2015 -...

2015-1 SOLVED PAPER - 2015

GENERAL APTITUDE

QUESTION 1 TO 5 CARRY ONE MARK EACH

1. Operators W , , ® are defined by : a W b = a ba b

-+

;

a b = a ba b

+-

; a ® b = ab.

Find the value of (66 W 6) ® (66 6).(a) –2 (b) –1(c) 1 (d) 2

2. Choose the most appropriate word from the options givenbelow to complete the following sentence.The principal presented the chief guest with a ________,as token of appreciation.(a) momento (b) memento(c) momentum (d) moment

3. Choose the appropriate word/phrase, out of the fouroptions given below, to complete the following sentence:Frogs __________.(a) Croak (b) Roar(c) Hiss (d) Patter

4. Choose the word most similar in meaning to the givenword: (EDUCE)(a) Exert (b) Educate(c) Extract (d) Extend

5. If logx (5/7) = –1/3, then the value of x is(a) 343/125 (b) 125/343(c) –25/49 (d) –49/25

QUESTION 6 TO 10 CARRY TWO MARKS EACH

6. A cube of side 3 units is formed using a set of smallercubes of side 1 unit; Find the proportion of the numberof faces of the smaller cubes visible to those which areNOT visible.(a) 1:4 (b) 1:3(c) 1:2 (d) 2:3

7. The following question presents a sentence, part of whichis underlined. Beneath the sentence you find four ways ofphrasing the underlined part. Following the requirementsof the standard written English, select the answer thatproduces the most effective sentence. Tuberculosis,

2015SET - 1

Duration: 3 hrs Maximum Marks: 100

INSTRUCTIONS

1. There are a total of 65 questions carrying 100 marks.

2. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject specificGATE paper for 85 marks. Both these sections are compulsory.

3. The GA section consists of 10 questions. Question numbers 1 to 5 are of 1-mark each, while question numbers 6 to10 are of 2-marks each. The subject specific GATE paper section consists of 55 questions, out of which questionnumbers 11 to 35 are of 1-mark each, while question numbers 36 to 65 are of 2-marks each.

4. Questions are of Multiple Choice Question (MCQ) or Numerical Answer Type. A multiple choice question will havefour choices for the answer with only one correct choice. For numerical answer type questions, the answer is a numberand no choices will be given.

5. Questions not attempted will result in zero mark. Wrong answers for multiple choice questions will result in NEGATIVE

marks. For all 1 mark questions, 13

mark will be deducted for each wrong answer. For all 2 marks questions, 23

mark

will be deducted for each wrong answer.

6. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.

2015 -2 SOLVED PAPER - 2015

together with its effects, ranks one of the leading causesof death in India.(a) ranks as one of the leading causes of death.(b) rank as one of the leading causes of death.(c) has the rank of the leading causes of death.(d) are one of the leading causes of death.

8. Humpty Dumpty sits on a every day while having lunch.The wall sometimes breaks. A person sitting on the wallfalls if the wall breaks.Which one of the statements below is logically valid andcan be inferred from the above sentences?(a) Humpty Dumpty always falls while having lunch.(b) Humpty Dumpty does not fall sometimes while

having lunch.(c) Humpty Dumpty never falls during dinner.(d) When Humpty Dumpty does not sit on the wall, the

wall does not break.9. Find the missing Value:

6 5

4 7 1

9 2

5

?

24

3

7 2

3

3

4

18 1 21

10. Read the following paragraph and choose the correctstatement.Climate change has reduced human security and threatenedhuman well being. An ignored reality of human progressis that human security largely depends upon environmentalsecurity. But on the contrary, human progress seemscontradictory to environment security. To keep up both atthe required level is a challenge to be addressed by oneand all. One of the ways to curb the climate change maybe suitable scientific innovations, while the other may bethe Gandhian perspective on small scale progress withfocus on sustainability.(a) Human progress and security are positively associated

with environmental security.(b) Human progress is contradictory to environmental

security.(c) Human security is contradictory to environmental

security.(d) Human progress depends upon environmental

security.

TECHNICAL SECTION

QUESTION 11 TO 35 CARRY ONE MARK EACH

11. Suppose A & B are two independent events withprobabilities P(A) ¹ 0 and P(B) ¹ 0. Let A & B be theircomplements Which of the following statement is FALSE?(a) P(A Ç B) = P(A)P(B)(b) P(A/B) = P(A)(c) P(A È B) = P(A) + P(B)

(d) P (A B)Ç P( A ).P( B)12. A region of negative differential resistance is observed in

the current voltage characteristics of a silicon PN junctionif(a) both the P-region and the N-region are heavily doped(b) the N-region is heavily doped compared to the

P-region(c) the P-region is heavily doped compared to the

N-region(d) an intrinsic silicon region is inserted between the

P-region and the N-region13. Consider a four bit D to A converter. The analog value

corresponding to digital signals of values 0000 and 0001are 0 V and 0.0625 V respectively. The analog value (inVolts) corresponding to the digital signal 1111 is ________.

