PAPER\ue000-2
1 Paper-2
Questions\ue000Q1.\ue000to\ue000Q20.\ue000carry\ue000one\ue000mark\ue000each. Q1.
Q2.
a is \u00b4 such\ue000th = 5B and Aa= t Aa 4\Bu,\ue000then If A and B are\ue000square\ue000matrices\ue000of\ue000order 40 0 d 7 (A)\ue0005
(B)\ue00025
(C)\ue000625
(D)\ue000None\ue000of\ue000these
d 3 -x x The\ue000particular\ue000integral\ue000of\ue000the\ue000differential\ue000equation is ( D - D ) y =e + e , D \u00ba dx 1 x -x (A) ( e +e ) 2 1 2 (C) x ( ex +e x ) 2
Q3.
(C)
2 1
x( e
x
+e
2
(D) x ( ex 2
1
s( s +1) e
-
(B)
2
s
(D)
s +1
-
-x
)
-
x
e )
s
( s + )1 e
-
2
s
( s +1) 2
Consider\ue000the ( ) as\ue000shown\ue000in\ue000fig.\ue000Q4.\ue000The\ue000FT\ue000of x t ( ) is x t x (t )
Fig\ue000Q4
1 1
-1
(A)
2 sin w- 2
t
(B)
w (C) 2 jwcos w Q5.
1
d -t If x( t) = { te u( t )},\ue000then\ue000laplace\ue000transform\ue000of ( ) is x t dt (A)
Q4.
(B)
2cos w- 2
j w (D) 2 jwsin w
If\ue000a\ue000resistor\ue000of\ue00010 W is\ue000placed\ue000in\ue000parallel\ue000with\ue000vo i
vs
Fig\ue000Q5 Linear Resistive Network
(A)\ue000increased
(B)\ue000decreased
(C)\ue000unchanged
(D)\ue000It\ue000is\ue000not\ue000possible\ue000to\ue000s
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
2 Paper-2
Q6.
Q7.
The\ue000current\ue000carrying\ue000capacity\ue000of\ue000a\ue0001\ue000W,\ue0004\ue000MW (A)\ue0000.5 kA
(B)\ue0002 kA
(C)\ue0002\ue000mA
(D)\ue0000.5\ue000mA
The\ue000equation\ue000governing\ue000the\ue000diffusion\ue000of\ue000neutral\ue000atom\ue000is
\u00b6\u00b6 (A) N =D N 2 \u00b6 t \u00b6 x
\u00b6\u00b6 (B) N =D N 2 \u00b6 x \u00b6 t
(C)
(D)
2
Q8.
2
2 \u00b6 \u00b6 N =D N 2 \u00b6 \u00b6 x t
2 \u00b6 \u00b6 N =D N 2 \u00b6 \u00b6 t x
The\ue000p-type\ue000substrate\ue000in\ue000a\ue000monolithic\ue000circuit\ue000should\ue000be\u (A)\ue000any\ue000dc\ue000ground\ue000point (B)\ue000the\ue000most\ue000negative\ue000voltage\ue000available\ue000in\ue000the\ue000circuit (C)\ue000the\ue000most\ue000positive\ue000voltage\ue000available\ue000in\ue000the\ue000circuit (D)\ue000no\ue000where, i.e.\ue000be\ue000floating
Q9.
Consider\ue000the\ue000List\ue000I\ue000and\ue000List\ue000II List\ue000I\ue000(\ue000Oscillator)
List\ue000II\ue000(\ue000Characteristic)
P.\ue000Colpitts\ue000Oscillator
1.\ue000RC\ue000Oscillator
Q.\ue000Phase\ue000shift\ue000Oscillator2.\ue000LC\ue000Oscillator R.\ue000Tunnel\ue000diode\ue000Oscillator 3.\ue000Negative\ue000resistance\ue000Oscillator S.\ue000Relaxation\ue000Oscillator
4.\ue000Sweep\ue000Circuits
The\ue000correct\ue000match\ue000is P
Q
R
S
(A)
1
2
3
4
(B)
2
1
3
4
(C)
1
2
4
3
(D)
2
1
4
3
Multiple Choice Questions
G AT E Electronics & Communications By
R. K. Kanodia Price 425.00
Pages 602
Published by NODIA & COMPANY
For\ue000previous\ue000years\ue000papers\ue000&\ue000sample\ue000chapter\ue000of\ue0 Feel\ue000free\ue000to\ue000call\ue000at\ue00009350292376\ue000for\ue000any\u
3 Paper-2
Q10.
For\ue000the\ue000circuit\ue000shown\ue000in\ue000fig.\ue000Q10, VCB = 05 . V\ue000and b = 100.\ue000The\ue000value\ue000of I Q is +5\ue000V
5\ue000kW
Vo
Fig\ue000Q10
IQ
-5\ue000V
Q11.
(A)\ue0001.68\ue000mA
(B)\ue0000.909\ue000mA
(C)\ue0000.134\ue000mA
(D)\ue000None\ue000of\ue000the\ue000above
A\ue000four-variable\ue000switching\ue000function\ue000has minterms m6 and m9 .\ue000If\ue000the literals complemented,\ue000the\ue000corresponding minterm numbers\ue000are (A) m3 and m0 (B) m9 and m6 (C) m2 and m0 (D) m6 and m9
Q12.
