gate 2011 q soln1

GATE 2011

1. The modes in a rectangular waveguide are denoted by TEmn/TMmn where m and n are the Eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?

(A) The TM10 mode of the waveguide does not exist

(B) The TE10 mode of the waveguide does not exist

(C) The TM10 and theTE10 modes both exist and have the same cut off frequency

(D) The TM10 and the TM01 modes both exist and have the same cut off frequencies.

Soln. (1) In a rectangular waveguide TEmn exists for all values of m and n except m = 0 and n = 0

i.e. TE00 does not exist

For TMmn to exist both values of m and n should be non-zero.

i.e. TMoo , TM01 , & TM10 do not exist.

Option (A) is correct

2. The Column 1 lists the attributes and the Column 2 lists the modulation systems. Match the attribute to the modulation system that best meets it.

Column 1 Column 2

P. Power efficient transmission of signals I. Conventional AM

Q. Most bandwidth efficient transmission of voice signals II. FM

R. Simplest receiver structure III. VSB

S. Bandwidth efficient transmission of signals with significant dc component IV. SSB - SC

(A) P IV, Q II, R I, S III

(B) P II, Q IV, R I, S III

(C) P III, Q II, R I, S IV

(D) P II, Q IV, R III, S I

GATE 2011

Soln. (2) Power efficient transmission FM Most bandwidth efficient SSBSC transmission

of voice signal

Simplest receiver structure conventional AM Bandwidth efficient transmission of VSB signals with significant DC component.

Option (B) is Correct

3. The differential equation 100 d2y/dt2 -20dy/dx +y = x(t) describes a system with an input x(t) and an output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform

GATE 2011

Soln. (3) Given :

Taking Laplace Transform on either side

Note So,

So the poles are with positive real part system is unstable. Option (A) is correct

4. For the transfer function G(j) = 5+j, the corresponding Nyquist plot for positive frequency has the form

GATE 2011

Soln. (4) In the given problem the transfer function has real part 5 which is fixed, only imaginary part increases with .

So option (A) is correct

GATE 2011

5. The trigonometric Fourier series of an even function does not have the

(A) dc term (B) cosine terms

(C) sine terms (D) odd harmonic terms

Soln. (5) f(t) is an even function thus bK the coefficients of sine terms will be zero.

Option (C) is correct

6. When the output Y in the circuit below is 1, it implies that data has

(A) changed from 0 to 1 (B) changed from 1 to 0

(C) changed in either direction (D) not changed

Soln.(6)

Data

Clock

D1 Q1

1

D2 Q2

2

As per the above diagram when data is 0 Q1 is 0 and (first FF)

Data is changed to 1

Q1 is 1

is connected to D2 so Q2 = 1

GATE 2011

So both the inputs of AND gate are 1 Y = 1 Option (A) is correct

7. The logic function implemented by the circuit below is (ground implies a logic 0)

(A) F = AND(P, Q) (B) F = OR(P, Q)

(C) F = XNOR(P, Q) (D) F = XOR(P, Q)

Soln. (7) From the given diagram O is connected to I0 and I3

1 is connected to I1 and I2

Therefore Option (D) is correct

8. The circuit below implements a filter between the input current i1 and the output voltage v0. Assume that the opamp is ideal. The filter implemented is a

(A) low pass filter (B) band pass filter

GATE 2011

(C) band stop filter (D) high pass filter

Soln. (8) In the given circuit

When = 0, inductor acts as short circuit

V0 = 0 When = , inductor acts as open circuit

V0 = i1 R1 So it acts as high pass filter.

Option (D) is correct

9. A silicon PN junction is forward biased with a constant current at room temperature. When the temperature is increased by 10C , the forward bias voltage across the PN junction

(A) increases by 60 mV (B) decreases by 60 mV

(C) increases by 25 mV (D) decreases by 25 mV

Soln. (9) For Si forward bias voltage changes by !"#$ %& For 100 increase, change will be

