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GATE EC 2012 Q. 1 - Q. 25 carry one mark each. MCQ 1.1 The current ib through the base of a silicon npn transistor is 1 + 0.1 cos (10000 t) mA At 300 K , the r in the small signal model of the transistor is (A) 250 (C) 25 SOL 1.1 (B) 27.5 (D) 22.5 Option (C) is correct. Given ib = 1 + 0.1 cos (1000 t) mA So, IB = DC component of ib = 1 mA In small signal model of the transistor VT r = IC = VT = VT IB IC / VT " Thermal voltage IC = I B = VT IB r = 25 mV = 25 VT = 25 mV, IB = 1 mA 1 mA The power spectral density of a real process X (t) for positive frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] , respectively, are So, MCQ 1.2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 2 GATE EC 2012 www.gatehelp.com (A) 6000/ , 0 (C) 6400/ , 20/ ( 2 ) SOL 1.2 (B) 6400/ , 0 (D) 6000/ , 20/ ( 2 ) Option (A) is correct. The mean square value of a stationary process equals the total area under the graph of power spectral density, that is E [X 2 (t)] = or, or, #S 3 3 E [X 2 (t)] = 1 2 X (f ) df #S 3 3 X ( ) d 3 (Since the PSD is even) E [X 2 (t)] = 2 # 1 # SX ( ) d 2 0 = 1 [area under the triangle + integration of delta function] = 1 ;2 b 1 # 1 # 103 # 6 l + 400E 2 = 1 66000 + 400@ = 6400 E [X (t)] is the absolute value of mean of signal X (t) which is also equal to value of X ( ) at ( = 0). From given PSD SX ( ) = 0 = 0 SX ( ) = X ( ) 2 = 0 X ( ) 2 = 0 = 0 X ( ) = 0 = 0 MCQ 1.3 In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is (A) 1750 (B) 2625 (C) 4000 Brought to you by: Nodia and Company PUBLISHING FOR GATE (D) 5250 Visit us at: www.nodia.co.in Page 3 SOL 1.3 MCQ 1.4 GATE EC 2012 www.gatehelp.com Option (C) is correct. For raised cosine spectrum transmission bandwidth is given as BT = W (1 + ) " Roll of factor BT = Rb (1 + ) Rb " Maximum signaling rate 2 3500 = Rb (1 + 0.75) 2 Rb = 3500 # 2 = 4000 1.75 A plane wave propagating in air with E = (8ax + 6ay + 5az ) e j ( t + 3x 4y) V/m is incident on a perfectly conducting slab positioned at x # 0 . The E field of the reflected wave is (A) ( 8ax 6ay 5az ) e j ( t + 3x + 4y) V/m (B) ( 8ax + 6ay 5az ) e j ( t + 3x + 4y) V/m (C) ( 8ax 6ay 5az ) e j ( t 3x 4y) V/m SOL 1.4 (D) ( 8ax + 6ay 5az ) e j ( t 3x 4y) V/m Option (C) is correct. Electric field of the propagating wave in free space is given as Ei = (8ax + 6ay + 5az ) e j ( t + 3x 4y) V/m So, it is clear that wave is propagating in the direction ( 3ax + 4ay). Since, the wave is incident on a perfectly conducting slab at x = 0 . So, the reflection coefficient will be equal to 1. i.e. Er = ( 1) Ei = 8ax 6ay 5az Again, the reflected wave will be as shown in figure. 0 0 i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the electric field of the reflected wave will be. Ex = ( 8ax 6ay 5az ) e j ( t 3x 4y) V/m MCQ 1.5 The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10 (ay + jaz ) e j 25x . The frequency and polarization of the wave, respectively, are (A) 1.2 GHz and left circular (B) 4 Hz and left circular (C) 1.2 GHz and right circular SOL 1.5 (D) 4 Hz and right circular Option (A) is correct. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 4 GATE EC 2012 www.gatehelp.com The field in circular polarization is found to be Es = E 0 (ay ! jaz ) e j x propagating in + ve x -direction. where, plus sign is used for left circular polarization and minus sign for right circular polarization. So, the given problem has left circular polarization. = 25 = c 2 f 25 = c 8 f = 25 # c = 25 # 3 # 10 2 2 # 3.14 = 1.2 GHz MCQ 1.6 Consider the given circuit In this circuit, the race around (A) does not occur (B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0 SOL 1.6 Option (A) is correct. The given circuit is Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between 0 and 1 checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 5 GATE EC 2012 www.gatehelp.com it won t oscillate for CLK = 0 . So, here race around doesn t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn t occur for the condition. So, Option (A) will be correct. MCQ 1.7 The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is (A) 4 (B) 6 (C) 8 SOL 1.7 (D) 10 Option (B) is correct. Y = 1, when A > B A = a1 a 0, B = b1 b 0 a1 a0 b1 b0 Y 0 1 0 0 1 1 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 Total combination = 6 MCQ 1.8 The i -v characteristics of the diode in the circuit given below are v 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 6 GATE EC 2012 www.gatehelp.com The current in the circuit is (A) 10 mA (C) 6.67 mA SOL 1.8 MCQ 1.9 (B) 9.3 mA (D) 6.2 mA Option (D) is correct. Let v > 0.7 V and diode is forward biased. By applying Kirchoff s voltage law 10 i # 1k v = 0 10 :v 0.7 D (1000) v = 0 500 10 (v 0.7) # 2 v = 0 10 3v + 1.4 = 0 (Assumption is true) v = 11.4 = 3.8 V > 0.7 3 So, i = v 0.7 = 3.8 0.7 = 6.2 mA 500 500 In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is (A) zero (B) a step function (C) an exponentially decaying function (D) an impulse function SOL 1.9 Option (D) is correct. The s -domain equivalent circuit is shown as below. vc (0) /s v (0) =c 1+1 1+1 C1 C 2 Brought to you by: Nodia and Company C1 s C2 s PUBLISHING FOR GATE I (s) = Visit us at: www.nodia.co.in Page 7 GATE EC 2012 www.gatehelp.com I (s) = b C1 C2 l (12 V) C1 + C 2 vC (0) = 12 V I (s) = 12Ceq Taking inverse Laplace transform for the current in time domain, i (t) = 12Ceq (t) MCQ 1.10 The average power delivered to an impedance (4 j3) by a current 5 cos (100 t + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W SOL 1.10 (Impulse) (D) 125 W Option (B) is correct. In phasor form Z = 4 j3 Z = 5 36.86c I = 5 100c A Average power delivered. Pavg. = 1 I 2 Z cos 2 = 1 # 25 # 5 cos 36.86c 2 = 50 W Alternate method: Z = (4 j3) I = 5 cos (100 t + 100) A 2 Pavg = 1 Re $ I Z . 2 = 1 # Re "(5) 2 # (4 j3), 2 = 1 # 100 = 50 W 2 MCQ 1.11 The unilateral Laplace transform of f (t) is 2 1 . The unilateral Laplace s +s+1 transform of tf (t) is (B) 2 2s + 1 2 (A) 2 s (s + s + 1) 2 (s + s + 1) s (D) 2 2s + 1 2 (s2 + s + 1) 2 (s + s + 1) Option (D) is correct. Using s -domain differentiation property of Laplace transform. (C) SOL 1.11 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 8 GATE EC 2012 If L f (t) F (s) dF (s) ds L [tf (t)] = d ; 2 1 ds s + s + 1E L tf (t) So, www.gatehelp.com 2s + 1 (s2 + s + 1) 2 With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (B) x = t 2 1 (A) x = t 1 2 2 = MCQ 1.12 SOL 1.12 2 (C) x = t (D) x = t 2 2 Option (D) is correct. t dx + x = t dt dx + x = 1 t dt dx + Px = Q (General form) dt IF = e # = e = e ln t = t Integrating factor, Pdt 1 # dt t Solution has the form x # IF = x#t = # ^Q # IF hdt + C # (1) (t) dt + C 2 xt = t + C 2 Taking the initial condition x (1) = 0.5 0.5 = 1 + C 2 C =0 2 xt = t & x = t 2 2 The diodes and capacitors in the circuit shown are ideal. The voltage v (t) across the diode D1 is So, MCQ 1.13 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 9 GATE EC 2012 www.gatehelp.com (A) cos ( t) 1 (C) 1 cos ( t) SOL 1.13 (B) sin ( t) (D) 