Q1. If A and B are square matrices of order 4 4 ´ such that A B =5 and A B =a × , then a is (A) 5 (B) 25 (C) 625 (D) None of these Q2. The particular integral of the differential equation ( ) , D D y e e D d dx x x 3 - = + º - is (A) 12( ) e e x x + - (B) 12 x e e x x ( ) + - (C) 12 2 x e e x x ( ) + - (D) 12 2 x e e x x ( ) - - Q3. If x t ddt te u t t ( ) { ( )} = - , then laplace transform of x t ( ) is (A) 11 2 s s ( ) + (B) s s ( ) +1 2 (C) e s s -+1 (D) e s s - + ( ) 1 2 Q4. Consider the x t ( ) as shown in fig. Q4. The FT of x t ( ) is (A) 2 2 sin w w - (B) 2 2 cosww- j (C) 2 jw w cos (D) 2 jw w sin Q5. If a resistor of 10W is placed in parallel with voltage source in the circuit of fig. Q5, the current i will be (A) increased (B) decreased (C) unchanged (D) It is not possible to say 1 Paper-2 x t ( ) -1 1 t 1 Fig Q4 Questions Q1. to Q20. carry one mark each. All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same. vs i Linear Resistive Network Fig Q5 PAPER -2Q6. The current carrying capacity of a 1 W, 4 MW resistor used in radio receiver is (A) 0.5 kA (B) 2 kA (C) 2 mA (D) 0.5 mA Q7. The equation governing the diffusion of neutral atom is (A) ¶¶ ¶¶ Nt D N x = 2 2 (B) ¶¶ ¶¶ Nx D N t = 2 2 (C) ¶¶ ¶¶ 2 2N t D Nx = (D) ¶¶ ¶¶ 2 2N x D Nt = Q8. The p-type substrate in a monolithic circuit should be connected to (A) any dc ground point (B) the most negative voltage available in the circuit (C) the most positive voltage available in the circuit (D) no where, i.e. be floating Q9. Consider the List I and List II List I ( Oscillator) List II ( Characteristic) P. Colpitts Oscillator 1. RC Oscillator Q. Phase shift Oscillator 2. LC Oscillator R. Tunnel diode Oscillator 3. Negative resistance Oscillator S. Relaxation Oscillator 4. Sweep Circuits The correct match is P Q R S (A) 1 2 3 4 (B) 2 1 3 4 (C) 1 2 4 3 (D) 2 1 4 3 2 Paper-2 Mul t ipl e Cho i c e Que s t i ons GATE Electronics & Communications By R. K. Kanodia Price 425.00 Pages 602 Published by NODIA & COMPANY For previous years papers & sample chapter of book please visit www.nodia.co.in Feel free to call at 09350292376 for any enquiry about book.Q10. For the circuit shown in fig. Q10,VCB =05 . V and b =100. The value of I Q is (A) 1.68 mA (B) 0.909 mA (C) 0.134 mA (D) None of the above Q11. A four-variable switching function has minterms m6 and m9 . If the literals in these minterms are complemented, the corresponding minterm numbers are (A) m3 and m0 (B) m9 and m6 (C) m2 and m0 (D) m6 and m9 Q12. The diode logic circuit of fig. Q12 is a (A) AND (B) OR (C) NAND (D) NOR Q13. The even part of a function x n u n u n [ ] [ ] [ ] = - - 4 is (A) 12 1 4 4 { [ ] [ ] [ ]} + - - - - - d n u n u n (B) 12 3 4 { [ ] [ ] [ ]} u n u n n + - - +d (C) 12 4 4 { [ ] [ ] [ ] [ ]} u n u n u n u n + - - - - - - (D) Above all 3 Paper-2 V1 V2 Vo D2 D1 Fig Q12 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same. -5 V IQVo +5 V5 kW Fig Q10Q14. The trigonometric Fourier series for the waveform shown in fig Q14 will be (A) A A t t t 2 2 13 3 15 5 + - + p (sin sin sin .... ) (B) A A t t t 2 2 12 2 13 3 + - + p (cos cos cos .... ) (C) A A t t t 2 2 13 3 15 5 + - + p (cos cos cos .... ) (D) A A t t t t 2 2 13 3 13 3 + + + + p (sin cos sin cos .... ) Q15. The poll–zero configuration of a phase–lead compensator is given by (A) (B) (C) (D) Q16. 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 4 Paper-2 s jw s jw s jw s jw 2 2 -p p x t ( ) p t -p A Fig Q14 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.Q17. Assertion (A): PSK is inferior to FSK. Reason (R): PSK require less bandwidth than FSK. 