!" ' "()! Option (D) is correct

10. In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

(A) 6.4 j4.8 (B) 6.56 j7.87

(C) 10 + j0 (D) 16 + j0

Soln. (10) When the terminals P & Q are short circuited, the circuit becomes

GATE 2011

16 00 25

j30

15

Isc

P

Q

From the current division rule

-. /'00012

/'0312

/'0312 4!5 65!7 Option (A) is correct

11. In the circuit shown below, the value of RL such that the power transferred to RL is maximum is

(A) 5 (B) 10

(C) 15 (D) 20

GATE 2011

Soln. (11) For maximum power transmission

89 8:; Hence the equivalent circuit is

For the calculation of RTH , replace the sources by their internal resistances

8:; ' 8:; 89 " Option (C) is correct

12. The value of the integral c (-3z+4)/(z2+4z+5) dz where c is the circle |z| = 1 is given by (A) 0 (B) 1/10

(C) 4/5 (D) 1

Soln. (12) 2>3>3>0 ?@ Denominator of the integrand can be written as z2+4z+5=(z+2)2+1=0

AB, @ 6 i.e. it will be outside the unit circle

thus the integration value will be zero


GATE 2011

13. A transmission line of characteristic impedance 50 is terminated by a 50 load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be /4 radians. The phase velocity of the wave along the line is

(A) 0.8*108 m/s (B) 1.2*108 m/s

(C) 1.6*108 m/s (D) 3*108 m/s

Soln. (13) We know

Phase difference EF ' path difference

E3 EF ' ' 2

AB,G 7 ' ' 2 G 4 ' 2( Given f = 10GHz = ' HI> Thus phase velocity of the wave along the line is

J KG ' H ' 4 ' 2 Or, J !4 ' L( M Option (C) is correct

14. Consider the following statements regarding the complex Pointing vector P for the power radiated by a point source in an infinite homogeneous and lossless medium Re(P) denotes the real part of P , S denotes a spherical surface whose centre is at the point source, and n denotes the unit surface normal on S. Which of the following statements is TRUE?

(A) Re (P) remains constant at any radial distance for the source

(B) Re (P) increases with increasing radial distance from the source

(C) s Re (P).n ds remains constant at any radial distance from the source (D) s Re (P).n ds decreases with increasing radial distance from the source

Soln. (14) The statement given in


GATE 2011

15. An analog signal is band limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is

(A) 1 bit/sec (B) 2 bits/sec

(C) 3 bits/sec (D) 4 bits/sec

Soln. (15) Since two samples are transmitted and each sample has 2 bits of information.

Therefore the information rate = 4bits/sec


16. The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

(A) G(s)H(s) = k[s(s+1)]/[(s+2)(s+3)]

(B) G(s)H(s) = k/[s(s-1)(s+2)(s+3)]

(C) G(s)H(s) = k[(s+1)]/[s(s+2)(s+3)2]

(D) G(s)H(s) = k[(s+1)]/[s(s+2)(s+3)] Soln. (16) In the figure given

X denotes poles

O denotes zero

From the root locus plot given in the figure we observe the following

One pole at zero

One zero at -1

GATE 2011

Three poles terminate to

Pole at -3 goes on both sides, it means two poles at -3.

Option (B) corresponds to the given plot.

17. A system is defined by its impulse response h(n) =2n u(n-2). The system is

(A) Stable and causal (B) Causal but not stable

(C) Stable but not causal (D) Unstable and non-causal

Soln. (17) h(n) = 2n u(n-2)

Since h(n) is existing for n>2

Thus h(n) = 0 for n

GATE 2011

19. The output Y in the circuit below is always 1 when

(A) Two or more of the inputs P, Q, R are 0

(B) Two or more of the inputs P, Q, R are 1

(C) Any odd number of the inputs P, Q, R is 0

(D) Any odd number of the inputs P, Q, R is 1 Soln. (19) Given circuit is

PR

PQPQ . QR PQ+QR

(PQ+QR).PR

PQ

Q

PR

RQR

Y = PQ + PR + RQ

GATE 2011

So that two or more inputs are 1 Y is always 1 Option (B) is correct

20. In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency.vi is a small signal input. The gain magnitude |vo/vi| at 10 Mrad/s is

(A) Maximum (B) Minimum

(C) Unity (D) Zero

Soln. (20) In the parallel RLC circuit

Z [I\R?] R ^ _9`

_'ab'ac

dB\?ef gB\?efs

So that for a tuned amplifier gain is maximum at resonant frequency. Option (A) is correct

21. Drift current in semiconductors depends upon

(A) Only the electric field

(B) Only the carrier concentration gradient

GATE 2011

(C) Both the electric field and the carrier concentration

(D) Both the electric field and the carrier concentration gradient

Soln. (21) Drift current h ij h kR[S l[mnoj So it depends on carrier concentration and electric field. Option (C) is correct

22. A Zener diode, when used in voltage stabilization circuits, is biased in

(A) Reverse bias region below the breakdown voltage

(B) Reverse breakdown region

(C) Forward bias region

(D) Forward bias constant current mode Soln. (22) For Zener diode

Voltage remains constant is breakdown region.

