(iii) Question B was answered correctly by 2,50,000 candidates

(iv) Question C was answered correctly by 2,60,000 candidates.

(v) Both questions A and B were answered correctly by 1,00,000 candidates.

(vi) Both questions B and C were answered correctly by 90,000 candidates.

(vii) Both questions A and C were answered correctly by 80,000 candidates.

y Total number of condidate attempted all questions correctly

Also, 630000 2 150000 100000 80000 60000 90000 90000y

630000 570000 2 y

30000y

The students failed to clear the test 150000 60000 90000 4 y 300000 4 30000 420000

90000 +y

80000 –y 90000 –y

100000 –y 60000 +y150000 +y

y

9000080000

100000

A(330000)

B(250000)

C(260000)


Since 2004

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6



.Technical Section.

Question 1

A piezoelectric pressure sensor has a bandpass characteristic with cut-off frequencies of 0.1 Hz and 1 MHz and a sensitivity of 100 mV/kPa. The sensor is subjected to a static constant pressure of 100 kPa. Its steady-state output will be

(A) 0 V (B) 0.1V (C) 1V (D) 10V

Ans. (A)

Sol. Given :

(i) Piezoelectric pressure sensor with cut-off frequencies of 0.1 Hz and 1 MHz

(ii) Sensitivity = 100 mV/kPa.

(iii) Sensor is subjected to a static constant pressure of 100 kPa.

Since, given input is static means 0 Hz signal and piezoelectric transducers is suitable only for dynamic inputs.

Hence, its steady state output will be zero.

Question 2

A voltage of 6cos(100 )t V is fed as y-input to a CRO. The waveform seen on the screen of the CRO is

shown in the figure. The Y and X axes settings for the CRO are respectively

(A) 1 V/div and 1 ms/div (B) 1 V/div and 2 ms/div

(C) 2 V/div and 1 ms/div (D) 2 V/div and 2 ms/div

Ans. (D)

Sol. Given :

6cos100yv t

peak to peak 12Vv

Number of vertical division = 6

Vertical sensitivity 12

(V/div) 2 V/div6



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7

Time period 2

( ) 20msec100

T

Number of horizontal division = 10

Horizontal sensitivity msec 20

2 msec /divdiv 10

Question 3

Let N be a 3 by 3 matrix with real number entries. The matrix is such that 2 0.N The eigen values of N are

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

Ans. (A)

Sol. Given : 2 0N

where N be a 3 by 3 matrix with real number entries.

Since, determinant of matrix = Product of the eigen value

2

0N

2 2 21 2 3 0

and trace of the matrix = Sum of the eigen value 1 2 3

Since, 2 0N

i.e., the given matrix is nilpotent matrix,

( ) 0kN

For nilpotent matrix, determinant and trace is zero.

So, 1 2 3 0 …(i)

2 2 21 2 3 0 …(ii)

From equation (i) and (ii)

The possible combination of the eigen values are, (0,0,0) and (0,1, 1)

Question 4

X and Y are two independent random variables with variances 1 and 2, respectively. Let Z = X– Y. The variance of Z is

(A) 0 (B) l (C) 2 (D) 3

Ans. (D)

Sol. Given : Two independent random variables X and Y with various 1 and 2

( ) ( )V z V X Y

2 2 2 2( ) (1) ( ) ( 1) ( ) 1 1 ( 1) 2 3V z V X V Y


Since 2004

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8



Question 5

A series R-C circuit is excited by a 1 0 V sinusoidal ac voltage source. The locus diagram of the phasor current ( )I x jy A, when C is varied, while keeping R fixed, is

(A) (B)

(C) (D)

Ans. (A)

Sol. Given : Series RC circuit by 1 0 V sinusoidal ac voltage source.

1 0

1I

Rj C

2 2

2 2 2 2 2 2

(1 )

1 1 1

j C j C j RC RC j CI x jy

j RC R C R C

2 2

2 2 21

RCx

R C

,

2 2 21

Cy

R C

1 1tan

RC

Question 6

Consider signal 1, | | 2

( ) .0 | | 2

tx t

t

Let ( )t denote the unit impulse (Dirac-delta) function. The value of

the integral 5

02 ( 3) ( 4)x t t dt is

(A) 2 (B) 1 (C) 0 (D) 3

Ans. (A)

Sol. Given : 1 2

( )0 2

tx t

t

5

02 ( 3) ( 4) 2 (1) 2x t t dt x

+jy

x x

+jy

x

+jy +jy

x

+jy

x1

R

0

t

( )x t

2

1

2



Since 2004

PAGE

9

Question 7

A 300 V, 5 A, 0.2 pf low power factor wattmeter is used to measure the power consumed by a load. The wattmeter scale has 150 divisions and the pointer is on the 100th division. The power consumed by the load (in Watts) is _________.

Ans. (200)

Sol. Given : 300 V, 5 A, 0.2 pf low power factor wattmeter

Wattmeter scale has 150 division and the pointer is on the 100th division.

Power ( ) 300 5 0.2 300P W

For 100th division 300

( ) 100150

P 200 W

Question 8

As shown in the figure, temperature is measured using a K type thermocouple. It has a sensitivity of 40 0V/ C. The gain (G) of the ideal instrumentation amplifier is 25. If the output 0V is 96 mV, then the value

of (in 0C ) is _______.

Ans. (96)

Sol. Given :

(i) Sensitivity 0( ) 40 V/ CS

(ii) Gain 25

(iii)Output 0( ) 96 mVV

(iv) 0ref

0 ref( )V S

096mV 25 40 μV/ C( 0)

3

6

96 10

40 25 10

096 C

Question 9

Consider the transfer function 2

( ) .( 1)( 2)

G ss s

The phase margin of ( )G s in degrees is ______.

