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EC-GATE-2016 GATE 2016 – A Brief Analysis (Based on student test experiences in the stream of EC on 31st January, 2016 – (Afternoon Session) Section wise analysis of the paper Section Classification 1 Mark Engineering Mathematics Networks Electronic Devices Analog Circuits Digital Circuits Signals and Systems Control Systems Communication Electromagnetics Verbal Ability Numerical Ability 5 1 3 3 3 3 2 3 2 2 3 30 2 Marks 4 3 3 4 3 4 3 4 2 3 2 35 Total No of Questions 9 4 6 7 6 7 5 7 4 5 5 65 Type of Questions asked from each section Digital Circuits Questions came from Boolean expression simplification, microprocessor, logic gates, Sequential circuits. Electronic Devices Questions came from Solar cell, pn junction diode, MOS capacitor, BJT Analog Circuits OP-amp, zener diode, diode, 555 timers, BJT, MOSFET Signals and Systems Questions came from Z-transforms, Sampling, Laplace transform, Filters, DFT. Networks Questions came from Basics of networks, Two port networks, Steady state Analysis. Control Systems Questions came from Root locus, R-H criterion, Block diagram reduction, Time Domain analysis Communications Questions came from Digital communication, Information theory. Electromagnetics Questions came from Wave guides, Polarization, Maxwells equations Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 1 EC-GATE-2016 Questions from the paper General Aptitude 1. Key: Exp: If y=mx+c curve passes through (0,0) and (2,6) then m= _________. 3 y=mx+c passing through (0,0) 0 0 c c 0 y=mx+c passing through (2,6) 6 2m m 3 2. Key: Exp: It takes 10s, 15s for two trains moving in same direction, to completely pass a pole. Length of first train is 120 m and other is 150m. The magnitude of the difference between speeds is m/s. (A) 2 (B) 10 (C) 12 (D) 22 (A) length Speed length speed time time 120 10 s 1 s1 12 150 15 s 2 s 2 10 s1 s 2 2 3. Key: 4. Key: Exp Four undergraduates are staying is a room. They agreed that older enjoys the more space. Manu is two months older than Sravan, who is one month younger than Trideep. Pavan is one month older than Sravan. Who will enjoy more space in room. (A) Manu (B) Sravan (C) Trideep (D) Pavan (A) The area bounded by 3x+2y=14 and 2x-3y=5 in the first quadrant is (A) 14.95 (B) 15.25 (C) 15.70 (D) 20.35 (B) 14 A ,0 3 B 0, 7 5 C , 0 2 5 D 0, 3 E 4,1 y B 3x 2y 14 2x 3y 5 E R O x C A Required area is area of OAB – area of CEA 1 14 7 15.25 sq.units 2 3 D Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 2 EC-GATE-2016 Technical 1. In below figure current ‘i’ is __________A. 1A 5 8V 1 1 1 8V C1 1 Key: Exp: -1 Nodal equation at V V 8 V V 8 V 0 1 1 1 1 4V 16 1A 5 8V i1 a 1 1 v V 4V By using KCL at node ‘a’. 84 1 i1 0 i1 5A 1 KCL at b 4 i1 i 0 4 5 i 0 1 i 1A 0V b 8V i 1 Px P1 2. The charge density profile shown in figure. The resultant potential distribution is best described by 0 Vx (A) (B) a b V(x) a (D) a x V(x) b 0 x b x V(x) a P1 0 (C) b 0 x Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 3 EC-GATE-2016 Key: (D) Exp: q q Electrical 0 N d x no N d x po x2 x 2w 0 Potential (x) 0 P(x) V(x) P1 b b a a P1 P1 3. Key: Exp: In the below circuit M1 is in saturation has transconductance gm = 0.01 seimens, Ignoring internal parasitic capacitance and assuming the channel length modulation to be zero the small signal input pole frequency (kHz) is ______. 