14. In the circuit shown, the switch SW is thrown fromposition A to position B at time t = 0. The energy (in mJ)taken from the 3V source to charge the 0.1 mF capacitorfrom 0V to 3V is

3V 120W SWB A

t = 0

0.1 Fm

(a) 0.3 (b) 0.45(c) 0.9 (d) 3

15. The result of the convolution x (–t) * d(–t – to) is(a) x(t+to) (b) x(t–to)(c) x(–t + to) (d) x(–t–to)

16. A function f(x) = 1 – x2 + x3 is defined in the closedinterval [–1,1]. The value of x, in the open interval (–1,1)for which the mean value theorem is satisfied, is

(a)12

- (b)13

-

(c)13 (d)

12


17. The electric field component of a plane wave traveling ina lossless dielectric medium is given by

y8 zE(z, t)a 2cos 10 t

2æ ö

-ç ÷è ø

ur$ V/m. The wavelength (in m)

for the wave is18. Negative feedback in a closed loop control system DOES

NOT(a) reduce the overall gain(b) reduce bandwidth(c) improve disturbance rejection(d) reduce sensitivity to parameter variation

19. For the circuit with ideal diodes shown in the figure, theshape of the output (Vout) for the given sine wave input(Vin) will be

+ +– –

V in V out0.5T T0

(a) 0.5T T0 (b) 0.5T T0

(c) 0.5T T0 (d) 0.5T T0

20. A silicon sample is uniformly doped with donor typeimpurities with a concentration of 1016/cm3. The electronand hole mobilities in the sample are 1200cm2/V-s and400cm2/V-s respectively. Assume complete ionization ofimpurities. The charge of an electron is 1.6 × 10–19 C. Theresistivity of the sample (in W-cm) is _______.

21. The waveform of a periodic signal x(t) is shown in thefigure.

x(t)3

0–4 –3–2

–1–3

1 2 34

t

A signal g(t) is defined by g(t) = t 12-æ ö

ç ÷è ø

. The average

power of g(t) is ______.

22. The polar plot of the transfer G(s) = 10(s 1)(s 10)

++ for

0 < w < ¥ will be in the(a) first quadrant (b) second quadrant(c) third quadrant (d) fourth quadrant

23. A unity negative feedback system has the open-loop

transfer function G (s) = K

s(s 1)(s 3)+ + . The value of

the gain K( > 0) at which the root locus crosses theimaginary axis is ______.

24. In the given circuit, the values of V1 and V2 respectivelyare

4W 4W

4W

V 5A2 2I V1

+ +– –

(a) 5V, 25V (b) 10V, 30V(c) 15 V, 35V (d) 0V, 20V

25. The value of p such that the vector

123

é ùê úê úê úë û

is an eigenvector

of the matrix

4 1 2p 2 114 4 10

é ùê úê úê ú-ë û

is –––––.

26. Consider a straight, infinitely long, current carryingconductor lying on the z – axis. Which one of thefollowing plots (in linear scale) qualitatively representsthe dependence of H f on r, where H f is the magnitudeof the azimuthal component of magnetic field outside theconductor and r is the radial distance from the conductor?

(a)Hf

r(b)

Hf

r

(c)Hf

r(d)

Hf

r27. In the network shown in the figure, all resistors are

identical with R = 300 W . The resistance Rab (in W ) ofthe network is __________.

RR

R RR R

RR

R

R

RR

R = 300W

28. In the circuit shown below, the Zener diode is ideal andthe Zener voltage is 6V. The output voltage Vo (in volts)is _______.

+

–V010 V

1kW

1kW


29. In an 8085 microprocessor, the shift registers which storethe result of an addition and the overflow bit are,respectively.(a) B & F (b) A & F(c) H & F (d) A & C

30. In the circuit, shown at resonance, the amplitude of thesinusoidal voltage (in Volts) across the capacitor is_____.

~+

–1 Fm

0.1mH4W

10 cos tw(Volts)

31. A sinusoidal signal of 2 kHz frequency is applied to adelta modulator. The sampling rate and stepsize D of thedelta modulator are 20,000 samples per second and 0.1 V,respectively. To prevent slope overload, the maximumamplitude of the sinusoidal signal (in Volts) is

(a)1

2p(b)

1p

(c)2p

(d) p

32. Consider the signal s(t) = m(t) cos (2 p fct) + µm (t) sin(2 p

fct) where µm (t) denotes the Hilbert transform of m(t) andband width of m(t) is very small compared to fc. The signals(t) is a.(a) High pass signal(b) Low pass signal(c) Band pass signal(d) Double side band suppressed carrier signal

33. Let Z = x + iy be a complex variable consider continuousintegration is performed along the unit circle in anticlockwise direction. Which one of the following statementsis NOT TRUE?

(a) The residue of 2z

z 1- at z = 1 is

12

(b)2

cz dz 0=òÑ

(c) c1 1 dz 1

2 i z=

p òÑ

(d) z (complex conjugate of z) is an analytical function34. A 16kb (=16,384 bit) memory array is designed as a square

with an aspect ratio of one (number of rows = number ofcolumns). The minimum number of address lines neededfor the row decoder is ______.

35. Consider the system of linear equations :x – 2y + 3z = –1x – 3y + 4z = 1 and–2x +4y – 3z = k,The value of k for which the system has infinitely manysolutions is _______.

QUESTION 36 TO 65 CARRY TWO MARKS EACH

36. Two sequences [a,b,c] and [A,B,C] are related as,

1 23 3

2 43 3

1 1 1A aB 1 bC c1

- -

- -

é ùé ù é ùê úê ú ê ú= w wê úê ú ê úê úê ú ê úë û ë ûw wê úë û Where

2j3

3 ep

w =

If another sequence [p, q, r] is derived as,

1 2 23 3 32 4 43 3 3

1 1 1 1 0 0p A / 3q 1 0 0 B/ 3r C / 31 0 0

é ù é ùé ù é ùê ú ê úê ú ê úê ú ê ú= w w wê ú ê úê ú ê úê ú ê úê ú ê úë û ë ûw w wë û ë û

,

then the relationship between the sequences [p, q, r] and[a, b, c] is(a) [p, q, r] = [b, a, c] (b) [p, q, r] = [b, c, a](c) [p, q, r] = [c, a, b] (d) [p, q, r] = [c, b, a]

37. The maximum area (in square units) of a rectangle whosevertices lie on the ellipse x2 + 4y2 = 1 is ______.

38. In the circuit shown, switch SW is closed at t = 0.Assuming zero initial conditions, the value of vc(t)(in Volts) at t = 1sec is _____.