Ans. (180)

Roomtemperature

INA= 25G

0V0

ref 0 C

M

N

N

M

Roomtemperature

INA= 25G

0V0

ref 0 C

M

N

N

M


Since 2004

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10



Sol. Given : 2

( )( 1)( 2)

G ss s

2

( )( 1)( 2)

G jj j

2 2

2( )

1 4G j

1 1( ) tan tan2

G j

We have to find phase margin, which is calculated over the gain cross over frequency ( )gc .

To find gc , ( ) 1gc

G j

2 2

21

1 4gc gc

2 24 ( 1)( 4)gc gc

4 2 24 4 4gc gc gc

4 25 0gc gc

2 2( 5) 0gc gc

2 0gc 2 5gc

0 /gc r s 5gc j

As frequency is a real quantity

Hence, gc Imaginary

0gc r/s

1 1( ) tan (0) tan (0)gc

G j

0( ) 0gc

G j

Phase margin 0180 0 0180 0 180

Short circuit to find gc

2

( )( 2)( 1)

G ss s

2

(0) 12 1

G

( ) 1G s at 0 r/s

Hence, 0gc r/s



Since 2004

PAGE

11

Question 10

The diodes given in the circuit are ideal. At 60ms,t pqV (in Volts) is _______.

Ans. (200)

Sol. Given figure,

During +ve half cycle of the line input, diodes 2D and 3D are reverse biased and can be replaced by open

circuit equivalent circuit can be drawn as,

From the figure, we can see that the, current can not flow as no closed loop path is available which can

connect inV .

During –ve half cycle of input, diodes 1D and 4D are reverse biased but 2D and 3D are forward

biased equivalent circuit can be drawn as,

pqV

p q

C

C

~100sin(2 50 ) Vt

inV

2D 4D

3D1D1C

2C

PQV P Q

inV

2D 4D

3D1D1C

2C

P Q

inV

2D

3D1D1C

V

2C

P Q

2CV

4D

PQV

Up to100 V


Since 2004

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12



Capacitors 1C and 2C charges through diodes 2D and 3D respectively upto 100 V as shown in figure.

Simplified circuit can be redrawn as,

KVL can be written as,

2 1

0in C PQ CV V V V

1 2PQ C C inV V V V

200PQ inV V …(i)

Also, given 100sin (2 50 )inV t V

Frequency of circuit 50Hzf

Time period 1 1

0.02 20ms50

Tf

From the waveform, it is clear that

60ms

0in tV

Hence, from equation (i), voltage PQV at 60 mst is given as,

200 200 0PQ inV V

60ms

200 VPQ tV

Question 11

An optical pulse containing 66 10 photons is incident on a photodiode and 64.5 10

electron-hole pairs are created. The maximum possible quantum efficiency (in %) of the photodiode is ________.

Ans. (75)

Sol. Given : Number of photons incident 6( ) 6 10N

Number of electron-hole pairs created 6( ) 4.5 10n

inV

2D

3D1

100 VC

V

2100 V

CV

PQV

20 ms 40 ms

+ 100 V

– 100 V

Vin

t60 ms



Since 2004

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13

Quantum efficiency of photo-diode is given by,

6

6

4.5 10100 100 75

6 10

n

N

Question 12

For the 3-bit binary counter shown in the figure, the output increments at every positive transition in the clock (CLK). Assume ideal diodes and the starting state of the counter as 000. If output high is 1 V and output low is 0 V, the current I (in mA) flowing through the 50 resistor during the 5th clock cycle is (up to one decimal place)

Ans. (10)

Sol. Given : 3 digit counter with starting state as 0000.

For 5th clock cycles the state is 101. The equivalent figure is,

By applying KCL at node A,

1 1

050 100 100

V V V

2 1 1 0V V V

0.5 VV

0.5

10mA50 50

VI

3-bit binarycounter

MSB

LSB

I 50 100

100

CLK

3-bit binarycounter

MSB

LSB

I 50 100

100

CLK

50

100

100

1 V

1 V

I

A


Since 2004

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14



Question 13 The representation of the decimal number (27.625)10 in base-2 number system is

(A) 11011.110 (B) 11101.101 (C) 11011.101 (D) 10111.110

Ans. (C)

Sol. Given : (27.625)10

10(27) (11011)

10 2(0.625) (0.101)

10 2(27.625) (11011.101)

Question 14

The approximate phase response of 2 0.01

2 2

100

0.2 100

se

s s

is

(A)

(B)

(C)

(D)

Ans. (A)

272

132

62

32

1 1

0

1

10.625 2

0.25 2

0.50 2

1.25

0.50

1.00

0

– 50

– 250

0 100

Ph

ase

(deg

ree)

Frequency (rad/s)200

0

– 180

0 100

Phas

e(d

egre

e)


100

– 150

0 100

Phas

e(d

egre

e)


0

180

0

0 100

Phas

e(d

egre

e)




Since 2004

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15

Sol. Given : 2 0.01

2 2

100( )

0.2 100

seF s

s s

( ) ( ) ( )F s G s T s

where, 2

2 2

100( )

0.2 100G s

s s

It has a quadratic pole having corner frequency

100r/sc n

and 0.01( ) sT s e

which represents the transportation lag in the system

0.01( ) jT j e

Total phase angle of ( )F s will be seen of angles of ( )G j and ( )T j .

Phase responses of ( )G j and ( )T j are drawn individually first and then their resultant phase response

that represents the overall phase response, are shown using equations (i) and (ii),

( ) 0.01T j …(i)

0

( ) 0 0

180

c

c

G j

…(ii)

Fig.1 Transportation lag phase plot Fig.2 Approximate Bode phase plot

Fig.3 Resultant phase response

Hence, the most approximate phase response is given as option (A).