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. 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 noise immunity can be obtained with (A) binary FSK (B) quaternary FSK (C) BPSK (D) QPSK Q19. 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 Q20. The beamwidth between first null of uniform linear array of N equally spaced ( element spacing = d) 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 5 Paper-2 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 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.Q21. If A = - éëêêêê ùûúúúú 0 2 2 0tan tan a a then ( ) cos sin sin cos I A - × - éëêê ùûúú a a a a2 is equal to (A) I A + (B) I A - (C) I A +2 (D) I A -2 Q22. For what value of x x 0 2 £ £ æè ç öø ÷ p , the function y x x = + ( tan ) 1 has a maxima ? (A) tan x (B) 0 (C) cot x (D) cos x Q23. The value of f x f x f a x dx a ( ) ( ) ( ) + - ò 2 0 2 is (A) 0 (B) 1 (C) a (D) 2a Q24. The integrating factor for the differential equation ( ) x xy dx y dy 3 4 3 2 0 + + = is given by (A) e x - (B) ex 2 (C) ex (D) e x - 2 Q25. cos ? pz z dz c - = ò 1 where c is the circle z =3 (A) i2p (B) - i2p (C) i6 2 p (D) - i6 2 p Q26. If 3 is the mean and 32 is the standard deviation of a binomial distribution, then the distribution is (A) 34 14 12 + æè ç öø ÷(B) 12 32 12 + æè ç öø ÷ (C) 45 15 60 + æè ç öø ÷(D) 15 45 5 + æè ç öø ÷ 6 Paper-2 Questions Q21. to Q75. carry two marks each. 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 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.Q27. For the differential equation dy dx x y = - 2 given that 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 (A) 0.2498 (B) 0.3046 (C) 0.4648 (D) 0.5114 Q28. The inverse Fourier transform of X j d d ( ) sin sin w w w w w = æè ç öø ÷ 4 4 2 is (A) t e t t 2 2 2 4 - - - rect { ( )} (B) t e t t 2 2 2 4 + + - rect ( ( )) (C) - + + - t t t { ( ) ( )} rect rect 2 8 2 8 (D) t t t { ( ) ( )} rect rect 2 4 2 8 + - - Q29. The time signal x t ( ) corresponding to X s s d ds s s ( ) = + æèçç öø÷÷ + + 22 21 9 1 3 is (A) e t t t t u t t - + + æèçç öø÷÷ 3 2 23 3 9 3 sin cos ( ) (B) ( sin cos ) ( ) e t t t t u t t - + + 3 2 2 3 3 (C) e t t t t u t t - + + æè ç öø ÷3 2 23 3 3 sin cos ( ) (D) ( sin cos ) ( ) e t t t t u t t - + + 3 2 3 2 3 Q30. The incidence matrix of a graph is as given below A = - - - - éëêêêê ùûúúúú 1 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0 The number of possible tree are (A) 40 (B) 70 (C) 50 (D) 240 7 Paper-2 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q31. In the fig Q31 the value of v1 is (A) 6 V (B) 7 V (C) 8 V (D) 10 V Q32. In the circuit of fig. Q32 the 30 V source has been applied for a long time. The switch is opened at t =1 ms. At t = 4 ms the vC ( ) 4ms is (A) 8.39 mV (B) 2.59 V (C) 1.13 mV (D) 2.77 V Q33. For a RLC series circuit R L = = 20 06 W , . H, the value of C will be [CD =critically damped, OD =over damped, UD =under damped]. CD OD UD (A) C =6mF C > 6 mF C < 6 mF (B) C =6mF C < 6 mF C >6mF (C) C > 6 mF C =6mF C < 6mF (D) C < 6 mF C = 6 mF C >6mF Q34. In the circuit shown in Fig. Q34 v( ) 0 8 - = V and i t t in ( ) ( ) = 4d . The v t C ( ) for t ³ 0 is (A)164e t - V (B) 208e t - V (C) 208 1 3 ( ) - - e t V (D)164 3 e t - V 8 Paper-2 1 W 1 W 3 W 2 W 6 W 8 V v1 –+ 2 W 6 W 18 VFig Q31 1 kW 6.25 kW 0.6 F m 25 kW t=1 ms 30 (-) V u t vC –+ Fig Q32 iin 20 mF 50 W vC –+ Fig Q34 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q35. In the circuit shown