MOCK TEST-4

MOCK TEST1/16 EC : ELECTRONIC AND COMMUNICATION ENGINEERING Duration : Three Hours Maximum Marks : 100 Read the following instructions carefully. 1. 2. Do not open the seal of the Question Booklet unlit you are asked to do so by the invigilator. Take out the Optical Response Sheet (ORS) from this Question Booklet without breaking the seal and read the instruments printed on the ORS carefully. If you find the Question Booklet Code printed at the right hand top corner of this page does not match with the Booklet Code on the ORS, exchange the booklet immediately with a new sealed Question Booklet. On the right half of the ORS, using ONLY a black ink ball point pen, (i) darken the bubble corresponding to your test paper code and the appropriate bubble under each digit of your registration number and (ii) write your registration number, your name and name of the examination center and put your signature at the specified location. This Question Booklet contain 16 pages including blank pages for rough work. After you are permitted to open the seal, please check all pages and report discrepancies, if any, to the invigilator. There are a total of 65 Question carrying 100 marks. All these questions are of objective type. Each question has only on correct answer. Question must be answered on the left hand side of the ORS by darkening the appropriate bubble (marked A, B, C, D) using ONLY a black ink ball point pen against the question number. For each question darken the bubble of the correct answer. More than on answer bubbled against a question will be treated as an incorrect response. Since bubbles darkened by the black ink ball point pen cannot be erased, candidates should darken the bubbles in the ORS very carefully. Question Q. 1 Q. 25 carry 1 mark each. Questions Q. 26 - Q. 55 carry 2 marks each. The 2 marks question include two pairs of common data questions and two pairs of linked answer questions. The answer of the second question of the linked answer questions depends on the answer to the first question of the pair. If the first question in the linked pair is wrongly answered or is unattemped, then the answer to the second question in the pair will not be evaluated. Question Q. 56 - Q. 65 belong to General Aptitude (GA) section and carry a total of 15 marks. Question Q. 56 - Q.60 carry 1 mark each, and questions Q. 61 -Q. 65 carry 2 marks each. Unattempted questions will result in zero mark and wrong answer will result in NEGATIVE marks. For all 1 mark questions 1/3 mark will be deducted for each wrong answer. For all 2 marks questions, 2/3 mark will be deducted for each wrong answer. However, in the case of the linked answer question pair, there will be negative marks only for wrong answer to the first question and no negative marks for wrong answer to the second question. Calculator is allowed whereas charts, graph sheets or tables are NOT allowed in the examination hall. Rough work can be done on the question paper itself. Blank pages are provided at the end of the question paper for rough work. Before the start of the examination, write your name and registration number in the space provided below using a blank ink ball point pen. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Names Registration Number EC For Answer Key and Full Solution mail to [email protected]. MOCK TEST FIRST WWW.GATEHELP.COM ρ = 4.25 rad/s.7 (C) 32. The natural 6 frequency and damping ratio of the system are respectively (A) 1 rad/s.27 (D) 16.95 cos 377t V EC-D www. Q.29 (B) 20. Compared to the uncompensated system.1 The unit impulse response of a second order system is 1 e−0.6t . 