( )T j

50 100 c

28.65

57.3

( )G j

100c

0180

( ) ( ) ( )F j T j G j

100c

57.3

237.3


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16



Question 15

An amplitude modulated signal is shown in the figure. The modulation index is (up to one decimal place)

Ans. (0.3)

Sol. The modulation index is given by,

13 7 6

0.313 7 20

Question 16

A coil having an impedance of (10 100)j is connected in parallel to a variable capacitor as shown in

figure. Keeping the excitation frequency unchanged, the value of the capacitor is changed so that parallel resonance occurs. The impedance across terminals p-q at resonance (in ) is _______.

Ans. (1010)

Sol. Given figure,

Condition for resonance, Img(Admittance) 0 and only real term exist.

1

10 100Y j c

j

2 2

10 100

10 100

jj c

2 2

10

10 100

2 21 10 100

101010

ZY

13

70

Time (ms)0.10

Vo

ltag

e(v

olt

s)

100j

10

P

Q

100j

10

P

Q

1

j c



Since 2004

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17

Question 17

Two periodic signals ( )x t and ( )y t have the same fundamental period of 3 seconds. Consider the signal

( ) ( ) (2 1).z t x t y t The fundamental period of ( )z t in seconds is

(A) 1 (B) 1.5 (C) 2 (D) 3 Ans. (D)

Sol. Given : ( ) ( ) (2 1).z t x t y t

Where, ( )x t and ( )x t we periodic signal of period 3.

Period of 3

(2 1) 1.52

y t

Period of ( ) 3x t

Note : Time shifting and time inversion do not have any effect on fundamental time period.

Now, period of ( ) ( ) (2 1)z t x t y t is given by, T LCM of (3 and 1.5) = 3.

Question 18

Consider two functions 2( ) ( 2)f x x and ( ) 2 1,g x x where x is real. The smallest value of x for

which ( ) ( )f x g x is _______.

Ans. (1)

Sol. Given : 2( ) ( 2)f x x , ( ) 2 1g x x

When, ( ) ( )f x g x

2( 2) 2 1x x

2 4 4 2 1x x x

2 6 5 0x x

1,5x

Thus, the smallest value of x is 1.

Question 19

An input ( ) sin( )p t t is applied to the system 1.

1

sG s

s

The corresponding steady state output is

si ,nty t where the phase (in degrees), when restricted to 0 00 360 , is _____.

Ans. (90)

Sol. Given : ( ) sin( )p t t , 1 , 1

1

sG s

s

( ) ( ) ( )y t G j p t 1 1

sin1 1

jt

j

sin( 90)t

( ) sin( )y t t where, 090

sin( )t ( )y t( )G s


Since 2004

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18



Question 20

An ideal square wave with period of 20 ms shown in the figure, is passed through an ideal low pass filter with cut-off frequency 120 Hz. Which of the following is an accurate description of the output?

(A) Output is zero.

(B) Output consists of both 50 Hz and 100 Hz frequency components.

(C) Output is a pure sinusoid of frequency 50 Hz.

(D) Output is a square wave of fundamental frequency 50 Hz.

Ans. (C)

Sol. Given : An ideal square wave with period of 20 ms,

Fundamental frequency of given signal ( )x t is 1 1

50Hz20ms

fT

The given signal ( )x t is half wave symmetric as it follows the equation

( )2

Tf t f t

i.e. ( 10) ( )f t f t

t

( )x t

1

1. . .. . .

t

( )x t

1

1. . .. . .

20 10 0 10 20 30

( )f t

1

t



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19

Since, the signal is half wave symmetric, it will contain only odd harmonics of fundamental frequency i.e.

50 Hz, 150 Hz, 250 Hz _________ etc. But if we pass it through ideal low pass filter with cut off frequency 120 Hz, so output will be pure sinusoid of frequency 50 Hz.

Question 21

In the given circuit, assume that the opamp is ideal and the transistor has a of 20. The current 0I (in

A ) flowing through the load LZ is _______.

Ans. (300)

Sol. Given figure,

Due to the virtual ground concepts, 2 VNI I CV V V

5 5 2 3

10.5 kΩ 10.5k 10.5C

c

VI

mA

3

10.5 20cI

I

mA

0

3 30.3 mA

10.5 10.5 20EI I

0 300 AI

20 10 0 10 20 30

( 10)f t

1

t20 10 0 10

20 30

( )f t

t

10.5 k

5V

20

LZ

0I

2V

10.5 k

5V

20

LZ

0I

2V

CV


Since 2004

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20



Question 22

The Thevenin equivalent circuit representation across terminals p-q of the circuit shown in the figure is

(A) 1 V source in series with 150 k (B) 1 V source in parallel with 100 k (C) 2 V source in series with 150 k (D) 2 V source in parallel with 200 k Ans. (C)

Sol. Given figure,

The Thevein’s equivalent circuit is given by,

For thV ,

100

4 2V100 100th pq ABV V V

Thevein’s resistance thR we remove the all independent source,

4 V

100k 100k

100k

p

q

4 V

100k 100k

100k

p

q

thRp

thV

q

4 V

100k 100k

100k

p

q

A

B

p

100k

100k100k

q



Since 2004

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21

100 100

100100 100thR

150k

Thus, the Thevenin’s equivalent circuit becomes,

Question 23

Consider a sequence of tossing of a fair coin where the outcomes of tosses are independent. The probability of getting the head for the third time in the fifth toss is

(A) 5

16 (B)

3

16 (C)

3

5 (D)

9

16

Ans. (B)

Sol. Given : Head appear third time in fifth toss.

The required probability ( )P1

2r

r n rCn p q

Where, 1 1

4, 2, ,2 2

n r p q

2 2

42

1 1 1 3

2 2 2 16P C

Question 24

The number of comparators required for implementing an 8-bit flash analog-to-digital converter is

(A) 8 (B) 128 (C) 255 (D) 256

Ans. (C)

Sol. Given : 8-bit flash analog-to-digital converter.