in fig. Q35, when the voltageV1 is 10 V, the current I is 1 A. If the applied voltage at port-2 is 100 V, the short circuit current flowing through at port 1 will be (A) 0.1 A (B) 1 A (C) 10 A (D) 100 A Q36. The network function of circuit shown in fig.Q36 is H VV j o ( ) . w w = = + 1 4 1 001 The value of the C and A is (A) 10 mF, 6 (B) 5 mF, 10 (C) 5 mF, 6 (D) 10 mF, 10 Q37. In germanium(ni = ´ - 2 4 1013 3 . cm ) semiconductor atT =300K, the acceptor concentrations is Na =1013 cm-3 and donor concentration is Nd =0. The thermal equilibrium concentration p0 is (A) 297 109 . ´ cm-3 (B) 268 1012 . ´ cm-3 (C) 295 1013 . ´ cm-3 (D) 2 4 . cm-3 Q38. A silicon (ni = ´ - 15 1010 3 . cm ) pn junction at T =300 K has Nd =1014 cm-3and Na =1017 cm-3 . The built-in voltage is (A) 0.63 V (B) 0.93 V (C) 0.026 V (D) 0.038 V Q39. The maximum electric field in reverse-biased silicon pn junction is | | Emax = ´ 3 10 5 V cm. The doping concentration are Nd = ´ 4 1016 cm-3 and Na = ´ 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 9 Paper-2 2 kW 15 kW C AvC ~ vi vo –+ vC –+ Fig Q36 V1 I Linear Resistive Network Fig Q35 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q40. A ideal n-channel MOSFET has parameters mn =525 cm V s 2 - , VTN =0 75 . V, tox = ° 400 A . When MOSFET is biased in the saturation region atVGS =5 V, the required rated current is I D sat ( ) =6 mA. The required ratioW L is (A) 14.7 (B) 11.2 (C) 9.61 (D) 7.2 Q41. For a n -channel enhancement-mode MOSFET the parameters areVTN =08 . V, ¢ = kn 8 mA V2and W L =5. If the transistor is biased in saturation region with I D =05 . mA, then required vGS is (A) 1.68 V (B) 2.38 V (C) 4.56 V (D) 3.14 V Q42. The cutin voltage for each diode in fig. Q42 isVg =06 . V. Each diode current is 0.5 mA. The value of R R 1 2 , and R3 will be respectively (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 Q43. In the circuit of fig. Q43 Zener voltage isVZ =5 V and b =100. The value of ICQ andVCEQ are (A)12 47 43 . , . mA V (B)12 47 5 7 . , . mA V (C)10 43 5 7 . , . A V (D)10 43 43 . , . A V 10 Paper-2 R1 +10 V -5 V +5 V 0 V R2 R3 Fig Q42 500 W +12 V Fig Q43 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q44. The transistors in the circuit of fig. Q44 have parameter VTN =08 . V, ¢ = kn 40mA/V2 and l =0. The width-to-length ratio of M2 is ( ) WL 2 1 = . IfVo =010 . V whenVi =5 V, then ( ) WL 1 for M1 is (A) 47.5 (B) 28.4 (C) 40.5 (D) 20.3 Q45. In the circuit of fig. Q45 the CMRR of the op-amp is 60 dB. The magnitude of the v o is (A) 1 mV (B) 100 mV (C) 200 mV (D) 2 mV Q46. If the X and Y logic inputs are available and their complements X and Y are not available, the minimum number of two-input NAND required to implement X Y Å is (A) 4 (B) 5 (C) 6 (D) 7 Q47. There are four Boolean variables x x x 1 2 3 , , and x 4 . The following function are defined on sets of them f x x x g x x x h x x ( , , ) ( , , ) ( , , ) ( , , ) ( , , 3 2 1 4 3 2 4 3 3 4 5 1 6 7 ==SSmm x x fg 2 1 , ) = Then h x x x x ( , , , ) 4 3 2 1 is (A) Sm(3, 12, 13) (B) Sm(3, 6) (C) Sm(3, 12) (D) 0 11 Paper-2 M1 Vo Vi +5 VM2 Fig Q44 R R R R 2 V vo 1 kW 1 kW 100 kW 100 kW Fig Q45 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 during his preparation.