Option (B) is correct

23. The circuit shown below is driven by a sinusoidal input vi = Vp cos(t/RC). The steady state output vo is:

(A) (Vp/3) cos(t/RC) (B) (Vp/3) sin(t/RC)

(C) (Vp/2) cos(t/RC) (D) (Vp/2) sin(t/RC)

GATE 2011

Soln. (23) In the given circuit

)p )m cos 8r&

+

R

R

C

Co

ZW@ 8 1s. tt1.s \R?@ 8 1s.

$u$v

>>w>

From the given sinusoidal input

^ tx

Then, @ tty1wz{ t

1

@ 8 1 t&

@ 8 t1 8 6

$u$v

zw|}

zw|}t1

2

Thus option (A) is correct

GATE 2011

24. Consider a closed surface S surrounding a volume V. IF r is the position vector of a point inside S, with the unit normal of S, the value of the integral

(A) 3V (B) 5V

(C) 10V (D) 15V

Soln. (24) Applying the divergence theorem

"B~!R! ? "! B~?J "3?J ") Note that ! B~ 3 Where B~is position vector Option (A) is correct

25. The solution of the differential equation dy/dx = ky, y(0) =c is :

(A) x= ce-ky (B) x= kecy

(C) y = cekx (D) y = ce-kx

Soln. (25) Differential equation

for y (0) = c

? Taking log on either side

S r W! W. WR r WAB r

GATE 2011

rW Option (C) is correct

26. The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative

permittivity r and relative permeability r = 1 is given by permittivity r and relative permeability r = 1 is given by

Assuming the speed of light in free space to be 3x108 m/s, the intrinsic impedance of free space to be 120, the relative permittivity r of the medium and the electric field amplitude Ep are

(A) r =3 , Ep = 120 (B) r =3 , Ep = 360

(C) r =9 , Ep = 360 (D) r =9 , Ep = 120

Soln. (26) Given j~ jmW1sLE>e( I~ 3W1sLE e( From the above expressions

7 EF AB,G 3 ( J KG 5 ' H ' 3 (eWr! J ' L(eWr ,J ._

AB, ' L 2'_ AB, Now we know

GATE 2011

2 ' 2 jm Option (D) is correct

27. A message signal m(t) = cos 2000t + 4cos 4000t modulates the carrier c(t)= cos2fct where fc = 1 MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy

(A) 0.5 ms < RC < 1 ms (B) 1 s 0.5 ms

Soln. (27) For proper demodulation of AM signal

The time constant should be much less than and much greater the

x.

! W! x 8]

; 8]

; [ 8] !"(

Option (B) is correct 28. The block diagram of a system with one input u and two outputs y1 and y2 is given below.

A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]

T is

GATE 2011

Soln. (28) From the given system diagram

w

State vector given

ww !

w

!

w \R? ww

\R?

Or And f And And And From the question

U V: UV

GATE 2011

Or

Or UV UV Only option (B) is satisfied.

29. Two system H1(z) and H2(z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H2(z) is

(A) (1-0.6z-1)/[z-1(1-0.4z-1)] (B) z-1(1-0.6z-1)/(1-0.4z-1)

(C) z-1(1-0.4z-1)/(1-0.6z-1) (D) (1-0.4z-1)/[z-1(1-0.4z-1)]

Soln. (29) In the given problem

Overall transfer function is z-1 since there is unit delay in transfer function

I@! I@ @ So I@ >aw;w> @!

k!/>awn!3>aw


30. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is

MVI A, 07H

RLC

MOV B, A

RLC

GATE 2011

RLC

ADD B

RRC

(A) 8H (B) 64H

(C) 23H (D) 15H

Soln. (30) MVIA, 07H A=07H (00000111)

RLC Rotate Acc. Left without carry 00001110

MOV B,A B 00001110

8Z]8Z]

A ADD B 00111000

88]