Number of comparators required for implementing an 8-bit flash analog-to-digital converter is given by,

2 1nN where, n = number of bit

82 1 255N

Question 25

Let 21( )f z z and 2( )f z z be two complex variable functions. Here z is the complex conjugate of z

. Choose the correct answer.

(A) Both 1( )f z and 2( )f z are analytic (B) Only 1( )f z is analytic

(C) Only 2( )f z is analytic (D) Both 1( )f z and 2( )f z are not analytic

Ans. (B)

150k

p

2 V

q


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Sol. Given : 21( )f z z and 2 ( )f z z

2 2 2 21( ) ( ) 2f z z x iy x ixy y

2 ( )f z z x iy x iy

We know that a function is said to be analytic if it satisfies Cauchy Riemann equation given as

x yu v and y xu v

Also, , , ,x y x yu u v v must be continuous functions.

Case-I : For 2 21( ) ( ) 2f z x y i xy

2 2 2u x y v xy

2 2x yu x v x x yu v

2 2y xu y v y y xu v

Hence, 1( )f z is analytic.

Case-II : For 2 ( )f z x iy

u x v y

1 1x yu v x yu v

0 0y xu v y xu v

As x yu v , hence 2 ( )f z is not analytic.

Hence, option (B) is correct.

Question 26

A multi-mode optical fiber with a large core diameter has a core refractive index 1 1.5n and cladding

refractive index 2 1.4142.n The maximum value of A (in degrees) for which the incident light from air

will be guided in the optical fiber is ± _______.

Ans. (30)

Sol. Given :

(i) Core refractive index 1( ) 1.5n

(ii) Cladding refractive index 2( ) 1.4142n

(iii) Air refractive index 0( ) 1n

Cladding

A

2 1.4142n

Core

2 1.4142n Cladding

Light

1 1.5n

Air



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23

Since, 2 20 1 2N.A. sin An n n

1 2 2sin (1.5) (1.4142)A

030A

Question 27

The product of sum expression of a Boolean function ( , , )F A B C of three variables is given by

( , , ) ( ).( ).( ).( )F A B C A B C A B C A B C A B C

The canonical sum of product expression of ( , , )F A B C is given by

(A) ABC ABC ABC ABC (B) ABC ABC ABC ABC

(C) ABC ABC ABC ABC (D ) ABC ABC ABC ABC

Ans. (B)

Sol. Given : ( , , ) ( )( )( )F A B C A B C A B C A B C

(001,011,100,111)M

(1,3,4,7)M (0,2,5,6)m

(000,010,101,110)m

ABC ABC ABC ABC

Hence, the correct option is (B).

Question 28

Two bags A and B have equal number of balls. Bag A has 20% red balls and 80% green balls. Bag B has 30% red balls, 60% green balls and 10% yellow balls. Contents of Bags A and B are mixed thoroughly and a ball is randomly picked from the mixture. What is the chance that the ball picked is red?

(A) 20% (B) 25% (C) 30% (D) 40%

Ans. (B)

Sol. Given :

For bag A For bag B

Red balls = 20 % Red balls = 30 %

Green balls = 80 % Green balls = 60 %

Yellow balls = 10 %

Let, both bags contains 10 balls each.

Cladding

A

2 1.4142n

Core

2 1.4142n Cladding

Light

1 1.5n

Air


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Hence,

In bag A In bag B

No. of red balls = 2 No. of red balls = 3

No. of green balls = 8 No. of Green balls = 6

No. of Yellow balls = 1

If balls of bags A and B are mixed, then,

Total no. of balls = 10 + 10 = 20

Total no. of red balls = 2 +3 = 5

Hence, if one ball is picked at random. Then the probability of this ball being of red color is

No. of red balls

(Red) 100%Total no. of balls

P 5

100% 25%20

Hence option (B) is correct.

Question 29

Given 2 22 4 4 ,F x y i yzj xz k

the value of the line integral .c

F dl

along the straight line c from

(0,0,0) to (1,1,1) is

(A) 3

16 (B) 0 (C)

5

12

(D) 1

Ans. (D)

Sol. Given : 2 22 4 4 ,F x y i yzj xz k

The equation form by, point (0,0,0) to (1,1,1)

0 0 0

1 0 1 0 1 0

x y zt

2 2( 2 ) 4 4C

F dl x y dx yz dy xz dz

1

2 2 3

0

( 2 ) 4 4t t dt t dt t dt

1 1 13 3 4

2

0 0 0

4 43 3 4C

t t tF dl t

1 41 1 1

3 3

Question 30

The average velocity v of flow of clear water in a 100 cm (inner) diameter tube is measured using the

ultrasonic flow meter as shown in the figure. The angle is 45°. The measured transit times are 1t

0.9950 ms and 2 1.0000t ms. The velocity v (in m/s) in the pipe is (up to one decimal place) ________.



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Ans. (5.025)

Sol. Given figure,

The measured transit times are 1t 0.9950 ms and 2 1.0000t ms.

From figure, 0 00sin 45

l

100 2 cml

1 0.9950mst and 2 1.0000msect

31

11.005025 10 Hz

0.9950msf

32

11 10 Hz

1 msecf

31 2 (1.005025 1)10 Hz 5.025f f f

2 cosfv

fd

. (5.025) 100 2 cm

502.5cm/sec12cos 22

f

f dv

5.025sec

Question 31

The voltage and current drawn by a resistive load are measured with a 300 V voltmeter of accuracy ± 1% of full scale and a 5 A ammeter of accuracy ± 0.5% of full scale. The readings obtained are 200 V and 2.5 A. The limiting error (in %) in computing the load resistance is (up to two decimal places) _______.