The\ue000diode\ue000logic\ue000circuit\ue000of\ue000fig.\ue000Q12\ue000is\ue000a D2
V1 V2
Q13.
Fig\ue000Q12
D1
Vo
(A)\ue000AND
(B)\ue000OR
(C) NAND
(D)\ue000NOR
The\ue000even\ue000part\ue000of\ue000a\ue000function xn [ ] = [ ]- [ - 4 ]is
un
un
1 (A) {1 +d[ n] - u[ n- 4]- u[- n- 4]} 2 1 (B) {u[n +3] -u[ n -4] 2
+ [dn ]}
1 (C) {u [n]+ u[ -n] - u[ n-4] -u[ -n -4]} 2 (D)\ue000Above\ue000all All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
4 Paper-2
Q14.
The trigonometricFourierseriesforthewaveformshowninfigQ14willbe
x( t)
Fig Q14
A
-p
(A)
(B)
(C)
(D) Q15.
+2A (sint 2 p
1
A 2A
(cost -
1
(cost -
1
A
2
+
p
A 2A 2
+
p
A 2A 2
+
p
-p 2
p
2
p
t
1 sin3t + sin5t .... ) 3 5 1 cos2t + cos3t .... ) 2 3
1 cos3t + cos5t .... ) 5 3
1 1 (sint +cost + sin3t + cos3t .... ) 3 3
Thepoll–zeroconfigurationofaphase–leadcompensatorisgivenby jw
jw
s
s
(A)
(B) jw
jw
s
s
(C) Q16.
(D)
While designing controller, the advantage of pole– zero cancellation is (A) The system order is increased (B) The system order is reduced (C) The cost of controller becomes low (D) System’s error reduced to optimum levels GATE ECE By R. K. Kanodia
MCQs : The book contains only solved Multiple choice questions (MCQ) which is the main requirement of the GATE exam. Each and every problem has its complete solution. We understand the fact that theoretical studies should be done from the standard book, that one has studied for the semester exams and thus one should use the same material to understand the concepts of the same. We have deliberately excluded theoretical matter in the book so as not to mislead the students. However, wherever needed, satisfactory explanation of the formula has been included in the solution.
5 Paper-2
Q17.
Assertion (A): PSKisinferiortoFSK. Reason (R): PSKrequirelessbandwidththanFSK.
Choose correct option: (A) Both A and R individually true and R is the correct explanation of A. (B) Both A and R individually true and but R is not the correct explanation of A. (C) A is true but R is false (D) A is false Q18.
Q19.
Q20.
In a certain telemetry system, the measured values are converted to digital form. The digital values can then be transmitted via FSK (binary or quaternary) , or BPSK or QPSK systems. Out of these the best noi immunity can be obtained with (A) binary FSK
(B) quaternary FSK
(C) BPSK
(D) QPSK
An antenna, when radiating, has a highly directional radiation pattern. When the antenna is receiving, its radiation pattern (A) is more directive
(B) is less directive
(C) is the same
(D) exhibits no directivity at all
The beamwidth between first null of uniform linear array of N equally spaced ( element spacing equally excited antenna is determined by (A) N alone and not by d
(B) d alone and not by N
(C) the ratio N d
(D) the product Nd
= d)
GATE ECE By R. K. Kanodia
Adherence to Pattern: All Multiple choice questions are strictly according to the GATE pattern. Every problem selected and included in the book is a model problem for the preparation of the exam which would thus prepare and equip the students better. Kindly not the standard of Multiple choice questions and their solution in every unit is much better than the ones available in a famous serie problems & solutions as far as GATE is concerned.
6 Paper-2
Questions Q21. to Q75. carry two marks each.
Q21.
Q22.
Q23.
é 0 ê If A =ê êtana êë 2
tan 0
aù 2 úthen ú ú ûú
(A) I
+A
(B) I
-A
(C) I
+ 2A
(D) I
-2 A
æ è
Forwhatvalueof x0ç £ x
p £ ö÷, the function=y x 2 ø (1 +tanx
(A) tan x
(B) 0
(C) cot x
(D) cos x 2a
Thevalueof
f ( x)
òf ( x) +f (2 a0
Q24.
-A
(I
écos a - sin a ù )×ê 2 úisequalto ê ú cosa û ësina
x)
dx is
(A) 0
(B) 1
(C) a
(D) 2a
Theintegratingfactorforthedifferentialequation (xy ) dx+2y dy=0 is given by x + 3
(A) e
-x
4
(B) e x
(C) e x Q25.
has a maxima ? )
(D) e
cospz
òz - 1 dz =? where
c is the circle z
-
3
2
2
x
=3
c
Q26.
(A) i2 p
(B) - i2 p
(C) i6 p2
(D) - i6 p2
If3isthemeanand
3 2
isthestandarddeviationofabinomialdistribution,thenthedistributionis
12
æ1 3 ö (B) ç + ÷ 2è 2 ø
60
æ1 4 ö (D) ç + ÷ 5è 5 ø
æ3 1 ö (A) ç + ÷ 4è 4 ø æ4 1 ö (C) ç + ÷ 5 è 5ø
12
5
GATE ECE By R. K. Kanodia
Levels of MCQs: The Multiple choice questions included in the guide are in a conceptually evolving method, allowing the student to progress from one level of complexity to another but always aiding in understanding the basic foundation of the s Thus, the MCQs gradually and scientifically advance from the basic level to a more complex level, helping in the systematic understanding of the problem rather than an abrupt one.