1 sin ( t) Option (A) is correct. The circuit composed of a clamper and a peak rectifier as shown. Clamper clamps the voltage to zero voltage, as shown The peak rectifier adds + 1 V to peak voltage, so overall peak voltage lowers down by 1 volt. So, vo = cos t 1 MCQ 1.14 In the circuit shown Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 10 GATE EC 2012 www.gatehelp.com (A) Y = A B + C (B) Y = (A + B) C (C) Y = (A + B ) C (D) Y = AB + C SOL 1.14 Option (A) is correct. Parallel connection of MOS & OR operation Series connection of MOS & AND operation The pull-up network acts as an inverter. From pull down network we write Y = (A + B) C Y = (A + B) + C = A B+C MCQ 1.15 A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount and decreases that of the second by . After encoding, the entropy of the source (A) increases (B) remains the same (C) increases only if N = 2 SOL 1.15 (D) decreases Option (D) is correct. Entropy function of a discrete memory less system is given as N 1 H = / Pk log b 1 l Pk k=0 where Pk is probability of symbol Sk . For first two symbols probability is same, so N 1 H = P1 log b 1 l + P2 log b 1 l + / Pk log b 1 l P1 P2 Pk k=3 = e P1 log P1 + P2 log P2 + = e 2P log P + Brought to you by: Nodia and Company PUBLISHING FOR GATE N 1 / P log P o k N 1 / P log P o k k=3 k k=3 k (P1 = P2 = P) Visit us at: www.nodia.co.in Page 11 GATE EC 2012 www.gatehelp.com Now, P1 = P + , P2 = P So, H l = =(P + ) log (P + ) + (P ) log (P ) + N 1 / P log P G k k k=3 By comparing, H l < H Entropy of source decreases. MCQ 1.16 A coaxial-cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given 0 = 4 # 10 7 H/m, 9 0 = 10 F/m , the characteristic impedance of the cable is 36 (A) 330 (C) 143.3 SOL 1.16 (B) 100 (D) 43.4 Option (B) is correct. Characteristic impedance. Z0 = ln b ba l b " outer diameter a " inner diameter 0 Z0 = ln b 0 r b a l 4 # 10 7 # 36 ln 2.4 b1l 10 9 # 10.89 = 100 = MCQ 1.17 The radiation pattern of an antenna in spherical co-ordinates is given by F ( ) = cos 4 ; 0 # # /2 The directivity of the antenna is (B) 12.6 dB (A) 10 dB (C) 11.5 dB SOL 1.17 (D) 18 dB Option (A) is correct. The directivity is defined as D = Fmax Favg Fmax = 1 Favg = 1 4 # F ( , ) d 2 = 1 ;# 4 0 = 1 ;# 4 0 Brought to you by: Nodia and Company PUBLISHING FOR GATE 2 2 # F ( , ) sin d d E 0 # 0 /2 cos 4 sin d d E Visit us at: www.nodia.co.in Page 12 GATE EC 2012 www.gatehelp.com /2 5 = 1 ;2 b cos lE 5 4 0 1 1 = 2 0 + D 5 4 # : = 1 # 2 = 1 5 10 4 D = 1 = 10 10 or, MCQ 1.18 SOL 1.18 D (in dB) = 10 log 10 = 10 dB If x [n] = (1/3) n (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (B) 1 < z < 1 (A) 1 < z < 3 3 3 2 (C) 1 < z < 3 2 Option (C) is correct. (D) 1 < z 3 n n x [n] = b 1 l b 1 l u [n] 3 2 n n n x [n] = b 1 l u [n] + b 1 l u [ n 1] b 1 l u (n) 3 3 2 Taking z -transform X 6z @ = 3 / n = 3 1 n n b 3 l z u [ n] + 3 / n = 3 1 n n b 3 l z u [ n 1] 1 3 3 / n = 3 1 n n b 2 l z u [ n] 3 n n n = / b 1 l z n + / b 1 l z n / b 1 l z n 3 3 2 n=0 n = 3 n=0 = 3 n=0 n 14 24 3 4I4 Series I converges if 3 / b 31z l + / b 1 z l 3 3 /b 1 l 2z m=1 n=0 14 24 3 14 24 3 4 II 4 4 III 4 m n Taking m = n 1 < 1 or z > 1 3z 3 Series II converges if 1 z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 MCQ 1.19 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 In the sum of products function f (X, Y, Z) = / (2, 3, 4, 5), the prime implicants are (A) XY, XY (B) XY, X Y Z , XY Z Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 13 GATE EC 2012 (C) XY Z , XYZ, XY SOL 1.19 