EC: ELECTRONICS AND COMMUNICATION · PDF fileEC: ELECTRONICS AND COMMUNICATION ENGINEERING...

GATE-2012

EC: ELECTRONICS AND COMMUNICATION ENGINEERING

Duration: Three Hours Maximum Marks: 100

Read the following instructions carefully.

1. Do not open the seal of the Question Booklet until you are asked to do so by the invigilator.

2. Take out the Optical Response Sheet (ORS) from this Question Booklet without breaking the seal and read the instructions printed on the ORS carefully. If you find that the Question Booklet Code printed at the right hand top corner of this page does not match with the Booklet Code on the ORS, exchange the booklet immediately with a new sealed Question Booklet.

3. On the right half of the ORS, using ONLY a black ink ball point pen, (i) darken the bubble

corresponding to your test paper code and the appropriate bubble under each digit of your registration number and (ii) write your registration number, your name and name of the examination center and put your signature at the specified location.

4. This Question Booklet contains 20 pages including blank pages for rough work. After you are

permitted to open the seal, please check all pages and report discrepancies, if any, to the invigilator.

5. There are a total of 65 questions carrying 100 marks. All these questions are of objective type.

Each question has only one correct answer. Questions must be answered on the left hand side of the ORS by darkening the appropriate bubble (marked A, B, C, D) using ONLY a black ink ball point pen against the question number. For each question darken the bubble of the correct answer. More than one answer bubbled against a question will be treated as an incorrect response.

6. Since bubbles darkened by the black ink ball point pen cannot be erased, candidates should

darken the bubbles in the ORS very carefully.

7. Questions Q.1 – Q.25 carry 1 mark each. Questions Q.26 – Q.55 carry 2 marks each. The 2 marks questions include two pairs of common data questions and two pairs of linked answer questions. The answer to the second question of the linked answer questions depends on the answer to the first question of the pair. If the first question in the linked pair is wrongly answered or is unattempted, then the answer to the second question in the pair will not be evaluated.

8. Questions Q.56 – Q.65 belong to General Aptitude (GA) section and carry a total of 15 marks.

Questions Q.56 – Q.60 carry 1 mark each, and questions Q.61 – Q.65 carry 2 marks each.

9. Unattempted questions will result in zero mark and wrong answers will result in NEGATIVE marks. For all 1 mark questions, ⅓ mark will be deducted for each wrong answer. For all 2 marks questions, ⅔ mark will be deducted for each wrong answer. However, in the case of the linked answer question pair, there will be negative marks only for wrong answer to the first question and no negative marks for wrong answer to the second question.

10. Calculator is allowed whereas charts, graph sheets or tables are NOT allowed in the examination

hall.

11. Rough work can be done on the question paper itself. Blank pages are provided at the end of the question paper for rough work.

12. Before the start of the examination, write your name and registration number in the space

provided below using a black ink ball point pen.

GATE-2012

Soln. (1) Given = 1 + 0.1 cos10000

Find in the small signal model

We know

= + 1

= + 1

= + 1

=

Where is the dc current through base

= 1

! = 25 at room temp.

So

= $%×'()×'() = 25Ω

Option (C) is correct

GATE-2012

Soln. (2) Spectral density plot is given

*+,$ - represents the total power in random signal.

So,

. = *+,$ - = 2 × 12 / 01232×')

4×')

= 5400 + $ × 6 × 2 × 1089 *+:$ - = ;<''

*+: - is the mean value of the random variable. Since there is an

impulse in the spectral density, the dc value corresponds to the mean

value *+: -

GATE-2012

*+: - = = 20>2?


Soln. (3) Signaling frequencies up to 3500 Hz.

Excess Bandwidth used is

@ = 0.75 × 3500CD = 2625CD

B.W. available is B then

E$ ≤ @

Where Rb is the data rate.

Thus B.W. required for transmission

G ≤ 2@

G ≤ 5250

GHIJ = 5250symbols/sec.