7 Paper-2
Q27.
Forthedifferentialequation dy = dx
- x2
giventhat y
x:
0
0.2
0.4
0.6
y:
0
0.02
0.0795
0.1762
Using Milne predictor–correction method, the y at next value of x is
Q28.
(A) 0.2498
(B) 0.3046
(C) 0.4648
(D) 0.5114
d æ4sin 4wsin2wö w) = TheinverseFouriertransformof( X j ç ÷is w dwè ø (A) t
2
- 2e- trect {2(t -
(B) t
2
+2e- trect (2( t +4))
4)}
(C) - t{rect (2t +8) + rect (2t - 8)} (D) t{rect (2t +4) -rect (2t - 8)} Q29.
2
d Thetimesignal (x )t correspondingto X ( s) = s 2 ds
æ1 ö 1 ç ÷ çs2 +9 ÷ +s +3 is è ø
2 æ- 3t 2t ö t + sin3t + cos3t ÷u( t) ç ÷ 3 9 è ø
(A) eç (B) ( e
- 3t
+ 2t sin3t +t2 cos3t )u (t
)
æ- 3t + 2t sin3t +t2 cos3t öu( t) ÷ 3 è ø
(C) eç (D) ( e Q30.
- 3t
+ t 2 sin3t +2t cos3t )u (t
)
Theincidencematrixofagraphisasgivenbelow
é1 ê0 ê A = ê0 ê0 ë
0
0
0
1
0
0
1
0
0
-1
1
0
0
1
0
0
-1
0
0
1
0
0
1
-1
1ù
ú ú - 1ú ú 0û 0
The number of possible tree are (A) 40
(B) 70
(C) 50
(D) 240
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
8 Paper-2
Q31.
InthefigQ31thevalueof v 1 is 2W
1W
3W
1W
Fig Q31
+ 2W
8 V
6W
v1
6W
18 V
–
Q32.
(A) 6 V
(B) 7 V
(C) 8 V
(D) 10 V
Inthecircuitoffig.Q32the30Vsourcehasbeenappliedforalongtime.Theswitchisopenedat t = 1 ( ) is ms. At t = 4 ms the 4 v Cms t=1 ms
1 kW
30u (-t ) V
Q33.
0.6 mF
6.25 kW
Fig Q32
+
25 kW
vC
–
(A) 8.39 mV
(B) 2.59 V
(C) 1.13 mV
(D) 2.77 V
Fora RLC seriescircuit= R20W ,L =06 . H,thevalueof C willbe [CD =critically damped, OD =over damped, UD =under damped]. CD
Q34.
OD
>6
UD
(A)
C
= 6 mF
C
C < 6 mF
(B)
C
= 6 mF
C < 6 mF
(C)
C
>6
mF
C
= 6 mF
C < 6 mF
(D)
C < 6 mF
C
=6
C
mF
mF -
InthecircuitshowninFig.Q340v(
C
> 6 mF
> 6 mF
= )8V
( ) tfor t ³ 0 is and iin ( t) = 4(dt) . TheC v Fig Q34
iin
(A) 164e
-t
V
(C) 208(1 - e
- 3t
50 W
20 mF
+ vC –
-
(B) 208e )V
t
V
-3 t
(D) 164e
V
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
9 Paper-2
Q35.
Inthecircuitshowninfig.Q35,whenthevoltage V1 is10V,thecurrent I is1A.Iftheappliedvoltage at port-2 is 100 V, the short circuit current flowing through at port 1 will be
Resistive
V1
Q36.
Fig Q35
I
Linear Network
(A) 0.1 A
(B) 1 A
(C) 10 A
(D) 100 A
V Thenetworkfunctionofcircuitshowninfig.Q36is H (w )= o V1 2 kW
vi
~
= 1
15 kW
C
4
+j 001 . w Fig Q36
+
+ vC –
vo
AvC
–
The value of the C and A is
Q37.
(A) 10 mF, 6
(B) 5 mF, 10
(C) 5 mF, 6
(D) 10 mF, 10
-3 In germanium(n i 2= .4 1´013 cm ) semiconductor at T
cm -3 anddonorconcentrationis N d
= 300 K, the acceptor concentrations is N a = 1013
= 0.Thethermalequilibriumconcentration
p 0 is
. ´10 9 cm -3 (A) 297 . ´ 1012 cm -3 (B) 268 . ´ 1013 cm -3 (C) 295 (D) 2.4 cm Q38.
-3
-3 Asilicon(n i = 15 . ´ 1010 cm ) pn junctionat T
= 300 Khas N d = 1014 cm -3 and N a = 1017 cm -3 .The
built-in voltage is
Q39.
