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GATE EC 2003 Q. 1-30 Carry One Mark Each MCQ 1.1 The minimum number of equations required to analyze the circuit shown in the figure is (A) 3 (B) 4 (C) 6 (D) 7 SOL 1.1 Hence (B) is correct option. Number of loops = b n + 1 = minimum number of equation Number of branches = b = 8 Number of nodes = n = 5 Minimum number of equation = 8 5+1 = 4 MCQ 1.2 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 SOL 1.2 For maximum power transfer * ZL = ZS = Rs jXs Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 2 GATE EC 2003 www.gatehelp.com Thus ZL = 1 1j Hence (C) is correct option. MCQ 1.3 A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100 . If each of R, L and C is doubled from its original value, the new Q of the circuit is (A) 25 (B) 50 (C) 100 SOL 1.3 Hence (B) is correct option. L Q=1 RC When R, L and C are doubled, 2L = 1 Q' = 1 2R 2C 2R Thus Q' = 100 = 50 2 MCQ 1.4 (D) 200 L =Q C 2 The Laplace transform of i (t) is given by 2 I (s) = s (1 + s) At t " 3 , The value of i (t) tends to (A) 0 (C) 2 SOL 1.4 (B) 1 (D) 3 From the Final value theorem we have 2 lim i (t) = lim sI (s) = lim s = lim 2 =2 t"3 s"0 s " 0 s (1 + s) s " 0 (1 + s) Hence (C) is correct answer MCQ 1.5 The differential equation for the current i (t) in the circuit of the figure is 2 i (A) 2 d 2 + 2 di + i (t) = sin t dt dt SOL 1.5 2 i (B) d 2 + 2 di + 2i (t) = cos t dt dt 2 i (C) 2 d 2 + 2 di + i (t) = cos t dt dt Applying KVL we get, di (t) 1 sin t = Ri (t) + L + dt C 2 i (D) d 2 + 2 di + 2i (t) = sin t dt dt Brought to you by: Nodia and Company PUBLISHING FOR GATE # i (t) dt Visit us at: www.nodia.co.in Page 3 GATE EC 2003 or sin t = 2i (t) + 2 di (t) + dt www.gatehelp.com # i (t) dt Differentiating with respect to t , we get 2di (t) 2d2 i (t) cos t = + i (t) + dt dt2 Hence (C) is correct option. MCQ 1.6 n -type silicon is obtained by doping silicon with (A) Germanium (B) Aluminium (C) Boron (D) Phosphorus SOL 1.6 Pentavalent make n type semiconductor and phosphorous is pentavalent. Hence option (D) is correct. MCQ 1.7 The Bandgap of silicon at 300 K is (A) 1.36 eV (B) 1.10 eV (C) 0.80 eV (D) 0.67 eV SOL 1.7 Hence option (B) is correct. For silicon at 0 K Eg0 = 1.21 eV At any temperature EgT = Eg0 3.6 # 10 - 4 T At T = 300 K, Eg300 = 1.21 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be remembered. MCQ 1.8 The intrinsic carrier concentration of silicon sample at 300 K is 1.5 # 1016 /m 3 . If after doping, the number of majority carriers is 5 # 1020 /m 3 , the minority carrier density is (A) 4.50 # 1011/m 3 (B) 3.333 # 10 4 /m 3 (C) 5.00 # 1020 /m 3 (D) 3.00 # 10 - 5 /m 3 SOL 1.8 By Mass action law np = ni2 2 16 16 .5 p = ni = 1.5 # 10 # 120 # 10 = 4.5 # 1011 n 5 # 10 Hence option (A) is correct. MCQ 1.9 Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photo diode are operated in X bias ad Y bias respectively (A) X: reverse, Y: reverse (B) X: reverse, Y: forward (C) X: forward, Y: reverse SOL 1.9 (D) X: forward, Y: forward Tunnel diode shows the negative characteristics in forward bias. It is used in forward Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 4 GATE EC 2003 www.gatehelp.com bias. Avalanche photo diode is used in reverse bias. Hence option (C) is correct. MCQ 1.10 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 (B) decrease (C) change polarity (D) increase SOL 1.10 Hence option (D) is correct. MCQ 1.11 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 (B) CB LO, CC HI, CE MO (C) CB MO, CC HI, CE LO (D) CB HI, CC LO, CE MO SOL 1.11 For the different combinations the table is as follows CE CE CC CB Ai High High Unity Av High Unity High Ri Medium High Low Ro Medium Low High Hence (B) is correct option. MCQ 1.12 The circuit shown in the figure is best described as a (A) bridge rectifier (C) frequency discriminator SOL 1.12 (B) ring modulator (D) voltage double This circuit having two diode and capacitor pair in parallel, works as voltage Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 5 GATE EC 2003 www.gatehelp.com doubler. Hence (D) is correct option. MCQ 1.13 If the input to the ideal comparators shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparators has a duty cycle of (A) 1/2 (C) 1/6 SOL 1.13 (B) 1/3 (D) 1/2 If the input is sinusoidal signal of 8 V (peak to peak) then Vi = 4 sin t The output of comparator will be high when input is higher than Vref = 2 V and will be low when input is lower than Vref = 2 V. Thus the waveform for input is shown below From fig, first crossover is at t1 and second crossover is at t2 where 4 sin t1 = 2V Thus t1 = sin - 1 1 = 2 6 t2 = = 5 6 6 5 6 Duty Cycle = 6 =1 2 3 Thus the output of comparators has a duty cycle of 1 . 3 Hence (B) is correct option. MCQ 1.14 If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is (A) 23 dB (B) 25 dB Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 6 GATE EC 2003 (C) 46 dB SOL 1.14 www.gatehelp.com (D) 50 dB Hence (C) is correct option. CMMR = Ad Ac 20 log CMMR = 20 log Ad 20 log Ac = 48 2 = 46 dB Where Ad " Differential Voltage Gain and AC " Common Mode Voltage Gain or MCQ 1.15 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 SOL 1.15 The gain of amplifier is gm Ai = gb + j C Thus the gain of a transistor amplifier falls at high frequencies due to the internal capacitance that are diffusion capacitance and transition capacitance. Hence (B) is correct option. MCQ 1.16 The number of distinct Boolean expressions of 4 variables is (A) 16 (B) 256 (C) 1023 SOL 1.16 (D) 65536 The number of distinct boolean expression of n variable is 22n . Thus 22 = 216 = 65536 Hence (D) is correct answer. 