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

1. Based on t he gi ven st at ement s, sel ect t heappropr iate opt ion with respect to grammar andusage.St at emen t s :I . The height of Mr. X is 6 feet .I I . The height of Mr. Y is 5 feet .(a) Mr. X is longer than Mr. Y.(b) Mr. X is more elongated than Mr. Y(c) Mr. X is tal ler than Mr. Y(d) Mr. X is lengthier than Mr. Y

2. The students _____ the teacher on teachers’ dayfor twenty years of dedicated teaching.(a) faci l i tated(b) fel ici tated(c) fantasized(d) faci l l i tated

3. After I ndia’s cr icket wor ld cup victory in 1985,Shrotr ia who was playing both tennis and cr ickett i l l then, decided to concentrate only on cr icket .And the rest is history.What does the under l ined phrase mean in thiscontext?(a) history wi l l rest in peace(b) rest is recorded in history books(c) rest is wel l known(d) rest is archaic

4. Given (9 inches)1/2 = (0.25 yards)1/2, which one ofthe fol lowing statements is TRUE?(a) 3 inches = 0.5 yards(b) 9 inches = 1.5 yards(c) 9 inches = 0.25 yards(d) 81 inches = 0.0625 yards

5. S, M, E and F are working in shifts in a team tofinish a project . M works with twice the efficiencyof others but for half as many days as E worked.S and M have 6 hour shifts in a day whereas Eand F have 12 hours shifts. What is the rat io ofcont r ibut ion of M to cont r ibut ion of E in t heproject?(a) 1 : 1 (b) 1 : 2(c) 1 : 4 (d) 2 : 1

(Q.6 – 10) : Car r y Tw o M ar k s Each6. The Venn diagram shows the preference of the

student populat ion for leisure act ivi t ies.

Readbooks

Watch TV

1912

713

44

15

17

Play sports

From the data given, the number of students wholike to read books or play spor ts is ____.(a) 44 (b) 51(c) 79 (d) 108

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

GATE – 2016 EC : ELECTRONICS

& COMMUNICATION ENGINEERINGSet - 2

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-2)

7. Social science discipl ines were in existence in anamorphous form unt i l the colonial per iod whenthey were inst i tut ional ized. I n varying degrees,t hey wer e in t ended t o fur t her t he col oni alinterest . I n the t ime of global izat ion and theeconomic r ise of postcolonial countr ies l ike India,convent ional ways of knowledge product ion havebecome obsolete.Which of the fol lowing can be logical ly infer redfrom the above statements?I . Soci al sci ence di sci pl i nes have become

obsolete.I I . Social science discipl ines had a pre-colonial

or igin.I I I .Social science discipl ines always promote

colonial ism.IV. Social science must maintain discipl inar y

boundar ies.(a) I I only (b) I and I I I only(c) I I and IV only (d) I I I and IV only

8. Two and a quar ter hours back, when seen in amir ror, t he reflect ion of a wal l clock wi thoutnumber markings seemed to show 1 : 30. What isthe actual cur rent t ime shown by the clock?(a) 8:15 (b) 11:15(c) 12:15 (d) 12:45

9. M and N star t from the same locat ion. M t ravels10 km East and then 10 km North-East. N travels5 km South and then 4 km South-East . What isthe shor test distance (in km) between M and Nat the end of their t ravel?(a) 18.60 (b) 22.50(c) 20.61 (d) 25.00

10. A wire of length 340 mm is to be cut into twopar ts. One of the parts is to be made into a squareand the other into a rectangle where sides are inthe rat io of 1:2. What is the length of the side ofthe square (in mm) such that the combined areaof the square and the rectangle is a MINIMUM?(a) 30 (b) 40(c) 120 (d) 180

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

1. The value of x for which the matr ix

A = 3 2 49 7 13

6 4 9 x

has zero as an eigen value

is _____.

2. Consider the complex valued funct ionf(z) = 2z3 + b| z| 3 where z is a complex var iable.The value of b for which the funct ion f(z) i sanalyt ic is _____

3. As x var ies fr om – 1 t o +3, which one of t hefol lowing descr ibes the behaviour of the funct ionf(x) = x3 – 3x2 + 1?(a) f(x) increases monotonical ly.(b) f(x) increases, then decreases and increases

again.(c) f(x) decreases, then increases and decreases

again.(d) f(x) increases and then decreasesThen the gi ven funct ion f(x) increases, t hendecreases and increase again.

4. How many dist inct values of x satisfy the equat ionsin(x) = x/2, where x is in radians?(a) 1 (b) 2(c) 3 (d) 4 or more

5. Consider the t ime-varying vector

I = ˆ ˆ15cos( ) 5sin( )x t y t i n Car t esi ancoordinates, where > 0 is a constant .When the vector magnitude | I | is at its minimumvalue, the angle that I makes with the x axis(in degrees, such that 0 180) is _____.

6. I n the circuit shown below, Vs is constant voltagesource and I L is a constant cur rent load.

R

Vs

+

–I L

The val ue of I L t hat maxi mi zes t he powerabsorbed by the constant cur rent load is

(a)V4R

s (b)V2R

s

(c)VR

s (d)

7. The switch has been in posit ion 1 for a long t imeand abrupt ly changes to posit ion 2 at t = 0.

+

–

3 1 2 4 2

2VC

t = 0

210V 5 A0.1F

+–

I f t ime t is in seconds, the capacitor voltage VC

(in volts) for t > 0 is given by(a) 4 (1 – exp (– t/0.5) (b) 10 – 6 exp (– t/0.5)(c) 4 (1 – exp (– t/0.6) (d) 10 – 6 exp (– t/0.6)

SOLVED PAPER 2016 (SET-2) 3

8. The figure shows an RLC circuit with a sinusoidalcur rent source.

I msin t

I R I L I C

R L

10 10 F

C

10 mH

At resonance, the rat io L

R

I,

I i .e., the rat io of the

magnitudes of the inductor cur rent phasor andthe resistor cur rent phasor, is_____

9. The Z-parameter matr ix for the two-port networkshown is

2,

3 2

j j

j j

Where the entr ies are in .