(A) 0.63 V
(B) 0.93 V
(C) 0.026 V
(D) 0.038 V
The maximum electric field in reverse-biased silicon pn junction is E max
concentration are N d4 =10´16 cm -3 and N a
|=3 ´|10
5
V cm. The doping
=4 ´1017 cm -3 . The magnitude of the reverse bias voltage is
(A) 3.6 V
(B) 9.8 V
(C) 7.2 V
(D) 12.3 V
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
10 Paper-2
Q40.
MOSFET is biased in the saturation required ratio W L is
Q41.
2 m n = 525 cm - ,V V TN s = .0 75V, t ox =400 A°. When = 6 mA. region at VGS = 5 V, the required rated ( current is The I D sat )
A ideal n-channel MOSFET has parameters
(A) 14.7
(B) 11.2
(C) 9.61
(D) 7.2
¢ =8 Fora n - channelenhancement-modeMOSFETtheparametersare V. TNV,=k08 n
mA V 2 and
05 required v GS is W L = 5. If the transistor is biased in saturation region w.ithmA, ID = then (A) 1.68 V (B) 2.38 V (C) 4.56 V (D) 3.14 V
Q42.
06 The cutin voltageforeachdiodeinfig.Q42is V g. = V.Eachdiodecurrentis0.5mA.Thevalueof R1 , R2 and R 3 will be respectively +10 V
R1
Fig Q42
+5 V R2 0 V R3
-5 V
Q43.
(A) 10 kW, 5 kW, 2.93 kW
(B) 6 kW, 3 kW, 3.43 kW
(C) 5 kW, 6 kW, 4.933 kW
(D) 6 kW, 8 kW, 6.43 kW
Inthecircuitoffig.Q43 Zener voltageis V Z
=5
Vand b = 100.Thevalueof I CQ and VCEQ are
+12 V
500 W
Fig Q43
(A) 12.47 mA , 43 . V
(B) 12.47 mA , 5.7 V
(C) 1 0. 4 3A , 5.7 V
(D) 10.43 A , 43 . V
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
11 Paper-2
Q44.
¢ =k08 . V, =4 0 mA / V 2 and l = 0.The Thetransistorsinthecircuitoffig.Q44haveparameter V TN n
width-to-length ratio of M 2W is L
(=1). If V
2
o
W . 010 Vwhen V i = 5 V,then ( = L
+5 V
) for M
1
1
is
Fig Q44
M1 Vo M2
Vi
Q45.
(A) 47.5
(B) 28.4
(C) 40.5
(D) 20.3
Inthecircuitoffig.Q45the CMRR oftheop-ampis60dB.Themagnitudeofthe v o is 2 V
Fig Q45 100 kW
R
R
1 kW
R
R
vo
1 kW
100 kW
Q46.
Q47.
(A) 1 mV
(B) 100 mV
(C) 200 mV
(D) 2 mV
Ifthe X and Y logicinputsareavailableandtheircomplements X and Y arenotavailable,theminimum Å is X Y number of two-input NAND required to implement (A) 4
(B) 5
(C) 6
(D) 7
TherearefourBooleanvariables ,x 1
2
x ,x 3 and x 4 . The following function are defined on sets of them
) =Sm (3, 4, 5) f ( x3 , x2 , x 1
g( x4 , x3 , x2 ) =Sm (1, 6, 7) h( x4 , x3 , x 2 , x1 ) = fg Then h( x4 , x3 , x2 , x1 ) is (A) Sm(3, 12, 13)
(B) Sm(3, 6)
(C) Sm(3, 12)
(D) 0 GATE ECE By R. K. Kanodia
Unit Division: Each unit has been further sub-divided into separate chapters and not clustered together. Thus the non-combination of all the problems in a single unit makes the reader, to remain focused and able to manage his time d preparation.
12 Paper-2
Q48.
The ideal inverter in fig. Q48 has a reference voltage of 2.5 V. The forward voltage of the diode is 0.75 V. The maximum number of diode logic circuit, that may be cascaded ahead of the inverter without produc logic error, is +5 V
+5 V
+5 V
Fig Q48 +5 V
A
Z
B C
n
Q49.
D
Stages of Diode Logic
(A) 3
(B) 4
(C) 5
(D) 9
Considerthefollowingsetof8085 mPinstruction MVI RLC MOV RLC RLC ADD
A, BYTE1 B, A B
If BYTE1 = 07H, then content of accumulator, after the execution of program will be
Q50.
(A) 46H
(B) 70H
(C) 38H
(D) 68H
Considertheexecutionofthefollowinginstructionby8085 mp MVI SHLD
H, 01FFH 2050H
After execution the contents of memory location 2050H, 2051H and registers H, L will be respe (A) 01H, FFH, FFH, (B) FFH,
01H, FFH,
(C) FFH,
01H,
(D) 01H, FFH,
01H 01H
01H, FFH 01H, FFH
GATE ECE By R. K. Kanodia
Includes Previous Exam Questions: This books contain questions on earlier IES, IAS & GATE exams that might be relevant to learn some concepts but we have purposely not mentioned them in our book. We believe and strongly advocate th year GATE contains new and unique problems.
13 Paper-2
Q51.
Thesystemshowninfig.Q51is
x[ n]
1 4
y[ n]
+
D
y[ n-2]
D
+
+
Fig Q51
1 4
+
-1 2
Q52.
