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Electronics and Communication Engineering Q. No. 1 to 25 Carry One Mark Each 1.
Consider the circuit shown in the figure. 0
Y
0
MUX
0
1
MUX
F
1 X
Z
The Boolean expression F implemented by the circuit is (A) XYZ XY YZ
(B) XYZ XZ YZ (D) XYZ XY YZ
Key
(C) XYZ XY YZ (B)
Exp:
F xyz z xy
y
0
xy
F xyz x y z F xyz xz yz
0
0
F
1
1
z
x
An LTI system with unit sample response h n 5 n 7 n 1 7 n 3 5 n 4 is a
2.
(A) Low – pass filter (B) high – pass filter (C) band – pass filter (D) band – stop filter Key:
(C)
Exp:
h n 5 n 7 n 1 7 n 3 5 n 4
Obtain h e j 5 7e j 7e3 j 54 j
At 0 and ; 2 j h e 0 For 0
at a frequency 0 maximum value of h e j is obtained 2 h e j
0 2 2 Thus Ideal behaviour of h[n] is Band pass filter.
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In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V. The ratio
amplitude of voltage across the capacitor is __________. amplitude of voltage across the resistor
5H
5
I 5F
V
Key: Exp:
(0.19 to 0.21) If I &V are in phase then the circuit is in resonance At resonance VC 1 L 1 5 Q 0.2 VR R C 5 5
4.
In a DRAM, (A) periodic refreshing is not required (B) information is stored in a capacitor (C) information is stored in a latch (D) both read and write operations can be performed simultaneously
Key: (B) 5.
Consider an n-channel MOSFET having width W, length L, electron mobility in the channel n and oxide capacitance per unit area Cox . If gate-to-source voltage VGS=0.7V, drain-to-
source voltage VDS=0.1V, n Cox 100A / V 2 , threshold voltage VTH=0.3 V and (W/L) =50, then the transconductance gm (in mA/V) is ___________. Key:
(0.45 to 0.55)
Exp:
Here, VDS VGS VPH , so n-channel MOSFET is working in linear region. I D n Cox
2 VDS W V V .V GS TH DS L 2
So, transconductance g m is in linear region and is given by
gm
6.
ID VGS
n Cox VDS const
W .VDS 100 106 50 0.1 5 104 0.5 mA V L
Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure. S2 S1 Wire
Radius a Radius b ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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For some charge placed on this structure, the potential and surface electric field on S1 are Va and Ea, and that on S2 are Vb and Eb, respectively, which of the following is CORRECT? (A) Va Vb and Ea Eb
(B) Va Vb and Ea Eb
(C) Va Vb and Ea Eb
(D) Va Vb and Ea Eb
Key:
(C)
Exp:
bc3 two spheres are joined with a conducting wire, the voltage on two spheres is same.
Va Vb The capacitance of sphere radius Ca a Cb b We know Q CV
Q a Ca a Q b Cb b 1 E a 4o 1 Eb 4o
a a2 b 1 b a b2
Ea E b 7.
For the circuit shown in the figure, P and Q are the inputs and Y is the output.
PMOS
Y NMOS
P Q
The logic implemented by the circuit is (A) XNOR
(B) XOR
(C) NOR
(D) OR
Key:
(B)
8.
An n-channel enhancement mode MOSFET is biased at VGS VTH and VDS VGS VTH , where VGS is the gate-to-source voltage, VDS is the drain-to-source voltage and VTH is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a (A) voltage source with zero output impedance (B) voltage source with non-zero output impedance (C) current source with finite output impedance (D) current source with infinite output impedance
Key: (C) Exp:
If channel length modulation is considered and significant it means 0
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V 1 and re A ID
Vgs
g m v gs
re
Vo
If VAS VTH and VDS VDS VTH then it indicates that MOSFET is working in saturation region and it can be used as an amplifier. So it can act as current source with finite output impedance. 9.
A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10, 5, 2 are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.
Key: Exp:
(2.12 to 2.16) The maximum resistance 5 10
R max 11.428 5
The minimum resistance 10
2
R min 5.333
R max 2.14 R min 10.
5
An npn bipolar junction transistor (BJT) is operating in the active region. If the reverse bias across the base – collector junction is increased, then (A) the effective base width increases and common – emitter current gain increases (B) the effective base width increases and common – emitter current gain decreases (C) the effective base width decreases and common – emitter current gain increases (D) the effective base width decreases and common – emitter current gain decreases
Key:
(C)
Exp:
If the reverse bias voltage across the base collector junction is increased, then their effective base width will decrease and collector current will increase, therefore their common-emitter current gain increases.