V (t)c+

–

t = 0

2W

3W

10 V

SW

5/6 F

39. A 3 input majority gate is defined by the logic functionM(a,b,c) = ab + bc + ca. Which one of the following gatesis represented by the function

( ) ( ) )M M(a, b,c) , M a,b,c ,c ?

(a) 3-Input NAND gate (b) 3-Input XOR gate(c) 3-Input NOR gate (d) 3-Input XNOR gate

40. A vector P is given by x y y3 2 2 2p x ya x y a x yza= - -

r r r r

.

Which one of the following statements is TRUE?

(a) pr

is solenoidal, but not irrotational

(b) pr

is irrotational, but not solenoidal

(c) pr

is neither solenoidal nor irrotational

(d) pr

is both solenoidal and irrotational


41. A lead compensator network includes a parallel combinationof R and C in the feed-forward path. If the transfer

function of the compensator GC (S) s 2s 4

++

The value of RC

is _____________.42. A MOSFET in saturation has a drain current of 1 mA for

VDS = 0.5V. If the channel length modulation coefficient is0.05 V–1, the output resistance (in kW) of the MOSFET is_____

43. The solution of the differential equation2

2d y 2 dy

dtdt+ + y = 0 with y(0) = y’ (0) = 1 is

(a) (2 –t)et (b) (1+2t)e-t

(c) (2 +t)e-t (d) (1 – 2t)et

44. The input X to the Binary Symmetric Channel (BSC)shown in the figure is ‘1’ with probability 0.8. The cross-over probability is 1/7. If the received bit Y = 0, theconditional probability that ‘1’ was transmitted is _______.

X 6/7 Y0

1/7

16/7

1

1/7

0P[X = 0] = 0.2P[X = 1] = 0.8

45. For the discrete time system shown in the figure, the polesof the system transfer function are located at(a) 2, 3 (b) 1/2 , 3(c) 1/2, 1/3 (d) 2, 1/3

+

16

- 56

z–1

+x(n)

z–1

46. The damping ratio of a series RLC circuit can be expressedas

(a)2R C

2L(b) 2

2LR C

(c)R C2 L

(d)2 CR L

47. For the NMOSFET in the circuit shown, the thresholdvoltage is Vth, where Vth> 0. The source voltage VSS isvaried from 0 to VDD. Neglecting the channel lengthmodulation, the drain current ID as a function of VSS isrepresented by

vss

vDD

(a)

ID

V –VDD thvss (b)

ID

Vth

vss

(c)

ID

V –VDD th

vss (d)

ID

V –VDD th

vss

48. In the circuit shown, assume that the opamp is ideal. Thebridge output voltage V0 (in mV) for d = 0.05 is ____.

+–

IV

100W

100W

50W

250 (1- )d W

250 (1+ )d

250 (1+ )d W

250 (1- )d W

V0– +

49. The electric field intensity of a plane wave traveling in freespace is given by the following expression E(x,t) = ay 24 pcos( tw – k0x) (V/m). In this field, consider a square area10 cm × 10 cm on a plane x + y = 1. The total time-averagedpower (in mW) passing through the square area is_______.

50. In the given circuit, the maximum power (in Watts) thatcan be transferred to the load RL is ____.

j2W

2W

4 0 VrmsÐ ~ RL


51. In the system shown in Figure (a), m(t) is a low-passsignal with bandwidth W Hz. The frequency response ofthe band-pass filter H(f) is shown in Figure (b).If it isdesired that the output signal z(t) = 10x(t), the maximumvalue of W (in Hz) should be strictly less than ________

H(f)Band PassFilter

Amplifierx(t)=m(t) cos(2400 f)p y(t)= 10x(t) + x (t)2 z(t)

f(in Hz)–1700 –700 0 700 1700

1

(a)H(f)

(b)

H(f)

52. The longitudinal component of the magnetic field insidean air-filled rectangular waveguide made of a perfectelectric conductor is given by the following expressionHZ (x,y,z,t) = 0.1 cos(25 px) cos(30.3 yp )

cos(12 p × 109t – b z) (A/m)The cross-sectional dimensions of the waveguide aregiven as a = 0.08 m and b = 0.033 m. The mode ofpropagation inside the waveguide is(a) TM12 (b) TM21(c) TE21 (d) TE12

53. The Boolean expression F(X,Y, Z) = X Y Z + X YZ + XY

Z + XYZ converted into the canonical product of sum(POS) form is

(a) F = (X + Y + Z)(X + Y + Z )(X + Y + Z )

( X + Y + Z )

(b) F = (X + Y + Z)( X + Y + Z )( X + Y + Z)

( X + Y + Z )

(c) F = (X + Y + Z)( X + Y + Z )(X + Y + Z)

( X + Y + Z )

(d) F = (X + Y + Z )( X + Y + Z)( X + Y + Z)(X + Y + Z)

54. For a silicon diode with long P and N regions, the accepterand donor impurity concentrations are 1 × 1017 cm–3 and1× 1015 cm–3, respectively. The lifetimes of electrons in Pregion and holes in N region are both 100 sm . Theelectron and hole diffusion coefficients are 49 cm2/s and36 cm2/s, respectively. Assume kT/q = 26mV, the intrinsiccarr ier concentration is 1 × 1010 cm–3 andq = 1.6 × 10–19 C. When a forward voltage of 208 mV isapplied across the diode, the hole current density (in nA/cm2) injected from P region to N region is _________.

55. Which one of the following graphs describes the function

f (x) = x 2e (x x 1)- + + ?