(A) Stable and causal
(B) Stable but not causal
(C) Causal but unstable
(D) unstable and not causal
Thetransferfunctionofasystemisgivenas
æ 1ö 2 zç + ÷ è 2ø . H( z ) = æz - 1 öæz -1 ö ç ÷ç ÷ è 2øè 3ø Consider the two statements Statement(i) :Systemiscausalandstable. Statement(ii) :Inversesystemiscausalandstable.
The correct option is
Q53.
(A) (i) is true
(B) (ii) is true
(C) Both (i) and (ii) are true
(D) Both are false
= x- t , Acausal LTI filterhasthefrequencyresponse H) sho j wn in fig. Q53. For the input si(gn)al ( w
e
jt
output will be H( j w) j2
-1
Fig Q53
1
w
-j2
(A) - 2 je(C) 4pje-
jt
jt
(B) 2 je
-
jt
(D) - 4pje-
jt
GATE ECE By R. K. Kanodia
Less Erroneous: The book has very few errors [less than 5%] compared to the other books available in the market which have upto 40% errors. This puts the students in a better and more com fortable situation as all the errors are traceable due to availability of the complete solution and moreover, the errors are never conceptual but data or typo mistakes. Kindly note t all the errata will be soon available at our website www.nodia.co.in
14 Paper-2
Q54.
Q55.
æ1 ö = a å n ç ÷will 4 è ø n=0 ¥
Thenumericvalueof
(A)
16
(C)
4
n
be
9
9
(B)
9
(D)
9
Eachoftwosequence[x ]nand y[ n] hasaperiod N
4
16
= 4.TheFScoefficientare
1 1 ]1 X [0] =X [3] = X [ 1] = X [2] =1 And Y[0 ],Y [1 ],Y [2 ],Y 3[ = 2 2 The FS coefficient [ ]Zforthesignal k z[ n] =x[ n] y [n ]will be (A) 6 (C) 6 Q56.
Q57.
(B) 6| k|
|k |
(D) e
j
p
k 2
A Routh tableisshownbelow.ThelocationofpoleonRHP, LHP andimaginaryaxisare s7
1
2
-1
-2
s5
1
2
-1
-2
s5
3
4
-1
s4
1
-1
-8
s3
7
8
s2
-18
-21
s1
-9
s0
-21
(A) 1,
2,
4
(B) 1,
6,
0
(C) 1,
0,
6
(D) None of the above
Theopen-looptransferfunctionofa ufb controlsystemis G( s) =
K (1 +2s)( 1+ 4s) 2
2
s (s
+2s + 8)
The position, velocity and acceleration error constants are respectively (A) 0,
0, 4K
(C) ¥ , K , 0 8
(B) 0, 4K ,
¥
(D) ¥ ,
K
¥,
8
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
15 Paper-2
Q58.
Theforward-pathtransferfunctionofa ufb systemis G( s) =
K ( s +1)( s + 2) ( s +5)( s +6)
The break points are
Q59.
Break-in
Breakaway
(A)
-1.563
-5.437
(B)
-5.437
-1.563
(C)
-1.216
-5.743
(D)
-5.743
-1.216
The Nyquist plotofasystemisshowninfig.Q59.Theopen-looptransferfunctionis 4s +1 G( s) H ( s)= 2 s ( s +1)( 2s +1) Im
Fig Q59
10.64 w
w=
=
¥
Re
0
The no. of poles of closed loop system in RHP are (A) 0
(B) 1
(C) 2
D) 4
GATE ECE By R. K. Kanodia
Time Management: Time is a very important factor in any competitive exam and the same applies for GATE too. It has been observed and concluded that if students can manage time, they can get a better score in GATE. The solutions provided a extremely logical yet tricky so that they save time when the student solve them in the examination, as they have already b used to solving difficult and tricky problems.
16 Paper-2
Q60.
( i). The state space representation is Forthenetworkshowninfig.Q60.Theoutputis t R
i1
1W
1H
v1
i3
iR
v2
Fig Q60
i2 vi
Q61.
4v1
1 F
1W
(A)
év&1 ù é 1 - 1ù év1 ù 1é ù = ê&i ú =ê- 3 1 ú êi ú + 0ê v iú R , i [4 û ë1 û ë û ë3 û ë
év1 1] ê ëi3
ù ú û
(B)
év&1 ù é 1- 1-ù év1 ù 1é ù = ê&i ú =ê- 3 - 1 ú êi ú + 0ê v iú R , i [4 û ë3 û ë û ë3 û ë
év1 1] ê ëi3
ù ú û
(C)
év&1 ù é1 êv& ú =1 ê ë2 û ë
év 4] ê 1 ëv 2
ù ú û
(D)
év&1 ù é 1 3ù év1 ù é1 ù êv& ú =ê 1- 6 ú êv ú + ê -1 vúi ,R i =[1 ûë 2 û ë û ë2 û ë
3-ù év1
ù é1 ù ú ê ú + ê vúi ,R i =[1 6 û ëv 2 û ë -1 û
év ù 4] ê 1 ú ëv 2 û
2 ( ) is Thepowerspectraldensityofabandpasswhitenoise nN t 2 asshowninfig.Q61.Thevalueof n
is SX( w )
4pB
Fig Q61
N
2
-wc
Q62.
