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GATE 2003 : Electronics And Communication Engineering

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For more files visit www.educationobserver.com/forum Q.1 Q.30 Carry One Mark Each 1. The minimum number of equations required to analyze the circuit shown in Fig.Q.1 is C C R ~ (a) 3 2. R (b) 4 R C R (c) 6 (d) 7 A source of angular frequency 1 rad/sec has a source impedance consisting of 1 resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is (a) 1 resistance (b) 1 resistance in parallel with 1 H inductance (c) 1 resistance in series with 1 F capacitor (d) 1 resistance in parallel with 1 F capacitor 3. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each R, L and C is doubled from its original value, the new Q of the circuit is (a) 25 4. (b) 50 The Laplace transform of i(t) is given by I ( s ) = As t (d) 200 2 s (1 + s ) , the value of i(t) tends to (a) 0 5. (c) 100 (b) 1 (c) 2 (d) The differential equation for the current i(t) in the circuit of Figure Q.5 is (a) 2 (b) d 2i dt d2i +2 +2 dt 2 (c) 2 2 d 2i dt 2 (d) d i dt 2 di + 2i ( t ) = cos t dt +2 2 +2 di + i ( t ) = sin t dt di + i ( t ) = cos t dt di + 2i ( t ) = sin t dt + i(t) 2 2H 1F sin t - For more files visit www.educationobserver.com/forum 6. n-type silicon is obtained by doping silicon with (a) Germanium 7. (b) Aluminum (d) Phosphorus (c) 0.80 eV (d) 0.67 eV The bandgap of silicon at 300 K is (a) 1.36 eV 8. (c) Boron (b) 1.10 eV The intrinsic carrier concentration of silicon sample of 300 K is 1.5 1016/m3. If after doping, the number of majority carriers is 5 1020/m3, the minority carrier density is (a) 4.50 1011/m3 (c) 5.00 1020/m3 9. (b) 3.33 104/m3 (d) 3.00 10-5/m3 Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photodiode are operated in X bias and Y bias respectively. (a) X: reverse, Y: reverse (c) X: forward, Y: reverse 10. (b) X: reverse, Y: forward (d) X: forward, Y: forward For an n-channel enhancement type MOSFET, if the source is connected at a higher potential than that of the bulk (i.e. VSB > 0), the threshold voltage VT of the MOSFET will (a) remain unchanged (c) change polarity 11. (b) decrease (d) increase Choose the correct match for input resistance of various amplifier configurations shown below. Configuration Input resistance CB: Common Base LO: Low CC: Common Collector MO: Moderate CE: Common Emitter HI: High (a) CB-LO, CC-MO, CE-HI (c) CB-MO, CC-HI, CE-LO 12. (b) CB-LO, CC-HI, CE-MO (d) CB-HI, CC-LO, CE-MO The circuit shown in figure is best described as a (a) bridge rectifier (b) ring modulator (c) frequency discriminatory (d) voltage doubler ~ output For more files visit www.educationobserver.com/forum 13. If the input to the ideal comparator shown in figure is a sinusoidal signal of 8V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of Input + Output Vref =2V (a) 14. 1 2 1 3 (c) 1 6 (d) q 12 If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then its common mode rejection ratio is (a) 23 dB 15. (b) - (b) 25 dB (c) 46 dB (d) 50 dB Generally, the gain of a transistor amplifier falls at high frequencies due to the (a) internal capacitances of the device (b) coupling capacitor at the input (c) skin effect (d) coupling capacitor at the output 16. The number of distinct Boolean expression of 4 variables is (a) 16 17. (c) 1024 (d) 65536 The minimum number of comparators required to build an 8 it flash ADC is (a) 8 18. (b) 256 (b) 63 (c) 255 (d) 256 The output of the 74 series of TTL gates is taken from a BJT in (a) totem pole and common collector configuration (b) either totem pole or open collector configuration (c) common base configuration (d) common collector configuration 19. Without any additional circuitry, an 8:1 MUX can be used to obtain (a) some but not all Boolean functions of 3 variables (b) all function of 3 variables but none of 4 variables (c) all functions of 3 variables and some but not all of 4 variables (d) all functions of 4 variables For more files visit www.educationobserver.com/forum 20. A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate(s). The combination circuit consists of (a) one AND gate (c) one AND gate and one OR gate 21. (b) one OR gate (d) two AND gates The Fourier series expansion of a real periodic signal with fundamental frequency ce f0 is given by gp ( t ) = n j 2 nfot it is given that C3 = 3 + j5. Then C-3 is n = (a) 5+j3 22. (b) -3-j5 (d) 3-j5 Let x(t) be the input to a linear, time-invariant system. The required output is 4x(t-2). The transfer function of the system should be (a) 4 e j 4 f 23. (c) -5+j3 (b) 2 e j 8 f (c) 4 e j 4 f (d) 2 e j 8 f A sequence x(n) with the z-transform X(z) = z 4 + z 2 2 z + 2 3z 4 is applied as an input to a linear, time-invariant system with the impulse response h(n) = 2 (n-3) where 1, n = 0 0, otherwise (n) = The output at n = 4 is (a) -6 24. (b) zero (c) 2 (d) -4 Figure shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right hand pole, the closed loop system is Im GH-plane Re =0 (-1,0) positive (a) always stable (b) unstable with one closed loop right hand pole (c) unstable with two closed loop right hand poles (d) unstable with three closed loop right hand poles For more files visit www.educationobserver.com/forum 25. A PD controller is used to compensate a system. uncompensated system, the compensated system has Compared (a) a higher type number 26. (d) larger transient overshoot The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is (a) the in-phase component (b) the quadrature-component (c) zero 27. (d) the envelope The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is (a) raised cosine 28. (b) flat (c) parabolic (b) 3 dB (c) 2 dB (d) 0 dB The unit of H is (a) Ampere (b) Ampere/meter (c) Ampere/meter 30. (d) Gaussian At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by (a) 6 dB 29. the (b) reduced damping (c) higher noise amplification to 2 (d) Ampere-meter The depth of penetration of electromagnetic wave in a medium having conductivity at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (a) 6.25 cm (b) 12.50 cm (c) 50.00 cm (d) 100.00 cm Q.31 Q.90 Carry Two Marks Each 31. Twelve 1 resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is (a) 32. 5 6 (b) 1 6 (c) 6 5 (d) 3 2 The current flowing through the resistance R in the circuit in figure has the form M=0.75H P cos 4t, where P is 1/10.24F (a) (0.18+j0.72) (b) (0.46+j1.90) 3 R=3.92 (c) -(0.18+j1.90) (d) -(0.192+j0.144) V=2cos4t ~ For more files visit www.educationobserver.com/forum The circuit for Q.33-34 is given in figure. For both the questions, assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0. 1 S C ii(t) i2(t) R L V R 33. At t = 0+, the current i1 is (a) 34. C V 2R (b) V R (c) V 4R (d) zero I1 ( s ) and I2 ( s ) are the Laplace transforms of i1 ( t ) and i2 ( t ) respectively. The equations for the loop currents I1 ( s ) and I2 ( s ) for the circuit shown in figure Q.33-34, after the switch is brought from position 1 to position 2 at t = 0, are 1 R + Ls + Cs (a) Ls 1 R + Ls + Cs (b) Ls Ls I s V 1( ) = s 1 I2 ( s ) 0 R+ Cs 1 R + Ls + Cs (c) Ls I1 ( s ) V = s 1 I s ( ) 2 R + Ls + 0 Cs 1 R + Ls + Cs (d) Ls 35. Ls I s V 1( ) = s 1 I s ( ) 2 R+ 0 Cs I1 ( s ) V = s 1 I s ( ) 2 R + Ls + 0 Cs Ls Ls An input voltage v(t) = 10 2 cos ( t + 10 ) + 10 3 cos ( 2t + 10 ) V is applied to a series combination of resistance R = 1 and an inductance L = 1H. The resulting steady state current i(t) in ampere is For more files visit www.educationobserver.com/forum ( (a) 10 cos ( t + 55 ) + 10 cos 2t + 10 + tan 1 2 (b) 10 cos ( t + 55 ) + 10 3 cos (2t + 55 ) 2 ( (c) 10 cos ( t 35 ) + 10 cos 2t + 10 tan 1 2 (d) 10 cos ( t 35 ) + 10 36. ) 3 cos (2t 35 ) 2 The driving point impedance Z(s) of a network has the