Suppose Zb(j) = Rb + j

Za

Zc

Zb

1 2

1 2

Then the value of Rb (in ) equals _____

10. The energy of the signal x(t) = sin(4 )

is________.4

tt

11. The Ebers-Moll model of a BJT is val id(a) only in a act ive mode(b) only in act ive and saturat ion modes(c) only in act ive and cut-off modes(d) in act ive, saturat ion and cut-off modes

12. A long-channel NMOS t ransistor is biased in thel inear region with VDS = 50 mV and is used as aresistance. Which one of the fol lowing statementsis NOT correct?(a) I f t he devi ce wi dt h W i s i ncr eased, t he

resistance decreases.(b) I f t he t hr eshol d vol t age i s r educed, t he

resistance decreases.(c) I f t he devi ce l engt h L i s i ncr eased, t he

resistance increase.(d) I f VGS is increased, the resistance increases.

13. Assume t hat t he di ode i n t he f i gur es hasVon = 0.7 V, but is otherwise ideal.

+–

R1

2 k i 2

R26 k2 V

The magnitude of the cur rent i 2 (in mA) is equalto_____

14. Resistor R1 in the circuit below has been adjustedso that I 1 = 1 mA. The bipolar t ransistors Q1 andQ2 are per fect ly matched and have very highcurrent gain, so their base currents are negligible.The supply voltage VCC is 6 V. The thermal voltagekT/q is 26 mV.

VCC

R1 I 2

I 1 R2

Q1 Q2

The value of R2 (in ) for which I 2 = 100 Ais __________.

15. Which one of the fol lowing statements is cor rectabout an ac-coupled commonemit ter ampl i fieroperat ing in the mid-band region?(a) The device parasit ic capacitances behave l ike

open circuits, whereas coupl ing and bypasscapacitances behave l ike shor t circuits.

(b) The device parasi t ic capacitances, coupl ingcapacitances and bypass capacitances behavel ike open circuits.

(c) The device parasi t ic capacitances, coupl ingcapacitances and bypass capacitances behavel ike shor t circuits.

(d) The device parasit ic capacitances behave l ikeshor t circuits, whereas coupl ing and bypasscapacitances behave l ike open circuits.


16. Transistor geometr ies in a CMOS inver ter havebeen adjusted to meet the requirement for worstcase charge and discharge t imes for dr iving a loadcapacitor C. This design is to be conver ted to thatof a NOR circuit in the same technology, so thati ts worst case charge and discharge t imes whiledr i ving the same capaci t or ar e simi lar. Thechannel lengths of al l t ransistors are to be keptunchanged. Wh i ch one of t he fol l owi ngstatements is cor rect?

VDD

I n Out

I n 1

Out

I n 2

C C

VDD

(a) Wi dt hs of PM OS t r ansi st or s shou l d bedoubled, whi le widths of NMOS t ransistorsshould be halved.

(b) Wi dt hs of PM OS t r ansi st or s shou l d bedoubled, whi le widths of NMOS t ransistorsshould not be changed.

(c) Widths of PMOS transistors should be halved,while widths of NMOS t ransistors should notbe changed.

(d) Wi dt hs of PM OS t r ansi st or s shou l d beunchanged, while widths of NMOS transistorsshould be halved.

17. Assume that al l the digital gates in the circuitshown in t he f i gur e ar e i deal , t he r esi st orR = 10 k and the supply voltage is 5 V. The Dfl ip-flops D1, D2, D3, D4 and D5 are ini t ial ized withlogic values 0, 1, 0, 1 and 0, respect ively. The clockhas a 30% duty cycle.

D QD1

D QD2

D QD3

D QD4

D QD5

ClockR = 10 k

The average power dissipated (in mW) in theresistor R is _____.

18. A 4 :1 mult iplexer is to be used for generat ingthe output car ry of a ful l adder. A and B are thebits to be added while Cin is the input car ry andCout is the output car ry. A and B are to be used asthe select bits with A being the more significantselect bit .

I 0

I 1

I 2

I 3 S1 S0

A B

Cout4 : 1Mux

Which one of the fol lowing statements cor rect lydescr ibes the choice of signals to be connected tothe inputs I 0, I 1, I 2 and I 3 so that the output isCout?(a) I 0 = 0, I 1 = Cin, I 2 = Cin and I 3 = 1(b) I 0 = 1, I 1 = Cin, I 2 = Cin and I 3 = 1(c) I 0 = Cin, I 1 = 0, I 2 = 1 and I 3 = Cin

(d) I 0 = 0, I 1 = Cin, I 2 = 1 and I 3 = Cin

19. The response of the system G(s) = 2

( 1)( 3)s

s s

to the unit step input u(t) is y(t). The value of dydt

at t = 0+ is ______.20. The number and di r ect ion of enci r clement s

around the point – 1 + j0 in the complex plane by

the Nyquist plot of G(s) = 1

4 2ss

is

(a) zero(b) one, ant i-clockwise(c) one, clockwise(d) two, clockwise

21. A discrete memoryless source has an alphabet{a1, a2, a3, a4} with cor responding probabi l i t ies

1 1 1 1, , , .

2 4 8 8

The minimum requi r ed aver age

codeword length in bits to represent this sourcefor er ror-free reconstruct ion is________.

22. A speech signal is sampled at 8 kHz and encodedinto PCM format using 8 bits/sample. The PCMdata is t ransmit ted through a baseband channelvia 4-level PAM. The minimum bandwidth (inkHz) required for t ransmission is _____.


23. A uniform and constant magnet ic field ˆB Bzexists in the z direct ion in vacuum. A par t icle ofmass m with a small charge q is int roduced into

this region with an ini t ial velocity ˆ ˆ .z zv xv zv

Given that B, m, q, vx and vz are all nonzero, whichone of t he fol l owing descr ibes t he event ualt rajectory of the par t icle?(a) Hel ical mot ion in the z direct ion.(b) Circular mot ion in the xy plane.(c) L inear mot ion in the z direct ion.(d) L inear mot ion in the x direct ion.

24. L et t he el ect r i c f i el d vect or of a pl aneel ect r omagnet i c wave pr opagat i ng i n ahomogenous medi um be expr essed as

( )ˆE E ,j t zxx e where the propagat ion constant

is a funct ion of the angular frequency . Assumethat () and Ex are known and are real. Fromt he informat ion avai lable, which one of t hefol lowing CANNOT be determined?(a) The type of polar izat ion of the wave.(b) The group velocity of the wave.(c) The phase velocity of the wave.(d) The power flux through the z = 0 plane.