Q63.
w c
w
(A) NB
(B) 2NB
(C) 2pNB
(D)
NB p
Inareceivertheinputsignalis100 mV,whiletheinternalnoiseattheinputis10 mV.Withamplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of receiver is (A) 2
(B) 0.5
(C) 0.2
(D) None of the above
12signalseachband-limitedto5kHzaretobetransmittedoverasinglechannelbyfrequencydivision multiplexing . If AM-SSB modulation guard band of 1 kHz is used, then the band width of the multiplexed signal will be (A) 131 kHz
(B) 81 kHz
(C) 121 kHz
(D) 71 kHz
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
17 Paper-2
Q64.
A carrier wave of 1 GHz and amplitude 3 V is frequency modulated by a sinusoidal modulating signal frequency of 500 Hz and of peak amplitude of 1 V. The frequency deviation is 1 kHz. The peak level of the modulating wave form is changed to 5 V and the modulating frequency is changed to 2 kHz. The expression for the new modulated wave form is (A) cos [2p´ 10 t +25 . cos (4p´ 10 t)] 6
3
(B) cos [2p´ 10 t +5 cos (4p´ 10 t)] 6
3
(C) 3cos 2 [ p´ 10 t +25 . cos(4p´ 10 t)] 6
3
(D) 3cos 2 [ p´ 10 t +5 cos (4p´ 10 t)] 6
Q65.
3
Let message signal m ( )t=cos(4p10 3t) and carrier signal ( )c=t 5cos (2p10 6t) are used to generate a FM
signal. It the peak frequency deviation of the generated FM signal is three times the transmissi p 1008 ´ (10)3( t) in the FM signal would bandwidth of the AM signal, then the coefficient of the 2term cos be
Q66.
Q67.
Q68.
Q69.
5
(A) 5J 4( 3)
(B)
5 (C) J 8( 4) 2
(D) 5J 4( 6)
Thecurlofvectorfield A=rz sinf u r
+3rz 2 cosf
2
J 8( 3)
ur
at point (5, 90°, 1) is
(A) 0
(B) 12u q
(C) 6u r
(D) 5u f
A 150 MHz uniform plane wave is normally incident from air onto a material. Measurements yield a SWR of 3 and the appearance of an electric field minimum at 0.3l in front of the interface. The impedance material is (A) 502 - j 641 W
(B) 641 - j 502
j (C) 641 + 502
j W (D) 502 + 641
W
W
The quarter-wave lossless 100 W line is terminated by load Z L 60 V, the voltage at the sending end is (A) 126 V
(B) 28.6 V
(C) 21.3 V
(D) 169 V
= 210 W. If the voltage at the receiving end is
An antenna can be modeled as an electric dipole of length 4 m at 3 MHz. If current is uniform over its length, then radiation resistance of the antenna is (A) 1.974 W
(B) 1.263 W
(C) 2.186 W
(D) 2.693 W
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
18 Paper-2
Q70.
Anarraycomprisestwodipolesthatareseparatedbyhalfwavelength.Ifthedipolesarefedbycurrents, that are 180° out of phase with each other, then array factor is (A) sin
æpcos q + pö ç ÷ 4 4 ø è
(B) cos
æpcos q + pö ç ÷ 4 2 ø è
(C) cos
æpcos q + pö ç ÷ 2 2 ø è
(D) sin
æpcos q + pö ç ÷ 2 2 ø è
Common Data Questions Common Data for Questions Q.71-73:
= 2/ V For the circuit shown in fig. Q71-73 transistor parameters are V. TNmA V, 2Kand = 0. n =l05 The transistor is in saturation. +10 V
Fig Q71-73
10 kW
vo
vi
~
VGG
Q71.
Q72.
Q73.
If I DQ istobe0.4mA,thevalueof VGSQ is (A) 5.14 V
(B) 4.36 V
(C) 2.89 V
(D) 1.83 V
Thevaluesof g m and ro are (A) 0.89 mS, ¥
(B) 0.89 mS, 0
(C) 1.48 mS, 0
(D) 1.48 mS, ¥
Thesmallsignalvoltagegain Av is (A) 14.3
(B) -14.3
(C) -8.9
(D) 8.9 GATE ECE By R. K. Kanodia
Variety: The book carries in it a large variety of problems. The words of one of the senior educators of a reputed coaching institute bear testimony to the fact wherein he comments that “We can’t expect so much variety of problems in a sin available in the market.”
19 Paper-2
Common Data for Questions Q74-75:
Consider three continuous-time periodic signals whose Fourier series representation are as follow x1 ( t) = Q74.
k æ 1ö å ç ÷e k =0 è3 ø 100
jk
2p 50
t
,
x 2 ( t) =
k
å
100
-
coskp e
jk
2p 50
t
,
=100 -
ækp öex 3 ( t) = å j sin ç ÷ è2 ø k = -100 100
jk
2p 50
t
Theevensignalsare (A) x 2 ( t ) only (B) x2 ( t) and x 3 ( t) (C) x1 ( t) and x3 ( t) (D) x1 ( t) only
Q75.