(a)x

f(x)1

(b)x

f(x)1

(c)x

f(x)1

(d)x

f(x)1

56. The pole –zero diagram of a causal and stable discrete-time system is shown in the figure. The zero at the originhas multiplicity 4. The impulse response of the system ish[n]. If h[0] = 1, we can conclude.(a) h[n] is real for all n(b) h[n] is purely imaginary for all n(c) h[n] is real for only even n(d) h[n] is purely imaginary for odd n

0.5

0.5

0.5

0.5

Re(z)× ×

××

Im(z)

57. The transmitted signal in a GSM system is of 200 kHzbandwidth and 8 users share a common bandwidth usingTDMA. If at a given time 12 users are talking in a cell, thetotal bandwidth of the signal received by the base stationof the cell will be at least (in kHz) _______.

58. All the logic gates shown in the figure have a propagationdelay of 20 ns. Let A = C = 0 and B =1 until time t = 0.At t = 0, all the inputs flip (i.e, A = C = 1 and B = 0) andremain in that state. For t > 0, output Z = 1 for a duration(in ns) of

ABC

Z

59. The built in potential of an abrupt p-n junction is 0.75 V.If its junction capacitance (CJ) at a reverse bias (VR) of1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25V is_________.

60. In the circuit shown, I1 = 80 mA and I2 = 4mA. TransistorsT1 and T2 are identical. Assume that the thermal voltageVT is 26 mV at 27°C. At 50°C, the value of the voltageV12 = V1 – V2 (in mV) is _____.

Q2

VS

I2 I1

V1V2

T2 T1

Q1

– +V12


61. Consider a uniform plane wave with amplitude (E0) of10 V/m and 1.1 GHz frequency travelling in air, andincident normally on a dielectric medium with complexrelative permittivity ( re ) and permeability ( rm ) as shownin the figure.

Dielectricµ = 1 – j2r

er = 1 – j2

10 cm |E| = ?

Air =

120 h

pW

|E | = 10 V/mFreq. = 1.1 Ghz

0

The magnitude of the transmitted electric field component(in V/m) after it has travelled a distance of 10 cm insidedielectric region is ________ .

62. A plant transfer function is given as

G(s) = 1p

K 1Ks s(s 2)

æ öæ ö+ ç ÷ç ÷ +è øè ø, When the plant operates in

a unity feedback configuration, the condition for thestability of the closed loop system is

(a) 1p

KK 0

2> > (b) 2K1 > Kp > 0

(c) 2K1 < Kp (d) 2K1 > Kp63. The open loop transfer function of a plant in a unity

feedback configuration is given as

G(s) = 2K(s 4)

(s 8)(s 9)

æ ö+ç ÷ç ÷+ -è ø

.

The value of the gain K (> 0) for which the point–1 + j2 lies on the root locus is _____ .

64. The circuit shown in the figure has an ideal opamp. Theoscillation frequency and the condition to sustain theoscillations, respectively, are

R2

R1

Vout

R

C

2C

+_

2R

(a) 1CR

and R1 = R2 (b) 1CR

and R1 = 4R2

(c) 12CR

and R1 = R2 , (d) 12CR

and R1 = 4R2

65. A source emits bit 0 with probability 13

and bit 1 with

probability 23 . The emitted bits are communicated to the

receiver. The receiver decides for either 0 or 1 based onthe received value R. It is given that the conditionaldensity functions of R are as

fR÷ 0 (r) = 3 x 1, 1 x 5,

R 14 6otherwise, otherwise,

1 1, and f (r) ,0 0

- £ £ - £ £ì ìï ïï ï=í íï ïï ïî î

|

The minimum decision error probability is(a) 0 (b) 1/12(c) 1/9 (d) 1/6


GENERAL APTITUDE

1. (c) (66 W 6) ® (66 6)

66 – 6 66 666 6 66 – 6

+æ ö æ ö= ®ç ÷ ç ÷è ø è ø+

60 7272 60

= ®

60 72 172 60

= ´ =

2. (b) Memento means something that serves as a reminder.3. (a) Croak is a hoarse sound made by a frog.4. (c) Educe means to draw out.

5. (a) 5 1

log –7 3xæ ö =ç ÷è ø

1–3 5

7xÞ =

–3 –35 77 5

x æ ö æ öÞ = =ç ÷ ç ÷è ø è ø

343125

xÞ =

6. (c) Total number of small cube 27 271

= =

Total surface area of 27 cubes = 27 × 6 × 12 = 162Surface area of cubes having side 3 units and visible= 6 × 32 = 54Surface area of not visible cubes = 162 – 54 = 108Required proportion = 54 : 108 = 1 : 2

7. (a)8. (b) The wall sometimes breaks, so Humpty Dumpty fall

sometimes while having lunch.9. 3

6 + 4 = 10, 10 52=

(7 + 4) + (2 + 1) = 14, 147

2=

(1 + 9 + 2) + (1 + 2 + 1) = 16, 16 82=

(4 + 1) + (2 + 3) = 10, 105

2=

3 + 3 = 6, 63

2=

Missing value is 3.10. (d) Line 3 and 4 clearly tells about this statement.

TECHNICAL SECTION

11. (c) We know that A and B are independent then( ) ( ) ( )P A B P A P BÇ =

A and B are independent then P(A/B) = P(A) andP(B/A) = P (B).

Also if A and B are independent then A and B arealso independent

i.e. (A B) (A). (B)Ç =P P P\ (a),(b), (d) are correct(c) is falseSince ( ) ( ) ( ) – ( )È = + ÇP A B P A P B P A B

( ) ( ) – ( ) ( )= +P A P B P A P B12. (a) It is the case of tunnel diode. In tunnel diode both

the P-region and the N-region are heavily doped &shows a region of negative differential resistance isobserved in current voltage characteristics of asilicon PN junction.