www.gatehelp.com (D) XY Z , XYZ, XY Z , XY Z Option (A) is correct. Prime implicants are the terms that we get by solving K-map F = XY + XY 14 24 3 4 4 prime implicants MCQ 1.20 A system with transfer function G (s) = (s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin ( t). The steady-state output of the system is zero at (B) = 2 rad/s (A) = 1 rad/s (C) = 3 rad/s SOL 1.20 (D) = 4 rad/s Option (C) is correct. G (s) = (s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) ( 2 + 9) (j + 2) G (j ) = (j + 1) (j + 3) (j + 4) The steady state output will be zero if G (j ) = 0 2 + 9 = 0 = 3 rad/s MCQ 1.21 The impedance looking into nodes 1 and 2 in the given circuit is (A) 50 (B) 100 (C) 5 k (D) 10.1 k Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 14 SOL 1.21 GATE EC 2012 www.gatehelp.com Option (A) is correct. We put a test source between terminal 1, 2 to obtain equivalent impedance ZTh = Vtest Itest By applying KCL at top right node Vtest + Vtest 99I = I b test 9 k + 1k 100 Vtest + Vtest 99I = I b test 10 k 100 ...(i) Ib = Vtest = Vtest 9k + 1k 10k But Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k 100Vtest + Vtest = I test 10 # 103 100 MCQ 1.22 2Vtest = I test 100 ZTh = Vtest = 50 Itest In the circuit shown below, the current through the inductor is (A) 2A 1+j Brought to you by: Nodia and Company PUBLISHING FOR GATE (B) 1 A 1+j Visit us at: www.nodia.co.in Page 15 GATE EC 2012 1A 1+j Option (C) is correct. (C) SOL 1.22 www.gatehelp.com (D) 0 A Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1 V1 (j 1 + 1) + j 1 + 1 0c = j 1 V1 = 1 1 + j1 1 V1 + 1 0c 1 + j + 1 = I1 = j1 j1 j = =1A (1 + j) j 1 + j Current MCQ 1.23 Given f (z) = 1 2. z+1 z+3 If C is a counter clockwise path in the z -plane such that z + 1 = 1, the value of 1 f (z) dz is 2 j # C (A) 2 (C) 1 SOL 1.23 (B) 1 (D) 2 Option (C) is correct. f (z) = 1 2 j 1 2 z+1 z+3 # f (z) dz C = sum of the residues of the poles which lie inside the given closed region. C & z+1 = 1 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 16 GATE EC 2012 www.gatehelp.com Only pole z = 1 inside the circle, so residue at z = 1 is. z + 1 f (z) = (z + 1) (z + 3) = lim (z + 1) ( z + 1) 2 = =1 2 (z + 1) (z + 3) z " 1 1 f (z) dz = 1 2 j # C Two independent random variables X and Y are uniformly distributed in the interval 6 1, 1@. The probability that max 6X, Y @ is less than 1/2 is (B) 9/16 (A) 3/4 So MCQ 1.24 (C) 1/4 SOL 1.24 (D) 2/3 Option (B) is correct. Probability density function of uniformly distributed variables X and Y is shown as P &[max (x, y)] < 1 0 2 Since X and Y are independent. P &[max (x, y)] < 1 0 = P b X < 1 l P bY < 1 l 2 2 2 P b X < 1 l = shaded area = 3 2 4 Similarly for Y : P bY < 1 l = 3 2 4 So P &[max (x, y)] < 1 0 = 3 # 3 = 9 2 4 4 16 Alternate method: Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 17 GATE EC 2012 www.gatehelp.com From the given data since random variables X and Y lies in the interval [ 1, 1] as from the figure X , Y lies in the region of the square ABCD . Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside the region of square AEFG . P &max 6X, Y @ < 1 0 = Area of 4AEFG 2 Area of square ABCD 3 3 2#2 = 9 = 2#2 16 x If x = 1, then the value of x is (A) e /2 (B) e /2 So, MCQ 1.25 (C) x SOL 1.25 (D) 1 Option (A) is correct. x= So, 1 = i = cos + i sin 2 2 x = ei 2 xx = ^ei 2 h & ^ei 2 h x i = e 2 Q. 26 to Q. 55 carry two marks each. MCQ 1.26 The source of a silicon (ni = 1010 per cm3) n -channel MOS transistor has an area of 1 sq m and a depth of 1 m . If the dopant density in the source is 1019 /cm3 , the number of holes in the source region with the above volume is approximately (A) 107 (B) 100 (C) 10 Brought to you by: Nodia and Company PUBLISHING FOR GATE (D) 0 Visit us at: www.nodia.co.in Page 18 SOL 1.26 GATE EC 2012 www.gatehelp.com Option (D) is correct. For the semiconductor n 0 p 0 = n i2 2 20 p 0 = n i = 1019 = 10 per cm3 n 0 10 Volume of given device, V = Area # depth = 1 m 2 # 1 m = 10 8 cm2 # 10 4 cm = 10 12 cm3 So total no. of holes is, p = p0 # V = 10 # 10 12 = 10 11 Which is approximately equal to zero. MCQ 1.27 A BPSK scheme operating over an AWGN channel with noise power spectral density of N 0 /2, uses equiprobable signals s1 (t) = 2E sin ( c t) and s2 (t) = 2E sin ( c t) T T over the symbol interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45c with respect to the received signal, the probability of error in the resulting system is (A) Q c 2E m (B) Q c E m N0 N0 (C) Q c SOL 1.27 E E (D) Q c 2N 0 m 4N 0 m Option (B) is correct. In a coherent binary PSK system, the pair of signals s1 (t) and s2 (t) used to represent binary system 1 and 0 respectively. s1 (t) = 2E sin c t T 2E sin t c T s2 (t) = where 0 # t # T , E is the transmitted energy per bit. General function of local oscillator 2 sin ( t), 0 # t < T 1 (t) = c T But here local oscillator is ahead with 45c. so, 2 sin ( t + 45c) 1 (t) = c T The coordinates of message points are s11 = Brought to you by: Nodia and Company PUBLISHING FOR GATE T # s (t) (t) dt 1 1 0 Visit us at: www.nodia.co.in Page 19 GATE EC 2012 = # T 2E sin t c T 0 = 2E T = = Similarly, Signal space diagram s21 = E 2 sin ( t + 45c) dt c T T # sin ( t) sin ( t + 45c) dt 2E T =1 T www.gatehelp.com c c 0 2 T # 0 T # 0 T 1 [sin 45c + sin (2 t + 45c)] dt c 2 1 dt + 1 E T (2 t + 45c) dt #0 sin c T 2 1 444444 2 444444 3 40 4 E 2 E 2 Now here the two message points are s11 and s21 . The error at the receiver will be considered. When : (i) s11 is transmitted and s21 received (ii) s21 is transmitted and s11 received So, probability for the 1st case will be as : P b s21 received l = P (X < 0) (as shown in diagram) s11 transmitted = P _ E/2 + N < 0i = P _N < E/2 i Taking the Gaussian distribution as shown below : Mean of the Gaussian distribution = E/2 Variance = N 0 2 Putting it in the probability function : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 20 GATE EC 2012 P bN < E= 2l = Taking, # `x + E/2 j 2 0 # www.gatehelp.com 0 1 3 2 N 0 2 e 2N 0 /2 dx 1 e `x + NE/2 j dx N 0 2 0 3 x + E/2 =t N 0 /2 N 0 dt 2 dx = So, P _N < E/2 i = Qc # 3 E/N 0 2 1 e t2 dt 2 E N0 m where Q is error function. Since symbols are equiprobable in the 2 nd case So, P b s11 received l = Q c E m N0 s21 transmitted So the average probability of error = 1 ;P b s21 received l + P b s11 received lE 2 s11 transmitted s21 transmitted MCQ 1.28 = 1 =Q c E m + Q c E mG = Q c E m 2 N0 N0 N0 A transmission line with a characteristic impedance of 100 is used to match a 50 section to a 200 section. If the matching is to be done both at 429 MHz and 1 GHz , the length of the transmission line can be approximately (b) 1.05 m (A) 82.5 cm (C) 1.58 cm SOL 1.28 (D) 1.75 m Option (C) is correct. Since Z 0 = Z1 Z 2 100 = 50 # 200 This is quarter wave matching. The length would be odd multiple of /4 . l = (2m + 1) 4 f1 = 429 MHz, f2 = 1 GHz , c= 3 # 108 = 0.174 m f1 # 4 429 # 106 # 4 8 l2 = c = 3 # 10 = 0.075 m f2 # 4 1 # 109 # 4 l1 = Only option (C) is odd multiple of both l1 and l2 . (2m + 1) = 1.58 = 9 l1 