Therealvaluedsignalsare (A) x 1 ( t ) and x2 ( t) (B) x2 ( t) and x 3 ( t) (C) x3 ( t) and x1 ( t) (D) x1 ( t) and x3 ( t)
Linked Answer Questions: Q76. to Q85. carry two marks each.
Statement for Linked
Answer Questions: Q76. and Q77:
08m, L = 4 mm, The parameters of an n - channel enhancement-mode MOSFET. are V, WV= 6= 4m TN 2 t ox =450 A°, m n = 650 cm -V . s Q76.
Theconductionparameter K n is (A) 0.8 mA V2 (B) 0.8 mA V 2 (C) 0.4 m A V2 (D) 0.4 mA V 2
Q77.
If VGS
=V DS = 3 V, then current I D is
(A) 1.94 mA (B) 2.87 mA (C) 5.68 mA (D) 3.84 mA
All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.
20 Paper-2
Statement for Linked
Answer Questions: Q78 and
Q79:
The 8-bit left shift register and D -flip-flop shown in fig. Q78-79 is synchronized with same clock. D flip-flop is initially cleared. b7
b6
D
b5
b4
Fig Q78-79
b3 b2 b1 b0
Q
CLK Q
Q78.
Thecircuitactas (A) Binary to 2’s complement converter (B) Binary to Gray code converter (C) Binary to 1’s complement converter (D) Binary to Excess–3 code converter
Q79.
IfinitiallyregistercontainsbyteB7,thenafter4clockpulsecontentsofregisterwillbe (A) 73
(B) 72
(C) 7E
(D) 74
Statement for Linked
Answer Questions: Q80 and
Q81:
A block diagram is shown in fig. Q80-81. 5 R 1(s )
+
1 s
Fig Q80-81 2s
+ +
C 2(s )
2
Q80.
Thetransferfunctionforthissystemis (A)
(C)
2s(2 s +1) 2s
2
+3s + 5
2s( 2s +1) 4s
2
+13s +5
(B)
(D)
2s( 2s +1) 2
2s
+13s +5
2s( 2 s +1) 2
4s
+3s + 5
GATE ECE By R. K. Kanodia
Attractive Format: We understand student psychology and the fact that if the book is in an attractive format, the student would feel good in reading the book. This fact also heightens the interest to study in a student. Thus the sty le of the book is so that it appeals to its readers, yet is expressive and detailed.
21 Paper-2
Q81.
Thepoleofthissystemare (A) - 075 . ± j 139 . (C) - 05 . ,
(B) - 041 . , - 609 . (D) - 025 . ± j 088 .
- 167 .
Statement for Linked
Answer Questions: Q82 and
Q83:
Ten telemetry signals, each of bandwidth 2 kHz, are to be transmitted simultaneously by binar The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The must be sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. Q82.
Q83.
Theminimumpossibledataratemustbe (A) 272.64 kbits/sec
(B) 436.32 kbits/sec
(C) 936.64 kbits/sec
(D)None of the above
Theminimumtransmissionbandwidthis (A) 218.16 kHz
(B) 468.32 kHz
(C) 136.32 kHz
(D) None of the above
Statement for Linked
Answer Questions: Q84 and
Q85:
In an air-filled waveguide, a TE mode operating at 6 GHz has Ey Q84.
Q85.
py =15sinæ2çpx ö÷cos æ ç bè èa ø
ö ø
sin÷ (wt - 12z) V m
Thecutofffrequencyis (A) 4.189 GHz
(B) 5.973 GHz
(C) 8.438 GHz
(D) 7.946 GHz
Theintrinsicimpedanceis (A) 35.72 W
(B) 3978 W
(C) 1989 W
. W (D) 7144
GATE ECE By R. K. Kanodia
Aim : The aim of the book is to provide quality material, a fact which can easily be seen in books available for the preparation of IIT-JEE, AIEEE, CPMT & CAT, but till date never observed in the material available GATE preparation. In other wor we want to provide ELITE material but which is also economical. E : Expressive L : Less Erroneous I : Individualistic T : Targeted approach E : Exhaustive content
22 Paper-2
Answers Paper-2 1.
(C)
2.
(B)
3.
(B)
4.
(B)
5.
(C)
6.
(A)
7.
(A)
8.
(B)
9.
(B)
10.
(B)
11.
(B)
12.
(B)
13.
(D)
14.
(C)
15.
(A)
16.
(B)
17.
(C)
18.
(B)
19.
(C)
20.
(D)
21.
(A)
22.
(D)
23.
(C)
24.
(B)
25.
(B)
26.
(A)
27.
(B)
28.
(C)
29.
(C)
30.
(A)
31.
(A)
32.
(D)
33.
(A)
34.
(B)
35.
(C)
36.
(C)
37.
(C)
38.
(A)
39.
(C)
40.
(A)
41.
(B)
42.
(A)
43.
(B)
44.
(D)
45.
(B)
46.
(A)
47.
(A)
48.
(A)
49.
(A)
50.
(C)
51.
(A)
52.
(C)
53.
(B)
54.
(C)
55.
(A)
56.
(A)
57.
(D)
58.
(A)
59.
(C)
60.
(B)
61.
(B)
62.
(A)
63.
(D)
64.
