gate –2016 - career launchermedia.careerlauncher.com/gate-exams/gate actual pa… · ·...

GENERAL APTI TUDE (GA)(Q.1 – 5) : Car r y One M ar k Each

1. An apple costs Rs. 10. An onion costs Rs. 8.Select the most suitable sentence with respect togrammar and usage.(a) The pr ice of an apple is greater than an onion.(b) The pr ice of an apple is more than onion.(c) The pr ice of an apple is greater than that of

an onion.(d) Apples are more cost l ier than onions.

2. The Buddha said, “Holding on to anger is l ikegrasping a hot coal with the intent of throwing i tat someone else; you are the one who gets burnt .”Select the word below which is closest in meaningto the word under l ined above.(a) burning (b) ignit ing(c) clutching (d) fl inging

3. M has a son Q and a daughter R. He has no otherchildren. E is the mother of P and daughter-inlawof M. How is P related to M?(a) P is the son-in-law of M.(b) P is the grandchi ld of M.(c) P is the daughter-in-law of M.(d) P is the grandfather of M.

4. The number that least fi ts this set : (324, 441, 97and 64) is ________.(a) 324 (b) 441(c) 97 (d) 64

5. I t takes 10 s and 15 s, respect ively, for two t rainst r avel l i ng at di f fer en t const an t speeds t ocompletely pass a telegraph post . The length ofthe first t rain is 120 m and that of the secondtrain is 150 m. The magnitude of the differencei n t he speeds of t he t wo t r ai ns (i n m/s) i s____________.(a) 2.0(b) 10.0(c) 12.0(d) 22.0

(Q.6 – 10) : Car r y Tw o M ar k s Each6. The velocity V of a vehicle along a st raight l ine is

measured in m/s and plot t ed as shown wi threspect to t ime in seconds. At t he end of the7 seconds, how much wil l the odometer readingincrease by (in m) ?

(a) 0 (b) 3(c) 4 (d) 5

No. of Quest i ons : 65 Maxi mum Mar k s : 100

GATE – 2016 EC : ELECTRONICS

& COMMUNICATION ENGINEERINGSet - 3

I NSTRUCTI ONS1. Total of 65 quest ions car rying 100 marks, out of which 10 quest ions car rying a total of 15 marks are in

General Apt i tude (GA)2. The Engineer ing Mathemat ics wi l l car ry around 15% of t he t ot al mar k s, the General Apt i tude sect ion

wil l car ry 15% of t he t ot al mar k s and the r emai n i ng 70% of t he t ot al mar k s.3. Types of Quest i ons

(a ) M u l t i p le Choi ce Quest i ons (M CQ) car rying 1 or 2 marks each in al l papers and sect ions. Thesequest ions are object ive in nature, and each will have a choice of four opt ions, out of which the candidatehas to mark the cor rect answer(s).

(b) Numer i cal Answ er Quest i ons of 1 or 2 marks each in al l papers and sect ions. For these quest ionsthe answer is a real number, to be entered by the candidate using the vir tual keypad. No choices wil lbe shown for these type of quest ions.

4. For 1-mar k mult iple-choice quest ions, 1/3 mar k s wil l be deducted for a wrong answer. L ikewise, for 2-mar k s mult iple-choice quest ions, 2/3 marks wil l be deducted for a wrong answer. There is no negat ivemarking for numer ical answer type quest ions.

2 SOLVED PAPER 2016 (SET-3)

7. The overwhelming number of people infectedwith rabies in India has been flagged by the WorldHealth Organizat ion as a source of concern. I t isest imated that inoculat ing 70% of pets and st raydogs against r abies can lead to a signi f icantreduct ion in the number of people infected withrabies.Which of the fol lowing can be logical ly infer redfrom the above sentences ?(a) The number of people in I ndia infected with

rabies is high(b) The number of people in other par ts of the

wor ld who are infected with rabies is low(c) Rabi es can be er adi cat ed i n I ndi a by

vaccinat ing 70% of st ray dogs(d) St ray dogs are the main source of r abies

wor ldwide8. A flat is shared by four fi rst year undergraduate

students. They agreed to al low the oldest of themto enjoy some extra space in the flat . Manu istwo months older than Sravan, who is t hreemonths younger t han Tr ideep. Pavan is onemonth older than Sravan. Who should occupy theextra space in the flat ?(a) Manu (b) Sravan(c) Tr ideep (d) Pavan

9. Find the area bounded by the l ines 3x + 2y = 14,2x – 3y = 5 in the first quadrant(a) 14.95 (b) 15.25(c) 15.70 (d) 20.35

10. A st raight l ine is fi t to a data set (ln x, y). Thisl ine intercepts the abscissa at ln x = 0.1 and hasa slope of – 0.02. What is the value of y at x = 5from the fi t?(a) – 0.030 (b) – 0.014(c) 0.014 (d) 0.030

TECH NI CAL SECTI ON(Q.1 – 25) : Car r y One M ar k Each

1. Consider a 2 2 square matr ix

A = ,x

where x is unknown. I f the eigenvalues of thematr ix A are ( + j) and ( – j), then x is equalto(a) + j (b) – j(c) + (d) –

2. For f(z) = 2

sin( ),

zz

the residue of the pole at z = 0

is _______.

3. The probabil ity of get t ing a "head" in a single tossof a bi ased coi n i s 0.3. The coi n i s t ossedrepeatedly t i l l a "head" is obtained. I f the tossesare independent , then the probabi l i ty of get t ing"head" for the first t ime in the fi fth toss is _______

4. The integral 1

0 (1 )

dx

x is equal to_______

5. Consider the first order ini t ial value problemy = y + 2x – x2, y(0) = 1, (0 x < )

with exact solut ion y(x) = x2 + ex. For x = 0.1, thepercentage difference between the exact solut ionand the solut ion obtained using a single i terat ionof the second-order Runge-Kut ta method wi thstep-size h = 0.1 is _______.