57.9 Cin 50PF 1 g m R 550PF VDD 1K VD 50pF Vin 5k M1 g m 0.015 R 1k R 1k 1 g m R 11 R in 5k f in 4. 1 57.9 kHz 2R in Cin z11 z12 The Z – parameter matrix for two port network shows z 21 z 22 3 IP OP 6 (A) 2 2 2 2 2 2 (B) 2 2 1 2 (C) 2 1 2 1 (D) 1 2 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 4 EC-GATE-2016 Key: Exp: (A) Since the given network is symmetric and reciprocal Z11 Z22 Z12 Z21 Z11 V1 I1 V Z21 2 I1 I2 0 3 5. Key: Exp: 6. output input 3 6 2 36 6 I1 I2 0 We know V2 V1 Z21 2 Z So 11 Z21 V1 Z12 2 2 Z22 2 2 V2 3 IL 0 If a right handed circularly polarized wave is incident on a place perfect conductor, then the reflected wave will be _______. (A) Right handed circularly polarized (B) left handed circularly polarized o (C) Elliptical with all angle 45 (D) horizontally polarized (B) If incident wave is right handed polarized then the reflected wave is left handed polarized. The block diagram of a feedback system is shown in figure the overall closed loop gain of the system is _________. x G2 y G1 H1 Exp: G2 G1 G1 1 G1H1 H1 G1G 2 Y x 1 G1H1 G1G 2 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 5 EC-GATE-2016 1 dx is equal to _______. 1 x 7. Key: 2 Exp: 8. f z Key: Exp: -1 0 1 0 1 1 x d 0 2 1 x 02 2 0 sin z , the residue of the pole at z = 0 is _______. z 2 z 0 is a simple pole as f z sin z z 2 1 z3 z5 z ...... 2 z 3! 5! z 0 z 0 ...... 1 z0 3! 5! Residue of f z at t 0 is 1 3 9. Key: Exp: The first two rows in the routh table for characteristics of a closed loop control system are given as s3 1 2k 3 4 s 2 2k The range of ‘k’ for which the system is stable is (A) 2.0 k 0.5 (B) 0 k 0.5 (C) 0 k 8 (D) 0.5 k (D) S3 1 2k 3 4 S2 2k From the table we can find characteristic equation s3 2ks2 2k 3 s 4 0 For stability 2k 2k 3 4 4k 2 6k 4 0 1 k k 2 0 2 1 So the conditions are k and k 2 combiningly k > – 2 2 10. Key: A triangle in the xy plane is bounded by the straight lines 2x =3y, y = 0 and x = 3. The volume above the triangle and under the plane x + y + z = 6 is _________. 10 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 6 EC-GATE-2016 Exp: Volume = z dx dy y R 3 2 x 3 y 0 6 x y dy dx x 0 3 2 y 6 x y z x 0 .dx y0 3 xb R 3 8 x3 8 2x 2 2 9 33 10 cubic units 9 3 27 0 11. B 2 x 3 2x 1 4 2 6 x 3 2 9 x dx 3 2 2 4x x 2 x 2 dx 3 9 2x 3y x O A x limits : 0 to 3 2 y limits: 0 to x 3 An analog baseband signal, band limited to 100MHz, is sampled at the nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities P1 P4 0.125 and P2 P3 . The information rate bits sec of the message source is _____. Key: Exp: 213.9 Information rate = rH H bits symbol r symbols sec If we assume each sample is mapped to symbol Then r = 200 MHz P1 P4 2P2 1 1 3 2P2 1 4 4 3 P2 8 3 P2 P3 8 2 6 8 H log 2 8 log 2 1.069 8 8 3 I.R 213.9 Mbps 12. Faraday’s law of electromagnetic s induction is mathematically described by which one of the followers. (A) .B 0 Key: B (C) E t (C) (B) .D V (D) H E D t Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 7 EC-GATE-2016 13. Key: Exp: The minimum number of 2 -input NAND gates required to implement a 2 – input XOR gate is (A) 4 (B) 5 (C) 6 (D) 7 (A) A Y AB AB A B B 14. For the unity feedback control system shown in below figure the open loop transfer function G(s) = 2 s s 1 x t et G s gt The steady state error due to unit step i p is _______. Key: 0.33 Exp: For the unit feedback system G S 2 S S 1 For unit step input the steady state error is 1 ess 1 kp Where k p limsG s lims s ess s 0 2 2 s s 1 1 1 0.33 1 2 3 Key: The bit error probability of a memory less BSc is 10-5. If 105 bits are sent over this channel, then the probability that not more than one bit will be in error is _________. 