3I Option (C) is correct

31. The first six points of the 8-point DFT of a real values sequence are 5, 1-j3, 0, 3-j4, 0 and 3+j4. The last two points of the DFT are respectively

(A) 0, 1-j3 (B) 1+j3, 5

(C) 0, 1+j3 (D) 1-j3, 5

Soln. (31) Given

R is a real sequence. DFT of X(k) is imaginary

(seen by the points)

DFT is odd so DFT points should be complex conjugate of each other

X(0) = X*(7)

X(1) = X*(6)

X(2) = X*(5)

GATE 2011

X(3) = X*(4)

Given first five DFT points

5, 1 - j3, 0, 3 - 4j and 3+4j

Thus

X(6) = X*(1) = 1+j3

X(7) = X*(0) = 5


32. For the BJT Q1 in the circuit shown below, = , VBEon= 0.7V, VCEsat =0.7V. The switch is initially closed. At time t=0, the switch is opened. The time t at which Q1 leaves the active region is

(A) 10 ms (B) 25 ms

(C) 50 ms (D) 100 ms

Soln. (32) The given circuit is as follows:

GATE 2011

5V

5V

0.5mA

10V

4.3K

5Ft=0

Q1

Applying KVL at the B-E junction

0!d

3!2 3!23!2 ( - ( Since A- -` When the switch is opened

-.m !" !"( )` )` ) )` )` ) )` ! 5!3 ' 2 ' ' 2 , ! !

Or, ' 'a!'a


33. In the circuit shown below, the network N is described by the following Y matrix:

GATE 2011

The voltage gain V2/V1 is

(A) 1/90 (B) -1/90

(C) -1/99 (D) -1/11

Soln. (33) From Y parameters

--

)) So - ) ) - ) ) Using equation (2)

- !) !) From the given figure

) -89

- AB,- Putting the value of I2 in above equation

!) !) !) !) !) !) !) Therefore

w

!!

AB, w


GATE 2011

34. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t=0. The current i(t) at a time t after the switch is

(A) i(t) = 15 exp(-2 x 103t) A (B) i(t) = 5 exp(-2 x 103t) A

(C) i(t) = 10 exp(-2 x 103t) A (D) i(t) = -5 exp(-2 x 103t) A

Soln. (34) Given: Initial charge on capacitor !" ' 2] Voltage on the capacitor

` !0'

a0'ab ")

(Note the direction of voltage)

Now switch is closed

DC

+

100V10

V() = 100V

Then

)

) ) ") Thus ) ) U) )VW &

"

W & AB, )

"W &

GATE 2011

r $ ]! 0'0'ab W'

"W' (l! Option (A) is correct

35. The system of equations

x+y+z =6

x+4y+6z=20

x+4y+z=

has NO solution for values of and given by

(A) = 6 , = 20 (B) = 6 , 20

(C) 6 , = 20 (D) 6 , 20

Soln. (35) Given equation are

@ 4 5 4@ And 5 G@ 3 If G 4\R? Then equation (3) becomes

4@ Equations (2) & (3) are same

i.e. infinite solutions

If G 4\R? 5 4@

Then equation (2) has same left hand side, so cannot have different right hand side. So will not have solution.

If G 4\R? It will have solution

GATE 2011

5 G@ G 4\R? will also give solution


36. A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is

(A) 2/36 (B) 2/6

(C) 5/12 (D) 1/2

Soln. (36) In the first toss, results can be 1, 2, 3, 4, 5, 6

For 1, the second toss results can be 2, 3, 4, 5, 6

For 2, the second toss results can be 3, 4, 5, 6

For3, the second toss results can be 4, 5, 6

For 4, the second toss results can be 5, 6

For 5, the second toss results can be 6

The required probability

! ! ! ! ! Option (C) is correct

37. A current sheet j = 10y A/m lies on the dielectric interface x=0 between two dielectric media with r1 = 5, r1 =1 in region-1 (x0). If the magnetic field in Region 1 at x =0 is H1 = 3x + 30y

A/m, the magnetic field in Region 2 at x= 0+ is :