6. Consider the signal x(t) = cos(6t) + sin(8t),where t is in seconds. The Nyquist sampling rate(in samples/second) for the signal y(t) = x(2t + 5)is(a) 8 (b) 12(c) 16 (d) 32

7. I f the signal x(t) = sin( ) sin( )

*t t

t t with * denot ing

the convolut ion operat ion, then x(t) is equal to

(a)sin( )t

t(b)

sin(2 )2

tt

(c)2sin( )t

t(d)

2sin( )t

t

8. A discrete-t ime signal x[n] = [n – 3] + 2 [n – 5]has z-t r ansfor m X(z). I f Y(z) = X(– z) i s t hez-t ransform of another signal y[n], then(a) y[n] = x[n] (b) y[n] = x[– n](c) y[n] = – x[n] (d) y[n] = – x[– n]

9. I n the RLC circuit shown in the figure, the inputvoltage is given by

vi(t) = 2 cos(200t) + 4 sin(500t).The output voltage vo(t) is

(a) cos(200t) + 2 sin(500t)(b) 2cos(200t) + 4 sin(500t)(c) sin(200t) + 2 cos(500t)(d) 2sin(200t) + 4 cos(500t)

SOLVED PAPER 2016 (SET-3) 3

10. The I -V character ist ics of three types of diodesat the room temperature, made of semiconductorsX, Y and Z, are shown in the figure. Assume thatthe diodes are uniformly doped and ident ical inal l respects except their mater ials. I f Egx, EgY andEgZ are the band gaps of X, Y and Z, respect ively,then

(a) EgX > EgY > EgZ

(b) Egx = EgY = Egz

(c) Egx < EgY < EgZ

(d) no relat ionship among these band gaps exists.11. The figure shows the band diagram of a Metal

Oxide Semiconductor (MOS). The sur face regionof this MOS is in

(a) inversion (b) accumulat ion(c) deplet ion (d) flat band

12. The figure shows the I -V character ist ics of a solarcel l i l luminated uni formly wi th solar l ight ofpower 100 mW/cm2. The solar cel l has an area of3 cm2 and a f i l l factor of 0.7. The maximumefficiency (in %) of the device is _______

13. The diodes D1 and D2 in the figure are ideal andthe capacitors are ident ical. The product RC isvery large compared to the t ime per iod of the acvol t age. Assumi ng t hat t he di odes do notbreakdown in the reverse bias, the output voltageVo (in volt ) at the steady state is _______

14. Consi der t he ci r cu i t shown i n t he f i gur e.Assuming VBE1 = VEB2 = 0.7 volt , the value of thedc voltage VC2 (in volt ) is _______

15. I n the astable mult ivibrator circuit shown in thefigure, the frequency of osci l lat ion (in kHz) at theoutput pin 3 is _______

16. I n an 8085 microprocessor, the contents of theaccumulator and the car ry flag are A7 (in hex)and 0, respect ively. I f the inst r uct ion RLC isexecuted, then the contents of the accumulator(in hex) and the car ry flag, respect ively, wi l l be(a) 4E and 0 (b) 4E and 1(c) 4F and 0 (d) 4F and 1

17. The logic funct ional i t y real ized by the ci r cui tshown below is

(a) OR

(b) XOR

(c) NAND

(d) AND18. The minimum number of 2-input NAND gates

required to implement a 2-input XOR gate is(a) 4 (b) 5(c) 6 (d) 7


19. The block diagram of a feedback control systemis shown in the figure. The overal l closed-loopgain G of the system is

(a)1 2

1 1

G GG

1 G H

(b)

1 2

1 2 1 1

G GG

1 G G G H

(c)1 2

1 2 1

G GG

1 G G H

(d)1 2

1 2 1 2 1

G GG

1 G G G G H

20. For me unity feedback control system shown inthe figure, the open-loop t ransfer funct ion G(s)is given as

G(s) = 2

.( 1)s s

The steady state er ror ess due to a unit step inputis

(a) 0(b) 0.5(c) 1.0(d)

21. For a superheterodyne receiver, the intermediatefrequency is 15 MHz and the local osci l latorfrequency is 3.5 GHz. I f the frequency of thereceived signal is greater than the local oscil latorfrequency, then the image frequency (in MHz)is_______.

22. An analog baseband signal, bandlimited to 100 Hz,is sampled at the Nyquist rate. The samples arequant ized into four message symbols that occurindependent ly with probabil it ies p1 = p4 = 0.125and p2 = p3. The informat ion rate (bits/sec) of themessage source is _______.

23. A binary baseband digital communicat ion systememploys the signal

p(t) =

1, 0 T

T

0, otherwise

SS

t

f or t r ansmi ssi on of bi t s. The gr aph i calrepresentat ion of the matched fi l ter output y(t )for this signal wi l l be

(a)

(b)

(c)

(d)

24. I f a r ight -handed ci rcular ly polar ized wave isincident normally on a plane per fect conductor,then the reflected wave wil l be(a) r ight -handed circular ly polar ized(b) left -handed circular ly polar ized(c) el l ipt ical ly polar ized with a t i l t angle of 45(d) hor izontal ly polar ized

25. Faraday's law of elect r omagnet ic induct ion ismathemat ical ly descr ibed by which one of thefol lowing equat ions?