1 Exp: Pebit 105 15. n 105 bits x no. of bits error P x 1 P x 0 P x 1 5 5 105 5 105 5 1 5 10 0 5 10 1 1 10 10 10 1 10 0 1 105 5 105 5 1 5 10 1 10 1 10 1 1 very close 105 1 105 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 8 EC-GATE-2016 16. For the circuit shown in figure R1 R 2 R 3 1, L 1H, C 1F. If input Vin cos 106 t , then the overall voltage gain of the circuit is _________. R1 R3 L C Vout R2 V in Key: Exp: -1 R 1 A1 1 1 1 6 2 6 jL 10 10 R3 1 1 1 A2 1 R 2 XC 11 2 1 6 10 106 V The overall voltage gain A v out A1 A 2 Vin Vout 1 2 1 Vin 2 17. Key: Exp: The forward path transfer function and the feedback path transfer function of single loop negative feedback control system is given as k s 2 G s 2 and H s 1 s 2s 2 If the variable perimeter k is real positive, then the location of the breakaway point on the root locus diagram of the system is _______. -3.414 To find break point, from characteristic equation we need to arrange k as function of s, then the dk 0 gives break point. root of ds Characteristic equation is given by s2 2s 2 k 2k 0 k s 2 s 2 2s 2 s 2 2s 2 k s2 2 d d 2 s 2 ds s 2s 2 s 2s ds s 2 dk ds s 2 2 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 9 EC-GATE-2016 s 2 2s 2 s 2 2s 2 dk ds s 2 2 dk 0 ds 2s 2 2s 4s 4 s 2 2s 2 0 j 1 2 s 2 4s 2 0 s 0.58 and 3.414 To find the valid break point we need to find that lies on root locus – 3.414 lies on root locus So break point – 3.414. 18. j Current I The I V characteristics of a solar cell is as shown in figure, when it is illuminated uniformly with solar light of power 100 mW/cm2. The solar cell has an area of 3 cm2 Isc 180mA and fill factor is 0.7. The maximum efficiency of the cell is _________%. Key: 21 Exp: Efficiency FF.Voc Isc 0.7 0.5 180 100% Pin (100 3) Voc 0.5V (0,0) Voltage V = 21%. 19. Consider the circuit shown below, Assuming VBE1 VEB2 0.7V . The value of D.C voltage Vc2 (V)is ____ . Vcc 2.5V 1 100 Q1 Q2 10 k 2 50 Vc2 1V 1k Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 10 EC-GATE-2016 Key: Exp: 0.5 VE2 2.5 VBE1 1.8V VB2 VE2 0.7 1.1V 0.1 105 A 10k 50 105 A I B2 I C2 VC2 5 104 103 0.5V 20. The I-V characteristics of three types of diodes at room temperature made of semiconductors x, y, z are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If E gx , E gy , E gz are the band gaps of x,y,z respectively then (A) E gx E gy E gt I (B) E gx E gy E g2 x z y (C) E gx E gy E gz (D) No relation V Key: (C) 21. The direct form structure of an FIR filter is shown in figure unity delay x(n) unity delay 5 5 Key: Exp: The filter can be used to approximate a (A) LPF (B) HPF (C) (C) BPF y[n] (D) BSF y n 5x n 5x n 2 H e j 5 1 e 2 j At 0, H e j 0 At ; H e j 0 Thus the filter is band pass filter Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 11 EC-GATE-2016 22. Two diodes D1 and D2 shown in figure are Ideal. The two capacitors are identical and the value of RC is very large than time period of ac signal. If the diodes don’t breakdown in reverse bias, the value of output voltage V0 (volts) in steady state is __________ D1 10sin t C R V0 C Key: Exp: 0 Vo 0volts D2 10V Vo 0V 10V 23. A discrete sequence x n n 3 2 n 5 has a z-transform X(z); if Y(z) = X(-z) is the ztransform for another sequence y[n], then, (A) y[n] x[n] (B) y[n] x[ n] (C) y[n] x[n] (D) y[n] x[ n] Key: (B) Exp: X(z) z3 2z5 Y(z) X(z) z3 2z5 On applying inverse z-transform, y[n] [n 3] 2[n 5] Consider x[ n] [ n 3] 2[ n 5] [n 3] 2[n 5] y[n] x[ n] 24. A continuous time signal x(t) =cos 6t sin 8t, the Nyquist sampling rate (samples/sec) of y(t)=x(2t+5) is (A) 8 (B) 12 (C) 16 (D) 20 Key: (C) Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 12 EC-GATE-2016 Exp: X(f ) Shifting doesn’t effects the sampling rate due to scaling by a factor ‘2’ spectral components are doubled. Thus maximum frequency of Y(f ) 8. 4 Nyquist rate 8 2 16Hz 3 3 4 2 t cos t is passed through an LTI system with impulse transfer 3 function H s es es . If ck represents kth coefficient of exponential Fourier series of output 25. An input signal x t 2cos Key: then the value of c3 is (A) 0 (C) Exp: H j 2cos (B) 1 (C) 2 (D) 3 2 2 For 2cos t , output 2k1 cos t 3 3 2 2cos k1 2cos 1 2 3 3 For cos t, output k 2 cos t k2 2 cos 2 2 output 2cos t 2 cos t 3 26. xt (A) Key: Exp: sin t sin t * , t t sin t t c3 2 If * is the convolution then the value of x t is (B) sin 2t 2 t (C) 2sin t t (D) sin 2 t t 2 (A) sin t 1 sa t t F sa t rect 2 1 F sa t rect 2 Convolution in time domain leads to multiplication in frequency domain 1 sin t rect rect X x t sa t t 2 2 Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 13 EC-GATE-2016 27. Figure I and II shows MOS capacitors of unit area. The capacitor in figure I has insulator X (with thickness t1 1m and r1 4 ) and Y (thickness t 2 3nm and r2 20 ). The capacitor in figure II has only insulator material X of thickness tq . Find tq _____________ nm. Metal t2 2 , Y t1 1 , X Metal 1 , X Figure II Figure I Key: Exp: 28. t eq 1.6 E1 E 2 . t1 t 2 E1E 2 4 20 CI 2.5 E1 E 2 E1t 2 E 2 t1 (4 2) (20 1) t1 t 2 E 4 CII 1 2.5 t Eq 1.6nm t Eq t Eq The injected electron concentration profile in the base region of npn BJT, biased in the active region is linear as shown in figure. If the area of the emitter – base junction is 0.001 cm2, n 800 cm2 V s in the base region, the collector current Ic mA at room temperature is __________ VT 26mV, q 1.6 1019 c . IB n IE Key: Exp: p n IC 0.5 m 6.65 Ic qADn 1014 cm3 1014 0 dn dn 1.6 1019 0.001 800 26 103 qAn V 4 dx dx 0.5 10 Ic 6.65mA Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 14 EC-GATE-2016 29. Frequency of oscillation at pin 3 is ___________kHz.. 4 8 R A 2.2k VCC 7 R B 4.7k 6 555 2 Out 3 1 C 0.022F Key: 5.64 Exp: f 30. The I-V characteristics of diodes D1 and D2 are given in figure. The simply voltage is varied from 0 to 100V. The breakdown occurs at 1.44 1.44 5.64kHz R A 2R B C [2200 24700] 0.022 106 ID D1 80V 70V V 0 to100V D2 D1 (A) Key: Exp: D2 D1 only (B) D2 only (C) Both D1 and D2 (D) None of D1 and D2 (D) When two diodes are connected in series, the effective breakdown voltages becomes equal to the sum of their individual breakdown voltage. Vb Vb1 Vb2 80 70 150V Since the applied voltage has a maximum of 100V (100<Vb), No diodes will be in breakdown region. Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 15 EC-GATE-2016 31. The surface region of the MOSFET is in (A) Accumulation (B) Inversion (C) Depletion (D) None of the above EC B B Key: Exp: EF Ei EV (B) The semiconductor used in the MOSFET is n-type. At the surface the intrinsic level is above EF as it is found at the distance of below EF, So, the surface is in inversion region. Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same. 16