13. 0.9375Analog output = [Resolution] × [Decimal equivalentof Binary]= (0.0625) × (15) = 0.9375V

14. (c) As the capacitor initially uncharged i.e. VC(0) = 0.When the switch is in position B then the capacitorwill be charged upto 3V for ¥ time.

– /( ) ( ) [ (0) – ( )] t= ¥ + ¥ tc c c cV t V V V e

– /3 – 3 te t=t = RC = 120 × 0.1 × 10–6 = 12m sec

ic = cdVC

dt

–6 – /(0.1 10 ) [3 – 3 ]t= ´ td edt

– / – /3 0.1 0.312 12

t t´= =t te e

Instantaneous power, p(t) = V(t) i(t)

– / – /0.3 0.9( ) 312 12

t tP t e et t= ´ =

– –

0 0

0.9 0.9 0.912 12 12

¥ ¥t t æ ö= = = = tç ÷è øò òt t

E Pdt e e dt

–612 10 0.912

´ ´=E

E = 0.9 mJ


15. (d) x(–t)*d(–t–t0) = x(–t)*d(t + t0)= x(–t (t + t0)) = x(–t – t0)

16. (b) f(x) = 1 – x2 + x3

By Mean value Theorem

( ) – ( )( )–

f b f af xb a

¢ = ....(1)

f ¢(x) = – 2x + 3x2 ....(2)f (–1) = –1 ....(3)f (1) = 1 ....(4)By putting value from eqn 2, 3, 4 in eqn 1

–2x + 3x2 = 1– (–1)1– (–1)

–2x + 3x2 = 13x2 –2x –1 = 0

x = 1, –13

So, –13 lies in closed interval [–1, 1]

17. 8.885

Given, 8ˆ( , ) .2 cos(10 – )2

=ur

yzE z t a t

Here, w = 108, b = 12

We know that

2pb =

t or

2pl =

b...(1)

Putting b = 12

in equation (1)

2 2 2 8.885 m12

pl = = p´ =

18. (b) In negative feedback, sensitivity to parameter variablewill be reduce. So, improve disturbance rejection isthere.Overall gain of negative feedback control systemgets reduce. Hence, bandwidth increases becauseproduct of gain and bandwidth will remains constanti.e. Gain × Bandwidth = constant

19. (c) Given circuit can be simplified as

+a

–b

D1

V1

D2

d

c

R–

+

+–

During positive half cycle both diodes D1 and D2 areforward biased. So output will be obtained in this

case but due to negative polarity negative half cyclewill be obtained.During negative half cycle both diodes D1 and D2 arereversed biased. So no output will be obtained.

20. 0.5208

1 1= =

s mN D nP

N q

16 –191

10 1.6 10 1200=

´ ´ ´= 0.5208 W–cm

21. 2The average power of g(t) is

22

233

(–3)1 0 1 13

3

´ + ´ + ´

3 3 23+

=

22. (a) G(s) = 0( 1)

10s

s1 +

+

0( 1)( )10

jG jj

1 w +w =

w +

2

210 1| ( ) |

100G j w +

w =w +

–1 –1( ) tan – tan10

G j wÐ w = w

| ( ) | ( )0 1 0

10 0

G j G jw w Ð w°

¥ °

jw

0w = 10 s

So, It is first quadrant.23. 12

For characteristics equation1 + G(s) H(s) = 0

1 0( 1)( 3)

+ =+ +

Ks s s

(s2 + s) (s + 3) + K = 0s3 + 3s2 + s2 + 3s + K = 0s3 + 4s2 + 3s + K = 0

2015 -10 SOLVED PAPER - 2015

3

2

1

0

1 3

412 – 0

2

s

s KKs

s KFor marginable stable root locus crosses the imaginaryaxis12 – 0

4K

= ; K = 12

24. (a)

4W 4W

4W

V 5A2 2I V1

+ +– –

I I

5 = I + I + 2I (By KCL)

4I = 5, 54

=I

154 54

= ´ =V V

V2 = 4 × 5 + V1 (By KVL)V2 = 25V

25. 17AX = l X

4 1 2 12 1 2 2

14 –4 10 3 3P

lé ù é ù é ùê ú ê ú ê ú= lê ú ê ú ê úê ú ê ú ê úlë û ë û ë û

4 2 64 3 2

14 – 8 30 3P

+ + lé ù é ùê ú ê ú+ + = lê ú ê úê ú ê ú+ lë û ë û

127 2

36 3P

lé ù é ùê ú ê ú+ = lê ú ê úê ú ê úlë û ë û

l = 12P + 7 = 2 × 12P = 17

26. (c) 2f =pr

IH

r is the distance from element

1Hrf µ

As r increases, value of Hf decreases. Hence, only(c) option is possible.

27. 100After parallel combination of adjacent branches, theequivalent circuit redraw as

R/2 R/2

R/2 R/2R RR R

R

Req

Remove the branches by bridge balance condition

R/2 R/2

R/2 R/2 R

R

Req

»

2 R2 R R R

Req

e

1 1 1 1 1R 2 2q R R R R

= + + +

e

1 1 2 2 1R 2q R

+ + +=

e

1 3R q R

= ; e

1 300 100R 3 3q

R= = =

28. 5

1 k W

1 k W+

–v010 V

1 k W

1 k W+

–v010 V

01 101 1

V ´=

+ (By Voltage Divider Rule)

V0 = 5 VIn this case zener diode is in cut off stateSo, V0 is 5V,

29. (b) In 8085 microprocessor, after performing the addition,result is stored in accumulator and in any carry(overflow bit) is generated, it updates the flags.