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 21 GATE EC 2012 www.gatehelp.com (2m + 1) = 1.58 - 21 l2 The input x (t) and output y (t) of a system are related as y (t) = # x ( ) cos (3 ) d 3 . The system is (A) time-invariant and stable (B) stable and not time-invariant t MCQ 1.29 (C) time-invariant and not stable SOL 1.29 (D) not time-invariant and not stable Option (D) is correct. y (t) = Time invariance : Let, # x ( ) cos (3 ) d t 3 x (t) = (t) y (t) = # (t) cos (3 ) d t 3 = u (t) cos (0) = u (t) For a delayed input (t t 0) output is y (t, t 0) = # (t t ) cos (3 ) d t 0 3 = u (t) cos (3t 0) Delayed output y (t t 0) = u (t t 0) y (t, t 0) ! y (t t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) = # t 3 =1 2 cos2 3t = # t 3 1 cos 6t 2 # 1dt 1 # cos 6t dt 2 t t 3 3 As t " 3, y (t) " 3 (unbounded) System is not stable. MCQ 1.30 The feedback system shown below oscillates at 2 rad/s when (A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5 SOL 1.30 (B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5 Option (A) is correct. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 22 GATE EC 2012 www.gatehelp.com K (s + 1) [R (s) Y (s)] s3 + as2 + 2s + 1 K (s + 1) K (s + 1) Y (s) ;1 + 3 E = s3 + as2 + 2s + 1 R (s) 2 s + as + 2s + 1 3 2 Y (s) [s + as + s (2 + k) + (1 + k)] = K (s + 1) R (s) Transfer Function K (s + 1) Y (s) H (s) = =3 2 R (s) s + as + s (2 + k) + (1 + k) Routh Table : Y (s) = For oscillation, a (2 + K) (1 + K) =0 a a = K+1 K+2 Auxiliary equation A (s) = as2 + (k + 1) = 0 s2 = k + 1 a s2 = k + 1 (k + 2) (k + 1) = (k + 2) = j k+2 = j k+2 (Oscillation frequency) = k+2 = 2 =2 and a = 2 + 1 = 3 = 0.75 2+2 4 The Fourier transform of a signal h (t) is H (j ) = (2 cos ) (sin 2 ) / . The value of h (0) is (B) 1/2 (A) 1/4 s2 s j k MCQ 1.31 (C) 1 SOL 1.31 (D) 2 Option (C) is correct. H (j ) = Brought to you by: Nodia and Company PUBLISHING FOR GATE (2 cos ) (sin 2 ) Visit us at: www.nodia.co.in Page 23 GATE EC 2012 www.gatehelp.com = sin 3 + sin We know that inverse Fourier transform of sin c function is a rectangular function. MCQ 1.32 So, inverse Fourier transform of H (j ) h (t) = h1 (t) + h2 (t) h (0) = h1 (0) + h2 (0) =1+1 =1 22 The state variable description of an LTI system is given by Jx1N J 0 a1 0NJx1N J0N o KO K OK O K O o Kx2O = K 0 0 a2OKx2O + K0O u Kx O Ka o3 0 0OKx 3O K 1O 3 LP L PL P L P Jx1N KO y = _1 0 0iKx2O Kx 3O LP where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 SOL 1.32 (D) a1 ! 0, a2 ! 0, a 3 = 0 Option (D) is correct. General form of state equations are given as o x = Ax + Bu o y = Cx + Du For the given problem R 0 a 0V R0V 1 S W SW A = S 0 0 a2W, B = S0W Sa S1W S 3 0 0W W SW T X TX Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 24 GATE EC 2012 www.gatehelp.com R 0 a 0VR0V R 0V 1 S WS W S W AB = S 0 0 a2WS0W = Sa2W Sa S 3 0 0WS1W S 0W WS W S W T XT X T X R0 0 a1 a2VR0V Ra1 a2V W WS W S S A2 B = Sa2 a 3 0 0WS0W = S 0W S 0 aa W WS W S S 0WS1W S 0W 31 T that followingXT X T has a tank of n = 3 . For controllability it is necessary matrix X U = 6B : AB : A2 B@ So, R0 0 a a V 1 2W S = S0 a2 0W S1 0 S W 0W T X a2 ! 