25. Light from free space is incident at an angle i tothe normal of the facet of a step-index large coreopt ical fibre. The core and cladding refract iveindices are n1 = 1.5 and n2 = 1.4, respect ively.

n2 (cladding)

n1 (core)

Freespace

L ight

i

The maximum value of i (in degrees) for whichthe incident l ight wi l l be guided in the core of thefibre is ________.

(Q.26 – 55) : Car r y Tw o M ar k s Each

26. The ordinary different ial equat ion

3 2, dy

ydx

with y(0) = 1

is to be solved using the forward Euler method.The largest t ime step that can be used to solvethe equat ion wi t hout mak ing t he numer icalsolut ion unstable is ________.

27. Suppose C is the closed curve defined as the circlex2 + y2 = 1 with C or iented ant i clockwise. The

value of 2 2( )xy dx x ydy over t he curve C

equals ________.

28. Two random var iables X and Y are dist r ibutedaccording to

fX,Y(x, y) = ( ) 0 1, 0 1

0 otherwisex y x y

The probabi l i ty P(X + Y 1) is ________

29. The matr ix A =

0 3 7

2 5 1 30 0 2 4

0 0 0

a

b

has det(A)=100 and

trace (A)=14. The value of | a – b| is ________.30. I n the given circuit , each resistor has a value

equal to 1 .

a

b

What is the equivalent r esistance across theterminals a and b?(a) 1/6 (b) 1/3 (c) 9/20 (d) 8/15

31. I n the circuit shown in the figure, the magnitudeof the cur rent (in amperes) through R2 is ___

+–

R1 R2

5

3

60 V

R3

Vx

+

–0.04Vx

5

32. A cont inuous-t ime fi l ter with t ransfer funct ion

H(s) = 2

2 6

6 8

s

s s

is converted to a discrete time filter

wi th t ransfer funct ion G(s) = 2

2

2 0.5032

0.5032

z z

z z k

so that the impulse response of the cont inuous-t ime fi l ter, sampled at 2 Hz, is ident ical at thesampling instants to the impulse response of thediscrete t ime fi l ter. The value of k is _____.


33. The Discrete Four ier Transform (DFT) of the4-point sequencex[n] = {x[0], x[1], x[2], x[3]} = {3, 2, 3, 4} isX[k] = {X[0], X[1], X[2], X[3]} = {12, 2j, 0, – 2j}.I f X1[k] i s t he DFT of t he 12-point sequencex1[n] = {3, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0}, the value of

1

1

X [8]X [11] is ________.

34. The switch S in the circuit shown has been closedfor a long t ime. I t is opened at t ime t = 0 andremains open after that . Assume that the diodehas zero reverse current and zero forward voltagedrop.

+–

t = 01

10 V 1 mH 10 F–

+VC

S

The steady state magni tude of t he capaci torvoltage VC (in volts) is ______.

35. A voltage VG is appl ied across a MOS capacitorwith metal gate and p-type si l icon substrate atT = 300 K . The inversion car r ier densi t y (innumber of car r iers per unit area) for VG = 0.8 V is2 1011 cm – 2. For VG = 1.3 V, the inversion car r ierdensity is 4 1011 cm – 2. What is the value of theinversion car r ier density for VG = 1.8 V?(a) 4.5 1011 cm – 2

(b) 6.0 1011 cm – 2

(c) 7.2 1011 cm – 2

(d) 8.4 1011 cm – 2

36. Consider avalanche breakdown in a si l icon p+njunct ion. The n-region is uniformly doped with adonor density ND. Assume that breakdown occurswhen the magnitude of the elect r ic field at anypoint in the device becomes equal to the cr i t icalfield Ecr i t. Assume Ecr i t to be independent of ND.I f the bui l t -in voltage of the p+n junct ion is muchsmal ler than the breakdown vol tage, VBR, therelat ionship between VBR and ND is given by

(a) BR DV N = constant

(b) D BRN V = constant

(c) ND VBR = constant(d) ND /VBR = constant

37. Consider a region of si l icon devoid of elect ronsand holes, wi t h an ionized donor densi t y of

17 3N 10 cm .d The electr ic field at x = 0 is 0 V/cm

and the elect r ic field at x = L is 50 kV/cm in theposit ive x direct ion. Assume that the electr ic fieldis zero in the y and z direct ions at al l points.

N = 10d+ 17 cm – 3

x = 0 x = L

Given q =1.6 10– 19 coulomb,0 = 8.85 10– 14 F/cm, r = 11.7 for si l icon, thevalue of L in nm is _____

38. Consider a long-channel NMOS t ransistor withsource and body connected together. Assume thatthe elect ron mobil i ty is independent of VGS andVDS. Given,gm = 0.5 A/V for VDS = 50 mV and VGS = 2 V,gd = 8 A/V for VGS = 2 V and VDS = 0 V,

where gm = D

GS

IV and gd =

D

DS

IV

The threshold voltage (in volts) of the t ransistoris ________.

39. The figure shows a hal f-wave rect i fier with a475 F fi lter capacitor. The load draws a constantcur rent I 0 =1 A from the rect i fier. The figure alsoshows the input voltage V i, the output voltage VC

and the peak-to-peak voltage r ipple u on VC. Theinput vol t age V i i s a t r i angle-wave wi t h anampli tude of 10 V and a per iod of 1 ms.

+–

+

–V i VC

I 0 = 1A475 F

+10 V V i

– 10 VVC

t

u

0

0

t

The value of the r ipple u (in volts) is ______


40. I n the opamp circuit shown, the Zener diodesZ1 and Z2 clamp the output voltage V0 to + 5 V or– 5 V. The switch S is ini t ially closed and is openedat t ime t = 0.

+

–

100 F+10 V

St = 0

470

– 10 V

V0

Z1

Z24 k

1 k– 10 V

10 k

0 V0 V

+10 V

The t ime t = t1 (in seconds) at which V0 changesstate is ________

41. An opamp has a fini te open loop voltage gain of100. I ts input offset voltage V ios (= + 5 mV) ismodeled as shown in t he ci r cui t below. Theampli fier is ideal in al l other respects. V input is25 mV.

+–

+–

V ios = 5 mV

V input

A0 = 100

1 k 15 k

The output voltage (in mil l ivolts) is ________42. An 8 Kbyte ROM with an act ive low Chip Select

i npu t (CS) i s t o be used i n an 8085microprocessor based system. The ROM shouldoccupy the address range 1000 H to 2 FFFH. Theaddress l ines are designated as A15 to A0, whereA15 is the most significant address bit .Which one of the fol lowing logic expressions wil lgenerate the cor rect CS signal for this ROM?