Ans. (2.5)

100 cm

/x x

T R

/x x

T R

045

v

100 cm

/x xT R

/x xT R

045

v l


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Sol. Given : Voltmeter ( ) 300 1%V and Ammeter ( ) 5 0.5%A

For 200V , 2.5AI

% limiting error 300

( ) 1 1.5%200

V

% limiting error 5

( ) 0.5 1%2.5

A

% liming error ( ) (1.5 1) 2.5%V

RI

Question 32

Consider a standard negative feedback configuration with 1

( )( 1)( 2)

G ss s

and ( ) .s

H ss

For

the closed loop system to have poles on the imaginary axis, the value of should be equal to (up to one decimal place) ________.

Ans. (9)

Sol. Given : Negative feedback system with

1

( ) , ( )( 1)( 2)

sG s H s

s s s

we have to find value of for which the poles of the system will be lying on imaginary axis.

We can find characteristic equation, which determines the location of closed loop poles as, 1 ( ) ( ) 0G s H s

( )

1 0( 1) ( 2)

s

s s s

3 23 2 0s s s s

3 23 3 0s s s

Hence we can form the Routh’s array using characteristic polynomial.

3 2( ) 3 3P s s s s

3

2

1

0

1 3 0

3 0

90 0

30 0

s

s

s

s

If closed loop poles will be lying on imaginary axis, then the system will be marginally stable and the conditions is satisfied by Routh’s array, if any complete row (except last) become zero.

Having seen the first column of Routh’s array, poles will on imaginary axis, when

9

03

9



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Question 33

For the sequence 1, 1,1, 1 ,x n with 0,1,2,3n the DFT is computed as 234

0

( ) ,j nk

n

X k x n e

for 0,1,2,3.k The value of k for which ( )X k is not zero is

(A) 0 (B) l (C) 2 (D) 3

Ans. (C)

Sol. Given : ( ) 1, 1,1, 1x n

The D.F.T. ( )X K is given as,

1

2 /

0

( )N

j Kn N

n

X K x n e

For the given problem 4N .

This equation can be re-written in matrix form if we represents DFT as linear transform as,

KnN N NX W x

where NX and Nx are column vectors and KnNW represents a K N matrix.

NW Twiddle factor 2 /j Ne

Here, / 24

jNW W e

So, we can write DFT ( )X K as,

0 0 0 04 4 4 40 1 2 3

4 4 4 40 2 4 6

4 4 4 40 3 6 9

4 4 4 4

(0) 0 (0)

(1) 1 (1)

(2) 2 (2)

(3) 3 (3)

X K xW W W W

X K xW W W W

X K xW W W W

X K xW W W W

From the periodicity property of Twiddle factor, we know that

K N KN NW W

Hence, we don’t need to find all the line values from 0 14 4, .....W W to 9

4W , but we need only to evaluate

values from 04W to 3

4W .

2

00 44 1

jW e

11 2

4 cos sin2 2

jW e j j

22 2

4 cos sin 1j

W e j

33 2

4

3 3cos sin

2 2

jW e j j

4 0 4 04 4 4W W W


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5 1 4 14 4 4W W W

9 1 8 1

4 4 4W W W

(0) 1 1 1 1 1

(1) 1 1 1

(2) 1 1 1 1 1

(3) 1 1 1

X

X j j

X

X j j

(0) 1 1 1 1 0X

(1) 1 1 0X j j

(2) 1 1 1 1 4X

(3) 1 1 0X j j

Hence, we can see that only

(2) 4 0X

Hence, for 2K

( ) 0X K

Question 34

A portion of an assembly language program written for an 8-bit microprocessor is given below along with explanations. The code is intended to introduce a software time delay. The processor is driven by a 5 MHz clock. The time delay (in s ) introduced by the program is _______.

MVIB, 64H : Move immediate the given byte into register B. Takes 7 clock periods.

LOOP : DCR B : Decrement register B. Affects Flags. Takes 4 clock periods.

JNZ LOOP : Jump to address with Label LOOP if zero flag is not set. Takes 10 clock periods when jump is performed and 7 clock periods when jump is not performed.

Ans. (280.8)

Sol. Given : 8-bit microprocessor of clock frequency 5 MHz,

MVIB, 64H

LOOP : DCR B

JNZ LOOP

Loop is executed for 100 times. In that 99 times condition is satisfied. 100th time condition is not satisfied.

Time delay ( )pdt is given by,

7 99(4 10 ) (4 7 ) 1404pdt T T T T T T

1 1

0.2 sCLK

Tf T

1404(0.2 s) 280.8 spdt



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29

Question 35

In the given circuit, the mesh currents 1 2,I I and 3I are

(A) 1 2 31A, 2A and 3AI I I (B) 1 2 32A, 3A and 4AI I I

(C) 1 2 33A, 4A and 5AI I I (D) 1 2 34A, 5A and 6AI I I

Ans. (A)

Sol. Given figure,

Applying KVL in loop 1,

1 2 38 7 2 0I I I …(i)

Applying KVL in loop 2,

1 2 36 7 5I I I …(ii)

By Supermesh, 3 1 2I I …(iii)

From equation (i), (ii) and (iii),

1 2 3A, I 2A, I 3AI I

Question 36

An opamp that is powered from a ± 5V supply is used to build a non-inverting amplifier having a gain of

15. The slew rate of the opamp is 60.5 10 V/s. For a sinusoidal input with amplitude of 0.3 V, the maximum frequency (in kHz) up to which it can be operated without any distortion is (up to 1 decimal place) ______.

Ans. (17.69)

+–

2I

6 1

2 11I 3I

5V

2 A

+–

2I

6 1

2 11I 3I

5V

2 A


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Sol. Given :

(i) A non-inverting amplifier of gain 15 is powered from a ± 5V supply

(ii) Slew rate 60.5 10 V/s.