pole-zero locations as shown in figure. If Z(0) =3, then Z(s) is (a) (b) (c) (d) 37. ) Im 3 ( s + 3) s2 + 2 s + 3 2 ( s + 3) X 2 1 s + 2s + 2 s-plane 3 ( s 3) 2 s 2s 2 -3 Re -1 2 ( s 3) 2 s 2s 3 X O denotes zero x denotes pole -1 The impedance parameters Z11 and Z12 of the two-port network in figure are (a) Z11 = 2.75 and Z12 =0.25 (b) Z11 = 3 and Z12 =0.5 2 2 3 1 (c) Z11 = 3 and Z12 =0.25 2 1 1 (d) Z11 = 2.25 and Z12 =0.5 1 38. An n-type silicon bar 0.1 cm long and m2 in cross-sectional area has a majority carrier concentration of 5 1020/m3 and the carrier mobility is 0.13m2/V-s at 300K. if the charge of an electron is 1.6 10-19 coulomb, then the resistance of the bar is (a) 106 ohm 39. 2 (b) 104 ohm (c) 10-1 ohm (d) 10-4 ohm The electron concentration in a sample of uniformly doped n-type silicon at 300 K varies linearly from 1017/cm3 at x = 0 to 6 1016/cm3 at x = 2 m. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 10-19 coulomb and the diffusion constant Dn = 35 cm2/s, the current density in the silicon, if no electric field is present, is (a) zero (b) -112 A/cm2 (c) +1120 A/cm2 (d) -1112 A/cm2 For more files visit www.educationobserver.com/forum 40. Match items in Group 1 with items in Group 2, most suitably. Group 1 Group 2 P LED 1 Heavy doping Q Avalanche photodiode 2 Coherent radiation R Tunnel diode 3 Spontaneous emission S LASER 4 Current gain (a) P 1 Q 2 R 4 S - 3 (c) P 3 Q 4 R 1 S - 2 41. (b) P 2 Q 3 R 1 S - 4 (d) P 2 Q 1 R 4 S - 3 At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435V, whereas a certain silicon diode requires a forward bias of 0.718V. Under the conditions stated above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (a) 1 42. (b) 1.98 eV (c) 1.17 eV (d) 0.74 eV (b) 2.0 mA (c) 3.5 mA (d) 4.0 mA If P is Passivation, Q is n-well implant, R is metallization and S is soruce/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process, is (a) P-Q-R-S 45. (d) 8 103 When the gate-to-source voltage (VGS) of a MOSFET with threshold voltage of 400mV, working in saturation is 900 mV, the drain current in observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation, the drain current for an applied VGS of 1400 mV is (a) 0.5 mA 44. (c) 4 103 A particular green LED emits light of wavelength 5490 A. The energy bandgap of the semiconductor material used there is (Planck s constant = 6.626 10-34J-s) (a) 2.26 eV 43. (b) 5 (b) Q-S-R-P (c) R-P-S-Q (d) S-R-Q-P An amplifier without feedback has a voltage gain of 50, input resistance of 1 K and output resistance of 2.5 K . The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is (a) 1 K 11 (b) 1 K 5 (c) 5 K (d) 11 K For more files visit www.educationobserver.com/forum 46. In the amplifier circuit shown in figure, the values of R1 and R2 are such that the transistor is operating at VCE= 3V and IC = 1.5mA when its is 150. For a transistor with of 200, the operating point (VCE, IC) is (a) (2V, 2 mA) VCC=6V (b) (3V, 2 mA) R2 R1 (c) (4V, 2 mA) (d) (4V, 1 mA) 47. The oscillator circuit shown in figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is C R (a) 48. 1 (2 6RC ) (b) C C R R 1 (2 RC ) (c) 1 ( 6RC ) (d) 6 (2 RC ) The output voltage of the regulated power supply shown in figure is + 1K 15 V DC Unregulated Power Source + Vz =3V - 40K 20K - (a) 3V (b) 6V (c) 9V Regulated DC Output (d) 12V For more files visit www.educationobserver.com/forum 49. The action of a JFET in its equivalent circuit can best be represented as a (a) Current Controlled Current Source (b) Current Controlled Voltage Source (c) Voltage Controlled Voltage Source (d) Voltage Controlled Current Source 5k 50. If the op-amp in figure is ideal, the output voltage Vout will be equal to 1k - 2V (a) 1V Vout + 3V 1k (b) 6V 8k (c) 14V (d) 17V 51. Three identical amplifiers with each one having a voltage gain of 50, input resistance of 1 K and output resistance of 250 , are cascaded. The