(a) 15 14 13 12 13 12A A (A . A A . A )

(b) 15 14 13 12A . A . (A A )

(c) 15 14 13 14 13 12A . A . (A A A . A )

(d) 15 14 13 12A A A . A

43. I n an N bit flash ADC, the analog voltage is fedsimultaneously to 2N – 1 comparators. The outputof the comparators is then encoded to a binaryformat using digi tal ci rcuits. Assume that theanalog voltage source V in (whose output is being

conver t ed t o di gi t al f or mat ) has a sour ceresistance of 75 as shown in the circuit diagrambel ow and t he i npu t capaci t ance of eachcomparator is 8 pF.The input must set t le to an accuracy of 1/2 LSBeven for a ful l scale input change for properconversion. Assume that the t ime taken by thethermometer to binary encoder is negl igible.

+

–

+

–+–

+

–V ref255

V ref2

V in

V ref1

Thermometercode tobinary

conversion

Digi talOutput

I f the flash ADC has 8 bit resolut ion, which oneof t he fol lowing al t ernat ives is closest t o themaximum sampling rate ?(a) 1 mega samples per second(b) 6 mega samples per second(c) 64 mega samples per second(d) 256 mega samples per second

44. The state t ransit ion diagram for a fini te statemachine with states A, B and C, and binary inputsX, Y and Z, is shown in the figure.Which one of the fol lowing statements is cor rect?

A B

C

X = 0, Y = 0, Z = 0

Y = 0, Z = 0

X = 1, Y = 1,Z = 1, Z = 1

X = 0, Z = 1

X = 1, Y = 0 Y = 0, Z = 1

Z = 0

Y = 1

Y = 1

(a) Transi t ions from State A are ambiguouslydefined.

(b) Transi t ions from State B are ambiguouslydefined.

(c) Transi t ions from State C are ambiguouslydefined.

(d) Al l of t he st at e t r ansi t i ons ar e def i nedunambiguously.


45. I n t he feedback syst em shown bel ow

G(s) = 2

1.

( 2 )s sThe st ep response of t he closed-loop syst emshould have minimum set t l ing t ime and have noovershoot .

K G( )s+

–

r y

The requi r ed value of gain k t o achieve thisis ________.

46. I n t he feedback syst em shown bel ow

G(s) = 1

( 1)( 2)( 3)s s s

K G( )s+

–

r y

The posit ive value of k for which the gain marginof the loop is exact ly 0 dB and the phase marginof the loop is exact ly zero degree is ________

47. The asymptot ic Bode phase plot of

G(s) = 1( 0.1)( 10)( )

ks s s p , with k and p1 both

posit ive, is shown below.

0.01 0.1 1 10 100 rad/s

0

– 45

– 135

– 225

– 270

The value of p1 is ________48. An i n for mat i on sour ce gener at es a bi nar y

sequence {n}. n can take one of the two possiblevalues – 1 and +1 with equal probabi l i ty and arest at i st i cal l y i ndependen t and i dent i cal l ydist r ibuted. This sequence is precoded to obtainanother sequence {n}, as n = n + kn– 3. Thesequence {n} is used to modulate a pulse g(t) togenerate the baseband signal

X(t) = ( T),nn

g t n

where g(t) = 1, 0 T

0, otherwise

t

I f there is a nul l at f = 1

3T in the power spectral

density of X(t), then k is ________49. An ideal band-pass channel 500 Hz- 2000 Hz is

depl oyed for commun i cat i on . A modem i sdesigned to transmit bits at the rate of 4800 bits/susing 16-QAM. The rol l-off factor of a pulse witha raised cosine spectrum that ut i l izes the ent irefrequency band is ________

50. Consider a random process X(t) = 3V(t) – 8, whereV(t ) is a zero mean stat ionary random processwith autocorrelat ion Rv() = 4e– 5| | . the power inX(t)is ________

51. A binary communicat ion system makes use of thesymbols “ zero” and “one” . There are channeler rors. Consider the fol lowing events:

x0 : a “zero” is t ransmit tedx1 : a “one” is t ransmit tedy0 : a “zero” is receivedy1 : a “one” is received

The fol lowing probabi l i t ies are given:

P(x0) = 1

,2

P(y0| x0) = 3

,4

and P(y0| x0) = 1

.2

The information in bits that you obtain when youlearn which symbol has been received (while youk now t hat a “ zer o” has been t r ansmi t t ed)is ________.

52. The paral lel-plate capacitor shown in the figurehas movable plates. The capacitor is charged sothat the energy stored in i t is E when the plateseparat ion is d. The capaci tor is then isolatedelect r ical ly and the plates are moved such thatthe plate separat ion becomes 2d.

d

At this new plate separat ion, what is the energyst or ed i n t he capaci t or, neglect ing fr i ngi ngeffects?

(a) 2E (b) 2E

(c) E (d) E/2


53. A lossless microst r ip t ransmission l ine consistsof a trace of width w. I t is drawn over a pract ical lyinfi ni t e ground plane and is separ at ed by adi el ect r i c sl ab of t h i ck ness t and r el at i vepermit t ivity r > 1. The inductance per unit lengthand the character ist ic impedance of this l ine areL and Z0, respect ively.

W = 0

t r r ; > 1 = 0

Which one of the fol lowing inequali t ies is alwayssat isfied?

(a) 00

LZ

r

tw

(b) 0

0

LZ

r

tw

(c) 00

LZ

r

wt

(d) 0

0

LZ

r

wt

54. A mi cr owave ci r cu i t consi st i ng of l ossl esst ransmission l ines T1 and T2 i s shown in t hefigure. The plot shows the magnitude of the inputreflection coefficient as a function of frequency f.The phase vel oci t y of t he si gnal i n t het ransmission

I nputT1 Length = 1 m T2 Length = L

Z0 = 50 Z0 = 50

50Open

1

0.8

0.6

0.4

0.2

0 0.5 1 21.5 2.5 f(in GHz)

| (r

efle

ctio

n c

oeff

icie

nt|

The length L (in meters) of T2 is ________55. A posit ive charge q is placed at x = 0 between two

infini t e metal plates placed at x = – d and atx = +d respect ively. The metal plates l ie in the yzplane.