30. 25At resonance

5–3 –6

1 1 10 rad / sec0.1 10 1 10LC

w = = =´ ´ ´

5 –3 5 –410 0.1 10 10 10 10LX L= w = ´ ´ = ´ =

5 –61 1

1010 10

= = =w ´

CXC

Z = R + j (XL – XC) = 4 + j (10 – 10) = 4104

VIZ

= =

1010 254C CV X I V= = ´ =

31. (a) fs = 20,000 samples per secondD = 0.1Vfm = 2 KHzTo prevent slope overload, condition given below is

m ms

2 f ATD

= p

D.fs = 2p fm Am0.1 × 20,000 = 2p × 2 × 103 × Am

m1A

2=

p32. (c) Given signal s(t) is the equation for single side band

modulation (lower side band). Thus it is a Band passfilter.

f

s(f)

33. (d) From option (a)

Residues of 2 –1z

z at z = 1 1

lim( –1).( –1) ( 1)®

=+z

zzz z

= 1 1

1 1 2=

+ (True)

From Option (b)2

cz dz 2 i= pòÑ (Residue = 0)

= 2pi × 0 = 0 (True)From option (c)

1òÑc

dzz

, Residue = 0

1lim . 1®

=z

zz

c1 dz 2 i (Residue 1) 2 i 1z

= p ´ = = p ´òÑ

c1 dz 2 iz

= pòÑ1 1 1

2=

p òÑc

dzi z

(True)

Hence only option (d) is false

34. 7Generally the structure of a memory chip = Numberof row (M) × Number of column (N) = M × NThe number of address line required for row decoderis n where

M = 2n

Taking log both sidesn = log2M

According to questionM = N

\ M × N = M × M = M2 = 16KM2 = 24 × 1024M2 = 24 × 210 = 214

M = 27 = 128n = log2M = 2 2log 27 7 log 2=n = 7

35. 2

Augmented Matrix1 –2 3 : –1

[A:B] 1 –3 4 : 1–2 4 –6 :

é ùê ú= = ê úê úë ûk

2 2 1–R R R¬ ; 3 3 12R R R¬ +

1 –2 3 : –10 –1 1 : 20 0 0 : – 2

é ùê úê úê úë ûk

Infinite many is possible,if P(A : B) = P(A) = r < number of variablesÞ k – 2 = 0Þ k = 2

36. (c) 2 43 3

1 2 2 3 63 3 3 3 32 1 4 4 53 3 3 3 3

1 1 1 1 0 0 1

1 0 0 1

1 0 0 1

é ùw wé ù é ùê úê ú ê úê úw w w = w wê ú ê úê úê ú ê úê úw w w w wë û ë û ë û

Q4 3 1 1 63 3 3 3 3. ;w = w w = w w 3 0 5 2

3 3 3 31;= w = w = w = w

2 13 3

1 23 3

11 1 1 13

1

ABC

é ùw w é ùê ú ê úê ú ê úê ú ê úë ûw wë û

2 13 3

–1 –23 3

1 2 –2 –13 3 3 3

1 1 111 1 1 1 13

1 1

abc

é ù é ùw w é ùê ú ê ú ê ú= w wê ú ê ú ê úê ú ê ú ê úë ûw w w wë û ë û

2 1 –1 0 03 3 3 3 3 3

–1 –2 –2 –13 3 3 3

1 2 0 0 –1 13 3 3 3 3 3

1 1 11 1 1 1 1 13

1 1 1

abc

é ù+ w + w + w + w + w + w é ùê ú ê úê ú= + + + w + w + w + w ê úê ú ê úë ûê ú+ w + w + w + w + w + wë û

2015 -12 SOLVED PAPER - 2015

23

3

j

ep

\ w = ; 2 1 –1 –1 –23 3 3 3 3 3 –1w + w = w + w = w + w =

0 0 31 3 0 03

0 3 0

p a cq b ar c b

é ù é ù é ù é ùê ú ê ú ê ú ê ú= =ê ú ê ú ê ú ê úê ú ê ú ê ú ê úë û ë û ë û ë û

37. 1

B

A

–1

12

1–1

2

O

x2 + 4y2 = 12 2

2 21

11 ( )2

x y+ =

Maximum Area= 4 × Area of DAOB

= 21 14 1 1m2 2

´ ´ ´ =

38. 2.528Vc (0) = 0 (Given)

V(t) = –

[ (0) – ( )] ( )t

RCc cV Vc e V¥ + ¥

10 2( )3 2

´¥ =

+cV (By voltage Divider)

( ) 4¥ =cV V

–

( ) [0 – 4] 4= +t

RcV t e

RC = 6 5 15 6

´ = ; 3 2 6R3 2 5

´= =

+

V(t) = 4 –t1(1– e )

V(t) = 4 (1 – e–t)at t = 1V(1) = 4 (1 – e–1) = 2.528 V

39. (b) M(a,b,c) = ab + bc + caLet Y = ( , , ) = + +M a b c ab bc ca

( , , )X M a b c ab bc ca= = + +Z = cM (X, Y, Z) = XY + YZ + ZX

( ) ( )= + + + +XY ab bc ca ab bc ca

( ) ( . . )= + +ab bc ca ab bc ca

( )[( )( ) ( )]= + + + + +ab bc ca a b b c c a

( )[( . . )( )]= + + + + + +ab bc ca a b ac b bc c a

( )[(= + + + + + +ab bc ca abc ab ac ac b c

]+ + +b a b c a b c

( )[ ]ab bc ca abc ab bc ac= + + + + +

= +ab c abc

YZ = ( ).+ + =ab bc ca c abc

ZX = abcAs, M (X, Y, Z) = XY + YZ + ZX

M(X, Y, Z) = + + +a b c abc abc abc

= ( ) ( )c ab ab c ab ab+ + +

( ) ( )= Å + Åc a b c a b= Å Åa b cThis is a X–OR gate expression.