Option (D) is correct

GATE-2012

Soln. (4) Given: plane wave propagating in air

*KL = M8OPJ + 6OPQ + 5OPDRSTUV8JW<Q /

(1) Electric field inside the perfect conductor = 0

i.e. E transmitted = 0

(2) Conducting slab is put at , = 0

So

* + *Y = 0

or *Y =−* = −M8OPJ + 6OPQ + 5OPDRSTUVW8JW<Q /


Soln. (5) Given

*KL = 10MOPQ + [OPDRSWT$%J

Wave is travelling in , – direction and

GATE-2012

= 25 MSWTJR or

$\ = 25

or $] ^_ = 25

or ` = $%×]$ = $%×8×'a$

= 1.19 × 104CD

` ≅ 1.2dCD

Real instantaneous form is

*KLef. =GM*KL × STUVR = 10gcos2 − 25,OPQ − sin2 − 25,OPDj At , = 0

*KLef. = 10gcos 2 OPQ − sin2 OPDj

As t increase wave traces the path as shown

So it is circularly polarized

Option (A) is correct

GATE-2012

Soln. (6) The given circuit is for the S – R flip flop

with A = S and B = R

In S – R flip flop race around condition does not occur


Soln. (7)

Following is given

GATE-2012

The output Y is logic 1 when the A input is higher than B input

A>B

01 00 1

10 00 01 2

11 00 01 10 3

6

Option (B) is correct

Soln. (8) Given

= klW'.m%'' n > 0.7 pn < 0.7 r For the diode in the given circuit.

Diode voltage 500 + 0.7

Applying KCL

GATE-2012

10 = 1000 + 500 + 0.7

⇒ = 4.8.% = 6.2


Soln. (9) C1 is charged to 12V switch is closed at = 0

Capacitor C1 will discharge and C2 will charge. Since both C1 and C2

are ideal and there is no resistance in the circuit, charging and

discharging time constant will be zero.

Thus current will exist like an impulse function.


GATE-2012

Soln. (10) Given

Impedance tu = 4 − [3Ω

Current

= 5 cos100 + 100

From the given impedance

Gu = 4Ω

General current equation is

= H cos2 + v So . = $ H$ Gu

= $ × 5$ × 4 = 50wO x


Soln. (11) Given: unilateral Laplace transform

of f(t) is fyf

GATE-2012

If ` ↔ x Then ` = − ||f x Thus if x = fyf

` = − ||f fyf~

= −5'.fyfW.$ffyfy 9

= $ffyfy Option (D) is correct

Soln. (12) Given : Differential equation is

|J|V + , = ⇒

|V|V + JV = 1

Integrating factor ` = S|V = S V = Solution is

, = . 3 , = 3 , = Vy$ +

GATE-2012

Given ,1 = 0.5

0.5 = $+

or = 0

Required solution is , = Vy$


Soln. (13) The given circuit consists of two parts .

First part is a clamper circuit and the second is peak rectifier. When

the circuit is excited by Cos2 , the clamping section clamps the

positive peak to 0 volt and –ve peak to -2volt. So whole cos2

GATE-2012

lowers by -1 volt.


Soln. (14) Series combination of N-MOS is equivalent to AND and parallel

combination is equivalent to OR

So = . + @ =. + @

GATE-2012

= + . @


Soln. (15) We know that the entropy is maximum when symbols are equi-

probable so if probability will change from equal to non equal entropy

will decrease.


Soln. (16) Given

Inner diameter a

Outer diameter b

Y = 10.89

t' = $ I~

= $ $.< ~

= $ $.< ~

GATE-2012

= $'$ '.4 = 60 × 8.8 × 0.875

t' = 15.9Ω

Alternative method

t' = 8> log |~

= 8>'.4 log $.< ~

88.8 × 0.380Ω

t' = 15.89Ω

No option seems to be falling in this range.

Soln. (17) Given

Ө = x<Ө0 ≤ Ө ≤ 2_

Ө is radiation intensity Ө, ∅ Directivity = <

YI| is radiated power.