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 producing logic error, is (A) 3 (B) 4 (C) 5 (D) 9 Q49. Consider the following set of 8085 mP instruction MVI A, BYTE1 RLC MOV B, A RLC RLC ADD B If BYTE1 = 07H, then content of accumulator, after the execution of program will be (A) 46H (B) 70H (C) 38H (D) 68H Q50. Consider the execution of the following instruction by 8085 mp MVI H, 01FFH SHLD 2050H After execution the contents of memory location 2050H, 2051H and registers H, L will be respectively (A) 01H, FFH, FFH, 01H (B) FFH, 01H, FFH, 01H (C) FFH, 01H, 01H, FFH (D) 01H, FFH, 01H, FFH 12 Paper-2 n Stages of Diode Logic Z C A B +5 V +5 V +5 V D +5 V Fig Q48 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 that every year GATE contains new and unique problems.Q51. The system shown in fig. Q51 is (A) Stable and causal (B) Stable but not causal (C) Causal but unstable (D) unstable and not causal Q52. The transfer function of a system is given as H z z z z ( ) = + æè ç öø ÷ - æè ç öø ÷- æè ç öø ÷ 2 12 12 13 . Consider the two statements Statement(i) : System is causal and stable. Statement(ii) : Inverse system is causal and stable. The correct option is (A) (i) is true (B) (ii) is true (C) Both (i) and (ii) are true (D) Both are false Q53. A causal LTI filter has the frequency response H j ( ) w shown in fig. Q53. For the input signal x t e jt ( ) = - , output will be (A) - - 2 je jt (B) 2 je jt - (C) 4pje jt - (D) - - 4pje jt 13 Paper-2 H j ( ) w w j2 -j2 1 -1 Fig Q53 + + ++ x n [ ] D y n [ ] y n-2 [ ] D -1214 14 Fig Q51 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 comfortable 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 that, all the errata will be soon available at our website www.nodia.co.inQ54. The numeric value of a n n n = æè ç öø ÷=¥å14 0 will be (A) 16 9 (B) 94 (C) 49 (D) 9 16 Q55. Each of two sequence x n [ ] and y n [ ] has a period N = 4. The FS coefficient are X X X X [ ] [ ] [ ] [ ] 0 3 12 1 12 2 1 = = = = And Y Y Y Y [ ], [ ], [ ], [ ] 0 1 2 3 1 = The FS coefficient Z k [ ] for the signal z n x n y n [ ] [ ] [ ] = will be (A) 6 (B) 6| | k (C) 6| | k (D) e j k p2 Q56. A Routh table is shown below. The location of pole on RHP, LHP and imaginary axis are 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 Q57. The open-loop transfer function of a ufb control system is G s K s s s s s ( ) ( )( ) ( ) = + + + + 1 2 1 4 2 8 2 2 The position, velocity and acceleration error constants are respectively (A) 0, 0, 4K (B) 0 4 , , K ¥ (C) ¥ , , K8 0 (D) ¥ ¥ , , K8 14 Paper-2 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q58. The forward-path transfer function of a ufb system is G s K s s s s ( ) ( )( ) ( )( ) = + + + + 1 2 5 6 The break points are Break-in Breakaway (A) -1.563 -5.437 (B) -5.437 -1.563 (C) -1.216 -5.743 (D) -5.743 -1.216 Q59. The Nyquist plot of a system is shown in fig. Q59. The open-loop transfer function is G s H s s s s s ( ) ( ) ( )( ) = + + + 4 1 1 2 1 2 The no. of poles of closed loop system in RHP are (A) 0 (B) 1 (C) 2 D) 4 15 Paper-2 Im Re 10.64 w ¥ = w= 0 Fig Q59 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 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.Q60. For the network shown in fig. Q60. The output is i t R ( ). The state space representation is (A) && vi vi v i 1 3 1 1 1 1 3 1 10 éë ê ùû ú = - - éë ê ùû ú éë ê ùû ú + éë ê ùû ú, i vi R = éë ê ùû ú[ ] 4 1 1 3 (B) && vi vi v i 1 3 1 3 1 1 3 1 10 éë ê ùû ú = - - - - éë ê ùû ú éë ê ùû ú + éë ê ùû ú, i vi R = éë ê ùû ú[ ] 4 1 1 3 (C) && vv vv v i 12 12 1 3 1 6 11 éë ê ùû ú = - éë ê ùû ú éë ê ùû ú + - éë ê ùû ú, i vv R = éë ê ùû ú[ ] 1 4 12 (D) && vv vv v i 12 12 1 3 1 6 11 éë ê ùû ú = - éë ê ùû ú éë ê ùû ú + - éë ê ùû ú, i vv R = éë ê ùû ú[ ] 1 4 12 Q61. The power spectral density of a bandpass white noise n t ( ) is N 2 as shown in fig. Q61. The value of n2 is (A) NB (B) 2NB (C) 2pNB (D) NB p Q62. In a receiver the input signal is 100 mV, while the internal noise at the input is 10 mV. With amplification 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 Q63. 