open circuit voltage gain of the combined amplifier is (a) 49 dB 52. (b) 51 dB (c) 98 dB (d) 102 dB An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3V is generated by charging a capacitor of 2 F in every cycle. The charging requires (a) constant voltage source of 3 V for 1 ms (b) constant voltage source of 3 V for 2 ms (c) constant current source of 3 mA for 1 ms (d) constant current source of 3 mA for 2 ms 53. The circuit shown in figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P Q R Q Z = RQ + PR + QP P P P Q R Z Y P Q R Z Y Output P Q R Z Y Q R Z Y For more files visit www.educationobserver.com/forum The circuit acts as a (a) 4 bit adder giving P + Q (c) 4 bit subtractor-giving Q - P 54. (b) 4 bit subtractor-giving P - Q (d) 4 bit adder giving P + Q + R If the functions W, X, Y and Z are as follows W = R + PQ + RS X = PQR S + P Q R S + P Q R S Y = RS + PR + PQ + P.Q Z = R + S + PQ + P.Q.R + PQ.S Then (a) W = Z, X = Z 55. (b) W = Z, X = Y (d) W = Y = Z (c) W = Y A 4 bit ripple counter and a 4 bit synchronous counter are made using flip-flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then (a) R = 10 ns, S = 40 ns (c) R = 10 ns, S = 30 ns 56. (b) R = 40 ns, S = 10 ns (d) R = 30 ns, S = 10 ns The DTL, TTL, ECL and CMOS families of digital ICs are compared in the following 4 columns (P) (Q) (R) (S) Fanout is minimum DTL DTL TTL CMOS Power consumption is minimum TTL CMOS ECL DTL Propagation delay is minimum CMOS ECL TTL TTL The correct column is (a) P 57. (b) Q The circuit shown in figure is a 4-bit DAC The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistances and the 5V inputs have a tolerance of 10%. The specification (rounded to the nearest multiple of 5%) for the tolerance of the DAC is (a) 35% (b) 20% (c) R (d) S R R 2R 4R 8R (c) 10% + R (d) 5% For more files visit www.educationobserver.com/forum 58. The circuit shown in figure converts INPUTS MSB MSB OUTPUTS (a) BCD to binary code (c) Excess 3 to Gray code 59. (b) Binary to excess 3 code (d) Gray to Binary code In the circuit shown in Figure, A is a parallel in, parallel-out 4-bit register, which loads at the rising edge of the clock C. The input lines are connected to a 4-bit bus, W. Its output acts as the input to a 16 4 ROM whose output is floating when the enable input E is 0. A partial table of the contents of the ROM is as follows Address 0 2 4 6 8 10 11 14 Data 0011 1111 0100 1010 1011 1000 0010 1000 W MSB C A 1 ROM E C: t1 t2 Time For more files visit www.educationobserver.com/forum The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is (a) 1111 60. (b) 1011 (c) 1000 (d) 0010 In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of register B. As a result (a) Carry flag will be set but Zero flag will be reset (b) Carry flag will be reset but Zero flag will be set (c) Both Carry flag and Zero flag will be reset (d) Both Carry flag and Zero flag will be set 61. Let X and Y be two statistically independent random variables uniformly distributed in the ranges (-1,1) and (-2,1) respectively. Let Z = X + Y. then the probability that [Z -2] is (a) zero 62. (b) 1 6 (c) 1 3 (d) 1 12 Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship, x ( n) , n 1 0, y ( n) = n=0 x ( n + 1) , n 1 where x(n) is the input and y(n) is the output. The above system has the properties (a) P, S but not Q, R (b) P, Q, S but not R (c) P, Q, R, S (d) Q, R, S but not P Data for Q.63-64 are given below. Solve the problems and choose the correct answers. The system under consideration is an RC low-pass filter (RC-LPF) with R = 1.0 k and C = 1.0 F. 63. Let H(f) denote the frequency response of the RC-LPF. Let f1 be the highest H ( f1 ) frequency such that 0 f f1 , 0.95. Then f1 (in Hz) is H (0) (a) 327.8 64. (b) 163.9 (c) 52.2 (d) 104.4 Let tg(f) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg(f2) in ms, is (a) 0.717 (b) 7.17 (c) 71.7 (d) 4.505 For more files visit www.educationobserver.com/forum Data for Q.65 66 are given below. Solve the problems and choose the correct answers. X(t) is a random process with a constant mean value of 2 and the autocorrelation 0.2 function Rx ( ) = 4 e + 1 . 