YI| = ʃ¡ʃ∅¢, ∅3Ω

So

GATE-2012

YI| = / / x< $_'

¢ sin ¢ 3¢3∅$'

Note that dΩ (solid angle) = sin ¢ 3¢3∅

So YI| = / / x<¢

$_'

sin ¢ 3¢3∅$'

2 = / x<¢

$_'

sin ¢ 3¢ =−2 5]£f¤¡% 9'

$_

YI| = $%

So = <y¥¤ ~ = 10HIJ

10+¢-HIJ

= 10

In dB

Directivity = 10¦' = 103@


GATE-2012

Soln. (18) Given

,+,- = §1 3_ ¨©e© −§1 2_ ¨e We have to find ROC

Above function can be written as

+,- = 8~e + 8~We − − 1 − $~$ 8~e ↔ W)D( Gp©t© > 8 8~We − − 1 ↔ WW)D( Gp©t© < 3

$~e ↔ WyD( Gp©t© > $ So overall ROC will be intersection of their ROCs

i.e. $ < ©t© < 3


Soln.(19) Given

`,, ª, t = «2,3,4,5 Find prime implicants.

GATE-2012

`,, ª, t = ,ª + ,ª

Above are prime implicants.


Soln. (20) Steady state output of the system is

ª = ©d[2© sin2 + ∡d[2 For ª to be zero

©d[2© can be zero

©d[2© = WUy4>Uy<>Uy>Uy4>Uy; ⇒ at 2 = 3O3/xS. ©d[2© = 0 thus ª = 0


GATE-2012

Soln. (21) G! = ®¯°¯ − − − −1

From the given fig.

£] = + 99 × 100

= 10< − − − −2 f] = 100 − − − −3 Thus

G! = ®¯°¯

= '±'' G! = 100Ω


GATE-2012

Soln. (22)

Applying KVL in loop EDABE

1∡ + 1. + + 1[. 1 − 1∡ = 0

Or + + 1[ = 0

1 + [ = −[ Or = − TT

GATE-2012

= + 1 = 1 − TT = T Option (C) is correct

Soln. (23) Given `t = D− $D8 = D8W$DDD8

`t = WDDD8 Poles are at -1 and -3

i.e. −1,0 and −3,0 to find

$ ∮]`t3t

To find

$ ∮] 5 D− D89 3t

Given ©t + 1© = 1

t = −1 is singularity in C and z=-3 is not in C

By Cauchy integral formula

= ∮] DD83t = 0

= $ ∮ `t3t

GATE-2012

GSx−1 = limD→W µDM–DRDD8 · =WWW8 = $$ = 1


Soln. (24) Random variables X and Y are uniformly distributed in the interval +−1, 1- i.e. −1 ≤ , ≤ 1 and −1 ≤ ª ≤ 1 is entire rectangle shown.

The region in which maximum of x , y is less than is $ is shown as

shaded.

¸ 5max »,, ª¼ < $9 = ½YI£^f¾I||Y¿£e½YI£^eVYY]VIe¿À = )y×)y$×$ = 4; Option (B) is correct

GATE-2012

Soln. (25) Given

, = >−1 = [ In polar coordinates

, = cos $ + [ sin $ = ST./$

Now ,J = [T = MST./$RT STy./$ = SW/$


Soln. (26) Given = 10'¸S$

n-channel MOS SO = 1xÁÂ & depth 1Â

Dopant Ã104/8

Volume (V) = × 3 = 1 × 10W8 = 10W$8

¸ = eÄyÅÆ = 'y'Ç = 10/8

So holes in volume = ∅ ×

= 10 × 10W$ = 10W

≅ 0


GATE-2012

Soln. (27) Given

x = $È! sin2] x$ = −$È! sin2]

Due to difference of 450 at coherent receiver the received signal

amplitude will be decreased by a factor of sin 45' = >$ So, x′ = >$$È! sin2] = È! sin2] x$′ = −È! sin2] Now noise power is

.Å = 2 × Å$ . @.É. = 2 × Å$ × 2G

.Å = 2Ã'G ……………………..(1)

*| = 1Ê/gx′ − x$′ j$!'