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Consider the state space realization x1 t 0 0 x1 t 0 u t , with the initial condition 0 9 x t 45 2 x t 2
x1 0 0 x 2 0 0
,
x12 t x 22 t is _______. where u(t) denotes the unit step function. The value of lim t
Key:
(4.99 to 5.01)
Exp:
x 1 t 0
...1
x .2 t 9x 2 t 45u t
... 2
Apply L.T to above equation
because initial conditions are zero
x1 t 0
Sx 2 s x 2 0 9X 2 s X 2 s s 9 X2 s
45 S
45 3
45 s s 9
S 5 5 s9 X 2 t 5u t 5e 9t u t X2 s
It
t
x12 t x 22 t It x 2 t 5 t
12.
1 0 The rank of the matrix 0 1 0
Key:
(4 to 4)
Exp:
1 1 0 0 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1
1 0 1 0 0
0 1 1 0 0
0 1 0 0 1
0 0 0 is ___________. 1 1
1 1 0 0 0 0 0 1 1 0 R 4 R 4 R 1 ~ 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1
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1 1 0 0 0 0 0 1 1 0 R 4 R 4 R 3 ~ 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1
1 1 0 0 0 0 1 1 0 0 R 2 R 3 ~ 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 R 4 R 4 R 3 ~ 0 0 1 1 0 0 0 0 1 1 0 0 0 1 1 1 1 0 0 0 1 1 0 R 5 R 5 R 4 ~ 0 0 1 1 0 0 0 1 0 0 0 0
0 0 0 1 0
Which is in Echelon form Rank No. of non zero rows 4 13.
A two – wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ______________
Key:
(48.0 to 51.0)
Exp:
Percentage of power reflected is 100 2
VSWR 1 5.8 1 4.8 0.7058 VSWR 1 5.8 1 6.8
% Power reflected = 100 49.82% 2
14.
The input x(t) and the output y (t) of a continuous-time system are related as yt
t
t T
x u du. The system is
(A) Linear and time-variant
(B) Linear and time-invariant
(C) Non-linear and time-variant
(D) Non-linear and time-invariant
Key:
(B)
Exp:
Given Input-output relationship describes integration over a fundamental period T. The integration over one period is linear and time-invariant.
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Which of the following statements is incorrect? (A) Lead compensator is used to reduce the settling time. (B) Lag compensator is used to reduce the steady state error. (C) Lead compensator may increase the order of a system. (D) Lag compensator always stabilizes an unstable system.
Key:
(D)
Exp:
The phase-lead controller adds zero and a pole, with the zero to the right of the pole, to the forward-path transfer function. The general effect is to add more damping to the closed-loop system. The rise time and settling time are reduced in general.
Reduces the steady state error Reduces the speed of response (i.e decreases) Increases the gain of original network without affecting stability Permits the increases of gain if phase margin is acceptable System becomes lesser stable Reduces the effect of noise Decrease the bandwidth
16.
The residues of a function f z (A)
1 1 and 27 125
(B)
1
z 4 z 1
1 1 and 125 125
Key:
(B)
Exp:
Z = 4 is a pole of order „1‟ (or) simple pole
3
are
(C)
1 1 and 27 5
(D)
1 1 and 125 5
1 1 1 Residue of f(z) at z= 4 = Res f z lim z 4 . and 3 3 z 4 z4 z 4 z 1 5 125
z = – 1 is a pole of order „3‟.
Res f z z 1
d 2 1 lim 2 3 1! z1 dz
1 3 z 1 . 3 z 4 z 1
d2 1 1 1 lim 2 2 z 1 dz z 4 125
17.
A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______
Key: Exp:
(7 to 7) For sinusoidal signal
SNR Q in dB 6.0 n 1.75 Given required SNR Q 40dB ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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6.0 n 1.75 40dB 6.0n 40 11.75 40 11.75 6.02 n 7 Since ' n ' must be an int eger n
18.
The general solution of the differential equation
d2 y dy 2 5y 0 in terms of arbitrary 2 dx dx
constants K1 and K2 is (A) K1e
1 6 x
1 6 x K2 e
(B) K1e
1 8 x
(C) K1e
2 6 x
2 6 x K2e
(D) K1e
2 8 x
Key:
(A)
Exp:
D2 2D 5 0 D 1 6 y k1 .e
1 8 x K2 e
2 8 x K2 e
roots are real and distinct
1 6 x
k 2e
1 6 x
Where k1, k2 are arbitrary constants. 19.
Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memory less binary symmetric channel with crossover probability P? (A) (B) 1 Capacity
1 Capacity
p
p 0
0
1
(C)
1
(D) 1 Capacity
1 Capacity
p
p 0
Key: Exp:
0
1
1
(C) For memory less binary Symmetric channel Channel capacity
C
C 1 H p H p p log 2
1 1 1 p log 2 p 1 p
1
p Cross over probability C 1 p log 2 p 1 p log 2 1 p At At At
p 0; C 1 p 1 C 1 p 1 2 C 0
0
1
P
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The output V0 of the diode circuit shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in Volts, neglecting the voltage drop across the diode, is ____________.