+

=0

q

x

at

= –

x

d

at

= +

x

d

The charge is at rest at t = 0, when a voltage +Vis appl ied to the plate at – d and voltage – V isappl ied to the plate at x = +d. Assume that thequant i ty of the charge q is small enough that i tdoes not per turb the field set up by the metalplates. The t ime that the charge q takes to reachthe r ight plate is propor t ional to

(a) d/V (b) /Vd

(c) / Vd (d) /Vd

ANSWERSGENERAL APTI TUDE

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

TECH NI CAL SECTI ON

1. (1) 2. (0) 3. (b) 4. (c) 5. (90) 6. (b) 7. (d) 8. (0.316)

9. (3) 10. (0.25) 11. (d) 12. (d) 13. (0.25) 14. (598.67) 15. (a) 16. (b)

17. (1.5) 18. (a) 19. (1) 20. (a) 21. (1.75) 22. (16) 23. (a) 24. (d)

25. (32.58) 26. (0.66) 27. (0) 28. (0.33) 29. (3) 30. (d) 31. (5) 32. (0.049)

33. (6) 34. (100) 35. (b) 36. (c) 37. (32.36) 38. (1.2) 39. (2.105) 40. (0.7985)

41. (413.8) 42. (a) 43. (a) 44. (c) 45. (1) 46. (60) 47. (1) 48. (– 1)

49. (0.25) 50. (100) 51. (0.405) 52. (a) 53. (a) 54. (0.1) 55. (c)


EXPL ANATI ONSGENERAL APTI TUDE

1. Height

Height

6 Feet5 Feet

Y(Statement I I )

X(Statement I )

Hence from the given figure Mr. X is tal ler thanMr. Y by 1 foot .

4. (9 inches)1/2 = (0.25 yards)1/2,Solving we get 9 inch = 0.25 yards

(since 1 inch = 0.028 yard)5. M is twice as efficient as E but worked for half as

many days. So in this case they will do equal workif their shifts had same t imings. But M’s shift isfor 6 hours, whi le E’s shift for 12 hours. Hence, Ewil l do twice the work as M.Rat io of contr ibut ion of M : E in work is 1 : 2.

6. Given Venn diagram is

Readbooks

Watch TV

12

713

44

15

17

Play sports

The number of students who l ike to read booksor play spor ts

= 13 + 12 + 44 + 7 + 15 + 17= 108

7. Social science disciplines had a pre-colonial or igin.

8.

10

11 12 1

2

Mirror image of 1 : 20 is 10 : 3010 : 30 was the t ime two and quar ter hour backso t ime now wil l be 12 : 45

9.

2 2

5 2

2 2

5 2

10 km

10 km

M

5 km 5 km

N O

4 km km

km

km

O

km

(Star t ing point)

From the given figure

MN = 2 2(O M) (O N)

OM = 5 2 5 2 2 5 7 2

ON = 10 5 2 2 2 10 3 2

MN = 2 2(5 7 2) (10 3 2)

= 25 98 70 2 100 18 60 2)

20.61 km

TECH NI CAL SECTI ON

1. From the given quest ion,A has an eigen value is zero

| A| = 0

3 2 49 7 13

6 4 9 x = 0

3(– 63 + 7x + 52) – 2(– 81 + 9x + 78) + 4(– 36 + 42) = 03(7x – 11) – 2(9x – 3) + 4(6) = 021x – 33 – 18x + 6 + 24 = 0

3x – 3 = 0x = 1

2. From the quest ion, the complex var iable funct ionis:

f(z) = 2z3 + b1| z| 3

Given f(z) is analyt ic.which is possible only when b = 0since | z3| is different iable at the or igin but notanalyt ic.2z3 is analyt ic everywhere f(z) = 2z3 + b | z3| is analyt iconly when b = 0


3. Given f(x) = x3 – 3x2 + 1f (x) = 3x2 – 6xf (x) = 0

– 1

1

0 1 2 3– 1

– 2

– 3

3x2 – 6x = 03x(x – 2) = 0

x = 0, 2f (x) = 6x – 6

At x = 0 f (0) = – 6 maximax = 2 f (2) = 6 minima

Then the gi ven funct ion f(x) increases, t hendecreases and increase again.

4. Here the given equat ion is

sin(x) = 2x

(where x is in radians)

y

y x = /2

x

y x = sin

Hence there are 3 dist inct values.5. Here the given t ime-varying vector

I = ˆ ˆ15cos 5sinx t y t

| I | = 2 2(15cos ) (5sin )t t

= 2 2225cos 25sint t

= 2 2225cos 25(1 cos )t t

= 225 200cos t

| I | is minimum when cos2t = 0cos2 t = cos 90

or = t = 90So, the value of is 90

6. I n maximum power t ransformat ion, half of thevoltage drop across source resistance, remaininghalf across the load. Voltage across source (R)

I L R = V2

s (from the given figure)

I L = V2R

s

7. At t = 0– , Switch is at posit ion-1

+–

3

10V 2

+

–

VC(0– )

where, Vc(0– ) =

10 22 3

= 4 V ...(1)

Vc (0– ) = Vc (0

+) = 4 VAt t =

VC() 2

24

5 A

+

–

Vc () = 5 2 = 10 V ...(2)The t ime constant of the circuit is

24

2

= Req Ceq

= (4 + 2) 0.1= 6 0.1 = 0.6 sec

Therefore Vc(t) = Vc() + [Vc(0+) – Vc()]e– t/

= 10 + (4 – 10)e– t/0.6

Vc(t) = (10 – 6e– t /0.6) V8. At resonance (for paral lel RLC circuit )

I R = II L = QI – 90I C = QI 90

For paral lel RLC circuit

L

R

II =

IQ CQ R

I L

=16

33

10 1010 10 10

10 10

= 0.316


9. For T - network

Za

Zc

Zb

1 2

1 2

Z11 = Za + Zc

Z22 = Zb + Zc

and Z12 = Z21 = Zc

Given [Z] =2

3 2

j j

j j

Therefore Z12 = jand Z22 = 3 + 2j

= 3 + j + j= Zb + Zc

= Rb + j + Zc

Rb = 3 Hence the value of Rb is 3.

10. Given the energy of the signal

x(t) = sin(4 )

4t

t

sin att

X( )

1

– a a

sin 44

tt

=

1 sin 44

tt

X( )

14

– 4 4

Energy, Ex(t) =21

X( )2

d

=24

4

1 12 4

d

=4

4

1 12 16

=1 1

82 16

=14

= 0.25 J

11. Ebers-Moll model of a BJT is val id for al l theregion of operat ion.

12. Here r ds =OX GS T

1W

C (V V )Ln

r ds = channel resistanceW r A cor rectVT r B cor rectL r C cor rectVGS r

D Wrong statementSo, statement (d) is not cor rect .