1 (C) 0.2/16 MOCK TEST Q.5 A conducting circular loop of radius 20 cm lies in the z = 0 plane in a field B = 20 cos 277tu z mWb/m2 .2 Consider the sketch shown below The root locus can be (A) (1) and (3) (C) (2) and (4) Q. The induced voltage in the loop is (A) − 0.com . 1.95 sin 377t V (C) − 0.4 Q.gatehelp.3 (B) (2) and (3) (D) (1) and (4) A PD controller is used to compensate a system. 0.64 Q.25 (D) 1.64 rad/s.8 rad/s.95 sin 377t V (B) 0. 0. 1. 25 carry one mark each.8t sin 0.95 cos 377t V (D) 0. the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot Q. The total area of the enclosing surface is (A) 34. φ = 135c.8 (B) 0.4 The surface ρ = 2. z = 3 and z = 4 define a closed surface. φ = 45c.Q. com .MOCK TEST 3/16 Q. The wrongly stored number is (A) − 37 as 1101 1011 (B) − 89 as 1010 0111 (C) − 48 as 1110 1000 (D) − 32 as 1110 0000 Q.9 dy Integrating factor of + 2y tan x = sin x is dx 2 (A) cos x (C) sec2 x (B) sin2 x (D) 2 tan x Q.8 A die is thrown 100 times.gatehelp.7 The residue of z cos 1 at z = 0 is z (A) 1/2 (C) 1/3 (B) −1/2 (D) − 1/3 Q.6 For a Hertz dipole antenna.11 A computer has the following negative numbers stored in binary form as shown. the half power beam width (HPBW) in the E-plane is (A) 360c (B) 180c (C) 90c (D) 45c Q.10 If A n # n is a triangular matrix then det A is (A) (C) % (− 1) a i=1 n i=1 n ii (B) (D) %a i=1 n ii / (− 1) a ii /a i=1 n ii Q. Getting an even number is considered a success. The variance of the number of successes is (A) 50 (B) 25 (C) 10 (D) None Q.12 Which of the following conversion is performed by the circuit shown in figure (A) Binary to gray code conversion (C) BCD conversion (B) Gray code to binary conversion (D) None of these EC-D www. The value of RF is (A) 1.6 kΩ (C) 800 Ω Q.14 (B) 8 kΩ (D) 1 kΩ For the circuit shown in fig.gatehelp.13 The step size of the following DAC 0.4/16 MOCK TEST Q. the minimum number of isolation regions are (A) 2 (C) 4 Q.com (B) 4 only (D) 3 and 4 EC-D .5 V.15 (B) 3 (D) 7 Consider the following graphs Non-planner graphs are (A) 1 and 3 (D) 3 only www. com .63 V (B) 0.19 (B) 60 Ω (D) 180 Ω The impulse response of an LTI system is h (t) = δ(2) (t). The value of R is (A) 40 Ω (C) 120 Ω Q.75 Ω The circuit shown below is critically damped.93 V (C) 0.gatehelp.17 In the circuit of figure the value of resistor Req is (A) 100 Ω (C) 60 Ω Q.20 Power of signal u [n] is (A) n (C) 1/2 (B) 1 (D) 3 EC-D www.5 Ω (D) 28.16 A silicon pn junction at T = 300 K has Nd = 1014 cm−3 and Na = 1017 cm−3 . The step response is (A) 1 (B) u (t) (C) δ(3) (t) (D) δ (t) Q.038 V Q.18 (B) 22.MOCK TEST 5/16 Q. The builtin voltage is (A) 0.026 V (D) 0. 7 V.5 V www.6 V and rf = 0 .22 4 (s + 3) s2 + s + 1 2 (s + 3) s + 2s + 2 2 (B) (D) 2 (s + 3) s + 2s + 2 2 4 (s − 3) s2 + s + 2 The diodes in the circuit shown below has parameters Vγ = 0.gatehelp.6 mA Q. If Z (0) = 3 .6/16 MOCK TEST Q.4 mA (C) 7.com (B) 7.8 V (C) 12.5 V (D) 8. the Z (s) is (A) (C) Q. The current iD is 2 (A) 8.2 V and VBE = 0.23 (B) 10 mA (D) 0 mA In the shunt regulator shown below.9 V EC-D . the VZ = 8.21 The driving point impedance Z (s) of a network has the pole zero location as shown below. The regulated output voltage Vo is (A) 11. 24 Consider the signal x (t) = rect (t) tri (t).31 (C) 4.5 kHz. 140 π rad (B) 6 kHz.26 If X (t) is a stationary process having a mean value E [X (t)] = 3 and auto correlation function RXX (τ) = 9 + 2e− τ . the magnitude of the network function can be approximated as H (ω) = 1/ω 2 . 