01

1 fi

RV V

R

0

1

15fV

i

RVA

V R

0 15 15 0.3sin 4.5siniV V t t

0 4.5 cosV

tt

Slew rate 0

max

4.5dV

dt

Given that slew rate 4.5 0.5 V/μs

6max4.5 2 0.5 10 V/sf

6

max

0.5 1017.69 kHz

4.5 6.28f

Question 37

Consider the following system of linear equations :

3 2 2x ky

6 2kx y

Here x and y are the unknowns and k is a real constant. The value of k for which there are infinite number of solutions is

(A) 3 (B) 1 (C) – 3 (D) – 6

Ans. (C)

Sol. Given : 3 2 2x ky and 6 2kx y

Condition for infinite solution,

( ) ( : ) No of unknownA A B

3 2 2

6 2

k

k

3 2 2

3 6 2 0

k

k k

3 0, 3k k

6 2 0, 3k k

14 kfR

1 1 kR

iV V

0V0.3siniV t



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31

Question 38

In the figure, an RLC load is supplied by a 230 V, 50 Hz single phase source. The magnitude of the reactive power (in VAr) supplied by the source is ________.

Ans. (67.3)

Sol. Given figure,

In the above figure,

(169.6 162.6) 230eqZ j j

300.477 22.485eqZ

The value of current is I is the given by,

00

2300.765 22.485

300.477 22.485I

The complex power supplied by source is given by, *S VI

0230(0.765 22.485 ) *S 162.57 67.29

The magnitude of reactive power supplied by source = 67.29.

Question 39

The Fourier transform of a signal, ( ),x t denoted by ( )X j , is shown in the figure.

162.6 j

162.6

230j 230V

50H

162.6 j

162.6

230j 230V

50 Hz

I

eqZ

0 1

( )X j

(rad/s)–1


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32



Let ( ) ( ) ( ).jty t x t e x t The value of Fourier transform of ( )y t evaluated at the angular frequency

0.5 rad/s is (A) 0.5 (B) 1 (C) 1.5 (D) 2.5 Ans. (C)

Sol. Given : ( ) ( ) ( )jty t x t e x t

Frequency shifting property of Fourier transform If ( ) ( )x t X j

Then, 00( ) ( )j tx t e X j

Apply Fourier transform,

0.5( ) 1 0.5 1.5y j

Hence, the correct option is (C).

Question 40

In the given relaxation oscillator, the opamps and the zener diodes are ideal. The frequency (in kHz) of the square wave output v0 is ________.

0 1

( )X j

(rad/s)–1

21 0 00.5 0.51 1

( )X j ( 1)X j

11

0.5

0.1 nF

50 k 50 k

100 k

50 k

25 k0v

5V

5V



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33

Ans. Marks to All

*** Given Question is wrong as the circuit can’t practically work as a relaxation oscillator.

Question 41

The sampling rate for Compact Discs (CDs) is 44,000 samples/s. If the samples are quantized to 256 levels and binary coded, the corresponding bit rate (in bits per second) is ________.

Ans. (352000)

Sol. Given : Symbol rate ( ) 44000samples/secr

Entropy for equally probable 256 levels

2log (256) 8bits/sampleH

Bit rate ( ) . bits/secbR r H 44000 8 bits/sec 352000 bits/sec

Question 42

A 1000 strain gage ( )gR has a gage factor of 2.5. It is connected in the bridge as shown in figure. The

strain gage is subjected with a positive strain of 400 m/m. The output 0V (in mV) of the bridge is (up to

two decimal places) _______.

Ans. (0.5)

Sol. Given : unloaded 1kR , Gage factor 2.5 , 400 m/m

For quarter bridge,

0 G.F.4 4

s sV V RV

R

60

2400 10 2.5

4V

60 200 2.5 10V

60 500 10V 500 V

Question 43

Consider the standard negative feedback configuration with 2

2

0.2 100( )

0.2 100

s sG s

s s

and 1

.2

H s The

number of clockwise encirclements of ( 1,0) in the Nyquist plot of the Loop transfer-function ( ) ( )G s H s

is ______. Ans. (0)

V02 V

1000

1000

1000

xR

1 k

2 V

xR

+

–

1 k

1 k

+ –0V


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Sol. Given : 2

2

0.2 100( )

0.2 100

s sG s

s s

, 1

.2

H s

The open loop transfer function is given by,

2

2

1 0.2 100( ) ( )

2 0.2 100

s sG s H s

s s

2

2

1 100 0.2( ) ( )

2 100 0.2

jG j H j

j

2 2 2

2 2 2

(100 ) (0.2 )1 1( ) ( )

2 2(100 ) ( 0.2 )G j H j

As magnitude has a constant value, hence it represents an all pass filter

1 12 2

0.2 0.2( ) ( ) tan tan

100 100G j H j

12

0.2( ) ( ) 2 tan

100G j H j

( ) ( )G j H j 12

0.2( ) ( ) 2 tan

100G j H j

0 1

2

00

100 1

2

0180

1

2

0 00 360

Hence, Nyquist plot can be drawn as

From the Nyquist plot shown, we can see that critical point ( 1 0)j lies outside the closed contour.