(a) B 0

(b) VD

(c)B

Et

(d)D

H Et


(Q.26 – 55) : Car r y Tw o M ar k s Each26. The par t i cu lar solut i on of t he in i t i al value

problem given below is2

212 36

d y dyy

dxdx = 0 with y(0) = 3

and 0

36x

dydx

(a) (3 – 18x) e– 6x (b) (3 + 25x) e– 6x

(c) (3 + 20x) e– 6x (d) (3 – 12x) e– 6x

27. I f the vectors e1 = (1, 0, 2), e2 = (0, 1, 0) ande3 = (– 2, 0, 1) form an orthogonal basis of the three-

dimensional r eal space 3 , t hen t he vect or

u = (4, 3,-3) 3 can be expressed as

(a) 1 2 32 11

35 5

u e e e

(b) 1 2 32 11

35 5

u e e e

(c) 1 2 32 11

35 5

u e e e

(d) 1 2 32 11

35 5

u e e e

28. A t r iangle in t he xy-plane is bounded by thestraight l ines 2x = 3y, y = 0 and x = 3. The volumeabove the tr iangle and under the plane x + y + z = 6is _______.

29. The values of the integral 1

2 2

z

c

edz

j z along a

closed contour c in ant i-clockwise direct ion for(i ) the point z0 = 2 inside the contour c, and(i i ) the point z0 = 2 outside the contour c,respect ively, are(a) (i ) 2.72, (i i ) 0(b) (i ) 7.39, (i i ) 0(c) (i ) 0, (i i ) 2.72(d) (i ) 0, (i i ) 7.39

30. A signal 2

2cos cos( )3

t t

is the input to an

LTI system with the t ransfer funct ionH(s) = es + e– s.I f Ck denotes the k th coefficient in the exponent ialFour ier ser ies of the output signal , then C3 isequal to(a) 0(b) 1(c) 2(d) 3

31. The ROC (r egi on of conver gence) of t hez-t r ansfor m of a di scr et e-t i me si gnal i srepresented by the shaded region in the z-plane.I f the signal x| n| = ( 2.0)| n| , – < n < + , thenthe ROC of i ts z-t ransform is represented by

(a)

(b)

(c)

(d)

32. Assume that the circuit in the figure has reachedthe steady state before t ime t = 0 when the 3 resistor suddenly burns out , result ing in an opencir cui t . The cur rent i (t) (in ampere) at t = 0+

is________.


33. I n the figure shown, the cur rent i (in ampere)is________.

34. The z-parameter matr ix 11 12

21 22

z z

z z

for the two-

por t network shown is

(a)2 22 2

(b)2 22 2

(c)9 36 9

(d)9 36 9

35. A cont inuous-t ime speech signal xa(t) is sampledat a r at e of 8 k H z and t he sampl es ar esubsequent ly grouped in blocks, each of size N.The DFT of each block is to be computed in realt ime using the radix-2 decimat ion-in-frequencyFFT algor i thm. I f the processor per forms al loperat ions sequent ial ly, and takes 20 (s forcompu t i ng each compl ex mu l t i pl i cat i on(including mult ipl icat ions by 1 and -1) and thet i me r equ i r ed for addi t i on /subt r act i on i snegl i gi bl e, t hen t he maxi mum val ue of Nis_______.

36. The di r ect for m st r uct ur e of an FI R (f i ni t eimpulse response) fi l ter is shown in the figure.

The fi l ter can be used to approximate a(a) low-pass fi l ter (b) high-pass fi l ter(c) band-pass fi l ter (d) band-stop fi l ter

37. The injected excess electron concentrat ion profi lein the base region of an npn BJT, biased in theact ive region, is l inear, as shown in the figure. I fthe area of the emit ter-base junct ion is 0.001 cm2,n = 800 cm2/(V-s) in the base region and deplet ionlayer widths are negl igible, then the col lectorcurrent Ic (in mA) at room temperature is _______.(Given: thermal vol tage VT = 26 mV at roomtemperature, elect ronic charge q = 1.6 10– 19 C)

38. Figures I and I I show two MOS capacitors of unitarea. The capaci tor in Figure I has insulatormater ials X (of thickness t1 = 1 nm and dielect r icconstant 1 = 4) and Y (of thickness t2 = 3 nm anddielect r ic constant 2 = 20). The capaci tor inFi gur e I I has onl y i nsul at or mat er i al X oft hi ckness tEq. I f t he capaci t or s ar e of equalcapacitance, then the value of tEq (in nm) is_______

39. The I -V character ist ics of the zener diodes Dl andD2 are shown in Figure I . These diodes are usedin the circuit given in Figure I I . I f the supplyvoltage is var ied from 0 to 100 V, then breakdownoccurs in

(a) D1 only(b) D2 only(c) both D1 and D2(d) none of D1 and D2


40. For the circuit shown in the figure, R1 = R2 = R3 = 1, L = 1 H and C = 1 F. I f t he i npu tVin = cos(106t), then the overall voltage gain (Vout/Vin)of the circuit is _______.

41. I n the circuit shown in the figure, the channellength modulat ion of al l t ransistors is non-zero( 0). Also, al l t ransistors operate in saturat ionand have negl igible body effect . The ac smal lsignal voltage gain (VO/V in) of the circuit is

(a) 1 1 2 3( )m o o og r r r

(b) 1 1 33

1( )m o o

m

g r rg

(c) 1 1 2 33

1( ( ( ) )m o o o

m

g r r rg

(d) 1 1 3 23

1( ( ) )m o o o

m

g r r rg

42. I n the circuit shown in the figure, t ransistor M1i s i n sat ur at i on and has t r ansconduct ancegm = 0.01 siemens. I gnor ing internal parasit iccapacitances and assuming the channel lengthmodulat ion to be zero, the small signal inputpole frequency (in kHz) is _______.