100 π rad (D) 7.28 An angle modulated signal is s (t) = cos 2π [2 # 106 t + 30 sin 150t + 40 cos 150t] The maximum frequency and phase deviations of s (t) are (A) 10. 80 π rad (C) 10.The variance of random variable Y = (A) 1 (B) 2. 26.gatehelp.25 In a certain frequency range.4 # 10−6 V (B) 9.27 A resistor R = 1000 Ω is maintained at 290 K and is shunted by a 10 µH inductor. The graph of x (t) is Q. 55 carry two mark each.2 # 10−8 V (C) 10.MOCK TEST 7/16 Q. The slope of the Bode plot in this range is (A) 40 dB/decade (B) − 40 dB/decade (C) 20 dB/decade (D) − 20 dB/decade Q.com EC-D .54 (D) 0 # X (t) dt will be 0 2 Q. The rms noise voltage across the inductor over a frequency bandwidth of 159 kHz is (A) 6.5 kHz. 100 π rad Q.4 # 10−12 V Q. Q. If the desired error probability is 10−5 and (εb /J 0) required to obtain an error probability of 10−5 for www.Q.29 A DS/BPSK spread spectrum signal has a processing gain of 500.4 # 10−10 V (D) 18.5 kHz. 31 A λ/2 dipole is kept horizontally at a height of λ0 /2 above a perfectly conducting v plane) looks infinite ground plane.gatehelp.5 dB. B = 1.0 dB In a material the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2 .4 dB Q.4 Wb/m2 . then the value of x22 u . The change in the magnetization M is (A) 164 kA/m (B) 326 kA/m (C) 476 kA/m (D) 238 kA/m Q.8/16 MOCK TEST binary PSK is 9.com EC-D .6 dB (B) 17. then the Jamming margin against a containers tone jammer is (A) 23. When H is reduced to 400 A/m. The radiation pattern in the lane of dipole ( E approximately as Q. is + 2xy 2 u + y22 u 2 x x dxdy 2y2 dx (A) 0 (B) u (C) 2u Q.33 (D) − u is equal to (B) cot x + cosec x + c (D) tan x + sec x + c is the circle x2 + y2 = 4 The value of f l (1 − i) is (B) 6 (2 + iπ) (D) 0 # 1 +dx sin x (A) − cot x + cosec x + c (C) tan x − sec x + c Q.34 f (z 0) = c + 7z + 1 dz . where c # 3z(z − z 0) 2 (A) 7 (π + i2) (C) 2π (5 + i13) www.30 (D) 109.5 dB (C) 117.32 2 2 2 y y If u = φ a k + xΨ a k. with an open loop gain of AOL = 105 and open loop input impedance Roi = 10 kΩ .02135 (D) 0.02199 Q. than closed loop input impedance (Rif ) at the inverting terminal of op-amp is (A) Rif = 10 kΩ (C) Rif = 40 kΩ EC-D (B) Rif = 1 Ω (D) Rif = 0.35 For dy/dx = x2 + y2 given that y = 0 at x = 0 .36 Consider the following circuit and input voltage to it. If output resistance of op-amp is zero.MOCK TEST 9/16 Q.com .02145 (C) 0.1 Ω www. If op-amp and diodes are ideal. the output voltage waveform is Q. The solution of differential equation for x = 0.37 Consider an op-amp circuit shown in figure.02193 (B) 0.gatehelp.4 using picard’s method is (A) 0. Assume transistor are matched.38 In the current mirror circuit shown below the transistor parameters are VBE = 0.cm) −1 (C) 2 # 10−5 (Ω .cm) −1 Q.cm) −1 (D) 6 # 10−3 (Ω .6 (C) − 6.8 Q. The signal frequency is sufficiently large.com .9 (D) − 3.68 mA (D) 432 µA Consider the common-source circuit with source bypass capacitor. The transistor parameters are VTN = 0.10/16 MOCK TEST Q.39 (B) 1. The voltage gain is (A) − 15.04 mA (C) 962 µA Q. Kn = 1 mA/V 2 and λ = 0 . The conductivity at T = 500 K is (A) 2 # 10−4 (Ω .gatehelp.cm) −1 at T = 300 K. β = 50 and the Early voltage is infinite.cm) −1 (B) 4 # 10−5 (Ω . µp = 600 cm2 /V-s and Nc = Nv = 1019 cm−3 .41 The parameters of n -channel depletion mode MOSFET are VTN =− 2 V and EC-D www.8 V.7 V.2 For a particular semiconductor material following parameters are observed: µn = 1000 cm2 /V-s. These parameters are independent of temperature. The