Hence, number of encirclement of ( 1 0)j point 0N

( 1 0)j 1

2 1

2

Im

Re



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35

Question 44

Consider the following equations,

2 2( , )2

V x ypx y xy

x

, 2 2( , )

2V x y

x qy xyy

Where p and q are constants, ( , )V x y that satisfies the above equation is

(A) 3 3

2 63 3

x yp q xy (B)

3 3

53 3

x yp q

(C) 3 3

2 2

3 3

x yp q x y xy xy (D)

3 32 2

3 3

x yp q x y xy

Ans. (D)

Sol. Given : 2 2( , )2

V x ypx y xy

x

…. (i)

2 2( , )2

V x yx qy xy

y

….(ii)

By integrating equation (i) w.r.t. x

3

2 2( , ) ( )3

pxV x y xy x y c y ….(iii)

Differentiating equation (iii), w.r.t. y

2( , )2

v x y cxy x

y y

….(iv)

From equation (ii) and (iv),

2 2 22 2c

xy x x qy xyy

2cqy

y

3

3

qyc

Thus, 3 3

2 2( , )3 3

x yV x y p q x y xy

Question 45

Unit step response of a linear time invariant (LTI) system is given by 2( ) (1 ) ( ).ty t e u t Assuming zero

initial condition, the transfer function of the system is

(A) 1

1s (B)

2

( 1)( 2)s s (C)

1

2s (D)

2

2s

Ans. (D)


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Sol. Given : Unit step response of a linear time invariant (LI) system is,

2( ) (1 ) ( )ty t e u t

Taking Laplace transform,

1 1

( )2

Y ss s

2

( 2)s s

The transfer function of the system is given by,

2( ) ( 2)

( )1( )

Y s s sH s

X ss

2

2s

Question 46

A Michelson interferometer using a laser source of wavelength 0 500nm, with both the mirrors

1 2( & )M M fixed and positioned equidistant from the splitter/combiner is shown in the figure. When a

dielectric plate of refractive index 1.5,n of thickness ,t is placed in front of the mirror 2,M a dark

fringe is observed on the detector. When the dielectric plate is removed without changing the position of

the mirrors 1 2&M M , a bright fringe is observed on the detector. The minimum thickness t (in mm) of

the dielectric is ________.

Ans. (250)

Sol. Given figure,

Wavelength 0( ) 500nm

Refractive index ( ) 1.5n

Laser

Detector

0 500 nm

n t

M

1M

Splitter-Combiner

Laser

Detector

0 500 nm

n t

M

1M

Splitter-Combiner



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37

Path difference 0

2

02( 1)2

t

Where, t thickness of dielectric.

9500 10

2 [1.5 1]2

t

250nmt

Question 47

Let * ,y n x n h n where * denotes convolution and x n and h n are two discrete time sequence.

Given that the z-transform of y n is 1 2( ) 2 3 ,Y z z z the z-transformer of * 2p n x n h n

is

(A) 22 3z z (B) 23z z (C) 22 3 1z z (D) 2 3 42 3z z z Ans. (D)

Sol. Given :

(i) 1 2( ) 2 3 ,Y z z z

(ii) [ ] [ ]* [ 2]P n x n h n

Apply z-transform on both sides

2( ) ( ) ( )P z X z H z z ….(i)

Also, [ ] [ ]* [ ]y n x n h n

Apply z-transform, ( ) ( ) ( )Y z X z H z ….(ii) Using equation (i) and (ii),

2( ) ( )P z Y z z 1 2 2(2 3 )z z z 2 3 42 3z z z

Question 48

Assuming ideal opamp, the RMS voltage (in mV) in the output 0V only due to the 230 V, 50 Hz

interference is (upto one decimal place) _________.

Ans. (10.84)

50 k

0.3 pF

3 pF1.3 pF

10 k1VdcV

230 V, 50 Hz

–

+ 0V


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38



Sol. Given figure,

From figure, 0 VNI IV V

When we remove dcV , capacitance of 3 pF and resistance of 10 k does not play any role in the value of

0V . So, the reduced equivalent circuit can be given as shown below,

0 1

1

50(50k)

1 i iV V j C V

j C

12 30(rms) (rms)(2 50) 3 10 50 10 iV V 72 75 10 230V

0(rms)

345mV 10.84 mV

100V

Question 49

A 2-bit synchronous counter using two -J K flip flops is shown. The expressions for the inputs to the

-J K flip flops are also shown in the figure. The output sequence of the counter starting from 1 2 00Q Q

is

50 k

0.3 pF

3 pF1.3 pF

10 k

230 V, 50 Hz

–

+ 0V0 V

50 k0.3 pF

–

+ 0V

3 pF230 V,

50 Hz

iV

230 V,

50 Hz

iV

1 2Q Q J QSET

1 2Q Q K QCLR

1Q J QSET

K QCLR

2Q1 2Q Q

1 2Q Q

Clock



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39

(A) 00 11 10 01 00... (B) 00 01 10 11 00... (C) 00 01 11 10 00... (D) 00 10 11 01 00... Ans. (C)

Sol. Given :

1 1 2 1 1 2J K Q Q

2 1 2 2 1 2J K Q Q

Clock J K 1nQ

0 X X nQ

1 0 0 nQ

1 0 1 0 1 1 1

nQ

We can find the output sequence using the truth table of J-K flip flop as shown, if we can identify inputs to even flip-flop property. Present state Inputs Next state

1Q 2Q 1 1 2J Q Q 1 1 2K Q Q 2 1 2J Q Q 2 1 2K Q Q 1Q 2Q

0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 1 1 1 1 1 0 0 0 0 1 1 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 1 1 1 0 0

Hence, output sequence for 1 2Q Q is 00, 01, 11, 10,00

Question 50

The inductance of a coil is measured using the bridge shown in the figure. Balance ( 0)D is obtained

with 1 1 2 41nF, 2.2 M , 22.2 k , 10 k .C R R R The value of the inductance xL (in mH) is ___.

1 2Q Q J QSET

1 2Q Q K QCLR

1Q J QSET

K QCLR

2Q1 2Q Q

1 2Q Q

Clock

~sV

1R

4R

xR

xL

2R

1C

D


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40



Ans. (222)

Sol. Given figure,

1 1 2 41nF, 2.2 M , 22.2 k , 10 k .C R R R

Above bridge is the example of Maxwell inductance capacitance bridge.