43. Following is the K-map of a Boolean funct ion offive var iables P, Q, R, S and X. The minimum sum-of-product (SOP) expression for the funct ion is

(a) PQSX PQSX QRSX QRSX

(b) QSX QSX

(c) QSX + QSX

(d) QS + QS44. For the circuit shown in the figure, the delays of

NOR gates, mult iplexers and inver ters are 2 ns,1.5 ns and 1 ns, respect ively. I f al l the inputs P,Q, R, S and T are applied at the same t ime instant,the maximum propagat ion delay (in ns) of thecircuit is _______.

45. For the circuit shown in the figure, the delay ofthe bubbled NAND gate is 2 ns and that of thecounter is assumed to be zero.

I f the clock (Clk) frequency is 1 GHz, then thecounter behaves as a(a) mod-5 counter(b) mod-6 counter(c) mod-7 counter(d) mod-8 counter


46. The fi r st two rows in the Routh table for thecharacter ist ic equat ion of a cer tain closed-loopcontrol system are given as

The range of K for which the system is stable is(a) – 2.0 < K < 0.5(b) 0 < K < 0.5(c) 0 < K < (d) 0.5 < K <

47. A second-order l inear t ime-invar iant system isdescr ibed by the fol lowing state equat ions

1 1( ) 2 ( ) 3 ( )d

x t x t u tdt

2 2( ) ( ) ( )d

x t x t u tdt

where x1(t) and x2(t) are the two state var iablesand u (t ) denot es t he i nput . I f t he ou t pu tc(t) = x1(t), then the system is(a) control lable but not observable(b) observable but not control lable(c) both control lable and observable(d) neither control lable nor observable

48. The for war d-pat h t r ansfer funct ion and thefeedback-path t ransfer funct ion of a single loopnegat ive feedback control system are given as

G(s) = 2

K ( 2)2 2s

s s

and H(s) = 1,

respect ively. I f the var iable parameter K is realposit ive, then the locat ion of the breakaway pointon the root locus diagram of the system is _______.

49. A wide sense stat ionary r andom process X(t)passes through the LTI system shown in t hefigure. I f the autocorrelat ion funct ion of X(t) isRX(), then the autocorrelat ion funct ion RY() ofthe output Y(t) is equal to

(a) 2RX() + RX( – T0) + RX( + T0)(b) 2RX() – RX( – T0) – RX( + T0)(c) 2RX() + 2RX( – 270)(d) 2RX() – 2RX( – 270)

50. A voice-grade AWGN (addit ive white Gaussiannoise) telephone channel has a bandwidth of4.0 kH z and t wo-sided noise power spect ral

density 2

= 2.5 10– 5 Watt per Hz. I f information

at the rate of 52 kbps is to be t ransmit ted overthis channel with arbit rar i ly small bi t er ror rate,t hen the minimum bi t -energy Eb (in mJ/bi t )necessary is _______.

51. The bit er ror probabi l i ty of a memoryless binarysymmetr ic channel is 10– 5. I f 105 bi ts are sent overthis channel, then the probabi l i ty that not morethan one bit wi l l be in er ror is _______.

52. Consider an air-fi l led rectangular waveguide withdimensions a = 2.286 cm and b = 1.016 cm.At 10 GHz operat ing frequency, the value of thepr opagat i on const an t (per met er ) of t hecorresponding propagat ing mode is _______

53. Consider an air-fi l led rectangular waveguide withdimensions a = 2.286 cm and b = 1.016 cm. Theincreasing order of the cut -off frequencies fordifferent modes is(a) TE01 < TE10 < TE11 < TE20

(b) TE20 < TE11 < TE10 < TE01

(c) TE10 < TE20 < TE01 < TE11

(d) TE10 < TE11 < TE20 < TE01

54. A r adar operat ing at 5 GHz uses a commonantenna for t r ansmission and recept ion. Theantenna has a gain of 150 and is al igned formaximum direct ional radiat ion and recept ion toa target 1 km away having radar cross-sect ion of3 m2. I f i t t ransmits 100 kW, then the receivedpower (in W) is _______.

55. Consider the charge profi le shown in the figure.The r esul t ant pot ent ial di st r i but ion i s bestdescr ibed by

(a) (b)

(c) (d)


ANSWERSGENERAL APTI TUDE

1. (c) 2. (c) 3. (b) 4. (c) 5. (a) 6. (d) 7. (a) 8. (c) 9. (b) 10. (a)

TECH NI CAL SECTI ON

1. (d) 2. (1) 3. (0.07203) 4. (2) 5. (0.06) 6. (c)7. (a) 8. (c) 9. (b) 10. (c) 11. (a) 12. (21)

13. (0) 14. (0.5) 15. (5.65) 16. (d) 17. (d) 18. (a)19. (b) 20. (a) 21. (3485) 22. (362.255) 23. (c) 24. (b)25. (c) 26. (a) 27. (d) 28. (10) 29. (b) 30. (b)31. (d) 32. (– 1) 33. (– 1) 34. (a) 35. (8) 36. (c)37. (6.656) 38. (1.6) 39. (a) 40. (– 1) 41. (c) 42. (57.8745)43. (b) 44. (7) 45. (d) 46. (d) 47. (a) 48. (– 3.41)49. (b) 50. (31.503) 51. (0.735) 52. (158.07) 53. (c) 54. (0.0122)55. (c)

EXPL ANATI ONSGENERAL APTI TUDE

2. The meaning of under l ined word grasping meansclutching (or holding something t ight ly).

3.