40. (a) 3 2 2 2– –=ur r r r

x y zP x ya x y a x yza

. .x y zP a a ax y z

æ ö¶ ¶ ¶Ñ = + +ç ÷¶ ¶ ¶è ø

ur r r r

( )3 2 2 2– –x y zx ya x y a x yzar r r

= 3x2 y –2x2y –x2y = 0Hence solenoidal because . 0Ñ =

ur

P

3 2 2 2

.

– –

¶ ¶ ¶Ñ =

¶ ¶ ¶

ur

x y za a a

Px y z

x y x y x yz

= ax (–x2z + 0) – ay (–2x y z) + az(–2x – y2 – x3) ¹ 0

0Ñ´ ¹ur

P , Hence P is solenoidal but not irrotational.41. 0.5

2( )4c

sG ss

+=

+Transfer function of lead compensator

(1 )( )1c

sG ss

a + t=

+ at.....(1)

2(1 )2( )

4(1 )4

c

s

G ss

+=

+


(1 )1 2( )2 (1 )

4

c

s

G ss

+=

+.....(2)

By comparing (1) & (2)

1 1,2 2

a = t =

1 0.52

RCt = = =

42. 20In case of Channel length modulationID = IDS (1 + l VDS)

–31 1

0.05 10o

DSr

I= =

l ´

ro = 20 kW

43. (b)2

22 0

d y dyy

dtdt+ + =

For C.F.D2 + 2D + 1 = 0D = –1, –1y(t) = (C1 + C2 t)e

–t

y(0) = 11 = C1y¢(t) = C2e

–t (C1 + C2 t) . e–t

y¢(0) = 1, 1 = C2 – C1, C2 = 2y(t) = (1 + 2t) . e–t

44. 0.4

P{X = 1/Y = 0} = { 0 / 1} { 1}

{ 0}P Y X P X

P Y= = =

=

1{ 0 / 1}7

P Y X= = =

P{X =1} = 0.8

6/7X Y

1/7

1/7

6/71 1

O O

6 1 2{ 0} 0.2 0.87 7 7

P Y = = ´ + ´ =

1 (0.8)7{ 1/ 0} 0.4

27

P Y Y= = = =

45. (c) H(Z) = 2

–2 2–1

( ) 15 1( ) 5 –1– 6 66 6

Y Z ZZ Z Z Z ZZ

= =++

2

1 1( – )( – )2 3

Z

Z Z=

So, poles are 12

and 13

46. (c) Quality factor(Q) given by

12

Q =e ....(1)

where e = Damping ratioQuality factor (Q) in RLC circuit given by

1 LQR C

= ....(2)

Comparing eqn.(1) and (2)

1 12

LR C

=e

2R C

Le =

47. (a) Here, VD = VG Þ VDS = VGSSo, VDs > VGS – VTHence, MOSFET is in saturationIn saturation,ID = K (VGS – VT)2 = K (VDD – VSS – VT)2

As, ss DV I ¯It means as value of Vss increases then the value ofID decreases and vice – versa

48. 250

+–

1V100W

100W

50W

250 (1- )d W250 (1+ )d W

250 (1- )d W

V0– +A

B

E F

C

D

VA = 1VVA = VB = 1 VSince, no current will flow in the Op-Amp

So, 100 501

50I I A= =

Sum of resistance in DEC branch = Sum of resistance

in DFC branch. So, total current 150 will equally

divide in branch DEC and DFC i.e. 1 1 12 50 100

æ ö =ç ÷è ø

2015 -14 SOLVED PAPER - 2015

IDEC = IDFC 1 A

100=

V0 = VF – VE = VFD – VED= (VF – VD) – (VE – VD) (1)

VFD = 250 (1 + d) × 1

100

VED = 250 (1 – d) × 1

100Putting value of VFD & VED in eqn (1)

0250(1 ) 250(1– ) 250 250 – 250 250–

100 100 100V + d d + d + d

= =

0500 5100

V d= = d = 5 × 0.05 = 0.25V

V0 = 0.25 × 103 mV = 250 mV49. 53.33

Given, E(x, t) = ay24p cos(wt – k0x)

Average power, Pavg = 201

2 xE

ah

21 (24 ) 7.532 120 x xa ap

= =p

Power (P) = .ò avgs

P d s

x + y = 1

2x ya a+

® unit vector normal to the surface

7.53 .2x

xs

aP a dy dz= ò

–2 –27.53 7.53 10 10 10 102 2S

dy dz= = ´ ´ ´ ´ò

P = 53.33 × 10–3w = 53.33mW

50. 1.67

j2W

I

~4 0ÐRL

For calculation of Thevenin’s resistance

j2

2 W

RthÞ2 2 42 || 22 2 2 2 45th

j jR jj

´= = =

+ Ð °

4 902 2 45thR Ð °

=Ð ° ; 2 45thR = Ð °

|Rth| = RL = 2 W

Maximum power transfer to RL

2max

2 2 45 2 1.67W2 1

Ð °= = ´ =

+ +LP I Rj

51. 350x(t) = m(t). cos(2400pt)spectrum of x(t)

–1200–w

X(f)

1200–w–1200 1200–1200+w 1200+w

y(t) = 10x(t) + x2(t)Let us draw the spectrum of having positivefrequenciesy(t) = 10 m(t) cos (2400 pt) + m2(t) cos2 (2400pt)

= 10 m(t) cos (2400 pt) + m2(t) 1 cos 4800

2t+ pé ù

ê úë û

= 10 m(t) cos(2400 pt) + 2 2( ) ( ) cos(4800 )2 2

m t m t t+ p

1200+W–2W –1200–W2W 2400–2W 2400+2W

1200

1200

2W < 7002400 – 2W > 1700W < 350

52. (c) 25 25 25 0.08 2m m aa

= Þ = = ´ =

30.3 30.3 30.3 0.033 1n n bb

= Þ = = ´ =

m = 2, n = 1Since Hz means TE mode, then TE21 mode becausem = 2 & n = 1.