For a Maxwell bridge the value of xL is given by,

2 4 1xL R R C

3 3 922.2 10 10 10 1 10xL 222 mH

Question 51

In the given circuit, superposition is applied. When 2V is set to 0 V, the current 2I is 6 A. When 1V is

set to 0 V, the current 1I is 6A. Current 3I (in A) when both sources are applied will be (up to two

decimal places) ______.

Ans. (1)

Sol. Given figure,

At 2 20, 6AV I

At 1 10, 6AV I

~sV

1R

4R

xR

xL

2R

1C

D

1V 2V

1I3 2 2I

6

3I

1V 2V

1I3 2 2I

6

3I



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41

Case 1 : When 2 0V , the figure becomes

Applying only 1V and taking 2 6AI as given condition,

1V Voltage drop in 2 6 2 12 Volt

Hence value of current 3I due to source 1V only

13

0 12' 2 A

6 6

VI

Case 2 : When 1 0V , the figure becomes

Now applying 2V only and taking 1 6I A

2V Voltage drop in 3 6 3 18V

Hence value of 3I due to source 2V only

23

0 18'' 3A

6 6

VI

Hence, using superposition theorem, value of 3I when both the sources are acting simultaneously

3 3 3' ''I I I 2 3 1 A

Question 52

The Boolean function ( , )F X Y realized by the given circuit is

(A) XY XY (B) XY XY (C) X Y (D) .X Y

Ans. (A)

3 2

6

1'I 2' 6AI

3'I

1V

1V

3 2

6

1 " 6AI

3 "I

2V

2V

F

X

Y


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42



Sol. Given figure,

F X XY Y XY X XY Y XY

( ) ( )F X X Y Y X Y XY XY

Question 53

A high Q coil having distributed (self) capacitance is tested with a Q-meter. First resonance at 61 10

rad/s is obtained with a capacitance of 990 pF. The second resonance at 62 2 10 rad/s is obtained with

a 240 pF capacitance. The value of the inductance (in mH) of the coil is (up to one decimal place) ________.

Ans. (1)

Sol. Given : 61 10 , 6

2 2 10 , 1 990 pFC , 2 240pFC

6

26

1

2 102

10n

The distributed capacitance is given by,

2

1 22 1d

C n CC

n

2

2

990 2 240

2 1dC

10pF

The inductance of the coil is given by,

21 1

1

( )coild

LC C

6 2 12

1

(10 ) (990 10) 10coilL

1mH

Question 54

The circuit given uses ideal opamps. The current I (in )A drawn from the source sv is (up to two decimal

places) ______.

F

X

Y

10.1k

20.0k

10kI1V

sV 20.0k

20.0k



Since 2004

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43

Ans. (1)

Sol. Given figure,

As we know that due to infinite input impedance ideal OP-AMP never draws any current.

Hence 1i and 1'i are equal and 2i and 2'i are equal.

From 1 1'i i , 1 0 0

10 20

V

k k

2V Volts

From 2 2'i i , 000

20 20

VV

k k

where 2V

0 2VV

Now the current I, that we have to evaluate is given as 1 3I i i

00

10 10.1s sV V V

Ik k

1 1 2

10 10.1k k

0.1mA 0.099 mAI 0.001mA 1 A

Question 55

Consider the linear system . 1 0

,0 2

x x

with initial condition

1(0) .

1x

The solution ( )x t for this

system is

(A) 2

2

1( )

10

t t

t

e tex t

e

(B)

2

10( )

10

t

t

ex t

e

(C) 2 2

2

1( )

10

t t

t

e t ex t

e

(D)

2

10( )

10

t

t

ex t

e

Ans. (D)

Sol. Given : . 1 0

,0 2

x x

with initial condition

1(0) .

1x

As the input is not given, hence we have to find the zero input response of the system which is given as the solution of homogeneous state equation i.e.,

X AX ( ) ( ) 0d

x t Ax tdt

10.1k

20.0k

10kI1VsV 20k

20.0k

0 V

3I

1i

0 V0 V

0 V

2i 0V

1'i

2'i

V


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44



The solution of this homogeneous equation takes a form, ( ) ( ) (0)x t t x ….(i)

Where, ( )t is state transition matrix and is given by,

1 1( ) [( ) ]t L sI A

0 1 0

[ ]0 0 2

ssI A

s

1 0

0 2

s

s

2 0

Adj[ ]0 1

ssI A

s

( 1)( 2) 0 ( 2)( 1)sI A s s s s

1 Adj[ ][ ]

sI AsI A

sI A

1

10

1[ ]1

02

ssI A

s

To find ( )t , taking inverse Laplace of 1[ ]sI A ,

1 1

2

0( ) [ ]

0

t

t

et L sI A

e

Thus, solution of ( )x t from equation (i), may be written as,

2

10( ) ( ) (0)

10

t

t

ex t t x

e



Since 2004

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45

Answer Key :

General Aptitude 1. D 2. A 3. C 4. C 5. B

6. A 7. C 8. C 9. A 10. D

Technical Section 1. A 2. D 3. A 4. D 5. A

6. A 7. 200 8. 96 9. 180 10. 200

11. 75 12. 10 13. C 14. A 15. 0.3

16. 1010 17. D 18. 1 19. 90 20. C

21. 300 22. C 23. B 24. C 25. B

26. 30 27. B 28. B 29. D 30. 5.025

31. 2.5 32. 9 33. C 34. 280.8 35. A

36. 17.69 37. C 38. 67.3 39. C 40. Marks to all

41. 352000 42. 0.5 43. 0 44. D 45. D

46. 250 47. D 48. 10.84 49. C 50. 222

51. 1 52. A 53. 1 54. 1 55. D

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GATE 2018 · 2018-02-19 · TM ® Since 2004 GATE 2018 Instrumentation Engineering (Afternoon Session - 04.02.2018) • GATE ACADEMY has taken due care in making questions and solutions