65. Let X be the Gaussian random variable obtained by sampling the process at t = ti and let Q ( ) = y2 e 2 dy. 1 2 The probability that x 1 is (a) 1 Q(0.5) 66. 1 (d) 1 Q 2 2 Let Y and Z be the random variables obtained by sampling X(t) at t =2 and t = 4 respectively. Let W = Y Z. The variance of W is (a) 13.36 67. 1 (c) Q 2 2 (b) Q(0.5) (b) 9.36 (c) 2.64 (d) 8.00 Let x(t) = 2cos(800 t) + cos(1400 t). x(t) is sampled with the rectangular pulse train shown in figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are p(t) 3 t -T0 -T0/6 0 T0/6 T0 T0=10-3sec (a) 2.7, 3.4 (c) 2.6, 2.7, 3.3, 3.4, 3.6 68. (b) 3.3, 3.6 (d) 2.7, 3.3 The signal flow graph of a system is shown in figure. The transfer function C (s) of the system is 1 1 R (s) 1 6 R(s) s s 1 C(s) -2 -4 -3 For more files visit www.educationobserver.com/forum (a) 69. 6 2 s + 29s + 6 6s (b) s ( s + 2) (c) 2 s + 29s + 6 The root locus of the system G ( s ) H ( s ) = s ( s + 27 ) (d) 2 s + 29s + 6 2 s + 29s + 6 K has the break-away point s ( s + 2) ( s + 3) located at (a) (-0.5,0) 70. (b) (-2.548,0) (c) (-4,0) (d) (-0.784,0) The approximate Bode magnitude plot of a minimum-phase system is shown in figure. The transfer function of the system is dB 160 140 20 0.1 10 100 (a) 10 ( s + 0.1)3 ( s + 10 )2 ( s + 100) (b) 10 ( s + 0.1)3 ( s + 10 ) ( s + 100) (c) 108 ( s + 0.1)2 ( s + 10 )2 ( s + 100) (d) 109 ( s + 0.1)3 ( s + 10 ) ( s + 100)2 8 A second-order system has the transfer function C (s) R (s) = 4 2 s + 4s + 4 . with r(t) as the unit-step function, the response c(t) of the system is represented by (a) (b) Step Response 1.5 Step Response 1 Amplitude 1 Amplitude 71. 7 0.5 0 0.5 0 0 2 4 Time(sec) Figure (a) 6 0 2 4 Time(sec) Figure (b) 6 For more files visit www.educationobserver.com/forum (c) (d) Step Response Step Response 1 2 Amplitude Amplitude 1.5 1 0.5 0.5 0 0 5 10 15 20 25 0 Time (sec) (a) Figure (a) 72. The gain G (s) H (s) = margin s ( s + 100)3 (a) 0 dB, 0 73. 10 Time(sec) Figure (d) (b) Figure (b) and 5 0 Figure (c) the phase (c) Figure (c) margin of a (d) Figure (d) feedback system with are (b) , (c) , 0 (d) 88.5 dB, The zero-input response of a system given by the state-space equation &1 1 0 x1 x1 ( 0 ) 1 x = is & = and x 0 ( ) x2 1 1 x2 2 0 tet (a) t 74. et (b) t et (c) t te t (d) t te A DSB-SC signal is to be generated with a carrier frequency fc = 1MHz using a nonlinear device with the input-output characteristic v0 = a0vi + a1v i3 where a0 and a1 are constants. The output of the nonlinear device can be filtered by an appropriate band-pass filter. cos (2 fc t ) + m ( t ) where m(t) is the message signal. Then the value of Let vi = Ac fc (in MHz) is (a) 1.0 (b) 0.333 (c) 0.5 (d) 3.0 For more files visit www.educationobserver.com/forum The data for Q.75 76 are given below. Solve the problems and choose the correct answers. 3 be m ( t ) = cos 4 10 t ( ) c ( t ) = 5 cos 2 10 ) t be the carrier. ( Let the message signal and 6 75. c(t) and m(t) are used to generate an AM signal. The modulation index of the Total sideband power generated AM signal is 0.5. Then the quantity is Carrier power (a) 76. 