(C)
65.
(D)
66.
(D)
67.
(C)
68.
(A)
69.
(B)
70.
(B)
71.
(C)
72.
(A)
73.
(C)
74.
(A)
75.
(B)
76.
(C)
77.
(A)
78.
(B)
79.
(C)
80.
(C)
81.
(C)
82.
(B)
83.
(A)
84.
(B)
85.
(B)
A Concern Publishers, who claim to be the partners in the progress of students, use cheap tactics to sell their material to them. A publis famous for providing books on GATE, which are incidentally also erroneous & copied, has also imitated our book this year. They have gone to the extent this year and introduced a revised edition of their book on GATE on Elect. & Comm., which also includes a large nu of Multiple Choice Questions [MCQ]. All the MCQ’s & their solutions have been copied from our book [GATE by RK Kanodia]. This is condemnable act of plagiarism. They are trying to restrict our book from reaching the students, by giving high incentives to retailers distributors. They are using all possible unfair means of the publishing field, to make money and monetary exploit students, yet the b fails to reach the students. An Appeal
We would like to make this Appeal to you against plagiarism and exploitation, keeping in mind the interest of stu publishing field, to support us in our fight. Our company has been started with very few resources and but have large vision. The greatest challenge for us is to reach e part of India. We realize that educators are busy people in times of change and thus they are not able to rate the books given to them able to suggest the same to the students. Thus students, who are dependent on such educators for the knowledge of books, are not a the resources available to the maximum. The restriction of erroneous and plagiarized work, in reaching the hands of students thus b a huge task. Therefore, we request and ask for your support against this condemnable act of plagiarism.
23 Paper-2
MCQ GATE-ECE by RK Kanodia Kindly note that our publication GATE-ECE by RK Kanodia, has the following features that make it an excellent study material in comparison to other books available on the GATE exam:
1. MCQs: The book contains only solved Multiple choice questions (MCQ) which is the main requirement of the GATE exam. Each and every problem has its complete solution. We understand that theoretical studies should be done from the standard book, that one has studied for the semester exams and thus one should use the same material to understand the concepts of the same. We have deliberately excluded theoretical matter in the guide book so as not to mislead the students. However, wherever needed, satisfactory explanation of the formula has been included in the solution. 2. Adherence to Pattern: All Multiple choice questions are strictly according to the GATE pattern. Every problem selected and included in the book is a model problem for the preparation of the exam which would thus prepare and equip the students better. Kindly note, that the standard of Multiple choice questions and their solution in every unit is much better than the ones available in a famous series of problems & solutions as far as GATE is concerned. 3. Levels of MCQs: The Multiple choice questions included in this book are in a conceptually evolving method, allowing the student to progress from one level of complexity to another but always aiding in understanding the basic foundation of the subject. Thus, the MCQs gradually and scientifically advance from the basic level to a more complex level, helping in the systematic understanding of the problem rather than an abrupt one. 4. Unit Division: Each unit has been further sub-divided into separate chapters and not clustered together. Thus the non-combination of all the problems in a single unit makes the reader, to remain focused and able to manage his time during his preparation. 5. Time Management: Time is a very important factor in any competitive exam and the same applies for GATE too. It has been observed and concluded that if students can manage time, they can get a better score in GATE. The solutions provided are extremely logical yet tricky so that they save time when the student solve them in the examination, as they have already been used to solving difficult and tricky problems.
6. Variety: The book carries in it a large variety of problems. The words of one of the senior educators of a reputed coaching institute bear testimony to the fact wherein he comments that “We can’t expect so much variety of problems in a single book available in the market.” 7. Includes Previous Exam Questions: This book contains questions on earlier IES, IAS & GATE exams that might be relevant to learn some concepts but we have purposely not mentioned them in our book. We believe and strongly advocate that every year GATE contains new and unique problems. 8. Less Erroneous: The book has very few errors [less than 5%] compared to the other books available in the market which have upto 40% errors. This puts the students in a better and more com fortable situation as all the errors are traceable due to availability of the complete solution and moreover, the errors are never conceptual but data or ty po mistakes. Kindly note that, all the errata will be soon available at our website www.nodia.co.in 9. Attractive Format: We understand student psychology and the fact that if the book is in an attractive format, the student would feel good in reading the book. This fact also heightens the interest to study in a student. Thus the style of the book is so designed that it appeals to its readers, yetisexpressiveanddetailed. 10. Aim : The aim of the book is to provide quality material, a fact which can easily be seen in books available for the preparation of IIT-JEE, AIEEE, CPMT & CAT, but till date never observed in the material available GATE preparation. In other words, we want to provide ELITE material but which is also economical.
E L I T E
: : : : :
Expressive Less Erroneous Individualistic Targeted approach Exhaustive content
We have received feedback, which state that the book fulfills more than what is stated above and thus it has been a great success last year, on all aspects. Everyone who got through, due to this book, has given excellent feedback. Reviews can be read at our website www.nodia.co.in. However, nothing in the world can be achieved without the help of constructive criticism and thus we would be obliged if you can send across your feedback to make our book, a GUIDE in true sense of the word.
Feel free to mail or call at following for any enquiry about b NODIA & COMPANY
09350292376
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