measured conductivity of the intrinsic material is σ = 10−6 (Ω . The output current Io is (A) 1.40 (B) − 9. The ratio W/L is (A) 7.56 mA (C) 9.3 V (C) 12. The state transition sequence is Q.5 mA at VGS = 0 and VDS = 3 V.7 µm when VBE = VBC = 0 .gatehelp.69 mA A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 # 1016 cm−3 and a heavily doped collector region with NC = 5 # 1017 cm−3 . The drain current is ID = 1.375 mA Q. Let the initial state be 00.3 V Q.42 (D) 4. The VBC at punch-through is (A) 26.43 Consider the following circuit The flip-flop are positive edge trigger red D FFs.2 V (D) 6.78 mA (B) 15. Each state is designated as a two bit string Q 0 Q1 .com .MOCK TEST 11/16 ' kn = 80 µA/V 2 .44 In the CMOS circuit shown below the output Y is EC-D www. The neutral base width is xB = 0.3 V (B) 18. 2. 0.4.46 For the system shown below the steady state error component due to unit step disturbance is 0.47 (B) 1250.4 (D) 463.com . 1684 (C) 125 # 103.12/16 MOCK TEST (A) (A + C) B (C) AB + C Q.016 Q.45 (B) (A + B) C (D) AB + C K The Nyquist plot is s (s + 1) (2s + 1) (3s + 1) For the certain ufb system G (s) = Q. 3CF 4 EC-D www. 3981 Following is the segment of a 8085 assembly language program LXI SP. The value of K1 and K2 are respectively (A) 16.003. EFFF H CALL 3000 H : : 3000 H LXI H.000012 and steady state error component due to unit ramp input is 0.gatehelp. − 3.04 sin (4t + 2.07c) A (D) 2.07c) A (C) 1.07c) A (B) 2.com .13c) A (D) − 2.13c) A EC-D www.51 In the circuit shown below the current i2 (t) is (A) 2. − 2.36 cos (4t + 41. − 2} (B) {1. − 4. 3. 50-51 : The circuit is as shown below Q. 4.48 DFT of the sequence g [n] = x [n − 2] is (A) {1. − 4} (C) {4.37 cos (4t − 41.gatehelp. 2.04 cos (4t + 2. − 3. 3. 4} Q.13c) A (C) 2. the contents of SP is (A) 3CF0 H (B) 3CF8 H (C) EFFD H (D) EFFF H Common Data for Q 48-49 : The discrete Fourier transform of a discrete sequence x [n] is XDFT [k] = {1. − 4} (D) {− 2. 2. 1. 4} (C) {3. − 4.04 sin (4t + 92.13c) A (B) − 2. 2} Common Data for Q.36 cos (4t + 41. − 4} (D) {1. 3} Q.07c) A Q.MOCK TEST 13/16 PUSH PSW SPHL POP PSW RET On completion of RET execution.36 cos (4t − 41. 3. 3. 2.50 In the circuit below the current i1 (t) is (A) 2. − 2.04 cos (4t + 92. − 2.49 The discrete Fourier transform of x [− n] is (A) {− 1. 1. 1} (B) {1. 3. 70 GHz (D) 4 GHz < f < 6 GHz EC-D www. Q. x1 = c = y . x 3 = d c . Choose the state variable as follows 2 d3 c = c p q o.25. x2 = dc = c = c x = 4 dt dt2 dt2 Q.14/16 MOCK TEST Statement for Linked Answer Q. y = 81 0 0 0B x W S 0 0 0 1W S 0 W S100 7 10 20W S100W R T T VX R V X 0 0 W 1 S 0 S 0 W S 0 S 0 W W 1 0 0 o= S x x + S W r y = 81 0 0 0B x W 0 0 1 W S 0 S 0 W − − − − 7 10 20 100 S S100W W R R VX V T T X S0 1 0 0 W S0W S0 0 1 0 W S0W o= S x x + S W r y = 8100 0 0 0B x W S0 0 0 1 W S0W S20 10 7 100W S1W R R V T T XV X 0 0 W 1 S 0 S0W S 0 S0W 1 0 W 0 o= S x x + S W r y = 8100 0 0 0B x W 0 0 1 W S 0 S0W S− 20 − 10 − 7 − 100W S1W T T X X Statement for Linked Answer Q.com .48 GHz < f < 7. 54-55: A 6 cm # 4 cm rectangular wave guide is filled with dielectric of refractive index 1.24 GHz < f < 3. 