1 2 (b) 1 4 1 3 (c) (d) 1 8 c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM 3 singal, then the coefficient of the term cos 2 1008 10 t in the FM signal (in ( ) terms of the Bessel coefficients) is (a) 5 J4 (3) 77. (b) 5 J8 ( 3) 2 (c) 5 J8 ( 4 ) 2 (d) 5 J4 ( 6 ) Choose the correct one from among the alternatives A, B, C, D after matching an item in Group 1 with the most appropriate item in Group 2. Group 1 Group 2 P Ring modulator 1 Clock recovery Q VCO 2 Demodulation of FM R Foster-Seely discriminator 3 Frequency conversion S Mixer 4 Summing the two inputs 5 Generation of FM 6 Generation of DSB-Sc (a) P 1 Q 3 R 2 S 4 (c) P 6 Q 1 R 3 S 2 78. (b) P 6 Q 5 R 2 S 3 (d) P 5 Q 6 R 1 S 3 A superheterodyne receiver is to operate in the frequency range 550 kHz 1650 Cmax kHz, with the intermediate frequency of 450 kHz. Let R = denote the Cmin required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then (a) R = 4.41, I = 1600 (b) R = 2.10, I = 1150 (c) R = 3.0, I = 1600 (d) R = 9.0, I = 1150 For more files visit www.educationobserver.com/forum 79. A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into 128 levels using a mid-rise uniform quantizer. The quantization noise power is (a) 0.768 V 80. (b) 10 dB (d) 3.072 V (c) 20 dB (d) 13 dB The input to a linear delta modulator having a step-size = 0.628 is a sine wave with frequency fm and peak amplitude Em. If the sampling frequency fs = 40 kHz, the combination of the sine-wave frequency and the peak amplitude, where slope overload will take place is Em fm (a) 0.3 V 8 kHz (b) 1.5 V 4 kHz (c) 1.5 V 2 kHz (d) 3.0 V 82. (c) 12 10-6V2 If Eb, the energy per bit of a binary digital signal, is 10-6 watt-sec and the onesided power spectral density of the white noise, N0 = 10-5 W/Hz, then the output SNR of the matched filter is (a) 26 dB 81. (b) 48 10-6V2 1 kHz If S represents the carrier synchronization at the receiver and represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is (a) = 0.5, S is required (c) = 0.5, S is not required 83. (b) = 1.0, S is required (d) = 1.0, S is not required A signal is sampled at 8 kHz and is quantized using 8-bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is (a) R = 32 kbps, SNRq = 25.8 dB (c) R = 64 kbps, SNRq = 55.8 dB 84. (b) R = 64 kbps, SNRq = 49.8 dB (d) R = 32 kbps, SNRq = 49.8 dB Medium 1 has the electrical permitivity 1=1.5 0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permitivity 2 = 2.5 0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = 2ux 3uy + 1uz volt/m, then E2 in medium 2 is (a) (c) 85. ( ) (2.0ux 7.5uy + 2.5uz ) volt/m (1.2ux 3.0uy + 1.0uz ) volt/m (b) (d) ( (2.0ux 2.0uy + 0.6uz ) volt/m (1.2ux 2.0uy + 0.6uz ) volt/m ) If the electric field intensity is given by E = xux + yuy + zuz volt/m, the potential difference between X(20,0) and Y(1,2,3) is For more files visit www.educationobserver.com/forum (a) +1 volt 86. (b) -1 volt (d) +6 volt A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with r = 4. The reflection coefficient for the normal incidence, is (a) zero 87. (c) +5 volt (b) 0.5 180 (c) 0.333 0 (d) 0.333 180 If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is give by ( ) E ( z, t ) = 10 cos 2 107 t = 0.1 z volt/m, then the velocity of the traveling wave is (a) 3.00 108 m/sec (c) 6.28 107 m/sec 88. (b) 2.00 108 m/sec (d) 2.00 107 m/sec A short-circuited stub is shunt connected to a transmission line as shown in Figure. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is /8 Z0 Z0 Z0 ZL=100ohm /2 Y (a) (0.01 j0.02) ohm (c) (0.04 j0.02) ohm 89. (b) (0.02 j0.01) ohm (d) (0.02 + j0) ohm A rectangular metal wave-guide filled with a dielectric material of relative permitivity r = 4 has the inside dimensions 3.0cm 1.2cm. The cut-off frequency for the dominant mode is (a) 2.5 GHz (b) 5.0 GHz (c) 10.0 GHz (d) 12.5 GHz For more files visit www.educationobserver.com/forum 90. Two identical antennas are placed in the = plane as shown in figure. The 2 elements have equal amplitude excitation with 180 polarity difference, operating at wavelength . The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for = 0, is s s 2 s (a) 2 cos 2 s (b) 2 sin s (c) 2 cos s (d) 2 sin

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Additional Info : Solved GATE exam paper study guide - gate 2003 : electronics and comm. engg. free online question paper
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