0 a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not. MCQ 1.33 Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is (A) 0.8 (C) 2 SOL 1.33 (B) 1.4 (D) 2.8 Option (A) is correct. We obtain Thevenin equivalent of circuit B . Thevenin Impedance : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 25 GATE EC 2012 www.gatehelp.com ZTh = R Thevenin Voltage : VTh = 3 0c V Now, circuit becomes as Current in the circuit, I1 = 10 3 2+R Power transfer from circuit A to B 2 P = (I 1 ) 2 R + 3I1 2 P = :10 3D R + 3 :10 3D 2+R 2+R P= 49R + 21 (2 + R) 2 (2 + R) P= 49R + 21 (2 + R) (2 + R) 2 P = 42 + 70R (2 + R) 2 2 dP = (2 + R) 70 (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 (42 + 70R) 2] 140 + 70R 84 140R 56 R MCQ 1.34 =0 =0 = 70R = 0.8 Consider the differential equation d 2 y (t) dy (t) dy +2 + y (t) = (t) with y (t) t = 0 = 2 and 2 dt dt dt Brought to you by: Nodia and Company PUBLISHING FOR GATE t = 0 =0 Visit us at: www.nodia.co.in Page 26 GATE EC 2012 dy The numerical value of dt (A) 2 is t = 0+ (B) 1 (C) 0 SOL 1.34 www.gatehelp.com (D) 1 Option (D) is correct. d 2 y (t) 2dy (t) + + y (t) = (t) dt dt 2 By taking Laplace transform with initial conditions dy 2 ;s Y (s) sy (0) dt E + 2 [sy (s) y (0)] + Y (s) = 1 t=0 & 2 6s Y (s) + 2s 0@ + 2 6sY (s) + 2@ + Y (s) = 1 Y (s) [s2 + 2s + 1] = 1 2s 4 Y (s) = 2 2s 3 s + 2s + 1 We know that, If, then, So, L y (t) dy (t) dt L Y (s) sY (s) y (0) ( 2s 3) s +2 (s2 + 2s + 1) 2 2 = 2s 3s + 2s + 4s + 2 2 (s + 2s + 1) 1 sY (s) y (0) = s + 2 2 = s + 1 2 + (s + 1) (s + 1) (s + 1) 2 sY (s) y (0) = = 1+ 1 s + 1 (s + 1) 2 By taking inverse Laplace transform dy (t) = e t u (t) + te t u (t) dt dy At t = 0+ , = e0 + 0 = 1 dt t = 0 The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d A = 0 is : (A) 2 (B) 2 + MCQ 1.35 (C) 1 SOL 1.35 (D) 0 Option (A) is correct. Divergence of A in spherical coordinates is given as d A = 12 2 (r 2 Ar ) : r 2r Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 27 GATE EC 2012 www.gatehelp.com = 12 2 (krn + 2) r 2r = k2 (n + 2) rn + 1 r = k (n + 2) rn 1 = 0 n+2 = 0 n = 2 MCQ 1.36 A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (B) 1/2 (A) 1/3 (C) 2/3 SOL 1.36 MCQ 1.37 (given) (D) 3/4 Option (C) is correct. Probability of appearing a head is 1/2 . If the number of required tosses is odd, we have following sequence of events. H, TTH, TTTTH, ........... 3 5 Probability P = 1 + b 1 l + b 1 l + ..... 2 2 2 1 P = 2 =2 3 1 1 4 In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if (A) Vin < 1.875 V (C) Vin > 3.125 V SOL 1.37 (B) 1.875 V < Vin < 3.125 V (D) 0 < Vin < 5 V Option (A) is correct. Given the circuit as below : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 28 GATE EC 2012 www.gatehelp.com Since all the parameters of PMOS and NMOS are equal. So, n = p W COX b l = COX bW l = COX bW l L M1 L M2 L Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0 Where I1 is drain current in M1 and I2 is drain current in M2 . C 2 Since, I1 = p OX bW l82VSD ^VSG VTp h V SDB 2 L & 0= p COX W 2 [2V V VTp h V SD] 2 b L l SD ^ SG Solving it we get, 2 ^VSG VTp h = VSD & 2 ^5 Vin 1h = 5 VD & Vin = VD + 3 2 For I1 = 0 , VD = 5 V So, Vin = 5 + 3 = 4 V 2 So for the NMOS VGS = Vin 0 = 4 0 = 4 V and VGS > VTn So it can t be in cutoff region. Case 2 : M2 must be in saturation region. So, I1 = I 2 p COX W C 2 2 (VSG VTp) VSD V SD@ = n OX W (VGS VTn) 2 2L 2 L6 & & & 2 2 (VSG VTp) VSD V SD = (VGS VTn) 2 2 (5 Vin 1) (5 VD) (5 VD) 2 = (Vin 0 1) 2 2 (4 Vin) (5 VD) (5 VD) 2 = (Vin 1) 2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 29 GATE EC 2012 www.gatehelp.com Substituting VD = VDS = VGS VTn and for N -MOS & VD = Vin 1 & 2 (4 Vin) (6 Vin) (6 Vin) 2 = (Vin 1) 2 & 48 36 8Vin = 2Vin + 1 & 6Vin = 11 & Vin = 11 = 1.833 V 6 So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V MCQ 1.38 A binary symmetric channel (BSC) has a transition probability of 1/8 . If the binary symbol X is such