13. Here Von = 0.7 V (assume the diode in the figure)

+–

i d

2 V

i 1

i 2

VA

2 K

6 K

Let diode : ON VA = 2 – 0.7 = 1.3 V

i 2 =1.3

;6K

i 1 =0.72K

i d = i 2 – i 1

=1.3 0.7

mA mA6 2

=1.3 2.1 0.8

6 6

= – ve

Which is not possibleWhen diode is OFF

+–

2 K

6 K2 V

i 2

i 2 =2V

2K 6K = 0.25 mA

Hene the magnitude of cur rent i 2 is 0.25 mA.


14. From the given figure

R2 =T 1

2 2

V Iln

I I

=3 3

6 6

26 10 1 10

100 10 100 10

= 598.67 17.

CLK Q1 Q2 Q3 Q4 Q5 Y = Q3 + Q5

0 0 1 0 1 0 01 0 0 1 0 1 12 1 0 0 1 0 03 0 1 0 0 1 14 1 0 1 0 0 15 0 1 0 1 0 0

The waveform of the gate outputY = Q3 + Q5

0 T 2T 3T 4T 5T

T1 = 5T

Average power dissipated

P =2

ONTVR T

=25 3T

10 5T

= 1.5 mW18. I n case of a ful l adder,

Cout (A, B, C ) = (3, 5, 6, 7)in

Appl ied atselect input

of MU X

Appl ied atdata input

of MU X

I 0 I 1 I 2 I 3

Cin

Cin

0 2 4 6

1 3 5 7

0 Cin Cin1

I 0 = 0I 1 = Cin

I 2 = Cin

I 3 = 119. Given the response of the system

G(s) =2

( 1)( 3)s

s s

Now Ldydt

= (sY(s) – y(0))

Y(s) =1 2

G( )( 1)( 3)

ss

s s s s

y(0) = Lt Y( )s

s s

(Applying ini t ial value theorem)

=2

Lt( 1)( 3)s

ss s

=

21

1 31 1

s

ss s

y(0) = 0

Ldydt

=( 2)

Y( )( 1)( 3)s s

s ss s s

=2

( 1)( 3)s

s s

0t

dydt

= Lt Ls

dys

dt

=( 2)

L t( 1)( 3)s

s ss s

=

21

1 31 1

s ss

s ss s

=1 0

1(1 0)(1 0)

20. Here the Nyquist plot is

G(s) =1

4 2ss

and G(s)| = – tan – 1 – tan – 1

2

at s = 0, G(s) =1 0 1

0.254 0 4

and G(s)= 0 = – tan – 1 0 – tan – 102

= 0


at s = , G(s) =

1 11 1

4 42 2

ss s

ss s

(0 1)(0 2)

=

10.5

2

and G(s)= = – tan – 1 180 – tan – 1 90= – 180Nyquist plot is

– 1 – 0.5 0.25Re

I m

Hence the number of encirclements of (– 1 + j0) = 021. The minimum required average codeword length

in bits for er ror free reconstruct ionLmin = H (Entropy)

H = 2 21 1

log 2 log 42 4

2 21 1

log 8 log 88 8

=2 3

2 2 21 1 1

log 2 log 2 log 22 4 8

32

1log 2

8

= 2 2 21 1 1

log 2 2 log 2 3 log 22 4 8

21

3 log 28

=1 1 3 32 2 8 8 = 1.75

Lmin = 1.75 bits/wordHence the minimum required average codewordlength is 1.75 bits/word.

22. From the given datafs = 8 kHz (speech signal)n = 8 bits/sample ; M = 4

The minimum band width

BWmin = 22 2 2

R R2log M 2log 4 2log 2

b b sf n

=2

8 84 log 2 4 4

s sf n f n = 16 kHz

Hence the minimum band width is 16 kHz.

23. Baz magnet ic field (magnet ic field exists in the zdirect ion in vacuum).

vxax + vzaz velocity

F = Q (v B) by Lorent ’s law

= Q (vxax + vzaz) Baz

Fy = Q vx . B (– ay)

This results in a circular path in the XY planewith vzaz component causing a l inear path.

Both result in a hel ical path in Z axis.

24. vp = / can be calculated. Polar izat ion can beident i fied.

r and r cannot be found, due to which powerflux cannot be calculated as power flux

P = 2

E12

, where = 120 r

r

Hence, the power flux through the z = 0 plane.

25. sin max = 2 2 2 21 2sin 1.5 1.4i n n

(Here i be the incident angle)

max = sin i = sin – 1 (0.5385) = 32.58 i = 32.58Hence, the maximum value of i is 32.58.

26. Given different ial equat ions

dydx

= – 3y + 2, y(0) = 1

I f 1 – 3h < 1, then solut ion of different ial equat ionis stable.

– 1 < 1 – 3h < 1– 2 < – 3h < 00 < h < 2/3

hmax =23

= 0.66

Hence, the largest t ime step that can be used tosolve the equat ion is 0.66.

27. By Green’s theorem

C = 2 2xy dx x ydy (where C = Closed curve)

= 2 2

R

( ) ( )d d

x y xy dxdydx dy

= R

(2 2 )xy xy = 0

So, the value of closed curve C is 0.


28. The probabi l i ty P(X + Y 1)

=

1 (1 )

0 0

( , )x

xy

x y

f x y dx dy

=

11 (1 ) 1 2

00 0 0

( )2

xx

x y x

yx y dx dy xy

=

1 2

0

(1 )(1 )

2x

xx x dx

=

1 1 22

0 0

(1 )( )

2x x

xx x dx dx

=

11 2 3

00

12 2 2 6

x

x x xdx

= 1 1 1

0.332 6 3

29. Trace of A = 14a + 5 + 2 + b = 14(Taking the diagonal element and then adding)

a + b = 7 ...(i )det (A) = 100

3 75 0 2 4

0 0

a

b = 100

5 2 a b = 10010 ab = 100

ab = 10 ... (i i )From equat ion (i ) and (i i )either a = 5, b = 2or a = 2 b = 5

| a – b| = | 5 – 2| = 3

30. Req a

b

1

1

1 1

1 1

1 1

1

By using delta to star conversion

a

b

1

1/3

1 1/3

1 1

1/3

1

1

a

b

1 4/3

4/3

1 4/3

1

Again by star to delta conversiona

b

1

1

4 1

4

4

Rab = {4 1) + (4 1)} {1 4)} 1 2

1 2

R . RR R

= 4 4 45 5 5

= 8

15

So, the equivalent resistance across the terminals

a and b is 8

15 .