4 MCQ 1.17 The minimum number of comparators required to build an 8-bits flash ADC is (A) 8 (B) 63 (C) 255 (D) 256 SOL 1.17 In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bit.s So, 2n - 1 = 28 1 = 255 Hence (C) is correct answer. MCQ 1.18 The output of the 74 series of GATE 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 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 7 GATE EC 2003 www.gatehelp.com (D) common collector configuration SOL 1.18 When output of the 74 series gate of TTL gates is taken from BJT then the configuration is either totem pole or open collector configuration . Hence (B) is correct answer. MCQ 1.19 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 functions of 3 variables but non of 4 variables (C) all functions of 3 variables and some but not all of 4 variables (D) all functions of 4 variables SOL 1.19 A 2n: 1 MUX can implement all logic functions of (n + 1) variable without andy additional circuitry. Here n = 3 . Thus a 8 : 1 MUX can implement all logic functions of 4 variable. Here (D) is correct answer. MCQ 1.20 A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate (s). The common circuit consists of (A) one AND gate (B) one OR gate (C) one AND gate and one OR gate SOL 1.20 (D) two AND gates Counter must be reset when it count 111. This can be implemented by following circuitry Hence (D) is correct answer. MCQ 1.21 The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = / cn e j2 f t . It is given that c3 = 3 + j5 . Then c 3 is n =- 3 (B) 3 j5 (A) 5 + j3 0 (C) 5 + j3 (D) 3 j5 SOL 1.21 Hence (D) is correct answer. HereC3 = 3 + j5 For real periodic signal * C k = Ck Thus C 3 = Ck = 3 j5 MCQ 1.22 Let x (t) be the input to a linear, time-invariant system. The required output is 4 (t 2). The transfer function of the system should be (A) 4e j4 f (B) 2e j8 f (C) 4e j4 f Brought to you by: Nodia and Company PUBLISHING FOR GATE (D) 2e j8 f Visit us at: www.nodia.co.in Page 8 SOL 1.22 GATE EC 2003 www.gatehelp.com Hence (C) is correct answer. y (t) = 4x (t 2) Taking Fourier transform we get Y (e j2 f ) = 4e j2 f2 X (e j2 f ) or Time Shifting property Y (e j2 f ) = 4e 4j f j 2 f X (e ) Thus H (e j2 f ) = 4e 4j f MCQ 1.23 A sequence x (n) with the z transform X (z) = z 4 + z2 2z + 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 (n) = ) 0, otherwise The output at n = 4 is (A) 6 (B) zero (C) 2 (D) 4 SOL 1.23 Hence (B) is correct answer. We have h (n) = 3 (n 3) or Taking z transform H (z) = 2z 3 4 2 4 X (z) = z + z 2z + 2 3z Now Y (z) = H (z) X (z) = 2z 3 (z 4 + z2 2z + 2 3z 4) = 2 (z + z 1 2z 2 + 2z 3 3z 7) Taking inverse z transform we have y (n) = 2[ (n + 1) + (n 1) 2 (n 2)+ 2 (n 3) 3 (n 7)] At n = 4 ,y (4) = 0 MCQ 1.24 Fig. 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 (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 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 9 GATE EC 2003 www.gatehelp.com SOL 1.24 Hence (A) is correct option. Z = P N N " Net encirclement of ( 1 + j0) by Nyquist plot, P " Number of open loop poles in right hand side of s plane Z " Number of closed loop poles in right hand side of s plane Here N = 1 and P = 1 Thus Z =0 Hence there are no roots on RH of s plane and system is always stable. MCQ 1.25 A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot SOL 1.25 PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot. Hence (C) is correct option. MCQ 1.26 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 (D) the envelope SOL 1.26 The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component nq (t) of the noise n (t) is completely rejected by the detector. Hence (A) is correct option. MCQ 1.27 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 (B) flat (C) parabolic (D) Gaussian SOL 1.27 The noise at the input to an ideal frequency detector is white. The PSD of noise at the output is parabolic Hence (C) is correct option. MCQ 1.28 At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by. (A) 6 dB (B) 3 dB (C) 2 dB SOL 1.28 (D) 0 dB Hence (B) is correct option. We have Pe = 1 erfc c Ed m 2 2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 10 GATE EC 2003 www.gatehelp.com Since Pe of Binary FSK is 3 dB inferior to binary PSK SOL 1.29 The unit of 4# H is (A) Ampere (B) Ampere/meter (C) Ampere/meter 2 MCQ 1.29 (D) Ampere-meter By Maxwells equations 4# H = 2D + J 2t Thus 4# H has unit of current density J that is A/m2 Hence (C) is correct option MCQ 1.30 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 dm (B) 12.50 cm (C) 50.00 cm SOL 1.30 (D) 100.00 cm Hence (B) is correct option. We know that \ 1 f f1 2 = Thus f2 1 2 = 1 4 25 1 # 25 = 12.5 cm or 2 = 4 Q.31-90 Carry Two Marks Each MCQ 1.31 SOL 1.31 Twelve 1 resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is (B) 1 (A) 5 6 (C) 6 (D) 3 5 2 For current i there is 3 similar path. So current will be divide in three path Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 11 GATE EC 2003 www.gatehelp.com so, we get Vab b i # 1l b i # 1l b 1 # 1l = 0 3 6 3 Vab = R = 1 + 1 + 1 = 5 eq i 6 363 Hence (A) is correct option. MCQ 1.32 The current flowing through the resistance R in the circuit in the figure has the form P cos 4t where P is (A) (0.18 + j0.72) (C) (0.18 + j1.90) SOL 1.32 (B) (0.46 + j1.90) (D) (0.192 + j0.144) Data are missing in question as L1 &L2 are not given The circuit for Q. 33 & 34 is given below. Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0 . At t = 0+ , the current i1 is (A) V 2R Brought to you by: Nodia and Company PUBLISHING FOR GATE MCQ 1.33 (B) V R Visit us at: www.nodia.co.in Page 12 SOL 1.33 MCQ 1.34 GATE EC 2003 www.gatehelp.com (C) V (D) zero 4R Data are missing in question as L1 &L2 are not given 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 the figure, after the switch is brought from position 1 to position 2 at t = 0 , are 1 V R + Ls + Cs Ls I1 (s) s (A) > H=I2 (s)G = = 0 G 1 R + Cs Ls 1 R + Ls + Cs Ls I1 (s) V s (B) > G== G 1 H= R + Cs I2 (s) Ls 0 1 R + Ls + Cs Ls I1 (s) V s (C) > = 1 R + Ls + Cs H=I2 (s)G = 0 G Ls SOL 1.34 1 V R + Ls + Cs Cs I1 (s) s (D) > G== G 1 R + Ls + Cs H=I2 (s) Ls 0 At t = 0 - circuit is in steady state. So inductor act as short circuit and capacitor act as open circuit. At t = 0 - , i1 (0 -) = i2 (0 -) = 0 vc (0 -) = V At t = 0+ the circuit is as shown in fig. The voltage across capacitor and current in inductor can t be changed instantaneously. Thus At t = 0+ , i1 = i2 = V 2R Hence (A) is correct option. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 13 MCQ 1.35 GATE EC 2003 www.gatehelp.com An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) V is applied to a series combination of resistance R = 1 and an inductance L = 1 H. The resulting steady-state current i (t) in ampere is (A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan 1 2) (B) 10 cos (t + 55c) + 10 3 2 cos (2t + 55c) (C) 10 cos (t 35c) + 10 cos (2t + 10c tan 1 2) (D) 10 cos (t 35c) + SOL 1.35 3 2 cos (2t 35c) Hence (C) is correct option v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) 1 4444 2 4444 3 1 4444 2 4444 3 4 4 v1 v2 Thus we get 1 = 1 and 2 = 2 Now Z1 = R + j 1 L = 1 + j1 Z2 = R + j 2 L = 1 + j2 v (t) v (t) i (t) = 1 + 2 Z1 Z2 = 10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 1+j 1 + j2 10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 12 + 22 + tan 1 1 12 + 22 tan 1 2 10 2 cos (t + 10c) 10 5 cos (2t + 10c) = + 2 + tan 1 45c 5 tan 1 2 i (t) = 10 cos (t 35c) + 10 cos (2t + 10c tan 1 2) = MCQ 1.36 The driving point impedance Z (s) of a network has the pole-zero locations as shown in the figure. If Z (0) = 3 , then Z (s) is 3 (s + 3) s + 2s + 3 3 (s + 3) (C) 2 s + 2s + 2 (A) SOL 1.36 2 2 (s + 3) s + 2s + 2 2 (s 3) (D) 2 s 2s 3 (B) 2 Hence (B) is correct option. Zeros = 3 Pole1 = 1 + j Pole 2 = 1 j Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 14 GATE EC 2003 www.gatehelp.com K (s + 3) (s + 1 + j)( s + 1 j) K (s + 3) K (s + 3) = = 2 2 (s + 1) j (s + 1) 2 + 1 From problem statement Z (0) = 0 = 3 Thus 3K = 3 and we get K = 2 2 2 (s + 3) Z (s) = 2 s + 2s + 2 Z (s) = MCQ 1.37 The impedance parameters z11 and z12 of the two-port network in the figure are (A) z11 = 2.75 and z12 = 0.25 (C) z11 = 3 and z12 = 0.25 SOL 1.37 (B) z11 = 3 and z12 = 0.5 (D) z11 = 2.25 and z12 = 0.5 Using 3 Y conversion 2 # 1 = 2 = 0. 5 2+1+1 4 R2 = 1 # 1 = 1 = 0.25 2+1+1 4 R3 = 2 # 1 = 0.5 2+1+1 R1 = Now the circuit is as shown in figure below. Now z11 = V1 I1 I2 = 0 = 2 + 0.5 + 0.25 = 2.75 z12 = R3 = 0.25 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 15 GATE EC 2003 www.gatehelp.com Hence (A) is correct option. MCQ 1.38 An n type silicon bar 0.1 cm long and 100 m2 i cross-sectional area has a majority carrier concentration of 5 # 1020 /m 2 and the carrier mobility is 0.13 m2 /V-s at 300 K. If the charge of an electron is 1.5 # 10 - 19 coulomb, then the resistance of the bar is (B) 10 4 Ohm (A) 106 Ohm (C) 10 - 1 Ohm SOL 1.38 (D) 10 - 4 Ohm Hence option (A) is correct. l We that R = , = 1 and = nqun A From above relation we have 1 R= nq n A = MCQ 1.39 0.1 # 10 - 2 = 106 - 19 - 12 20 5 # 10 # 1.6 # 10 # 0.13 # 100 # 10 The electron concentration in a sample of uniformly doped n -type silicon at 300 K varies linearly from 1017 /cm 3 at x = 0 to 6 # 1016 /cm 3 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 cm 2 /s, the current density in the silicon, if no electric field is present, is (A) zero (B) -112 A/cm 2 (C) +1120 A/cm 2 SOL 1.39 (D) -1120 A/cm 2 Hence option (D) is correct. dn = 6 # 1016 1017 dx 2 # 10 - 4 0 = 2 # 1020 Now Jn = nq e E + Dn q dn dx Since no electric field is present, E = 0 and we get So, Jn = qDn dn dx = 1.6 # 10 - 19 # 35 # ( 2 # 1020) = 1120 A/cm 2 MCQ 1.40 Match items in Group 1 with items in Group 2, most suitably. Group 1 Group 2 P. LED 1. Heavy doping Q. Avalanche photo diode 2. Coherent radiation R. Tunnel diode 3. Spontaneous emission S. LASER 4. Current gain Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 16 GATE EC 2003 www.gatehelp.com (A) P - 1, Q - 2, R - 4, S - 3 (B) P - 2, Q - 3, R - 1, S - 4 (C) P - 3 Q - 4, R - 1, S - 2 (D) P - 2, Q - 1, R - 4, S - 3 