(A) H2 = 1.5x+30y - 10z A/m (B) H2 = 3x+30y - 10z A/m

(C) H2 = 1.5x+40y A/m (D) H2 =3x+30y + 10z A/m

Soln. (37) Given

Current sheet ~ e( Magnetic field in region I at

GATE 2011

I~ 3 3e( For magnetic fields boundary condition are

R~ R~ and I I h ' \S Here (Normal component) I I ' 3 ' I or, I !" I I ' > I I > I 3 > So the total field is

I !" 3 > Option (A) is correct

38. A transmission line of characteristic impedance 50 is terminated in a load impedance ZL. The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of /4 from the load. The value of ZL is :

(A) 10 (B) 250

(C) (19.23 + j46.15) (D) (19.23 j46.15)

Soln. (38) Voltage maximum at the load is observed at Ge5 Therefore @9should be real )f8 >u

@9 00 Voltage minimum is at the load


GATE 2011

39. X(t) is a stationary random process with auto correlation function Rx()=exp(- 2). This process is passed through the system shown below. The power spectral density of the output process Y(t) is

(A) (42f2 + 1) exp(-f2) (B) (42f2 - 1) exp(-f2)

(C) (42f2 + 1) exp(-f) (D) (42f2 - 1) exp(-f)

Soln. (39)

PSD fK P6K PfK fK 5K fK fK U8V fK WE Thus U V Option (A) is correct

40. The output of a 3 stage Johnson (twisted ring) counter is fed to a digital to analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is

GATE 2011

Soln. (40) The truth table for Johnson counter and DA counter output is given below:

GATE 2011

Q2 Q1 Q0 D2 D1 D0 V0

0 0 0 0 0 0 0

1 0 0 1 0 0 4

1 1 0 1 1 0 6

1 1 1 1 1 1 7

0 1 1 0 1 1 3

0 0 1 0 0 1 1

0 0 0 0 0 0 0

Thus the output at option (A) matches the V0 output

41. Two D flip flops are connected as a synchronous counter that goes through the following QB QA sequence 00 11 01 10 00 .....

The connections to the inputs DA and DB are

(A) DA = QB, DB = QA

(B) DA =, DB = (C) DA = (QA + QB), DB =QA (D) DA = (QAQB + ), DB =

Soln. (41) D FFs are connected as a synchronous counter. The sequence goes through the following sequence

Present state Next state

QB QA QB QA

0 0 1 1

1 1 0 1

0 1 1 0

1 0 0 0

0 0 1 1

Now using the excitation table of D Flip-flop

GATE 2011


42. In the circuit shown below, for the MOS transistor nCox = 100mA/V2 and the threshold

voltage VT = 1V. The voltage Vx at the source of the upper transistor is

(A) 1 V (B) 2 V

(C) 3 V (D) 3.67 V

Soln. (42) For upper MOS

) 4 ) ) ): " ) 5 ) Upper MOS will be in saturation because

) ) ): For lower MOS

) ) ) ): ) So ) ) ): Hence both MOS will be in saturation

-w [S ] y9 { ) ): -w [S]5) Similarly, - [S])

GATE 2011

But -w - Thus [S]55) [S]5) On solving ) 3) Option (C) is correct

43. An input x(t) = exp(-2t)u(t) + (t-6) is applied to an LTI system with impulse response h(t) = u(t).The output is

(A) [1 - exp(-2t)]u(t) + u(t+6) (B) [1 - exp(-2t)]u(t) + u(t-6)

(C) 0.5[1 - exp(-2t)]u(t) + u(t+6) (D) 0.5[1 - exp(-2t)]u(t) + u(t-6)

Soln. (43) Given W 4 W/ I I '

ab

! ab

!" W 4 Option (D) is correct

44. For a BJT, the common base current gain = 0.98 and the collector base junction reverse bias saturation current Ico = 0.6A. This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20A. The collector current Ic for this mode of operation is

(A) 0.98 mA (B) 0.99 mA

(C) 1.0 mA (D) 1.01 mA

Soln. (44) For common emitter configuration

-. - -`

GATE 2011

XX !HL!HL 5 -. 5 ' " ' !4 7 3 [ , ! Option (D) is correct

45. If F(S) = L[f(t)] = 2s(s+1)/(s2+4s+7) then the initial and final values of f(t) are respectively

(A) 0, 2 (B) 2, 0

(C) 0, 2/7 (D) 2/7, 0

Soln. (45) K 3d


46. In the circuit shown below, the current I is equal to

(A) 1.4 0 A (B) 2.0 0 A

GATE 2011

(C) 2.8 0 A (D) 3.2 0 A Soln. (46) Applying delta to star conversion circuit becomes