P

M Q

E

R Daughter-in-law

Daughter

(+ or –)

(+ or –)(–) Female(+) Male

Son

GrandChild

Husband

Mother

(+)

So, from the above relat ion diagram it is clearthat P is the grandchi ld of M.

5. Here the speed of fi rst t rain (S1) = 120

12 m/s10

and the speed of second train (S2) = 150

10 m/s15

The magnitude of difference in the speeds oftwo t rains = (S1 – S2) = (12 – 10) m/s = 2 m/s

6. From the given figure.The odometer reading increases from star t ingpoint to end point .Odometer reading = Area of the given diagramArea of the velocity and t ime graph per second

1st sec t r iangle

= 1 1

1 12 2

2nd sec square= 1 1 = 1

3rd sec square + t r iangle

= 11 1 1 1

2 =

11

24th sec t r iangle

= 1

1 2 12

5th sec st raight l ine= 0

6th sec t r iangle

= 1 1

1 12 2

7th sec t r iangle

= 1 1

1 12 2

Total Odometer reading at 7 seconds

=

1 1 1 11 1 1 0 m

2 2 2 2

= 5 mSo, the odometer reading is increased by 5 m.

8. Manu age = sravan age + 2 monthsManu age = Tr ideep age – 3 monthsPavan age = Sravan’s age + 1 monthFrom the above statementTr ideep age > Man > Pavan > SravanHence, Tr ideep can occupy the ext ra space in theflat .

9.


A = 14

,03

B = [0, 7],

C = 5

,02

D = 50,

3

E = [4, 1],F = [0, 1]

Requ i r ed ar ea = Ar ea of BFE + Ar ea ofquadr i lateral FEOC

= 1 1

4 6 (4 2.5) 12 2

= 12 6 6.5

2

= 12 + 3.25= 15.25 sq. units

So, t he ar ea bounded by t he gi ven l i nes i s15.25 sq. units.

10. The equat ion of a st raight l ine isy = mx + c

m = slope = – 0.02set (log x, y)Now, the l ine intercepts the abscissa at

log x = 0.1So, at abscissa y = 0 y = mx + c

0 = – 0.02 0.1 + c c = 0.002

y = mx + cy = – 0.02 1og x + c

At x = 5 y = – 0.02 log 5 + 0.002 = – 0.030So, the value of y is – 0.030.

TECH NI CAL SECTI ON

1. From the given square matr ix det (A) = 2 – x

( + j) ( – j) = 2 – x()2 – (j)2 = 2 – x

2 – (j2) (2) = 2 – x2 + 2 = 2 – x (since j2 = – 1)

2 = – x x = –

2. Given f(z) = 2

sin z

z

= 3 5

2

1...........

3! 5!z z

zz

= 31

............3! 5!z z

z

Residue of f(z)= 1 (at z = 0)

3. The probabil ity of gett ing ‘‘head’’ for the first t imein fi fth loss

(P) = (0.7)4 (0.3)= 0.07203

4. Let I = 1

01

dx

x

= 1

02 1 x

= – 2 [(0) – 1] I = 2

5. Given the first order ini t ial value problem is

dydx

= 22y x x

y(0) = 1

0 x < Given f(x, y) = y + 2x – x2,

xo = 0,

yo = 1,

h = 0.1

k1 = hf (xo, yo)

= 0.1 (1 + 2(0) – 02)

= 0.1

k2 = hg (x0 + h, y0 + k1)

= 0.1 ((y0 + k1) + 2(x0 + h) – (x0 + h)2)

(since h = 0.1)

= 0.1 ((1 + 0.1) + 2(0.1) – (0.1)2)

= 0.1 (1.1 + 0.2 – 0.01)

= 0.129

y1 = 0 1 212

y k k

= 11 0.1 0.129

2

(since k1 = 0.1 and k2 = 0.129)

= 1 + 0.1145 = 1.1145

For exact solut ion,

y(x) = x2 + ex

y(0.1) = (0.1)2 + e0.1

= 0.01 + 1.1050= 1.1152

error = 1.1152 – 1.1145= 0.00069

Relat ive er ror = error 0.00069

( ) 1.1152y x

= 0.00062

Percentage Error = (0.00062 100)% = 0.06%


6. Given signal

x(t) = cos (6t) + sin (8t)

Where t is in seconds

and y(t) = x(2t + 5)

y(t) = cos (12 t + 30) + sin (16 t + 40)

fm1= 6

fm2= 8

fm = 8 Hz

(fS)min = 2fm

= 16 Hz

Thus the Nyquist sampling rate is 16 Hz.

7. H er e t he convol ut i on of t wo si nc pul ses i ssinc pulse.

So, x1(t) = sin t

tNow x(t) = x1(t) * x1(t)

X() = X1() · X1()= X1()

x(t) = x1(t) = sin t

t

So, the value of x(t) is sin t

t8. Here (a)nx(n) X(z/a)

a = – 1(– 1)n x(n) X(– z)

but x(n) = [n – 3] + 2[n – 5]y(n) = (– 1)n x (n)

= (– 1)n [(n – 3) + 2[n – 5] y(n) = – (n – 3) – 2 (n – 5)

= – x(n)Hence, the value of signal y(n) is – x(n).