53. (a) Given, F(X, Y, Z) = X Y Z X Y Z X Y Z X Y Z+ + +

This is a minterm expression where minterm is (2, 4,6, 7)So, max term is pm(0, 1, 3, 5)POS form or max term expression as

(X Y+ Z) (X Z) ( ) ( )Y X Y Z X Y Z+ + + + + + +

54. 28.617Using formula

v/vTn0. . (e 1)r -

¶ = pp

p

q DL

19 5 208/26

31.6 10 36 10 ( 1)

60 10

-

-

´ ´ ´ ´ -=

´e

= 28.617 A/cm2

55. (b)2 –( ) ( 1)

0 10.2 1.0150.4 1.04570.6 1.07560.8 1.09631.0 1.10361.2 1.09631.4 1.0751.6 1.0411.8 0.9982 0.947

= + + xx f x x x e

So, only (b) option is possible.

56. (c)4

2 2( )(( 0.5) (0.25)(( – 0.5) 0.25)

=+ + +

zH zz z

4 4

4 4 –40.25 (1 0.25 )z z

z z z= =

+ +

–4 –1(1 0.25 )z= +

By using Binomial Theorem1–0.25z–4 + 0.0625z–8– 0.015625z–12 + ...h(x) = d(n) – 0.25d(n – 4) + 0.0625 d(n – 8) + .....

57. 400For 8 users we need 200 kHz and for 16 users weneed 400 kHz. Here, 12 users are using so 400 kHz isrequired. Since 12 calls are taking so 200 kHz extrabandwidth is required. So total bandwidths is 400kHz.

58. 40

A

C

B

B

B

(with 0 delay)

(with 20ns delay)t = 20 n sec

t < 0 t = 0 t > 0

A

A

B

B

(without delay)

(20 n sec delay) t = 20 n sec t = 60 n sec

t = 20 n sec(without delay)

(20 n sec delay)(X-OR-ing) t = 40 n sec

C

t = 40 n sec(AB C)Z = Å

(AB C)Z = Å

So, Z =1 for 40 nanosecond duration.59. 2.5

1j

R OC

V Vµ

+

2

1

1

2

R O

R O

V VCC V V

+=

+

2

5 7.25 0.75 21.25 0.75 1

+= =

+CC2 = 2.5 pF

60. 83.52VT at 50° C

11600TTV =

T = 273 + 50 = 323 k

VT = 323 27.84

11600=

=BE

T

VV

C SI I eHere, VE = 0

2015 -16 SOLVED PAPER - 2015

=B

T

VV

C SI I e

1 1=BV V and 2 2=BV V

127.841 =

V

sI I e ; 2

27.842 =V

sI I e

1

2

27.841

227.84

=

V

VI eI

e;

1 2–27.84 27.8480

4

æ öç ÷è ø=

V V

e

1 2–27.8420

æ öç ÷è ø=

V V

e

1 20

–3 83.52

27.84= Þ =

V VV mV (Since, V0 = V1– V2)

61. 0.1Medium 1 Medium 2Air Dielectrich1 = 120pW mr = 1 – j2

er = 1 – j2E1 = 10 V/m h2 = 120 pWSince,

h1 = h2So, E1 = E2 = 10 V/mE3 ® Electric field in the dielectric after travelling10 cmE3 = E2 e

–yz

g = a + jb = ( )wm s + wej j

0 0 r rj ja + b = w m e m e

0 0 [1– 2]j j ja + b = w m e

0 0 0 02j= w m e + w m e

9

0 0 82 2 1.1 102 46.07

3 10´ p´ ´

a = w m e = =´

For, z = 10 cmWe have, E3 = 10 × e–10 × 10–2 × 46.07

= 10e–4.6 = 0.1 V/m

62. (a) Given, G(s) = 1 1( 2)p

KKs s s

æ öæ ö+ ç ÷ç ÷ +è ø è øFor characteristic equation1 + G(s) H(s) = 0

1 + 1 1( 2)p

KKs s s

æ öæ ö+ ç ÷ç ÷ +è ø è ø = 0

12

( )1 0

( 2)

++ =

+psK K

s ss2(s + 2) + sKp + K1 = 0s3 + 2s2 + sKp + K1 = 0

3

21

11

01

1

22

2-

p

p

s K

s KK K

s

s K

For stability of a closed loop systemK1 > 02Kp – K1 > 0

12p

KK > ; 1 0

2pK

K > >

63. 25.54For any point lies on the root locus must satisfy thecondition as.1 + G(s) or H(s) = 0or, G(s) H(s) = –1

2( 4) –1

( 8)( – 9)K s

s s+

=+

2(–1 2 4) –1

(–1 2 8)[(–1 2) – 9]+ +

=+ + +

K jj j

(3 2)–1

(7 2) (1– 4 – 4 – 9)+

=+

K jj j

(3 2)–1

(7 2) (–4 –12)+

=+

K jj j

9 4 149 4 16 144

+=

+ +

K

13 153 160

=K

K = 25.5464. (d) The given Op-Amp circuit is Wein bridge Oscillator

circuit.Operating frequency of Wein bridge’s oscillator isgiven as

1 2 1 2

1w =

R R C CHere, R1 = R, C1 = 2CR2 = 2R, C2 = C

1 122 2

w = =´ ´ ´ RCR R C C

Condition For Sustain Oscillation

3³F

i

RR

Here RF = R1Ri = R2

1

23³

RR

; 1

24=

RR

; R1 = 4R2

65. (b)

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