14 00

V2 2

2

j4 j4

Net impedance

65PP 65 3/3

- 3ud Option (B) is correct

47. A numerical solution of the equation f(x) = x+x -3 = 0 can be obtained using Newton Raphson method. If the starting value is x=2 for the iteration, the value of x that is to be used in the next step is

(A) 0.306 (B) 0.739

(C) 1.694 (D) 2.306

Soln. (47) The given function is

K _ 3 As per the Newton Raphson method

S S Starting values is for iteration. K _ 3 _ K _

GATE 2011

K _ Then

uwu

_ w_

!45 Option (C) is correct

Common Data questions (48-49):

48. The channel resistance of an N channel JFET shown in the figure below is 600 when the full channel thickness (tch) of 10m is available for conduction. The built in voltage of the gate P+N junction (Vbi) is -1V. When the gate to source voltage (VGS) is 0 V, the channel is depleted by 1 m on each side due to the built in voltage and hence the thickness available for conduction is only 8m.

The channel resistance when VGS = 0V is

(A) 480 (B) 600

(C) 750 (D) 1000

Soln. (48) Depletion layer width

) )p Given )p ) and for ) ), [( and for ) 3), can be calculated

GATE 2011

So w

2

[( Channel width increases channel resistance increases For 10[ m channel resistance is 600 for 7[( channel resistance /'L = 750


49. The channel resistance when VGS = -3V is

(A) 360 (B) 917

(C) 1000 (D) 3000

Soln. (49) For 4[( Channel resistance

/'/


Common Data questions (50-51):

The input output transfer function of a plant H(S) = 100/[s(s+10)2]. The plant is placed in a unity negative feedback configuration as shown in the figure below.

50. The signal flow graph that DOES NOT model the plant transfer function H(S) is

GATE 2011

Soln. (50) Phase cross over frequency

ys{ 7

\R ys{

Since \R ys{ 5" Or, ^ B\?eWr! PI6^P P1P1

' g o & o 4? Option (C) is correct

GATE 2011

51. The gain margin of the system under closed loop unity negative feedback is

(A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

Soln. (51) Option D is considered

is forward gain ! ! !

Z

ww

Z Z ZZ L1 L2 non touching loop

wuu 'wuu

ywuu {

ywuu {

Option (D) does not match

GATE 2011

LINKED ANSWER QUESTIONS (52-53):

A four phase and an eight phase signal constellation are shown in the figure below.

52. For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r1 and r2 of the circles are

(A) r1 = 0.707d, r2 =2.782d (B) r1 = 0.707d, r2 =1.932d

(C) r1 = 0.707d, r2 =1.545d (D) r1 = 0.707d, r2 =1.307d

Soln. (52) For M-ary

? s yE{j Where jis distance of any point from the origin For 4 ary B jw 7 ary B j For 4 ary M = 4 ? R yE3{ B

For 7 ary M = 8 ? R yEL{ B If ? ? ? R yE3{ B ? AB,B _ !?

R yEL{ B ?

GATE 2011

y{ !


53. Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is

(A) 11.90 dB (B) 8.73 dB

(C) 6.79 dB (D) 5.33 dB

Soln. (53) Probability of error

j So

w

w

!dd!2d

w y!2d!dd{

3!5 To achieve same error, 2nd must have 3.42 time than Ist

The value in dB = 10log (3.42) = 5.33 dB


Statement for Answer Questions (54-55):

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV. The small signal input vi= Vp cos(t) where Vp= 100mV.

GATE 2011

54. The bias current IDC through the diodes is

(A) 1 mA (B) 1.28 mA

(C) 1.5 mA (D) 2 mA

Soln. (54) -` !d!LHH ( Option (A) is correct

55. The ac output voltage Vac is

(A) 0.25 cos(t) mV (B) cos(t) mV

(C) 2cos(t) mV (D) 22 cos(t) mV

Soln. (55) For a.c. analysis diodes will be replaced by its dynamic resistance

B '0## " ). vHH '

v, '

v ) cos^

'a cos^ ) rA^()

GATE 2011


GENERAL APTITUDE (GA) QUESTIONS

Q.(56-60) Carry one mark each.

56. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair:

Gladiator : Arena

(A) dancer : stage (B) commuter : train

(C) teacher : classroom (D) lawyer : courtroom

Soln. (56) Gladiator : Arena

Gladiator - A person (often a slave or captive) who was armed with a sword or other weapon and compelled to fight to death in public arena against another person or a wild animal, for the entertainment of spectators.

Best choice is Lawyer : courtroom

Option D is best choice

57. There are two candidates P and Q in an election. During the campaign, 40% of the voters promised to vote for P, and rest for Q. However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q. 25% of the voters went back on their promise to vote for Q and instead voted for P. Suppose, P lost by 2 votes, then what was the total number of voters?

(A) 100 (B) 110

(C) 90 (D) 95

Soln. (57) Lets take total number of voters be 100

P Q

(40) (60)

40% 60%

GATE 2011

(-)15% of 40(-6) -25% of 60(-15)

+25% of 60 (+15) +15% of 40(+6)

49 51

P lost by 2 votes.


58. Choose the most appropriate word from the options given below to complete the following sentence:

Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which _____________ treatments are unsatisfactory.

(A) Similar (B) Most

(C) Uncommon (D) Available

Soln. (58) Option (D) Available is the most appropriate choice.

59. Choose the word from the options given below that is most nearly opposite in meaning to the given word:

Frequency

(A) Periodicity (B) Rarity

(C) Gradualness (D) Persistency

Soln. (59) The best choice is rarity which means shortage or scarcity

Option (C)

60. Choose the most appropriate word from the options given below to complete the following sentence:

It was her view that the countrys problems had been ___________ by foreign technocrats, so that to invite them to come back would be counterproductive.

(A) identified (B) ascertained

(C) exacerbated (D) analysed

GATE 2011

Soln. (60) The clue is that foreign technocrats did something negatively to the problem so it is counterproductive to invite them. All other options are non negative. The best choice is exacerbated which means aggravated.


Q.(61-65) carry two marks each.

61. The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood. Serums to fight with diphtheria and tetanus were developed this way.

It can be inferred from the passage, that horses were

(A) given immunity to diseases

(B) generally quite immune to diseases

(C) given medicines to fight toxins

(D) given diphtheria and tetanus serums

Soln. (61) Here option B is most appropriate

62. The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below.

The distances covered during four laps of the journey are listed in the table below

Lap Distance (kilometers) Average speed (kilometers per hour)

P 15 15

GATE 2011

Q 75 45

R 40 75

S 10 10

From the given data, we can conclude that the fuel consumed per kilometer was least during the lap

(A) P (B) Q

(C) R (D) S

Soln. (62) Fuel consumed per Km. will be least when mileage (kilometre per litre) mentioned in the graph on y axis will be maximum.

From the graph we observe that mileage is maximum when vehicle is driven at 45Km/hour. Thus the stretch which Q covered at 45 kmph mileage was highest and fuel consumption per litre lowest.


63. Three friends, R, S and T shared toffee from a bowl. R took 1/3rd of the toffees, but returned four to the bowl. S took 1/4th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl?

(A) 38 (B) 31

(C) 48 (D) 41

Soln. (63) R S T

8 3& 5 5&

3 &S

Remaining 17

Let the total no. of toffees in the bowl With R 2 5 Remaining toffees in the bowl

2 5

GATE 2011

No. of toffees with S 3 2 5 3 Remaining is bowl 23 2 5 5

No. of toffees with T 23 2 5 5

Remaining toffees in bowl 23 2 5 5

Given 23 2 5 5

Or 23

2 5 4


64. Given that f(y) = |y|/y, and q is any non zero real number, the value of |f(q)-f(-q)| is

(A) 0 (B) -1

(C) 1 (D) 2

Soln. (64) Given K PP &

q is non zero real no.

to find

PKo KoP K PP Ko

PP

Ko PP PP

PKo KoP PP PP

PP


GATE 2011

65. The sum of n terms of the series 4 + 44 + 444 + . is

(A) (4/81) [10n+1 -9n -1]

(B) (4/81) [10n-1 -9n -1]

(C) (4/81) [10n+1 -9n -10]

(D) (4/81) [10n -9n -10]

Soln. (65) Sum of series f 5 55 555 5 3H 3H 2 3H R

3H ! H R

3L S R


gate 2011 q soln1

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