9. The input voltage is given byV i(t) = 2cos 200t + 4 sin 500t

Let us apply Superposit ion theorem only consider2cos 200t, then circuit becomes

So, 0V ( )t = 2 cos 200 t

Now consider only 4 sin 500t, then circuit becomes

So, again 0V ( )n t = 4 sin 500 t

Final ly (according to superposit ion theorem)

V0(t) = 0 0V ( ) V ( )nt t

V0(t) = 2cos (200t) + 4 sin (500 t)10. From the given fig :

Vr3> Vr2

> Vr1

Vr Eg

Where Vr is cut -in voltageSo, EgZ > EgY > EgX

12. Fil l factor = max max

T SC OC

P PP I V

where Pmax = Maximum powerPT = Theoret ical power

= I SC · VOC

0.7 = max3

P

180 10 0.5

Pmax = 63 10– 3 WNow Maximum efficiency

(max) = max

in

PP

= 3

3 22

63 10 W100 %

W100 10 3cm

cm

= 21%Thus the maximum efficiency (in %) of the deviceis 21%.


13.

Diodes are ideal therefore dur ing Posit ive cycleof output voltage

(V0) = 10 – 10 = 0 VDur ing Negat ive cycle, the diodes are Reversebiased so output

(V0) = 0 V V0 = 0 V (always)So, the output voltage V0 (in volt ) at steady stateis 0 V.

14. Here VE1= (2.5 – 0.7)= 1.8 V

VB2= VE1 – VEB2

= (1.8 – 0.7)= 1.1 V

I B2=

2BV 1

10k

= 1.1 1 0.110 10k k

I C2 = 20.1

I 5010k

VC2 = I C2 (1K)

= 50(0.1)

(1 )10

kk

= 0.5 VHence, the value of the dc voltage VC2 is 0.5V.

15.

From the above figureT1 = 0.693 (RA + RB)C

= 0.693 (2.2k + 4.7k) 0.022 = 0.693 6.9 103 0.022 10– 6

= 0.1052 m secT2 = 0.693 RBC

= 0.693 (4.7 k) 0.022 = 0.693 4.7 103 0.022 10– 6

= 0.0716562 m sec

Total per iod T = T1 + T2

= 0.1768562 m secFrequency of osci l lat ions

(f) = 1T

= 5.65 kHz

16. Given

After execut ing inst ruct ion

17. From given fig. i t is a AND gate

Input Output

A B Y

0 0 0

0 1 0

1 0 0

1 1 1

18. The minimum no. of NAND gates required for2-input EX-OR gate = 4

19. From the given block diagram, we have

Y( )X( )

ss

= G(s) = 1 2

1 1 2 2

G G1 G H G G

20. Given G(s) = 2

and H( ) 1( 1)

ss s

Type-1 System, to the unit step input the ess = 0So, the steady state er ror ess due to a unit stepinput is 0.

21.

f I f = 15 MHzfLo = 3500 MHz

fs – fLo = f I f

fs = fLo + f I f

= 3515 MHzfsi = image frequency

= fs – 2 f I f

= 3515 – 2 15= 3485 MHz

So, the image frequency is 3485 MHz.22. Here the samples are quant ized into four message

symbols that occur independently with probabilitiesare given by


P1 = 0.125P4 = 0.125P2 = 0.375P3 = 0.375

H = 2 21 1

0.125log 0.125 log0.125 0.125

2 21 1

0.375 log 0.375 log0.375 0.375

= 2 2 28

0.125 log 8 0.125 log 8 0.375 log3

28

0.375 log3

I nformat ion rate = H 200= 1.811 200= 362.255

23. The graphical representation of the matched fi lteroutput y(t) for this signal is given by

24.

Here RHCP r ight handed circular ly polar ized LHCP Left -handed circular ly polar izedI f the wave is incident on per fect conductor thenreflect ion coefficient is given by

0

0

E1

Er

i

Er0= E i0

180I f incident wave is t ravel ing along + Z direct ionthen the reflected wave wil l be t ravel ing along– Z direct ion.Thus, the reflected wave is left hand circular lypolar ized (LHCP).

25. Faraday’s law of elect r omagnet ic induct ion ismathematically described by the equat ion is givenby

E

=

Bt

or

E

= Ht

26. The different ial equat ion is given by

2

212 36

d y dyy

dxdx = 0

D2 + 12D + 36 = 0(D + 6)2 = 0

D = – 6, – 6The solut ion is given by

y = 6 61 2C Cx xe xe ...(i )

y(0) = 3 3 = C1

C1 = 3

(i ) y = 6 623 Cx xe xe

dydx

= 6 6 6218 C 6x x xe xe e

0x

dydx

= – 18 + C2

– 36 = – 18 + C2

C2 = – 18 The solut ion of y = 3 e– 6x – 18 x e– 6x

(Since C1 = 3 & C2 = – 18)27. The equat ion is given by

u = x1e1 + x2e2 + x3e3 ...(i )Put t ing the value of u, e1, e2 & e3 in the givenequat ion (i ), we get

(4, 3, – 3) = x1 (1, 0, 2) + x2 (0, 1, 0)+ x3(– 2, 0, 1)

x1 – 2x3 = 4 (a)x2 = 3 (b)

2x1 + x3 = – 3 (c)On solving these equat ions, we get

x1 = 2,

5

x2 = 3,

x3 = 115

u = 1 2 32 11

35 5

e e e

28. Volume = zdx dy

=

23 3

0 0

6

x

x y

x y dx dy

= 10

2since 2 = 3 , = & 3

3x y y x x

x + y + z = 6

z = 6 – x – y


29. (i ) When the point z0 = 2 inside the contour c.

12 ( 2)

z

c

edz

j z = 1

2 (2)2

j fj

e2 = 7.39(i i ) When the point z0 = 2 outside the contour c.