12 signals each band-limited to 5 kHz are to be transmitted over a single channel by frequency division 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 16 Paper-2 wc -wc 4pB SX( ) w w N2 Fig Q61 i1 iR v2 1 F 1 W 1 H 1 W vi i3 4v1 v1i2 Fig Q60 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.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 [ . cos ( )] 2 10 25 4 10 6 3 p p ´ + ´ t t (B) cos [ cos ( )] 2 10 5 4 10 6 3 p p ´ + ´ t t (C) 3 2 10 25 4 10 6 3 cos [ . cos ( )] p p ´ + ´ t t (D) 3 2 10 5 4 10 6 3 cos [ cos ( )] p p ´ + ´ t t Q65. Let message signal m t t ( ) cos ( ) = 4 103 p and carrier signal c t t ( ) cos ( ) =5 2 106 p are used to generate a FM signal. It the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of theAMsignal, then the coefficient of the term cos ( ( ) ) 2 1008 103 p ´ t in the FM signal would be (A) 5 3 4 J ( ) (B) 52 3 8 J ( ) (C) 52 4 8 J ( ) (D) 5 6 4 J ( ) Q66. The curl of vector field A u u = f + f r r r r z z sin cos 3 2 at point (5, 90°, 1) is (A) 0 (B)12uq (C) 6ur (D) 5uf Q67. 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 of material is (A) 502 641 - j W (B) 641 502 - j W (C) 641 502 + j W (D) 502 641 + j W Q68. The quarter-wave lossless 100Wline is terminated by load ZL =210W. If the voltage at the receiving end is 60 V, the voltage at the sending end is (A) 126 V (B) 28.6 V (C) 21.3 V (D) 169 V Q69. 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) 1.974W (B) 1.263W (C) 2.186W (D) 2.693W 17 Paper-2 All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.Q70. An array comprises two dipoles that are separated by half wavelength. If the dipoles are fed by currents, that are 180° out of phase with each other, then array factor is (A) sin cos p q p 4 4 + æè ç öø ÷ (B) cos cos p q p 4 2 + æè ç öø ÷ (C) cos cos p q p 2 2 + æè ç öø ÷ (D) sin cos p q p 2 2 + æè ç öø ÷ Common Data Questions Common Data for Questions Q.71-73: For the circuit shown in fig. Q71-73 transistor parameters areVTN =2 V, Kn =05 . mA /V2 and l =0. The transistor is in saturation. Q71. If I DQ is to be 0.4 mA, the value ofVGSQ is (A) 5.14 V (B) 4.36 V (C) 2.89 V (D) 1.83 V Q72. The values of g m and ro are (A) 0.89 mS, ¥ (B) 0.89 mS, 0 (C) 1.48 mS, 0 (D) 1.48 mS, ¥ Q73. The small signal voltage gain Av is (A) 14.3 (B) -14.3 (C) -8.9 (D) 8.9 18 Paper-2 vo vi VGG 10 kW +10 V ~ Fig Q71-73 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 single book available in the market.”Common Data for Questions Q74-75: Consider three continuous-time periodic signals whose Fourier series representation are as follows. x t e k k jk t 1 0 100 250 13 ( ) = æè ç öø ÷= - å p , x t k e k jk t 2 100 100 250 ( ) cos = = - - å p p , x t j k e k jk t 3 100 100 250 2 ( ) sin = æè ç öø ÷= - - å p p Q74. The even signals are (A) x t 2 ( ) only (B) x t 2 ( ) and x t 3 ( ) (C) x t 1 ( ) and x t 3 ( ) (D) x t 1 ( ) only Q75. The real valued signals are (A) x t 1 ( ) and x t 2 ( ) (B) x t 2 ( ) and x t 3 ( ) (C) x t 3 ( ) and x t 1 ( ) (D) x t 1 ( ) and x t 3 ( ) Statement for Linked Answer Questions: Q76. and Q77: The parameters of an n -channel enhancement-mode MOSFET are VTN =08 . V,W =64mm, L = 4mm, tox = ° 450 A , mn =650 cm V s 2 - . Q76. The conduction parameter Kn is (A) 0.8m V A 2 (B) 0.8 mA V2 (C) 0.4m V A 2 (D) 0.4 mA V2 Q77. IfV V GS DS = =3 V, then current I D is (A) 1.94 mA (B) 2.87 mA (C) 5.68 mA (D) 3.84 mA 19 Paper-2 Linked Answer Questions: Q76. to Q85. carry two marks each. All questions are drawn from the bookGATEECE by R. K. Kanodia published by NODIA&COMPANY. For full solution refer the same.