3

GATE-2012

= ! <È!!' x$2] 3 = $È!y 2!' x$2] 3

*| = $È! ~

Now

. = Ë µ È <Å£fÌ£ÍY· = Ë ÎÏ yÐ<×$ÅE Ñ

= Ë ÎÏ 24Ã'ÊGÑ As, Ê = E ⇒ Ê. G = 1

So, . = Ë µ È<Å· Option (D) is correct

Soln. (28) Characteristic impedance of line used for matching =100 Ω

Required to match 50 Ω section with 200 Ω

GATE-2012

Frequency = 429 MHz & 1 GHz

For 429 MHz = \< = ]< = 8×'a<×<$4×'Ò = 0.175

For 1GHz $ = \y< = ]< y = 8×'a<××'Ç = 0.075

Matching sections are of length λ/4. For these sections to match at

both the above frequencies they should be multiple of LCM of and $ i.e. 0.013125 m

Multiple of this which matches the given option is 1.05 m.


Soln. (29) Given that input and output are related as

ª = /,Ó. cos3Ó3ÓVW∞

We will first see whether the system is time invariant or not output at − ' is

GATE-2012

ª − ' = / ,Ó. cos3Ó3ÓVWVW∞

Let y’ (t) is the output for the input , − '

ª ′ = /,Ó − '. cos3Ó3ÓVW∞

ª ′ = / ,Ó. cos 3Ó + '3Ó

VWVW∞

ª ′ ≠ ª − ' so system is not time invariant

Check for stability.

ª = /x$3ÓV

W∞→ ∞

As → ∞

So for bounded input output is not bounded system is not stable


GATE-2012

Soln. (30) The characteristic equation can be written as

1 + dxCx = 0

i.e. 1 + Õff)Ify$f = 0

or, x8 + Ox$ + 2 + Öx + 1 + Ö = 0

Routh array for the above characteristic equation can be written as

S3 1 (2+K)

s2

a (1+K)

s1 a(2+K) – (1+K)

a

s0 (1+K)

for oscillations

I$ÕWÕI = 0

⇒ O = Õ$Õ~

Now

GATE-2012

Ox$ + 1 + Ö = 0

−O2$ + 1 + Ö = 0

Given 2 = 2O3/xS

−4O + 1 + Ö = 0

−4. Õ$Õ+ 1 + Ö = 0

−41 + Ö + 2 + Ö + 1 + Ö = 0

1 + Ö+2 + Ö − 4- = 0

Or, Ö = −1, 2

But Ö = −1 is not possible as system will not oscillate for this as O = 0

So, Ö = 2

Or, O = Õ$Õ = 8< = 0.75


Soln. (31) Given the Fourier transform of ℎ C[2 = $ØÙUÙÚÛ $UU

= 2 cos2 5$ ÙÚÛ $U$U 9 = C[2.C$[2 C[2 = 2 cos2 = MSTU + SWTUR C$[2 = $ ÙÚÛ $U$U = 20I22

GATE-2012

0I22 is the Fourier transform of Gate function

Comparing it with the given function Ó/2 = 2 or, Ó = 4

ÖÓ = 2 or, Ö = 2/Ó = 2/4 = 1/2

C[2 = 2 cos2C$[2 = MSTU + SWTURC$[2 C[2 = STUC$[2 + SWTUC$[2 ⇒ ℎ = ℎ$ + 1 + ℎ$ − 1

So, ℎ0 = 1 Option (C) is correct

GATE-2012

Soln. (32) LTI system is given by

Î,,$,8ÜÜÜ Ñ = Ý 0 O 00 0 O$O8 0 0 Þ Ý

,,$,8Þ +Ý001Þ

Output,

ª = +1 0 0-Ý,,$,8Þ Controllable matrix

Ë] = +@ @ $@- ……………… (1)

@ = Ý 0 O 00 0 O$O8 0 0 ÞÝ001Þ = Ý 0O$0 Þ

$@ = @ = Ý 0 O 00 0 O$O8 0 0 Þ Ý0O$0 Þ

$@ = ßOO$00 à

GATE-2012

Ë] = Ý0 0 OO$0 O$ 01 0 0 Þ For system to be controllable

©Ë]© ≠ 0

⇒ 0 − OO$$ ≠ 0

⇒ O ≠ 0

O$ ≠ 0

a3 can be zero matrix should not be zero


Soln. (33) In the given circuit

= 'W8$E = m$E = 3 + G = 3 + mE$E

= ;'E$E Power transferred from circuit A to circuit B =V.I.