12 ( 2)

z

c

edz

j z = 0 ( z = 2 l ies out side c)

30. The t ransfer funct ionH(ej) = ej + e– j = 2cos

Here x(t) = 22cos

3t

0 =23

H(j0) =2 1

2cos 2 13 2

y(t) =2

2cos 1803

t

x(t) = cos t0 =

H(j0) = 2 cos () = – 2y(t) = 2 cos(t + 180)

y(t) =2

2cos3

t

– 2 cos(t + )

1 =23

2 =

T1 = 3 T2 = 2T0 = 6

0 =0

2T 3

y(t) = 2cos(20t + ) – 2cos(30t + )

y(t) = 0 0(2 ) (2 )j t j te e 0 0(3 ) (3 )j t j te e

y(t) = 0 0(2 ) (2 )j t j te e 0 0(3 ) (3 )j t j te e

C3 = 1Thus the value of C3 is 1.

31. x(y) = 1

(2) ( ) ( 1)2

nn u n u n

ROC = ( 2) ( ½)z z

So, the ROC of z-tansform is nul l .

32. Circuit at t = 0– , the di rect ion of cur rent andcorrect component was not given.

Not t = 0+

So, i (0+) =4

1A4

The magni tude of he cur rent is 1 ampere(negat ive).

33.

From above fig.Nodal Analysis

(V 8) V (V 8) V0

1 1 1 1

4V = 16V = 4 Volts

Now from KCLi – 4 + 4 + 1 = 0

i = – 1AHence the cur rent (i ) in t he given ci r cui t i s– 1 Amp.

34. Given network is in lat t ice formWhere Za = 3

Zb = 0Zc = 0Zd = 6

But i t is not symmetr ical & balanced


So, Z11 = 2

1

1 I 0

V3 // 6 2

I

Z21 = 2

2

1 I 0

VI

KVL– V2 – 2I 1 + 0 = 0

V2 = – 2I 1

Z21 = – 2Also

Z22 = 1

2

2 I 0

V3 // 6 2

I

Z21 = 1

1

2 I 0

VI

KVL– V1 – 2I 2 = 0

V1 = – 2I 2

Z12 = – 2So, t he Z-parameter mat r ix for the two por tnetwork is given by

Z =2 2

2 2

35. The number of complex mult ipl icat ions required

for DIF-FFT = 2

Nlog N

2

2

Nlog N

2

(20 sec) = 125 sec

Solving the above equat ion, we get N = 836. From the given figure

y(n) = 5[x(n) – x(n – 2)]Y(ej) = 5[1 – e– 2j] X(ej)H(ej) = 5[1 – e– 2j]

Thus, i t is a Band pass fi l ter

37. We know that

Collector cur rent (I C) = AeDndndx

= TAe Vndndx

WhereA = Area for the emit ter-base funct ione = Elect ronic chargeVT = Thermal voltage

I C = 0.001 1.6 10– 19 800 0.026 4

4

10 00.5 10

I C = 6.656 mAHence the col lector cur rent (I C) is 6.656 mA.

38. We known that

Now C =Ad

Where C = CapacitanceA = Area of a capacitor = dielect r ic constant

Now C =1 2

1 2

C CC C

=9 9

12

9 9

4 201 10 3 10 8.8521 10

4 101 10 3 10

= 2.5 109 0

C =0r

eqt

teq =0 0

9 90 0

4

2.5 10 2.5 10r

= 1.6 10– 9 m= 1.6 nm

So, the value of thickness (teq) is 1.6 nm.

39.

From the above fig., i t is clear thatboth zener diodes are in reverse biased

VBZ1 = 80 V

VBZ2 = 70 V

D1 have l ist saturat ion cur rent .When we wil l vary the voltage above 80VD1 get breaks down & wil l replaced by 80V. &through i t ‘’ cur rent can flow through i t .But because of D2 we wil l take minimum currenti .e. net cur r ent equals t o r ever se saturat ioncurrent of D2 as we know.


The diode have least saturat ion wil l break downfirst and it wil l replaced by i ts break down voltageand t he net cur rent equal upt o ot her diodereverse saturat ion cur rent .

40.

From above fig.6 6

6

V 1 10 101 1

V 10x

in

ss ss

V0 = 06 6

V1V

V10 101

xx

ss

s

0V

Vin =

60

6

V V 10. 1

V V 10x

x in

s sss

overal l voltage gain 0

in

VV

= – 1

41. The ac small signal voltage gain

0VVin

= 1 01 02 032

1m

m

g r r rg

42. CM1 = 50PF [1 – AV]AV = – gm RD

= – 0.01 1AV = – 10

CMi = 50PF [1 + 10]= 0.55 10– 9 F= 0.55 nF

fp =1

2 R Ci mi

fp =1

2 5K 0.55mF

= 3 9

12 5 10 0.55 10

= 57.8745 kHzSo, t he smal l signal i nput pole fr equency is57.8745 kHz.

43. The minimum sum-of-product (SOP) expressionfor the funct ion

= Q.S.X Q.S.X

44. The maximum propagat ion delay of the circuit= tpd, NOR + tpd, mux + tpd, NOR + tpd,mux

= (2 + 1.5 + 2 + 1.5) ns= 7 ns

45. At 6th Clock pulseQ2Q1Q0 = 110

I t makes NAND gate output ‘0’ at 8 ns due to i tsdelay. At that t ime counter receives 7th, 8th Clockpulse and counts 111, 000. Hence, i t is mod– 8counter.