31. From the given figure

+–

R1 R2

5

3

60 V Vx

+

–

Vx

5

5 0.16Vx

0.4 Vx

Using KVL (kirchhoff’s voltage law) in the outer loop

60 – 5(0.16 Vx) V3 V

5x

x = 0

or Vx = 25 V The cur rent flowing through

R2 =V 255 5

x = 5 A

So, the magnitude of the cur rent is 5 A.


32. Given a cont inuous-t ime fi l t er wi t h t r ansferfunct ion

H(s) = 2

2 6 1 12 46 8

ss ss s

h(t) = e– 2t u(t) + e– 4t u(t)Given sampling frequency (fs) = 2 HzFor discrete t ime,

t = nTs = 2n

h[n] = (e– n + e– 2n) u[n]

H(z) = 1 1 2 1

1 1

1 1e Z e Z

= 1 2

Z Z

Z Ze e

=2

2

2Z 0.5032Z

Z 0.5032 0.049Z

k = 0.049Hence the value of k is 0.049.

33. Discrete Four ier Transform (DFT) of the 4-pointsequence is

x1[n] = 3n

x

X1[K] = {12, 2j , 0, – 2j , 12, 2j , 0, – 2j , 12, 2j , 0, – 2j}X1[8] = 12; X1(11) = – 2j

1

1

X [8]X [11] =

122 j = 6

34. At t = 0–

+–

1

10 V L10

(0 ) 10A1

i

i L(0– ) =

101

= 10 A

for t > 0 (using Laplace t ransform)

+–

VC(s)

–

+

I ( )s

10 10 – 3

10– 3s

106

10 s

I (s) =3

63

10 10

1010

10s

s

Vc(s) =610

I ( )10

ss

Vc(s) =6

2 8

10

10s Taking inverse Laplace t ransform, we get

Vc(t) = 100 sin104t V ( Vc(t) = V0 sin t) St eady st at e magn i t ude vol t age acr osscapacitor is 100 V.

35. I n a MOS - Capacitor(VG – VT) Q

Wher e VG = vol t age appl i ed acr oss a M OScapacitor with metal gate and p-type si l icon

Q = Inversion car r ier density

G1 T

G2 T

(V V )(V V )

=

1

2

QQ

T

T

0.8 V1.3 V

=

11

11

2 10

4 10

On solving, we get VT = 0.3 V

Now ConsiderG2 T

G3 T

V VV V

=

2

3

QQ

1.3 0.31.8 0.3

=11

3

4 10Q

I nversion car r ier density with VG = 1.8 VQ3 = 6 1011/cm2

Hence, the value of inversion car r ier density is6 1011/cm2

36. I n any type of PN junct ion

VBR 1

Doping Concent rat ion

i .e. VBR D

1N (or ) VBR =

2

D

E2 Nq

(Here VBR = break down voltage, ND = donor density) VBR ND is a constant

37. | | = D N0

N Xq

50 103 = 19

17N14

1.6 1010 X

8.85 10 11.7

(Since = 0 r)

On solving XN = 3.2356 10– 6 cm= 3.236 10– 8 m

XN = L = 32.36 nmHence, the value of L is 32.36 nm.


38. Since VGS > VDS, MOSFET is in l inear operat ionI D = K N [VGS – VT] VDS

D

GS

IVd

d = K N VDS

gm = K N VDS

0.5 10– 6 = K N [50 10– 3] K N = 10– 5

I D = K N [VGS – VT] VDS

D

S

IVd

dd = K N [VGS – VT]

gd = K N [VGS – VT]8 10– 6 = 10– 5 [2 – VT]

6

5

8 10

10

= 2 – VT

8 10– 1 = 2 – VT

0.8 = 2 – VT

VT = 1.2 VHence threshold voltage of the transistor is 1.2 V.

39. Vr ipple =DCI .TC

Here T = Time per iodC = Fil ter capacitor

Vr ipple =3

6

1 1 10

475 10

= 2.105 Volts

40. I ni t ial ly switch is closed and VB = 10 V V01 = – 10 V V0 = – V2 = – 5 V

VA = 0V 51K K

4K 1K 5K

= – 1 V

+

–

10 V

V01V0VB

VA

470 k4K

1K

100 F

10K

– 10 V

At t = 0;The switch is opened and as t , VB approaches– 10 V.Let at t = T1,VB exceeds VA (– 1 V) so that V01 changes from – 10 Vto 10 V

V0 changes from – 5 V to 5 VVB = Vf + (V i – V f)e

– t/

= – 10 + [10 – 1 – 10]e– t/RC

(Since = RC)At t = T1 VB = – 1

– 1 V = – 10 + 20e– T1/RC

T1 =20

RC ln9

= 10 103 100 10– 6 0.798= 0.798 sec

Hence, the required t ime at which V0 changesstate is 0.798 s.

41. Overal l input = V ios + V input

(Here V ios = input offset voltage)= 5 mV + 25 mV = 30 mV

V0 =

F

1

F

OL 1

R1

ROveral l input

R11+ 1

A R

= 3

15K1

1K 30 101 15K

1+ 1100 1K

= 3

16K1K 30 10

1 16K1+ 1

100 K

= 316

30 10 100116

= 348000

10 V116

= 413.79 mVHence, the output V0 is 413.79 mV.