= ;'EE$ ~ mE$~

GATE-2012

. = <$m'EE$y

|á|E = E$y.m'W<$m'E$E$E$y = 0

Or, 70G + 2$ = 42 + 70G2G + 2 ⇒ 5G + 2 = 23 + 5G ⇒ 5G + 10 = 6 + 10G

⇒ 4 = 5G

G = 0.8


Soln. (34) Given differential equation is

|yQ|Vy + 2 |Q|V + ª = â

With ª Vã'( = −2 and |Q|V |t = 0 = 0

We know that

ℒ 5|y^|Vy9 = x$`x − x`0 − |^|V 0 & ℒ 5|^|V9 = x`x − `0

GATE-2012

Taking Laplace transform

x$ªx + 2x + 2xªx + 4 + ªx = 1

Or x$ + 2x + 1ªx = −2x + 3 ªx = − $f8fy ªx = − 5 $f+ fy9 ª = −+2SWV + SWV-

|Q|V = −+−2SWV + SWV − SWV-

|Q|V |t = 0+ = −+2 + 1 − 0- = 1


Soln. (35) Given

©© = åe

L = åeOPY since it is

radially outward .

∇. L in spherical coordinates is

∇. L = Yy . ççY $Y + Y ÙÚÛ¡ ççY ¡ sin ¢ + Y ÙÚÛ¡ çç∅ ∅

∇. L = Yy . ççY $åe + 0 + 0

GATE-2012

∇. L = èYy ççY e$

So ∇. L will be zero if ççY e$ = 0

i.e. + 2 is constant, that is possible only if + 2 = 0

⇒ = −2


Soln. (36) P (odd tosses) = P(H)+P(TTH)+P(TTTTH)+……………

= $+ $~8 + $~% +⋯ = $ µ1 + $~$ + $~< +⋯· = $ µ1 + <+ <~$ +⋯· = $ ß W±à = $ × <8 = $8 Option (C) is correct

GATE-2012

Soln. (37) For P MOS

êë = ê − ë = 5 − e

For P MOS to be ON

êë > © !á© ⇒ 5 − e > 1

Or, e < 4

So option C and D get eliminated. For small Vin output is high and P

MOS is in linear region and N MOS is cutoff region. For high Vin P

MOS is in cutoff and N MOS is in linear region and for Vin in

between both are in saturation.

So, P MOS in linear region for e < 1.875


Soln. (38) Given transitional probability of

GATE-2012

¸S = ¸0. ¸ §1 0_ ¨ + ¸1¸ ì0 1_ í

= 4' × m+ ' × m = ;8' + m' = m'' = m There is no option as 7/8

Soln. (39) For phase modulator

∅ = 2 ] + åÌ Maximum phase deviation is

∅HIJ =åÌmax+ - = 2åÌ—1

GATE-2012

For frequency modulator

∅ = 2 ] + 2å^ / 3 V'

M∅′ RHIJ = 2å^ Î/ 3 V

'ÑHIJ

M∅′ RHIJ = 2å^ / 3 $

'

= 2å^ / 23 $

'

M∅′ RHIJ = 8å^

Given

M∅′ RHIJ = ∅HIJ

8G^ = 2åÌ

Or, èïèð = 4


GATE-2012

Soln. (40) Given the magnetic field

CD = 3 cos2.094 × 10$, cos2.618 × 10$ª Cos6.283 × 10' − t

Since magnetic field component is in the direction of propagation so

the mode is transverse i.e. TE

The standard Hz is given by

CD = C' cos ,O ~cos ,ñ ~cos2 − t Comparing the two equation we get

HI = 2.094 × 10$

Or, = 2.094 × I × 10$

= 2.094 × 8 × 10$

= 2.000 × 10$

= 2.618 × × 10$

= 2.618 × .$ × 10$

= 1.0005 × 10$

GATE-2012

So mode is TE21

2 = 6.283 × 10'