10sin t f 50Hz
1k
V0
Key:
(3.15 to 3.21)
Exp:
Vo
21.
Consider the random process X t U Vt, where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is ____________
Key:
(2)
Exp:
Given x t U Vt
Vm 10 3.1847V
x 2 U 2V E x 2 E U 2V E U 2E V 0 2 1 2
22.
For the system shown in the figure, Y (s) / X (s) = __________.
X s
Key: Exp:
(0.95 to 1.05) Y s X s
G s 2
X s
Y s
2 1 1 1 2
G s
Y s
23.
The smaller angle (in degrees) between the planes x + y + z =1 and 2x – y + 2z = 0 is ________.
Key:
(54.0 to 55.0)
Exp:
x y 2 1 2x y 2z 0 We have angle between two planes a1x b1 y c1 y d1 0 a x x b2 y c2 y d 2 0
is cos cos
a1a 2 b1b2 c1c2 a12 b12 c12 a 22 b 22 c22 2 1 2 111 4 1 4
3 1 54.73 3 9 3
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Consider the circuit shown in the figure. Assume base-to- emitter voltage VBE=0.8 V and common base current gain of the transistor is unity. 18V
44k
16k
4k
2k
The value of the collector- to – emitter voltage VCE (in volt) is _______. Key:
(5.5 to 6.5)
Exp:
Given VBE 0.8V; 1 As 1; is very large So IE IC
18 16 4.8V 60 4.8 08 IC 2mA 2 103 VCE 18 6 103 2 103 VB
18 12 6V 25.
In the figure, D1 is a real silicon pn junction diode with a drop of 0.7V under forward bias condition and D2 is a zener diode with breakdown voltage of -6.8 V. The input Vin(t) is a periodic square wave of period T, whose one period is shown in the figure. 10F
Vin t 14V
D1 10
0 14V
t sec onds
Vout t
D2
Assuming 10 T. where is the time constant of the circuit, the maximum and minimum values of the output waveform are respectively, (A) 7.5 V and –20.5V (B) 6.1 V and –21.9V (C) 7.5 V and –21.2 V Key:
(A)
Exp:
When Vi 14V, the equivalent circuit is
(D) 6.1 V and –22.6 V
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VC
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0.7V
Vo
10
Vi
6.8V
Vi VC Vo VC Vi Vo
VC 14 6.8 0.7 14 7.5 6.5V Maximum Vo 7.5V When Vi 14V, the equivalent circuit is VC
Vi
10
Vo
Vo Vi VC 14V 6.5 20.5V Minimum Vo 20.5V
Q. No. 26 to 55 Carry Two Marks Each
26.
If the vector function F ax 3y k1z ay k 2 x 2z az k 3 y z is irrotational, then the values of the constants k1, k2 and k3 respectively, are (A) 0.3, –2.5, 0.5 (B) 0.0, 3.0, 2.0 (C) 0.3, 0.33, 0.5 (D) 4.0, 3.0, 2.0
Key:
(B)
Exp:
curl F 0
i j k 0 x y z 3y k1z k 2 x 2z k 3 y z i k 3 2 j 0 k1 k k 2 3 0 k1 0, k 2 3, k 3 2
27.
The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________
Key:
(5.19 to 5.23)
Exp:
Total power when 50% is 2 PT PC 1 2
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0.5 2 PT 5 1 5 1 0.125 5 1.125 2 PT 5.625
When 40% Total power remains 5.625 0.4 2 5.625 PC 1 5.625 PC 1 0.08 2 PC 5.22
28.
Consider an LTI system with magnitude response |f | , | f | 20 1 | H f | 20 0, | f | 20
And phase response Arg H f 2f . If the input to the system is x t 8cos 20t 16sin 40t 24 cos 80t . 4 8 16
Then the average power of the output signal y (t) is _________. Key:
(7.95 to 8.05)
Exp:
Consider an LTI system with magnitude response |f | , | f | 20 1 | H f | 20 0, | f | 20
And phase response Arg H f 2f . If the input to the system is x t 8cos 20t 16sin 40t 24 cos 80t . 4 8 16
Then the average power of the output signal y (t) is _________. 28. Obtain X f for the given x(t) X f 4 f 10 f 10 8 f 20 f 20 12 f 40 f 40 12
Xf
Hf
8
1
multiply
4
40
20
10
10
20
40 f
40
1
20
1 2
10
2
10
20
f
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4 f 10 f 10 2 1 f 10 f 10 2
Y f
y t 4cos 2t Thus max power
29.
16 8 2
A MOS capacitor is fabricated on p-type Si (silicon) where the metal work function is 4.1 eV and electron affinity of Si is 4.0 eV. EC-FF=0.9 eV, where EC and EF are the conduction band minimum and the Fermi energy levels of Si, respectively. Oxide r 3.9, 0 8.85 1014 F / cm. oxide thickness tox = 0.1 m and electronic charge q = 1.6×10-19 C. If the measured flat band voltage of the capacitor is –1V, then the magnitude of the fixed charge at the oxidesemiconductor interface, in nC/cm2, is __________.