SOL 1.40 LED works on the principal of spontaneous emission. In the avalanche photo diode due to the avalanche effect there is large current gain. Tunnel diode has very large doping. LASER diode are used for coherent radiation. Hence option (C) is correct. MCQ 1.41 At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions state above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (B) 5 (C) 4 # 103 SOL 1.41 (D) 8 # 103 Hence option (C) is correct. V We know that I = Io `e V 1j where = 1 for germanium and = 2 silicon. As per question V V Io `e e 1j = Io `e V 1j D1 T si Dsi hVT DGe n T Ge VDsi or 0.718 Io = eVV 1 = e 2 #026 # 10 1 = 4 # 103 .1435 Io e 26 # 10 1 e V 1 si T -3 DGe -3 si T MCQ 1.42 A particular green LED emits light of wavelength 5490 Ac. The energy bandgap of the semiconductor material used there is (Plank s constant = 6.626 # 10 - 34 J s ) (A) 2.26 eV (B) 1.98 eV (C) 1.17 eV (D) 0.74 eV SOL 1.42 Hence option (A) is correct 34 3# 8 Eg = hc = 6.626 # 10 # 10 10 = 3.62 J 54900 # 10 19 E (J) In eV Eg (eV) = g = 3.62 # 10 19 = 2.26 eV e 1.6 # 10 Alternatively 1.24 Eg = 1.24 eV = = 2.26 eV ( m) 5490 # 10 4 m MCQ 1.43 When the gate-to-source voltage (VGs) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 17 GATE EC 2003 www.gatehelp.com 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 (B) 2.0 mA (C) 3.5 mA (D) 4.0 mA SOL 1.43 We know that ID = K (VGS VT ) 2 2 ID2 = (VGS2 VT ) Thus ID1 (VGS1 VT ) 2 Substituting the values we have 2 ID2 = (1.4 0.4) = 4 ID1 (0.9 0.4) 2 or ID2 = 4IDI = 4 mA Hence option (D) is correct. MCQ 1.44 If P is Passivation, Q is n well implant, R is metallization and S is source/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 (B) Q S R P (C) R P S Q (D) S R Q P SOL 1.44 In n well CMOS fabrication following are the steps : (i) n well implant (Q) (ii) Source drain diffusion (S) (iii) Metalization (R) (iv) Passivation (P) Hence option (B) is correct. MCQ 1.45 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 (B) 1 k (A) 1 k 11 5 (C) 5 k SOL 1.45 Hence (A) is correct option. We have Ri = 1k , = 0.2, A = 50 Thus, MCQ 1.46 (D) 11 k Rif = Ri = 1 k (1 + A ) 11 In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3 V and IC = 1.5 mA when its is 150. For a transistor with of 200, the operating point (VCE , IC ) is Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 18 GATE EC 2003 www.gatehelp.com (A) (2 V, 2 mA) (C) (4 V, 2 mA) SOL 1.46 (B) (3 V, 2 mA) (D) (4 V, 1 mA) The DC equivalent circuit is shown as below. This is fixed bias circuit operating in active region. In first case VCC IC1 R2 VCE1 = 0 or 6 1.5mR2 3 = 0 or R2 = 2k IB1 = IC1 = 1.5m = 0.01 mA 1 150 In second case IB2 will we equal to IB1 as there is no in R1. Thus IC2 = 2 IB2 = 200 # 0.01 = 2 mA VCE2 = VCC IC2 R2 = 6 2m # 2 k = 2 V Hence (A) is correct option. MCQ 1.47 The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is (A) 1 (2 6 RC) (B) 1 (2 RC) (C) 1 ( 6 RC) (D) 6 (2 RC) Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 19 GATE EC 2003 www.gatehelp.com SOL 1.47 The given circuit is a R C phase shift oscillator and frequency of its oscillation is 1 f= 2 6 RC Hence (A) is correct option. MCQ 1.48 The output voltage of the regulated power supply shown in the figure is (A) 3 V (C) 9 V SOL 1.48 (B) 6 V (D) 12 V If we see th figure we find that the voltage at non-inverting terminal is 3 V by the zener diode and voltage at inverting terminal will be 3 V. Thus Vo can be get by applying voltage division rule, i.e. 20 V = 3 20 + 40 o or V0 = 9 V Hence (C) is correct option. MCQ 1.49 The action of 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 SOL 1.49 For a JFET in active region we have 2 IDS = IDSS c1 VGS m VP From above equation it is clear that the action of a JFET is voltage controlled current source. Hence option (D) is correct. MCQ 1.50 If the op-amp in the figure is ideal, the output voltage Vout will be equal to Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 20 GATE EC 2003 www.gatehelp.com (A) 1 V (C) 14 V SOL 1.50 (B) 6 V (D) 17 V The circuit is as shown below 8 (3) = 8 k 1+8 3 V+ = V- = 8 V 3 V+ = Now applying KCL at inverting terminal we get V- 2 + V- Vo = 0 1 5 or Vo = 6V- 10 = 6 # 8 10 = 6 V 3 Hence (B) is correct option. MCQ 1.51 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 opened circuit voltages gain of the combined amplifier is (A) 49 dB (B) 51 dB (C) 98 dB SOL 1.51 (D) 102 dB The equivalent circuit of 3 cascade stage is as shown in fig. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 21 GATE EC 2003 www.gatehelp.com 1k 50V1 = 40V1 1k + 0.25k 1k 50V2 = 40V2 V3 = 1k + 0.25k V2 = Similarly or or V3 = 40 # 40V1 Vo = 50V3 = 50 # 40 # 40V1 AV = Vo = 50 # 40 # 40 = 8000 V1 or 20 log AV = 20 log 8000 = 98 dB Hence (C) is correct option. MCQ 1.52 An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3 V 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 voltage source of 1 mA for 1 ms (D) Constant voltage source of 3 mA for 2 ms SOL 1.52 If a constant current is made to flow in a capacitor, the output voltage is integration of input current and that is sawtooth waveform as below : t VC = 1 # idt C0 The time period of wave form is T = 1 = 1 = 2 m sec f 500 -3 20 # 10 1 # idt 2 # 106 0 or i (2 # 10 - 3 0) = 6 # 10 - 6 or i = 3 mA Thus the charging require 3 mA current source for 2 msec. Hence (D) is correct option Thus MCQ 1.53 3= The circuit in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P 5 Q 5 R and Z = RQ + PR + QP The circuit acts as a Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 22 GATE EC 2003 www.gatehelp.com (A) 4 bit adder giving P + Q (B) 4 bit subtractor giving P Q (C) 4 bit subtractor giving Q-P (D) 4 bit adder giving P + Q + R SOL 1.53 Hence (B) is correct answer. We have Y = P 5 Q 5 R Z = RQ + PR + QP Here every block is a full subtractor giving P Q R where R is borrow. Thus circuit acts as a 4 bit subtractor giving P Q . MCQ 1.54 If the function W, X, Y and Z are as follows W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + P .Q Z = R + S + PQ + P .Q .R + PQ .S Then, (A) W = Z, X = Z (B) W = Z, X = Y (C) W = Y SOL 1.54 (D) W = Y = Z Hence (A) is correct answer. W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + PQ = RS + PR $ PQ $ PQ = RS + (P + R )( P + Q)( P + Q) = RS + (P + PQ + PR + QR )( P + Q) = RS + PQ + QR (P + P ) + QR = RS + PQ + QR Z = R + S + PQ + PQR + PQS = R + S + PQ $ PQR $ PQS = R + S + (P + Q )( P + Q + R)( P + Q + S) = R + S + PQ + PQ + PQS + PR + PQR + P RS + PQ + PQS + PQR + QRS = R + S + PQ + PQS + PR + PQR + PRS Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 23 GATE EC 2003 www.gatehelp.com + PQS + PQR + QRS = R + S + PQ (1 + S) + PR (1 + P ) + PRS + PQS + PQR + QRS = R + S + PQ + PR + PRS + PQS + PQR + QRS = R + S + PQ + PR (1 + Q ) + PQS + QRS = R + S + PQ + PR + PQS + QRS Thus W = Z and X = Z MCQ 1.55 A 4 bit ripple counter and a 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 (B) R = 40 ns, S = 10 ns (C) R = 10 ns S = 30 ns (D) R = 30 ns, S = 10 ns SOL 1.55 Propagation delay of flip flop is tpd = 10 nsec Propagation delay of 4 bit ripple counter R = 4tpd = 40 ns and in synchronous counter all flip-flop are given clock simultaneously, so S = tpd = 10 ns Hence (B) is correct answer. MCQ 1.56 The DTL, TTL, ECL and CMOS famil GATE 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 (B) Q (C) R (D) S SOL 1.56 The DTL has minimum fan out and CMOS has minimum power consumption. Propagation delay is minimum in ECL. Hence (B) is correct answer. MCQ 1.57 The circuit shown in the figure is a 4 bit DAC Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 24 GATE EC 2003 www.gatehelp.com The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistance and the 5 v inputs have a tolerance of ! 10%. The specification (rounded to nearest multiple of 5%) for the tolerance of the DAC is (A) ! 35% (B) ! 20% (C) ! 10% SOL 1.57 (D) ! 5% Hence (A) is correct answer. Vo = V1 :R bo + R b1 + R b2 + R b 3D R 2R 4R 4R Exact value when V1 = 5 , for maximum output VoExact = 5 :1 + 1 + 1 + 1 D = 9.375 248 Maximum Vout due to tolerance Vo max = 5.5 :110 + 110 + 110 + 110 D 90 2 # 90 4 # 90 8 # 90 Tolerance MCQ 1.58 = 12.604 = 34.44% = 35% The circuit shown in figure converts (A) BCD to binary code (C) Excess -3 to gray code SOL 1.58 (B) Binary to excess - 3 code (D) Gray to Binary code Hence (D) is correct answer. Let input be 1010; output will be 1101 Let input be 0110; output will be 0100 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 25 GATE EC 2003 www.gatehelp.com Thus it convert gray to Binary code. MCQ 1.59 In the circuit shown in the figure, A is 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 at input to a 16 # 4 ROM whose output is floating when the input to a partial table of the contents of the ROM is as follows Data 0011 1111 0100 1010 1011 1000 0010 1000 Address 0 2 4 6 8 10 11 14 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 (C) 1000 SOL 1.59 (B) 1011 (D) 0010 After t = t1, at first rising edge of clock, the output of shift register is 0110, which in input to address line of ROM. At 0110 is applied to register. So at this time data stroed in ROM at 1010 (10), 1000 will be on bus. When W has the data 0110 and it is 6 in decimal, and it s data value at that add is 1010 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 26 GATE EC 2003 www.gatehelp.com then 1010 i.e. 10 is acting as odd, at time t2 and data at that movement is 1000. Hence (C) is correct answer. MCQ 1.60 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 rest but Zero flag will be set (C) Both Carry flag and Zero flag will be rest (D) Both Carry flag and Zero flag will be set SOL 1.60 CMP B & Compare the accumulator content with context of Register B If A < R CY is set and zero flag will be reset. Hence (A) is correct answer. MCQ 1.61 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 # 1) is (A) zero (B) 1 6 SOL 1.61 (C) 1 (D) 1 3 12 The pdf of Z will be convolution of pdf of X and pdf of Y as shown below. Now p [Z # z] = p [Z # 2] = #- 3 fZ (z) dz z #- 3fZ (z) dz -2 = Area [z # 2] = 1 # 1 #1 = 1 26 12 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 27 GATE EC 2003 www.gatehelp.com Hence (D) is correct option. MCQ 1.62 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 y (n) = *0, 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 SOL 1.62 (D) Q, R, S but not P System is non causal because output depends on future value For n # 1 y ( 1) = x ( 1 + 1) = x (0) Time varying y (n n0) = x (n n0 + 1) Depends on Future y (n) = x (n + 1) i.e. None causal y (1) = x (2) For bounded input, system has bounded output. So it is stable. y (n) = x (n) for n $ 1 = 0 for n = 0 = x (x + 1) for n # 1 So system is linear. Hence (A) is correct answer. Common data for Q 63 & 64 : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 28 GATE EC 2003 www.gatehelp.com The system under consideration is an RC low-pass filter (RC-LPF) with R = 1 k and C = 1.0 F. MCQ 1.63 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) 324.8 (B) 163.9 (C) 52.2 SOL 1.63 (D) 104.4 The frequency response of RC-LPF is 1 H (f) = 1 + j2 fRC Now or or or or or or H (0) = 1 H (f1) 1 = $ 0.95 H (0) 1 + 4 2 f12 R2 C2 1 + 4 2 f12 R2 C2 # 1.108 4 2 f12 R2 C2 # 0.108 2 f1 RC # 0.329 f1 # 0.329 2 RC f1 # 0.329 2 RC f1 # 0.329 2 1k # 1 or f1 # 52.2 Hz Thus f1 max = 52.2 Hz Hence (C) is correct answer. MCQ 1.64 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 SOL 1.64 (D) 4.505 Hence (A) is correct answer 1 H ( ) = 1 + j RC ( ) = tan 1 RC d ( ) RC tg = = d 1 + 2 R2 C2 10 3 = = 0.717 ms 2 1 + 4 # 10 4 # 10 6 Common Data for Questions 65 & 66 : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 29 GATE EC 2003 www.gatehelp.com X (t) is a random process with a constant mean value of 2 and the auto correlation function Rxx ( ) = 4 (e - 0.2 + 1). MCQ 1.65 Let X be the Gaussian random variable obtained by sampling the process at t = ti and let 3 Q ( ) = # 1 e dy 2 The probability that 6x # 1@ is (A) 1 Q (0.5) (B) Q (0.5) x2 2 SOL 1.65 MCQ 1.66 (C) Q c 1 m (D) 1 Q c 1 m 22 22 Hence (D) is correct option. We have RXX ( ) = 4 (e - 0.2 + 1) RXX (0) = 4 (e - 0.2 0 + 1) = 8 = 2 or = 2 2 Given mean =0 Now P (x # 1) = Fx (1) X = 1 Qc m = 1 Qc 1 0 m = 1 Qc 1 m 22 22 Let Y and Z be the random variable obtained by sampling X (t) at t = 2 and t = 4 respectively. Let W = Y Z . The variance of W is (A) 13.36 (B) 9.36 (C) 2.64 SOL 1.66 MCQ 1.67 at x = 1 (D) 8.00 Hence (C) is correct option. W = Y Z E [W2] = E [Y Z] 2 = E [Y2] + E [Z2] 2E [YZ] 2 = w We haveE [X2 (t)] = 4 [e - 0.2 0 + 1] = 4 [1 + 1] = 8 E [Y2] = E [X2 (2)] = 8 E [Z2] = E [X2 (4)] = 8 E [YZ] = RXX (2) = 4 [e 0.2 (4 2) + 1] = 6.68 2 E [W2] = w = 8 + 8 2 # 6.68 = 2.64 = Rx (10) Let x (t) = 2 cos (800 ) + cos (1400 t). x (t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 30 GATE EC 2003 www.gatehelp.com (A) 2.7, 3.4 (C) 2.6, 2.7, 3.3, 3.4, 3.6 SOL 1.67 (B) 3.3, 3.6 (D) 2.7, 3.3 Hence (D) is correct option. The frequency of pulse train is f 1- 3 = 1 k Hz 10 The Fourier Series coefficient of given pulse train is T /2 Cn = 1 # Ae jn t dt To T /2 o o o =1 To # To /6 To /6 Ae j t dt o A [e j t] T /6 T /6 To ( j o) A = (e j t e j T /6) ( j2 n) = A (e j /3 e j /3) j2 n Cn = A sin ` n j n 3 = o o o or o o o From Cn it may be easily seen that 1, 2, 4, 5, 7 , harmonics are present and 0, 3, 6, 9,.. are absent. Thus p (t) has 1 kHz, 2 kHz, 4 kHz, 5 kHz, 7 kHz,... frequency component and 3 kHz, 6 kHz.. are absent. The signal x (t) has the frequency components 0.4 kHz and 0.7 kHz. The sampled signal of x (t) i.e. x (t)* p (t) will have 1 ! 0.4 and 1 ! 0.7 kHz 2 ! 0.4 and 2 ! 0.7 kHz 4 ! 0.4 and 4 ! 0.7 kHz Thus in range of 2.5 kHz to 3.5 kHz the frequency present is 2 + 0.7 = 2.7 kHz 4 0.7 = 3.3 kHz MCQ 1.68 The signal flow graph of a system is shown in Fig. below. The transfer function C (s)/ R (s) of the system is Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 31 GATE EC 2003 6s s + 29s + 6 s (s + 27) (D) 2 s + 29s + 6 6 s + 29s + 6 s (s + 2) (C) 2 s + 29s + 6 (A) SOL 1.68 www.gatehelp.com (B) 2 2 Mason Gain Formula pk 3 k T (s) = 3 In given SFG there is only forward path and 3 possible loop. p1 = 1 31 = 1 + 3 + 24 = s + 27 s s s L1 = 2 , L2 = 24 and L3 = 3 s s s where L1 and L3 are non-touching C (s) p1 3 1 This = 1 (loop gain) + pair of non touching loops R (s) ^ s +s27 h = 2 1 ^ 3 24 s h + 2 . 3 s s s s s (s + 27) =2 s + 29s + 6 Hence (D) is correct option. = MCQ 1.69 s + 27 K The root locus of system G (s) H (s) = has the break-away point s (s + 2)( s + 3) located at (A) ( 0.5, 0) (B) ( 2.548, 0) (C) ( 4, 0) SOL 1.69 ^sh 1 + 29 + s62 s (D) ( 0.784, 0) We have or 1 + G (s) H (s) = 0 K 1+ =0 s (s + 2)( s + 3) or which gives K = s (s2 + 5s2 + 6s) dK = (3s2 + 10s + 6) = 0 ds s = 10 ! 100 72 = 0.784, 2.548 6 The location of poles on s plane is Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 32 GATE EC 2003 www.gatehelp.com Since breakpoint must lie on root locus so s = 0.748 is possible. Hence (D) is correct option. MCQ 1.70 The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is (A) 108 (C) SOL 1.70 (s + 0.1) 3 (s + 10) 2 (s + 100) (s + 0.1) 2 (s + 10) 2 (s + 100) (B) 107 (D) (s + 0.1) 3 (s + 10)( s + 100) (s + 0.1) 3 (s + 10)( s + 100) 2 The given bode plot is shown below At = 0.1 change in slope is + 60 dB " 3 zeroes at = 0.1 At = 10 change in slope is 40 dB " 2 poles at = 10 At = 100 change in slope is 20 dB " 1 poles at = 100 K ( 0s.1 + 1) 3 Thus T (s) = s s ( 10 + 1) 2 ( 100 + 1) Now 20 log10 K = 20 or K = 10 10 ( 0s.1 + 1) 3 108 (s + 0.1) 3 Thus T (s) = s = s ( 10 + 1) 2 ( 100 + 1) (s + 10) 2 (s + 100) Hence (A) is correct option. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 33 MCQ 1.71 GATE EC 2003 www.gatehelp.com A second-order system has the transfer function C (s) =2 4 R (s) s + 4s + 4 With r (t) as the unit-step function, the response c (t) of the system is represented by SOL 1.71 The characteristics equation is s2 + 4s + 4 = 0 Comparing with 2 s2 + 2 n + n = 0 2 we get 2 n = 4 and n = 4 Thus =1 ts = 4 = 4 = 2 1#2 n Critically damped Hence (B) is correct option. MCQ 1.72 The gain margin and the phase margin of feedback system with 8 are G (s) H (s) = (s + 100) 3 (B) 3, 3 (A) dB, 0c (C) 3, 0c (D) 88.5 dB, 3 SOL 1.72 Hence (B) is correct option. MCQ 1.73 The zero-input response of a system given by the state-space equation 1 0 x1 x1 (0) 1 xo1 =xo G = =1 1G=x G and =x (0)G = = 0 G is 2 2 2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 34 GATE EC 2003 www.gatehelp.com tet (A) = G t SOL 1.73 et (B) = G t et (C) = t G te We have t (D) = t G te 1 xo1 =xo G = =1 2 1 A == 1 s (sI A) = = 0 0 x1 x1 (0) 1 G=x G and =x (0)G = = 0 G 12 2 0 1G 0 10 s 1 0 G =1 1G = = 1 s 1G s 1 0 s 1 1 >(s 1) (sI A) = H = > +1 (s 1) 2 + 1 (s 1) (s 1) et 0 L 1 [(sI A) 1] = eAt = = t t G te e et 0 1 et At x (t) = e # [x (t0)] = = t t G= G = = t G te e 0 te 0 1 2 1 s 1 H Hence (C) is correct option. MCQ 1.74 A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input-output characteristic V0 = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band-pass filter. i Let Vi = Ac cos (2 fi ct) + m (t) is the message signal. Then the value of fci (in MHz) is (A) 1.0 (B) 0.333 (B) 0.5 SOL 1.74 (D) 3.0 Hence (C) is correct option. 1 vi = Ac cos (2 fc t) + m (t) v0 = ao vi + avi3 ' ' v0 = a0 [Ac cos (2 fc' t) + m (t)] + a1 [Ac cos (2 fc' t) + m (t)] 3 ' ' = a0 Ac cos (2 fc' t) + a0 m (t) + a1 [(Ac cos 2 fc' t) 3 ' ' + (Ac cos (2 fc') t) 2 m (t) + 3Ac cos (2 fc' t) m2 (t) + m3 (t)] ' ' ' = a0 Ac cos (2 fc' t) + a0 m (t) + a1 (Ac cos 2fc' t) 3 + 3a1 Ac2 ; 1 + cos (4 fc' t) E m (t) 2 ' = 3a1 Ac cos (2 fc' t) m2 (t) + m3 (t) ' The term 3a1 Ac ( cos 4 f t ) m (t) is a DSB-SC signal having carrier frequency 1. MHz. 2 Thus 2fc' = 1 MHz or fc' = 0.5 MHz ' c Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 35 GATE EC 2003 www.gatehelp.com Common Data for Question 75 & 76 : Let m (t) = cos [(4 # 103) t] be the message signal & c (t) = 5 cos [(2 # 106 t)] be the carrier. MCQ 1.75 SOL 1.75 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) 1 (B) 1 2 4 (C) 1 3 Hence (D) is correct option. 2 PT = Pc c1 + m 2 or MCQ 1.76 SOL 1.76 (D) 1 8 2 P (0.5) 2 Psb = Pc = c 2 2 Psb = 1 Pc 8 c (t) and m (t) are used to generated an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2 (1008 # 103 t)] in the FM signal (in terms of the Bessel coefficients) is (A) 5J4 (3) (B) 5 J8 (3) 2 (C) 5 J8 (4) (D) 5J4 (6) 2 Hence (D) is correct option. AM Band width = 2fm Peak frequency deviation = 3 (2fm) = 6fm 6f Modulation index = m = 6 fm The FM signal is represented in terms of Bessel function as xFM (t) = Ac 3 /Jn ( ) cos ( c n n) t n =- 3 c + n m = 2 (1008 # 103) 2 106 + n4 # 103 = 2 (1008 # 103), n = 4 Thus coefficient = 5J4 (6) MCQ 1.77 Choose the correct one from among the alternative A, B, C, D after matching an item in Group 1 with most appropriate item in Group 2. Group 1 Group 2 P. Ring modulator 1. Clock recovery Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 36 GATE EC 2003 Q. VCO R. Foster-Seely discriminator S. Mixer 2. 3. 4. 5. 6. www.gatehelp.com (A) P 1; Q 3; R 2; S 4 Demodulation of FM Frequency conversion Summing the two inputs Generation of FM Generation of DSB-Sc (B) P 6; Q = 5; R 2; S 3 (C) P 6; Q 1; R 3; S 2 (D) P 5; Q 6; R 1; S 3 SOL 1.77 Hence (B) is correct option. Ring modulation $ Generation of DSB - SC VCO $ Generation of FM Foster seely discriminator $ Demodulation of fm mixer $ frequency conversion MCQ 1.78 A superheterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = Cmax /Cmin denote the 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 (B) R = 2.10, I 1150 (A) R = 4.41, I = 1600 (C) R = 3.0, I = 600 (D) R = 9.0, I = 1150 SOL 1.78 Hence (A) is correct option. fmax = 1650 + 450 = 2100 kHz fmin = 550 + 450 = 1000 kHz 1 or f= 2 LC frequency is minimum, capacitance will be maximum f2 R = Cmax = max = (2.1) 2 2 Cmin fmin or R = 4.41 fi = fc + 2fIF = 700 + 2 (455) = 1600 kHz MCQ 1.79 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 (B) 48 # 10 - 6 V2 (B) 12 # 10 - 6 V2 SOL 1.79 (D) 3.072 V Hence (C) is correct option. Step size = 2mp = 1.536 = 0.012 V L 128 2 (0.012) 2 Quantization Noise power = = 12 12 = 12 # 10 6 V2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 37 MCQ 1.80 GATE EC 2003 If Eb , the energy per bit of a binary digital signal, is 10 - 5 watt-sec and the one-sided power spectral density of the white noise, N0 = 10 - 6 W/Hz, then the output SNR of the matched filter is (A) 26 dB (B) 10 dB (C) 20 dB SOL 1.80 www.gatehelp.com (D) 13 dB Hence (D) is correct option. Eb = 10 - 6 watt-sec No = 10 - 5 W/Hz 6 (SNR) matched filler = Eo = 10 - 5 = .05 N 2 # 10 2 (SNR)dB = 10 log 10 (0.05) = 13 dB o MCQ 1.81 The input to a linear delta modulator having a step-size 3 = 0.628 is a sine wave with frequency fm and peak amplitude Em . If the sampling frequency fx = 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 1 kHz SOL 1.81 Hence (B) is correct option. For slopeoverload to take place Em $ 3 fs 2 fm This is satisfied with Em = 1.5 V and fm = 4 kHz MCQ 1.82 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 (B) = 1.0, S is required (C) = 0.5, S is not required (D) = 1.0, S is not required SOL 1.82 Hence (A) is correct option. If s " carrier synchronization at receiver " represents bandwidth efficiency then for coherent binary PSK = 0.5 and s is required. MCQ 1.83 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 (B) R = 64 kbps, SNRq = 49.8 dB (C) R = 64 kbps, SNRq = 55.8 dB SOL 1.83 (D) R = 32 kbps, SNRq = 49.8 dB Hence (B) is correct option. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 38 GATE EC 2003 www.gatehelp.com Bit Rate = 8k # 8 = 64 kbps (SNR)q = 1.76 + 6.02n dB = 1.76 + 6.02 # 8 = 49.8 dB MCQ 1.84 Medium 1 has the electrical permittivity 1 = 1.5 0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity 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) (2.0ux 7.5uy + 2.5uz ) volt/m (B) (2.0ux 2.0uy + 0.6uz ) volt/m (C) (2.0ux 3.0uy + 1.0uz ) volt/m SOL 1.84 Hence (C) is correct option. We have E1 = 2ux 3uy + 1uz E1t = 3uy + uy and E1n = 2ux Since for dielectric material at the boundary, tangential component of electric field are equal (x = 0 plane) E1t = 3uy + uy = E2t E1n = 2ux At the boundary the for normal component of electric field are D1n = D2n or 1 E1n = 2 E2n or 1.5 o 2ux = 2.5 o E2n or E2n = 3 ux = 1.2ux 2.5 Thus MCQ 1.85 (D) (2.0ux 2.0uy + 0.6uz ) volt/m E2 = E2t + E2n = 3uy + uz + 1.2ux If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m, the potential difference between X (2, 0, 0) and Y (1, 2, 3) is (B) 1 volt (A) + 1 volt (C) + 5 volt SOL 1.85 (D) + 6 volt Hence (C) is correct option. We have E = xux + yuy + zuz t t t dl = ux dx + uy dy + uz dz VXY = # E.dl Y X = # 1 2 t xdxux + t # ydyutz + # zdzuz 2 3 0 0 0 22 20 y2 = = x + +z G 21 22 23 = 1 [22 12 + 02 22 + 02 32] = 5 2 MCQ 1.86 A uniform plane wave traveling in air is incident on the plane boundary between Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 39 GATE EC 2003 www.gatehelp.com air and another dielectric medium with r = 4 . The reflection coefficient for the normal incidence, is (A) zero (B) 0.5+180c (B) 0.333+0c SOL 1.86 (D) 0.333+180c Hence (D) is correct option. = Reflection coefficient 1 = 2 2 + 1 Substituting values for 1 and 2 we have = = 1 r = 1 1 + r 1+ + = 1 = 0.333+180c 3 o o o o or MCQ 1.87 0 or o 4 4 since r = 4 If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2 107 t 0.1 z) V/m, then the velocity of the traveling wave is (A) 3.00 # 108 m/sec (B) 2.00 # 108 m/sec (C) 6.28 # 107 m/sec (D) 2.00 # 107 m/sec SOL 1.87 Hence (B) is correct option. We have E (z, t) = 10 cos (2 # 107 t 0.1 z) where = 2 # 107 t = 0.1 7 Phase Velocity u = = 2 # 10 = 2 # 108 m/s 0.1 MCQ 1.88 A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 40 GATE EC 2003 www.gatehelp.com (A) (0.01 j0.02) mho (C) (0.04 j0.02) mho SOL 1.88 (B) (0.02 j0.01) mho (D) (0.02 + j0) mho The fig of transmission line is as shown below . [Z + jZo tan l] We know that Zin = Zo L [Zo + jZL tan l] For line 1, l = and = 2 , ZL1 = 100 2 [Z + jZo tan ] Thus Zin1 = Zo L = ZL = 100 [Zo + jZL tan ] For line 2, l = and = 2 , ZL2 = 0 (short circuit) 8 [0 + jZo tan ] 4 Thus Zin2 = Zo = jZo = j50 [Zo + 0] Y = 1 + 1 = 1 + 1 = 0.01 j0.02 Zin1 Zin2 100 j50 Hence (A) is correct option. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 41 MCQ 1.89 GATE EC 2003 www.gatehelp.com A rectangular metal wave guide filled with a dielectric material of relative permittivity r = 4 has the inside dimensions 3.0 cm # 1.2 cm. 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 SOL 1.89 Hence (A) is correct option. 8 u = c = 3 # 10 = 1.5 # 108 2 0 In rectangular waveguide the dominant mode is TE10 and fC = v ` m j2 + ` n j2 2 a b 8 1 2 + 0 2 = 1.5 # 108 = 2.5 GHz = 1.5 # 10 ` 0.03 j ` b j 2 0.06 MCQ 1.90 Two identical antennas are placed in the = /2 plane as shown in Fig. The elements have equal amplitude excitation with 180c 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 (A) 2 cos b 2 s l (C) 2 cos a s k SOL 1.90 (B) 2 sin b 2 s l (D) 2 sin a s k Hence (D) is correct option. Normalized array factor = 2 cos d Now 2 cos 2 2 = d sin cos + = 90c, = 2 s, = 45c, = 180c d sin cos + = 2 cos ; E 2 = 2 cos 8 2 . 2 2 s cos 45c + 180 B 2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 42 GATE EC 2003 www.gatehelp.com = 2 cos 8 s + 90cB = 2 sin ` s j Answer Sheet 1. (B) 19. (D) 37. (A) 55. (B) 73. (C) 2. (C) 20. (D) 38. (A) 56. (B) 74. (C) 3. (B) 21. (D) 39. (D) 57. (A) 75. (D) 4. (C) 22. (C) 40. (C) 58. (D) 76. (D) 5. (C) 23. (B) 41. (C) 59. (C) 77. (B) 6. (D) 24. (A) 42. (A) 60. (A) 78. (A) 7. (B) 25. (C) 43. (D) 61. (D) 79. (C) 8. (A) 26. (A) 44. (B) 62. (A) 80. (D) 9. (C) 27. (C) 45. (A) 63. (C) 81. (B) 10. (D) 28. (B) 46. (A) 64. (A) 82. (A) 11. (B) 29. (C) 47. (A) 65. (D) 83. (B) 12. (D) 30. (B) 48. (C) 66. (C) 84. (C) 13. (B) 31. (A) 49. (B) 67. (D) 85. (C) 14. (C) 32. (*) 50. (B) 68. (D) 86. (D) 15. (B) 33. (*) 51. (C) 69. (D) 87. (B) 16. (D) 34. (A) 52. (D) 70 (A) 88. (A) 17. (C) 35. (C) 53. (B) 71 (B) 89. (A) 18. (B) 36. (B) 54. (A) 72 (B) 90. (D) Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in

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