Or, ` = ;.$8$ × 10' = 10dCò

So the wave will propagate in the waveguide with

nÌ >


Soln. (41) For the given circuit

Êx = ff = −ß EyE °à Êx = Wóyó~ff ó¯

It is transfer function of high pas filter with cutoff frequency

GATE-2012

2 = E] O3/xS Option (B) is correct

Soln. (42) Given

ℎ = $~e ª+- = ℎ+- ⊗ ¦+- ª+- = « ℎ+ − å-¦å∞èãW∞

Then

ª+0- = « ℎ+−å-+¦å∞èãW∞ - = ℎ+0-¦+0- 1 = 1¦+0- Or, ¦+0- = 1

ª+1- = « ℎ∞èãW∞ +1 − å-¦+å- = ℎ+1-¦+0- + ℎ+0-¦+1- ℎ+1 − å- will be zero for å > 1

¦+å- will be zero for å < 0 (causal sequence)

$ = $ × 1 + 1. ¦+1-

Or, ¦+1- = 0


GATE-2012

Soln. (43) From the given circuit when select signal of MUX i.e. A=1, then Q

will be selected and it will be fed back to D flip-flop and gives the

output Q again.

So, at A=1, it holds state.

When A=0, the Ë will be selected by MUX and fed back to D flip-

flop and output will be inverted.

Thus for A=1 it hold the state and for A=0 it interchanges the state.

i.e. if Q=0and then it will go to Q=1

and if Q=1then it will go to Q=0


GATE-2012

Soln. (44) AC equivalent circuit for the given circuit can be drawn as the

following:

Applying KVL in input loop

13.7 − õ + ö12Ö − 100Ö. ö − 0.7 = 0

ö = 9.9Â õ = ö = 100 × 9.9Â

= 0.99

È ≅ 1 = $;HÐ = 26Ω

GATE-2012

t = = 2.6Ω

l = ''Õǁ$Õ$; = 412

t′ = tǁ ''Õ<$ = 221Ω

lf = l DÄDÄ′E° = 412 $$$$'Õ~

©lf© = 10


Soln. (45) Given ½ − ö = 6

To find õ −

In the given circuit

½ − ö = 6 so, current through 2Ω resister is 3A.

The current entering any network should be same as leaving the

network, so the branch õ − should have the same current of 3A.

GATE-2012

So, = õ + 3 + 2

= 5 + õ

So, ] − = −5


Soln. (46) The function is given

`, = ,8 − 9,$ + 24, + 5

In the interval [1, 6]

` ′, = 3,$ − 18, + 24 = 0

i.e. , = 2, 4

`”, = 6, − 18

`"2 = 12 − 18 < 0

`”4 = 24 − 18 > 0

Hence `, has maximum value at , = 2

The value is

28 − 9 × 2$ + 24 × 2 + 5 = 25

GATE-2012


Soln. (47) Given = 5−5 −32 0 9 & = 51 00 19 Characteristic equation of A is © − ù© = 0

⇒ 5−5 − ù −32 0 − ù9 = 0

⇒ −5 − ù−ù + 6 = 0

⇒ ù$ + 5ù + 6 = 0

ù$ = −5ù − 6

ù$ = −5ù$ − 6ù

= −5−5ù − 6 − 6ù

ù8 = 25ù − 6ù + 30

= 19ù + 30

Thus the value of

8 = 19 + 30 Option (B) is correct

GATE-2012

Soln. (48) Using ABCD parameters.