Key:
(6.85 to 6.95)
Exp:
VFB MS
QF Cox
q MS qM qS
qM q xo Ec E f 4.1 4.0 0.9 0.1 0.9 0.8eV
0.8 q 1V 0.8V q E Cox ox 34.5 109 F cm 2 t ox MS
QF 34.5 109 QF 0.2 34.5 109 1 0.8
Q F 6.9 nc cm 2 30.
An electron (q1) is moving in free space with velocity 105 m/s towards a stationary electron (q2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is ___________ ×10-8m. [Given, mass of electron m = 9.11×10-31kg, charge of electron e = -1.6×10-19 C , and permittivity 0 1/ 36 109 F / m]
Key: Exp:
(4.55 to 5.55) Work done due to field and external agent must be zero
1 qV MV 2 2 1.6 1019
2 1.6 1019 1 m 105 4o 2
5.058 108 m ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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The values of the integrals 11 1 xy xy dy dx dx and 0 0 x y 3 0 0 x y 3 dy are 1
Key: Exp:
(A) same and equal to 0.5
(B) same and equal to -0.5
(C) 0.5 and – 0.5, respectively
(D) - 0.5 and 0.5, respectively
(C) 11 xy dx dy To find 3 0 0 x y
xy
1
Consider
1
x y
3
0
x. x y
31 1
3 1
dy x 0
1
1
x y
3
dy 0
y
x y
3
dy
x y x dy 3 0 x y
1
0
1 1 x 1 1 1 1 dy x dy 2 2 3 2 2 x 1 x 0 x y 0 x y 1
2 1 x 1 1 x y x 1 1 2 x 12 x 2 2 1 2 x 12 x 2 0
1 1 1 1 x 2 2 x 1 x 1 x x 1 2 x 1 1 11 1 xy 1 1 1 dy dx dx 3 2 x 1 0 2 0 0 x y 0 x 1 11 xy 1 dx dy Similarly 3 2 0 0 x y
32.
Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is _________.
Key: (2.4 to 2.6) Exp: Let „X‟ is a random variable which takes number of attempts Given probability of any attempts to be successful, p 40%
X pX
1
40 2 2 3 , q 1 100 5 5 5
2
3
4
2 3 2 3 3 2 3 3 3 2 5 5 5 5 5 5 5 5 5 5
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3
2 3 2 3 2 3 2 E X Xp X 1 2 3 4 ... 5 5 5 5 5 5 5 2 3 2 2 3 2 2 3 3 3 1 2 3 4 ..... 1 1 x 1 2x 3x 2 4x 3 .... 5 5 5 5 5 5
2 25 2.5 5 4 Average number of attempts that passengers need to make to get seat reserved is „2.5‟
33.
Figure I shows a 4-bits ripple carry adder realized using full adders and Figure II shows the circuit of a full-adder (FA). The propagation delay of the XOR, AND and OR gates in Figure II are 20 ns, 15 ns and 10 ns respectively. Assume all the inputs to the 4-bit adder are initially reset to 0.
Z4
FA
Z3
S3
FA
Z2 FA
S2
Figure I
S1
Xn Yn
Y0 X 0
Y1 X1
Y2 X 2
Y3 X 3
Z1
FA
Z0
Sn Zn 1
Zn
S0
Figure II
At t=0, the inputs to the 4-bit adder are changed to X3X2 X1X0 1100, Y3Y2 Y1Y0 0100 and Z0 1.
The output of the ripple carry adder will be stable at t (in ns) = ___________ Key:
(70.0 to 70.0)
34.
The permittivity of water at optical frequencies is 1.75 0 . It is found that an isotropic light source at a distance d under water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is c . 5m
Air
C
Water
d
Light Source
The value of d (in meter) is _____________ Key:
(4.2 to 4.4)
Exp:
2 1 C sin 1 sin 1 49.106 1 1.75 tan C
5 5 d 4.33m d tan C
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A unity feedback control system is characterized by the open-loop transfer function G s
10K s 2 s3 3s 2 10
The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below. j
j5.43K
s plane
j
G s plane
j
s Re j R
0
0
K
j
2K
j
Re G
j5.43K
Nyquist path for G s
Nyquist Plot of G s
If 0 < K < 1, then the number of poles of the closed-loop transfer function that lie in the right – half of the s-plane is (A) 0 (B) 1 (C) 2 (D) 3 Key: Exp:
(C) N=0, Because 0 < L < 1 There are no encircles around (Y, 0) And G S So,
10K S 2 S 3S 10 3
2
10K S 2
S 3.72 S 0.31 1.598i
P2 NPZ Z2
OR
C.E S3 3S2 10KS 20K 10 If stable 30 K > 20K+10 K>1 Here, in the question asking 0 < K < 1 So, System is unstable 36.