46.3

2

1

0

1 2 32 4

2 (2 3) 40 for stabi l i ty

24

s ks k

k ks

ks

4k2 + 6k – 4 > 04k2 + 8k – 2k – 4 > 04k(k + 2) – 2(k + 2) > 0(4k – 2) (k + 2) > 0(4k – 2) > 0, k + 2 > 0k > – 2, k > 0.50.5 < k <

So the range of k for which the system is stable is0.5 < k <

47. Given 1 1( ) 2 ( ) 3 ( )d

x t x t u tdt

1x = – 2x1 + 3U

and 2 2( ) ( ) ( )d

x t x t u tdt

2x = – x2 + U

c = x1


1

2

x

x

=

1

2

2 0 3U

0 1 1

x

x

[c] = 1

2

1 0x

x

Hence by applying Gilber t ’s test , the system iscontrol lable but not observable.

48. Given t ransfer funct ion

G(s) = 2

( 2),

( 2 2)k s

s s

H(s) = 1

At break away point dkds

= 0

2

22 2

d sds s s

= 0

2 2

2 2

( 2 2) ( 2) ( 2) ( 2 2)

( 2 2)

d ds s s s s s

ds dss s

2

2 2

1( 2 2) ( 2)(2 2)( 2 2)

s s s ss s

= 0

– s2 – 4s – 2 = 0 – 0.58, – 3.41Valid BAP is – 3.41

49. Y(t) = X(t) – X(t – To)Autocorrelat ion funct ion for o/p

= Ry () = E[y(t) Y(t + )]Ry () = E [(X(t) – X (t – To)] [X (t + )

– X (t + – To)]Ry () = E [(X(t) X (t + ) – X(t) X (t + – To)

– X (t – To) X(t + ) + X (t – To) X (t + – To)]Ry () = [Rx () – Rx ( – To) – Rx ( + To) + Rx ()]Ry () = 2 Rx () – Rx ( – To) – Rx ( + To)

So, the autocor rect ion funct ion of Ry() of theoutput Y(t) is 2Rx() – Rx( – To) – Rx( + To).

50. From given dataChannel t ransmission rate (C) = 52 kbpsChannel band width B = 4 kHz

2

= 2.5 10– 5

N = 4 103 2.5 10– 5 2

C = 2

SB log 1

N

S = 1638.2

Eb =S J / sec

R bits / secb

= 31.503

CB

= log2 (1 + S/N)

log2(1 + S/N) = CB

(1 + S/N) = 2C/B = 252/4 = 213 = 8192 S/N = 8191 S = 8191 N S = 8191 4 103 2.5 10– 5 2

= 819.1 2

Eb = 819.1 2

Rb

= 31.503

So the minimum bit energy (Eb) is 31.503 mJ/bit .51. P = 10– 5 N = 105

Given quest ion can be solved by two methods.M et hod 1 :Binomial : nCx p

x qn– x

P[x = 0] + P[x = 1] = 5 510 5 0 5 100C (10 ) (1 10 )

5 510 5 1 5 10 11C (10 ) (1 10 )

= (1) (1) 0.367 + 0.367 = 0.735M et hod 1 :

Poisson =!

xex

= np = 10– 5 105 = 1(since n = 105, p = 10– 5)

=1 1

1(1)1!

ee

2 e– 1 = 0.735So, the probabi l i ty that not more than one bitwi l l be in er ror is 0.735.

52. GivenAir fi l led rectangular waveguidea = 2.286 cmb = 1.016 cmf = 10 GHzAssume dominant mode (TE10) is propagat ing inthe waveguide,So cut-off frequency of TE10 mode is given by

fc(TE10) = 2ca

= 103 10

2 2.286

fc = 6.56 GHz


propagat ion constant is given by

= j

= 2

0 0 1 cfif

= 2

98

1 6.562 10 10 1

103 10j

= j158.07 m – 1

Therefore the value of propagat ion constant isgiven by

= 158.07 m – 1

53. Givena = 2.286 cmb = 1.016 cmair fi l led rectangular waveguide

11(TE )cf = 2 2

1 12c

a b (Since m = 1, n = 1)

=10

2 2

3 10 1 12 (2.216) 1.016

11(TE )cf = 16.15 GHz

01(TE )cf =103 10

2 2 1.016cb

= 14.76 GHz

20(TE )cf =103 10

2.286ca

= 13.12 GHz

10(TE )cf =103 10

2 2 2.286ca

= 6.56 GHz

I ncreasing order of the cut-off frequency isgiven by

TE10 < TE20 < TE01 < TE11

54. Givenfrequency, f = 5 GHz = 5 109 Hz

wave length, = 8

9

3 105 10

cf

= 0.06m

gain of antenna, G = 150Range of target , Rmax = 1 km = 103 m,radar cross-sect ion, = 3m2,t ransmit ted power, Pt = 100 kWRadar range equat ion is given by

Rmax =

12 4

2

2R

P G4 Since A G

4(4 ) P

t

e

G

The received power, PR is given by

PR = 3 2

3 3 4

100 10 150 150 (0.06) 3(4 ) (10 )

= 1.22 10– 8 = 0.0122 WHence, the received power is 0.0122 W.

55. Let us consider a = 1 and b = – 1For l ine (1) :Now the equat ion of the l ine to point (– 1, 0) and(0, – 1) is given by

y – 0 =1

( 1)1

t

y = – t – 1

1

( 1)t

t dt

=2( 1)

1 02

tt

For l ine (2) :The equat ion of the l ine to the point (0, – 1) and(1, 0) is given by

y – (– 1) =1

( 0)1

t

y + 1 = ty = t – 1

0

( 1)t

t dt =

2( 1)

0 12

tt

At t = 0 :

y =2( 1)

2t

y =12

At t = 0+ :

y =2( 1)

2t

y =12

gate –2016 - career launchermedia.careerlauncher.com/gate-exams/gate actual pa… · ·...

Documents