of 49
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
Option (A) is correct
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
Option (D) is correct
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
Option (D) is correct
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
Option (B) is correct
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
Option (C) is correct
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
Option (C) is correct
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
Option (D) is correct
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
Option (B) is correct
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
Option (A) is correct
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
Option (D) is correct
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
Option (B) is correct
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
Option (C) is correct
49. The channel resistance when VGS = -3V is
(A) 360 (B) 917
(C) 1000 (D) 3000
Soln. (49) For 4[( Channel resistance
/'/
Option (C) is correct
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{ !
Option (D) is correct
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
Option (D) is correct
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
Option (B) is correct
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.
Option (A) is correct
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.
Option (C) is correct
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.
Option (B) is correct
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
Option (C) is correct
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
Option (D) is correct
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
Option (C) is correct