The signal x t sin 14000t , where t is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows : 1, H f 0,
| f | 12kHz | f | 12kHz.
What is the number of sinusoids in the output and their frequencies in kHz? (A) Number = 1, frequency = 7 (C) Number = 2, frequencies = 2, 7 Key:
(B) Number = 3, frequencies= 2,7,11 (D) Number = 2, frequencies = 2, 11
(B)
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Given input signal x t sin 1400t Input signal spectrum
Xf 1 2
7
1 2
7
f kHz
Sampled signal spectrum is the spectrum of X(f) which repeats with integer multiples of 9 kHz.
Xs f
Sampled signal spectrum:
16
7
11 9
2 2
7
9
11 16
f
The sampled signal spectrum is passed through a LPF of cutoff frequency 12 KHz. Thus the filtered out sinusoids are of 2 KHz 7 KHz and 11 KHz frequency. 37.
A unity feedback control system is characterized by the open-loop transfer function G s
2 s 1 s ks 2 2s 1 3
The value of k for which the system oscillates at 2 rad/s is ________. Key:
(0.74 to 0.76)
Exp:
G s
2 s 1 s ks2 2s 1 3
2 rad sec K ?? 1 G s H s 0
1 3
s s2 s1 s0
2 k 1 s ks 2s 1 3
2
s3 ks2 4s 3 0
1 4 k 3 4k 3 0 k 3
For marginal stable 4k 3 3 0 k 0.75 k 4
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Take auxiliary equation
ks 2 3 0 3 2 s 30 4 s j2 2 rad sec k
38.
3 4
Consider the circuit shown in the figure.
3i 0 P 1
10V
i0
1
1 1 Q
The Thevenin equivalent resistance (in ) across P – Q is _________. Key:
(-1.01 to -0.99)
Exp:
To find R th
V I
io
V V 1 Nodal at V Here i o
1
V 3i o V I0 I I V 3V V I 0
V
1
R th 1
39.
The transfer function of a causal LTI system is H(s) = 1/s. If the input to the system is x(t) = [sin(t) / t]u(t), where u(t) is a unit step function, the system output y(t) as t is ______.
Key:
(0.45 to 0.55)
Exp:
Given x t
sin t u t
t By using frequency integration property, xt t
s
L X1 u du
L Consider x1 t sin t u t
1 X1 s s 1 2
1 1 2 du tan s u 1 2 s
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L x t
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1 tan 1 s X s 2
Y s X s H j
1 1 tan 1 s 2s s
By using final value theorem,
1 1 1 lim y t lim sY s lim tan 1 s t s 0 s 0 2 2 2
40.
An integral I over a counter clock wise circle C is given by I C
z2 1 z e dz. z2 1
If C is defined as |z| = 3, then the value of I is (A) i sin 1 Key:
(D)
Exp:
z2 1 z I C z 2 1 e dz
(B) 2i sin 1
(C) 3i sin 1
(D) 4i sin 1
z2 1 z z2 1 Consider f z ez 2 e z 1 z i z i z i are simple poles of f(z) which lie inside |z|=3 z 2 1 iei Residue of f(z) at z i lim z i e z z i z i z i
z 2 1 z ie i & Residue of f(z) at z i lim z i e z i z i z i
z2 1 z i i i i e dz 2i ie ie 2 e e 2 z 1 C
By residue theorem, I
ei e i 4i 2i
41.
4isin 1
Consider a binary memory less channel characterized by the transition probability diagram shown in the figure. 0
0.25
0
0.25
0.75
1
The channel is (A) Lossless Key:
0.75
(B) Noiseless
1
(C) Useless
(D) Deterministic
(C)
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It is a useless channel as MAP criteria cannot decide anything on receiving „0‟ we cannot decide what is transmitted.
42.
An abrupt pn junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the pn junction? Ex
(A)
(B)
p side
n side
0, 0
0, 0
x
x W
W
(C)
n side
W
Ex p side
n side
Ex p side
(D) x
0,0
n side
W
Ex p side
0,0
x
Key:
(A)
Exp:
If left side is p-region and right side is n-region then electric field triangle will be down warded and if the left side is n-region and right side is p-region, then electric field triangle will be upward.
43.
A second – order LTI system is described by the following state equations, d x1 t x 2 t 0 dt d x 2 t 2x1 t 3x 2 t r t dt
Where x1 t and x 2 t are the two state variables and r(t) denotes the input. The output c(t) = x1(t). The system is. (A) Undamped (oscillatory)
(B) Under damped
(C) Critically damped
(D) Over damped
Key:
(D)
Exp:
x 1 (t) x 2 (t)
sX1 (s) X 2 (s) 1 x 2 (t) 2 x1 (t) 3x 2 (t) r(t) sX 2 (s) 2X1 (s) 3X 2 (s) R(s) s2 2 3s X1 (s) R(s) R(s) C(s) X1 (s) 2 s 3s 2 C(s) 1 R(s) (s 1)(s 2) system is over damped
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Consider the parallel combination of two LTI systems shown in the figure. h1 t x t
yt
h2 t
The impulse responses of the systems are h1 t 2 t 2 3 t 1 h2 t t 2.