Given = 10

(i) $ = 3 (since 1Ω connected to port B draws 1AMP

current) $ = −3 (since it is the current coming out of port)

= $ − @$

10 = 3 + 3@

(ii) $ = 5

$ = −2

10 = 5 + 2@

= '4

@ = $'4

Given if = 8

$£] =? $ = 0

GATE-2012

= $ − @$

8 = $£] − 0

$£] = ½ = Ç = 7.2


Soln. (49) = 10

$ = −7$ = $ − @$

10 = −7$ − @$

10 = − m4 $ − $'4 $

$ = −1 (current is drawn from the input)


GATE-2012

Soln. (50) Source body junction capacitance

T = ∈½|

∈Y= 11.7`0 = 1Â × 0.2Â = 0.2 × 10W$$

3 = depletion width of ¸ − junction

3 = 10 = 10W

T = .4×'(y×.m×'.$×'(y'(a

= 2.082 × 10W%

T = 2`


GATE-2012

Soln. (51) Gate source capacitance ¿ is

¿ = ∈½| =∈∈Y½|

= 1Â × â = 1 × 10W; × 20 × 10W4

= 2 × 10W<$

3 = 1 =10W4

So, ¿ = .4×'(y×8.4×$×'(±'(Ç

¿ = 69.42 × 10Wm

= 0.69 × 10W%

¿ = 0.7`


Soln. (52) Magnetic flux density at a distance from the wire

©@© = $Y

GATE-2012

For < O → = ü. $

So @ = ý$

©@© ∝ ` < O

For > O

= ü × O$

So ©@© = Iy$Y

©@© ∝ Y ` > O


Soln. (53) It is uniform and defends on both b and d.


GATE-2012

Soln. (54) From the given transfer function

dx = fIf

So ∅ = OW UI − OW U

For phase lead ∅ should be positive

⇒ OW UI > OW U

⇒ O < ñ

Both options A & B are satisfied but option C will put poles and zeros

on RHS of S-plane thus not possible


Soln. (55) For phase to be maximum

ç∅çU = 0

yy+ yy

= 0

GATE-2012

Or, Uy − y± = 0

Uy − $Uy< = 0

2$ + 4 − 2 − 22$ = 0

2$ = 2

Or, 2 = >2O3/xS. Option (A) is correct

Soln. (56) Let , = 1.001

Given ,$%4 = 3.52 and ,$';$ = 7.85

So, ,88$ = ,$%4. ,$';$ = 3.52 × 7.85 = 27.64


Soln. (57) Option (C) is correct

GATE-2012

Soln. (58) Most appropriate word is non chalance meaning casual/non serious.


Soln. (59) Latitude refers to freedom of action, freedom of expression etc.

Option (B) is most appropriate choice.

Soln. (60) Option (B) is correct

GATE-2012

Soln. (61) In the given passage “Discipline on the battlefield kept units obedient,

intact and fighting, even when the odds and conditions were against

them”

Choice ‘A’ comes out as the best answer option.

Soln. (62) Let the number of Rs. 20/- notes be ,

and Rs. 10/- notes be y

20, + 10ª = 230

, + ª = 14

After soling we get , = 9O3ª = 5

Thus the numbers of Rs. 20/- notes is 9 and Rs. 10/- notes be 5


Soln. (63) Divide the 8 bags into three parts.

GATE-2012

3, 3 and 2 respectively

Case –I If we compare the 3, 3 bags on pans of balance. We can

identify on which side is the heavier bag placed. Then we will need

only one more weighing for identifying faulty bag ,so only two

weighing’s are required.

Case – II If 3, 3 compared bags are equal, the faulty bag is one

among the remaining two bags. So one more weighing is required to

find out faulty bag from these two bags. Hence overall two weighing’s

are required.

Option (A) correct

Soln. (64) As per the given table

Total expenses are Rs. 10,500/-

expenses excluding savings are Rs. 9000/-

Hence, required percentage

= 4''''%'' × 100

= 85.714% ≅ 86%


GATE-2012

Soln. (65) Probability that A & B will meet will be given by the graphical

representation. where shaded region represents favorable area

Total sample space = 60 × 60 = 3600

Favorable cases = 3600 − 2 × $ × 45 × 45~ = 1575

Therefore the required probability = %m%8;'' = m; Option (C) is correct

EC: ELECTRONICS AND COMMUNICATION · PDF fileEC: ELECTRONICS AND COMMUNICATION ENGINEERING...

Documents

Transcript of EC: ELECTRONICS AND COMMUNICATION · PDF fileEC: ELECTRONICS AND COMMUNICATION ENGINEERING...