52-53 Determine the state-space representation for the transfer function given in question.54 The range of frequencies over which single mode operation will occur is (A) 2.53 (A) (B) (C) (D) C (s) 100 = 4 3 R (s) s + 20s + 10s2 + 7s + 100 R R V V S 0 1 0 0W S 0 W S 0 0 1 0W S 0 W o= S x x + S W r .gatehelp.33 GHz (B) 2 GHz < f < 3 GHz (C) 4.52 C (s) 24 = 3 2 R (s) s + 9s + 26s + 24 Ro V R R V R V 1 0V Sx1W S 0 WSx1W S 0 W o2W = S 0 (A) Sx 0 1WSx2W + S 0 W r S W S o3W S− 24 − 26 − 9W Sx Sx 3W S24W WS W S W To V X R T XT XV T X R R R V V Sx1W S0 1 0 WSx1W S 0 W o2W = S0 0 1 WSx2W + S 0 W r (C) Sx S o3W Sx W S S9 26 24W Sx 3W S24W WS W S W T X T XT X T X Ro V R VR V R V Sx1W S 0 1 0WSx1W S 0 W o2W = S 0 0 1WSx2W + S 0 W r (B) Sx S o3W Sx W S S24 26 9W Sx 3W S24W WS W S W T XT X VT X RoX V T R R V V 1 0 WSx1W R Sx1W S 0 S0W o2W = S 0 (D) Sx 0 1 WSx2W + S 0 W S W S o3W S− 9 − 26 − 24W Sx Sx 3W S24W WS W S W T X T XT X T X Q. 6 GHz (C) 3 GHz < f < 3. over which guide support both TE10 and TE 01 modes and no other. The best alternative among the four. (A) Show their virility (B) reaffirm their commitment of the company (C) bring down the government (D) demonstrate their strength.56. is In pursuance of their decision of resist what they saw as anti-labour policies.60 In the following sentence....5 GHz < f < 4.02 GHz General Aptitude(GA) Questions Q.Q 60 carry one mark each Q..gatehelp...... The odd word from the group.57 Which one of the following is the synonym of the word ADMONISH ? (A) warn (B) escape (C) worship (D) distribute Q.55 The range of frequencies. the company employee’s union launched agitation to.MOCK TEST 15/16 Q.56 Which one of the following is the Antonym of the word SILENCE ? (A) attune (B) babble (C) achromatic (D) aurora Q.59 A pair of CAPITALIZED words shown below has four pairs of words. Four different ways of completing the sentence are indicated. is (A) Swim (B) Swill (C) Ablution (D) Bathe Q. a part of the sentence is left unfinished.58 One of the four words given in the four options does not fit the set of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair. is NUTS : BOLTS (A) Nitty : Gritty (B) Bare : Feet (C) Naked : Surprise (D) Hard : Soft Q. EC-D www..5 GHz < f < 3.6 GHz (D) 2.35 GHz < f < GHz (B) 2.com .... is (A) 3.. 61-65 carry two marks each Q. then one km east and then one km north to arrive at the destination. The groups A and C are combined to form group D .gatehelp. then one km east then one km north.62 There are 60 students in a class.com EC-D . 20 and 25 students each. In how many ways can you start from any city and come back to it without travelling on the same road more than once ? (A) 8 (C) 16 Q.61 Let N = 1421 # 1423 # 1425 . These students are divided into three groups A .65 A man starting at a point walks one km east.63 Four cities are connected by a road network as shown in the figure.16/16 MOCK TEST Q. and C of 15.64 (B) 12 (D) 20 The roots of the equation ax2 + 3x + 6 = 0 will be reciprocal to each other if the value of a is (A) 3 (B) 4 (C) 5 (D) 6 Q. then two km north. What is the shortest distance from the starting point to the destination ? (B) 7 km (A) 2 2 km (C) 3 2 km (D) 5 km END OF THE QUESTION PAPER www. What is the average weight of the students in group D ? (A) more than the average weight of A (B) more than the average weight of C (C) less than the average weight of C (D) cannot be determined Q. B . What is the remainder when N is divided by 12 ? (A) 0 (B) 9 (C) 3 (D) 6 Q. com with Subject MOCK Test Solutions and provide following details in your email • • • • • • • Name: College: City: Mode of Preparation: Coaching/self study Phone no: How have you heard about GATHELP? .gatehelp.MOCK TEST 17/16 ATTENTION! For Detailed Solutions Please Mail to [email protected] . Are you referring any book by NODIA for your GATE Preparation: Yes/No (If Yes.. Mention Books) EC-D www. 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