If the input x(t) is a unit step signal, then the energy of y(t) is ____________. Key:
(7.0 to 7.0)
Exp:
Since h1 t and h 2 t are connected in parallel the resultant system can be given as follows.
xt
yt
h1 t h 2 t
y t x t * h1 t h 2 t
From the given h1 (t) & h 2 (t)
y(t)
h1 t h 2 t 2 t 2 3 t 1 t 2
2
xt ut
2
y t 2u t 2 3u t 1 u t 2
Energy of y t
1
2
2
45.
2 1
1
2
2
t
dt 1 dt 4 1 13 7 2
1
Assuming that transistors M1 and M2 are identical and have a threshold voltage of 1V, the state of transistors M1 and M2 are respectively. 3V (A) Saturation, Saturation (B) Linear, Linear
2.5V
M2
2V
M1
(C) Linear, Saturation (D) Saturation, Linear Key:
(C)
Exp:
If VD VG VTH , then transistor is working in saturation region. So, For M2 transistor
VD2 VG2 VTH 3V 2.5 1 V Assume that M1 is working in saturation, so that ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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ID1 ID2 VGS1 VTH VGS2 VTH 2V VG 2 VS2 2.5V VS2 VS2 VD1 0.5V.
Now, for M1, transistor to work in saturation VD1 VG1 VTH , but it is not satisfied by M1 transistor and VG1 VTH , so, transistor M1 is ON but working in linear region. 46.
A programmable logic array (PLA) is shown in the figure. P P Q Q R R * * * *
* *
F
* * *
P
Q
R
The Boolean function F implemented is (A) P Q R P Q R P Q R
(B) P Q R P Q R P Q R
(C) P Q R P Q R P Q R
(D) P Q R P Q R P Q R
Key:
(C)
Exp:
PQR PQR PQR
47.
A modulating signal given By x(t) = 5 sin 4103 t 10 cos 2103 t V is fed to a phase modulator with phase deviation constant kp=0.5rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is __________
Key:
(69.9 to 70.1)
Exp:
x t 5sin 4 103 t 10 cos 2 102 t
Transform theorem frequency
f1 t f c
1 d kp. m t 2 dt
d m t 5cos 4 103 t 10 cos 2 103 t 4 103 10sin 2 103 t .2 103 dt
d mt 5cos 2 10 4 103 0 20 103 cos 12 20 103 dt t 0.5 f i t 20
1 5 20 70 kHz 2
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The minimum value of the function f x x x 2 3 in the int erval 100 x 100 occurs at x = 1 3
________. Key:
(-100.01 to -99.99)
Exp:
1 x3 f x x x 2 3 x 3 3 f x
3x 2 1 x2 1 3
x2 1 0 x 1 f x 2x f 1 2 0 at x 1, f x has local minimum. f 1 2 0 at x 1, f x has local maximum
1 2 For x = 1, local minimum value f 1 1 3 3 Finding f 100 333433.33
f 100 333233.33
x 100, 100 are end points of interval Minimum occurs at x 100 49.
The switch in the circuit, shown in the figure, was open for a long time and is closed at t = 0. The current i(t) (in ampere) at t = 0.5 seconds is ________ it
5
t0
10A
5 2.5H
Key: Exp:
(8.0 to 8.3) At t=05
10A
5
iL 0 5A
i L O
At t 0 it
5 2.5H
5
5A
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L 2.5 1 iL 0 R 5 2 2t i L 5e
i t 10 5e 2t
At t 0.5s i 0.5 10
50.
5 8.16A e
In the voltage reference circuit shown in the figure, the op-amp is ideal and the transistors Q1 Q2….., Q32 are identical in all respects and have infinitely large values of common – emitter current the relation IC=IS exp ((VBE/VT), where Is is the saturation current. Assume that the voltage VP shown in the figure is 0.7 V and the thermal voltage VT=26mV.