42. Here 8 kbyte ROM is given 2n = 8 kB = 23(210) = 213 (am an = am+n)

D0

A0 CS RD

8 kB

ROMA12

D7


13 Address l ines are required for memory chip.But the address range given as 1000 H - 2FFF H

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0A A A A A A A A A A A A A A A A0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0

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

I n order to get CS as low, the condit ion is

A15 = A14 = 0 and A13 = 0/1, A12 = 1/0.43. The total capacitance = (2N – 1) C = (28 – 1) 8 pF

= 2.04 nF (Here the input capacitance of eachcomparator = 8 pF)

~

75

2.04 nF

The t ime constant = = RC = 153 ns

Sett l ing t ime (Ts) = 5 = 5 RC = 765 ns

Sampling rate =1

Set t l ing t ime (T )s

1 M samples/secH ence, t he max i mum sampl i ng r at e i s1 mega samples per second.

44. For St at e A

X Y Z0 0 0 A B0 0 1 A A0 1 0 A A0 1 1 A A1 0 0 A C1 0 1 A C1 1 0 A A1 1 1 A A

Binary inputPresent State Next State

Pr esen t St at e B

X Y Z0 0 0 B A0 0 1 B C0 1 0 B B0 1 1 B B1 0 0 B A1 0 1 B B1 1 0 B B1 1 1 B B


Pr esen t St at e C

X Y Z0 0 0 C C0 0 1 C C0 1 0 C C0 1 1 C B1 0 0 C C1 0 1 C A1 1 0 C C1 1 1 C AB


In state ‘C’ when XYZ = 111; the ambiguity occurs.Because from state ‘C’When, X = 1, Z = 1 Next state = AWhen, Y = 1, Z = 1 Next state = BSo, t r ansi t ions fr om state ‘C’ are ambiguousdefined.

45. G(s) = 2

1

2s s

Y( )R( )

ss

=2

2

2

KK2

K 2 K12

s ss s

s s

Minimum set t l ing t ime and no overshoot implies

= 1

n = K2 n = 2

n = 1

K = 1 or K = 1

Hence, the required value of gain K to achievethis is 1.

46. 1 + G(s)H(s) = K

1 0( 1)( 2)( 3)s s s

(s + 1)(s2 + 5s + 6) + K = 0s3 + 5s2 + 6s + s2 + 5s + 6 + K = 0

s3 + 6s2 + 11s + 6 + K = 0Gain margin = 0 dB and phase margin = 0I t implies marginal stable systemBy Routh Array

3

2

0

1 11

6 (6 K )66 6 K

06

6 K

s

s

s

s


For marginal stable system,60 – K = 0

K = 60Hence, the posit ive value of K is 60.

47. G(s) = 1

K( 0.1)( 10)( )s s s p

G(s) = 1 1 1

1

tan tan tan0.1 10 p

G(s)=1 = 1 1 1

1

1 1 1tan tan tan

0.1 10 p

= – 135

– tan – 1 10 – tan – 1 0.1 – tan – 1 1

1p = – 135

– 84.28 – 5.72 – tan – 1 1

1p = – 135

1

1

1tan

p = – 135 + 90

1

1

1tan

p

= 45

1

1p = 1 p1 = 1

Hence, the value of p1 is 1.48. Power spectral density of X(t) = SX(f)

SX(f) =

22( )

R ( ) ,T

j fnb

n

G fe

Rb() = E[nn – ]

= E[(n + kn– 3)(n– + k n– – 3)]= E[nn– ] + kE[n– 3 n– ] + kE[nn– – 3]

+ k2E[n– 3 n– – 3]= E[nn– ] + kE[n– 3 n– – 3+3] + kE[nn– – 3]

+ k2E[n– 3 n– – 3]= R() + kR( – 3) + kR( + 3) + k2R()= (1 + k2)R() + kR(+ 3) + kR( – 3)

Hence autocorrelat ion funct ion can be defined as:

Rb() =

21 0

3

0 otherwise

k

k

Power spectral densitysb(f) = 1 + k2 + 2k cos(2f3T)

Null wi l l occur at f =1

3T

At f =1

3Tsb(f) = 1 + k2 + 2k cos(2) = 0

1 + k2 + 2k = 0 (k + 1)2 = 0 k = – 1Hence the value of k is – 1.

49. H er e t he r ange of Channel spect r um500 Hz – 2000 Hz (Here Rb = rate of t ransmission)Hence Bandwidth (BW) = 1500 Hz

BW =2

R(1 )

log Mb

1500 =2

R(1 )

log 16b

1500 =2

4800(1 )

log 16

1500 = 1200 (1 + ) = 0.25Hence, the rol l-off factor of a-pulse is 0.25.

50. Given random processX(t) = 3V(t) – 8 and E[V(t)] = 0

Rv() = 4e– 5| |

Power of X(t ) = E[X 2(t)]= E [9V2 (t)] + 64 – 48 E [V(t)]= 9E [V2 (t)] + 64 – 48 E [V(t )]

E [V2 (t)] = Rv(0) = 4Power of X(t) = ((9 4) + 64)

= 100Hence the power in X(t) is 100.

51. 3/4x0 y0

1/4

1/2

1/2x1 y1

P(x0) =1

;2

P(x1) = 12

0

0

Pyx

=3

;4

0

1

Pyx

= 12

[P (x, y)] = [P(x)]d [P(y| x)]

=1/2 0 3/4 1/4

0 1/2 1/2 1/2


[P(x, y)] =3/8 1/8

1/4 1/4

P(y| x0) =1

0 200

( , ) log kk

k

yP x y P

x

= – {P(x0, y0)log2 P(y0| x0)+ P(x0, y1)log2 P(y1| x0)}

= 2 23 3 1 1

log log8 4 8 4

= 0.40552. Let E = E1,

Energy E1 =21

1

Q2C

(Here Q = charge across capacitor )For elect r ical ly isolated Q2 = Q1

d2 = 2d1

C2 =1C

2 (Here d1 = d2 = plate separat ion)

E2 =2 2 22 1 1

12 1

Q Q Q2

2C2C 2C2

= 2E1 = 2EHence the energy stored in the capacitor is 2E.

53. Z0 =LC

Here Z0 = Character ist ics impedanceL = Inductance per unit length

Z0 =L

Weff

t

as eff < 0r

Z0 >0

LWr

t

54. At frequency of 1 GHz the l ine behaves as 2

length there by giving open circuit at the inputof T2 l ine. (From figure)With 50 terminat ion at T1 the load is matchedand | | = 0 as seen in the graph.Here | | = input reflect ion coefficientT1 – T2 = lossless t ransmission l ine

f = 1 GHz

=10

9

2 1020 cm

1 10

Now L =2

= 10 cm = 0.1 m

Hence, the length (L) of T2 is 0.1 m.55. For velocity being free,

Since K.E. = 21mV

2

V =Eq

E = EnergyV = Applied voltageq = elect r ic charge of a metal plate

21mV

2 = qV

v =2 Vd q

t m

dt

=2 Vqm

t =2 V

d

qm

t V

d

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