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. The D flip-flop is initially cleared. Q78. The circuit act as (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. If initially register contains byte B7, then after 4 clock pulse contents of register will be (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. Q80. The transfer function for this system is (A) 2 2 1 2 3 5 2s s s s ( ) + + + (B) 2 2 1 2 13 5 2s s s s ( ) + + + (C) 2 2 1 4 13 5 2s s s s ( ) + + + (D) 2 2 1 4 3 5 2s s s s ( ) + + + 20 Paper-2 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 style of the book is so designed that it appeals to its readers, yet is expressive and detailed. b7 b6 b5 b4 b3 b1 b2 b0 D QQ CLK Fig Q78-79 ++ 1s2 5 2s + R s 1( ) C s 2( ) Fig Q80-81Q81. The pole of this system are (A) - ± 0 75 139 . . j (B) - - 0 41 609 . , . (C) - - 05 167 . , . (D) - ± 0 25 088 . . j Statement for Linked Answer Questions: Q82 and Q83: Ten telemetry signals, each of bandwidth 2 kHz, are to be transmitted simultaneously by binary PCM. The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The signals must be sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. Q82. The minimum possible data rate must be (A) 272.64 kbits/sec (B) 436.32 kbits/sec (C) 936.64 kbits/sec (D)None of the above Q83. The minimum transmission bandwidth is (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 E x a y b t z y = æè ç öø ÷ æè ç öø ÷- 15 2 12 sin cos sin ( ) p p w V m Q84. The cutoff frequency is (A) 4.189 GHz (B) 5.973 GHz (C) 8.438 GHz (D) 7.946 GHz Q85. The intrinsic impedance is (A) 35.72W (B) 3978W (C) 1989W (D) 7144 . W 21 Paper-2 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 words, we want to provide ELITE material but which is also economical. E : Expressive L : Less Erroneous I : Individualistic T : Targeted approach E : Exhaustive contentAnswers 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)22 Paper-2 A Concern Publishers, who claim to be the partners in the progress of students, use cheap tactics to sell their material to them. A publisher, 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 number 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 book 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 students & 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 every 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, to be able to suggest the same to the students. Thus students, who are dependent on such educators for the knowledge of books, are not able to use the resources available to the maximum. The restriction of erroneous and plagiarized work, in reaching the hands of students thus becomes a huge task. Therefore, we request and ask for your support against this condemnable act of plagiarism.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 comfortable 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 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, yet is expressive and detailed. 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 : Expressive L : Less Erroneous I : Individualistic T : Targeted approach E : 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 book: NODIA & COMPANY 09350292376 pk.goel@nodia.co.in 23 Paper-2