20k
20k
Q2
Vout 15V
5k Vp
Q1
15V
Q3
Q32
The output voltage Vout(in volts) is __________. Key: Exp:
(1.1 to 1.2) KCL at node „a‟ Vo Vi Vx 0.7 20 5
20k 5k
VC Vi 4Vx 2.8
VS 31I
Vo 5Vx 2.8
Now, Is e
VX VT
31 s e
a Vx
Vo
VS VT
V Vx n31 S VT VT
20k
Vx VS n31 VT
Vx 0.789V From equation (i) Vo 5 0.789 2.8 1.145V
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The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input „In‟ and an output „Out‟. The initial state of the FSM is S0. In 0 Out 0
00 In 1 Out 0
In 0 Out 0
S0
In 1 Out 0
01 In 0 S1 Out 0
In 0 Out 0
10 S2 In 1 Out 1
In 1 Out 0
11
S3
If the input sequence is 10101101001101, starting with the left-most bit, then the number times „Out‟ will be 1 is __________. Key: Exp:
(4 to 4) From the state diagram, let us obtain the transition of states and out when IN channel. Initial state is So, the input sea is 10101101001101 IN 1 then S0 S1 , with out 0 When Next
IN 0 then S1 Sz with out 0 IN 1 then S1 Sn with out 1 IN 0 then S3 S2 with out 0 IN 1, then S2 S3 with out 1 IN 1, then S3 S1 with out 0 IN 0, then S1 S2 with out 0 IN 1, then S2 S3 with out 1 IN 0, then S3 S2 with out 0 IN 0, then S2 Su with out 0 IN 1, then S0 S1 with out 0 IN 1, then S1 S1 with out 0 IN 0, then S1 S2 with out 0 IN 1, then S2 S3 with out 1
The ticketed mark now corresponding to output = 1. So output will be 1 „4‟ times.
52.
Standard air – filled rectangular waveguides of dimensions a = 2.29 cm and b= 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominatnt TE10 mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is. (A) 8.19 GHz f 13.1 GHz (B) 8.19 GHz f 12.45 GHz (C) 6.55 GHz f 13.1 GHz
Key:
(D) 1.64 GHz f 10.24 GHz
(B)
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Cut off frequency of TE10 is f c Since, b
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c 1 3 108 1 65.5 108 Hz 2a 2 2.29 10 2
a next higher mode is TE 20 2
c 2 13.1GHz 2 a f 0.95 13.1 12.45GHz fC
TE 20
53.
For a particular intensity of incident light on a silicon pn junction solar cell, the photocurrent density (JL) is 2.5 mA/cm2 and the open-circuit voltage (Voc) is 0.451 V. consider thermal voltage (VT) to be 25mV. If the intensity of the incident light is increased by 20 times, assuming that the temperature remains unchanged. Voc (in volts) will be ______.
Key:
(0.51 to 0.54)
Exp:
3.6 1011 A cm 2 0.451 VOC 0.025 e VT 1 e 1 Now if the intensity of the light is increased by 20 times it means their photocurrent will also increased by 20 times. JS
JL
2.5 103
KT J L n 1 q JS 20 2.5 103 25 103 n 1 11 3.6 10 0.5262 Volt. Voc
54.
In the circuit shown, transistors Q1 and Q2 are biased at a collector current of 2.6mA. Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain V0/Vs in the mid-band frequency range is _____________ (up to second decimal place). 5V
1k VO Q1 VS
Q2 R B2 5V
Key: Exp:
(49.0 to 51.0) a.c equivalent circuit for the given figure
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RC
I 2.6 10 3 gm C 100 m VT 26 10 3 R C 1k; AV
RE
1 ; gm
g m R E 100 1 50 1 gmR E 11
VS
~ RE
A V 50
55.
1k
1 gm
Two n-channel MOSFETs, T1 and T2, are identical in all respects except that the width of T2 is double that of T1. Both the transistors are biased in the saturation region of operation, but the gate overdrive voltage (VGS-VTH) of T2 is double that of T1, where VGS and VTH are the gate – to – source voltage and threshold voltage of the transistors, respectively. If the drain current and transconductance of T1 are ID1 and gm1 respectively, the corresponding values of these two parameters for T2 are (A) 8ID1 and 2gm1
Key: Exp:
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(B) 8ID1 and 4gm1
(C) 4ID1 and 4gm1
(D) 4ID1 and 2gm1
(B) Drain current in saturation is 1 W 2 I D n Cox VGS VTH 2 L For transistor T1 I ID ID1 and g m g m1 D1 n Cox VGs VTH VGS L For transistor T2 W2 2W1 2W
VGS VTh 2 2 VGS VTh 1 2 VGS VTh 2 1 2W I D2 n Cox 2 VGS VTH 8I D1 2 L ID2 2W g m2 n Cox 2 VGs VTh 4g m 1 VGS2 L
General Aptitude Q. No. 1 - 5 Carry One Mark Each 1.
The ninth and the tenth of this month are Monday and Tuesday ___________. (A) figuratively
(B) retrospectively
(C) respectively
(D) rightfully
Key:
(C)
2.
500 students are taking one or more courses out of Chemistry, Physics, and Mathematics. Registration records indicate course enrolment as follows: Chemistry (329). Physics (186). ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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Mathematics (295). Chemistry and Physics (83), Chemistry and Mathematics (217), and Physics and Mathematics (63). How many students are taking all 3 subjects? (A) 37 Key:
(D)
Exp:
Given
(B) 43
(C) 47
A x 2 83
... 1
A y 2 63
... 2
A x 3 217
... 3
(D) 53
C 329
x1
And x1 x 2 A x 3 329
...(4)
x 2 A y1 y 2 186
...(5)
x 3 A y 2 z1 295
... 6
x1 x 2 x 3 y1 y 2 z1 A 500
...(7)
P 186
x3
x2 A
y1 y2
z1
M 295
1 2 3 x 2 y 2 x 3 363 3A ...(8) 4 5 6 3A 2 363 3A x1 y1 z1 810 3A 2 363 3A x1 y1 z1 810 3A 726 500 x 2 x 3 y1 A 810
From 8 From 7
3A 726 500 363 3A A 810 863 A 810 A 53 Alternate method n C 329, n P 186, n M 295, n C P 83;
n C M 217, P M 63 n P C M n C n P n M n C P n C m n P M n P C M . 500 329 186 295 83 217 63 n P C m n P C m 500 447 53.
It is _________ to read this year‟s textbook __________ the last year‟s.
3.
(A) easier, than
(B) most easy, than
(C) easier, from
(D) easiest, from
Key:
(A)
4.
Fatima starts from point P, goes North for 3 km, and then East for 4km to reach point Q. She then turns to face point P and goes 15km in that direction. She then goes North for 6km. How far is she from point P, and in which direction should she go to reach point P? (A) 8km, East
Key:
(B) 12 km, North
(C) 6km, East
(D) 10km, North
(A)
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The required distance FP x 100 36 64 x 8, East
4 km 3 km
'x '
F
Q
5 km P
15 km 6 km
10
O
5.
A rule states that in order to drink beer one must be over 18 years old. In a bar, there are 4 people. P is 16 years old, Q is 25 years old, R is drinking milkshake and S is drinking beer. What must be checked to ensure that the rule is being followed? (A) Only P‟s drink
(B) Only P‟s drink and S‟s age
(C) Only S‟s age
(D) Only P‟s drink, Q‟s drink and S‟s age
Key:
(B)
Exp:
For rules to be followed, we need to check P's drink and S's age.
Q. No. 6- 10 Carry Two Marks Each
6.
Each of P, Q, R, S, W, X, Y and Z has been married at most once. X and Y are married and have two children P and Q. Z is the grandfather of the daughter S of P. Z and W are married and are parents of R. Which one of the following must necessarily be FALSE? (A) X is the mother-in-law of R
(B) P and R are not married to each other
(C) P is a son of X and Y
(D) Q cannot be married to R
Key:
(D)
7.
The number of 3-digit numbers such that the digit 1 is never to the immediate right of 2 is (A) 781
(B) 791
Key:
(C)
Exp:
Total no. of 3 digit no‟s = 91010 = 900
(C) 881
(D) 891
The no. of 3-digit numbers in which „1‟ is to the immediate right of 2 = 19 2
1 2
10 choices 1
9 choices 19 choices The no. of 3-digit no‟s such that the digit 1 is never to immediate right of 2 is 900 19 881 ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
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Alternate method Total no. of 3 digit numbers are 9 10 10 9 10 10 900. Numbers with digit 1 is to the immediate right of 2 are
2 1 x
x 2 1 19
1 1 10
9 11
So, no. of 3 digit numbers such that the digit 1 is never to the immediate right of 2 are = 90019= 881
8.
A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25m intervals in this plot. R
425 450
550 575 P
Q
S
575 550 500
T
500
475 0
1
2km
Which of the following is the steepest path leaving from P? (A) P to Q Key: Exp:
(B) P to R
(C) P to S
(D) P to T
(B) Closer lines represents steepest path
Alternate method The steepest path will be the path which is deepest from sea level. So, P to R is the steepest path.
9.
1200 men and 500 women can build a bridge in 2weeks. 900men and 250 women will take 3 weeks to build the same bridge. How many men will be needed to build the bridge in one week? (A) 3000 (B) 3300 (C) 3600 (D) 3900
Key: Exp:
(C) Given 1200 Men + 500 Women can build a bridge in 2 weeks. And 900 Men + 250 Women will take 3 weeks to build the same bridge
To complete in a week; there are 2400 Men + 1000W required in the first equation and 2700 Men + 750 Women required in the second equation. 2400 M + 1000W = 2700M + 750W 1W
6M 5
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The no. of men required to build the bridge in one week 6M 2400M 1000 3600 Men 5 Alternate method Let a man can build the bridge in x weeks and a woman can build the bridge in y weeks.
So,
120 500 1 2 x y 900 250 1 3 x y
By equations i and ii ; weget x 3600; y 3000 A man build the bridge3600 weeks Required men 3600 to build in a week.
10.
“If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective section, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.” Which of the following statements best reflects the author‟s opinion? (A) An intimate association does not allow for the necessary perspective. (B) Matters are recorded with an impartial perspective. (C) An intimate association offers an impartial perspective. (D) Actors are typically associated with the impartial recording of matters.
Key:
(A)
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