that P (X = 0) = 9/10, then the probability of error for an optimum receiver will be (A) 7/80 (B) 63/80 (C) 9/10 (D) 1/10 SOL 1.38 MCQ 1.39 The signal m (t) as shown is applied to both a phase modulator (with k p as the phase constant) and a frequency modulator (with k f as the frequency constant) having the same carrier frequency. The ratio k p /k f (in rad/Hz) for the same maximum phase deviation is (B) 4 (A) 8 (C) 2 SOL 1.39 (D) Option (B) is correct. General equation of FM and PM waves are given by FM (t) = Ac cos ; c t + 2 k f # m ( ) d E t 0 PM (t) = Ac cos [ c t + k p m (t)] For same maximum phase deviation. k p [m (t)] max = 2 k f ; # m ( ) d E t 0 Brought to you by: Nodia and Company PUBLISHING FOR GATE max Visit us at: www.nodia.co.in Page 30 GATE EC 2012 www.gatehelp.com k p # 2 = 2 k f [x (t)] max where, x (t) = # m ( ) d t 0 [x (t)] max = 4 k p # 2 = 2 k f # 4 kp = 4 kf The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t z) So, MCQ 1.40 The phase velocity v p of the wave inside the waveguide satisfies (B) v p = c (A) v p > c (C) 0 < v p < c SOL 1.40 (D) v p = 0 Option (D) is correct. Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t z) x = 2.094 # 102 y = 2.618 # 102 = 6.283 # 1010 rad/s For the wave propagation, 2 ( 2 + 2) = x y c2 Substituting above values, 10 2 = c 6.283 # 10 m (2.0942 + 2.6182) # 10 4 3 # 108 - j 261 is imaginary so mode of operation is non-propagating. vp = 0 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 31 MCQ 1.41 GATE EC 2012 The circuit shown is a (A) low pass filter with f3dB = 1 rad/s (R1 + R2) C (B) high pass filter with f3dB = (C) low pass filter with f3dB = 1 rad/s R1 C 1 rad/s R1 C (D) high pass filter with f3dB = SOL 1.41 www.gatehelp.com 1 rad/s (R1 + R2) C Option (B) is correct. First we obtain the transfer function. 0 Vi (j ) 0 Vo (j ) =0 + 1 +R R2 1 j C Vo (j ) Vi (j ) = 1 +R R2 1 j C Vo (j ) = Vi (j ) R2 R1 j 1 C At " 0 (Low frequencies) 1 " 3, so V = 0 o C At " 3 (higher frequencies) 1 " 0, so V (j ) = R2 V (j ) o R1 i C Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 32 GATE EC 2012 www.gatehelp.com The filter passes high frequencies so it is a high pass filter. H (j ) = Vo = R2 Vi R1 j 1 C H (3) = R2 = R2 R1 R1 2 times of maximum gain 6H (3)@ H ^ j 0h = 1 H (3) 2 R2 = 1 b R2 l 1 2 2 R1 R1 + 2 2 0 C At 3 dB frequency, gain will be So, 2 2 2R 1 = R 1 + 2 R1 = 1 C2 2 0 1 2C 2 1 R1 C Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 0 = MCQ 1.42 (C) 1 SOL 1.42 (D) 3/2 Option (A) id correct. Convolution sum is defined as y [n] = h [n] * g [n] = For causal sequence, y [n] = 3 / h [n] g [n k] k = 3 3 / h [n] g [n k] k=0 y [n] = h [n] g [n] + h [n] g [n 1] + h [n] g [n 2] + ..... For n = 0 , y [0] = h [0] g [0] + h [1] g [ 1] + ........... y [0] = h [0] g [0] g [ 1] = g [ 2] = ....0 ...(i) y [0] = h [0] g [0] For n = 1, y [1] = h [1] g [1] + h [1] g [0] + h [1] g [ 1] + .... y [1] = h [1] g [1] + h [1] g [0] 1 = 1 g [1] + 1 g [0] 2 2 2 1 h [1] = b 1 l = 1 2 2 1 = g [1] + g [0] g [1] = 1 g [0] From equation (i), Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 33 GATE EC 2012 g [0] = So, www.gatehelp.com y [0] 1 = =1 h [0] 1 g [1] = 1 1 = 0 MCQ 1.43 The state transition diagram for the logic circuit shown is SOL 1.43 Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure. D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn X 0 = Q , X1 = Q If A = 0, (toggle of previous state) Qn + 1 = Qn If A = 1, Qn + 1 = Qn So state diagram is MCQ 1.44 The voltage gain Av of the circuit shown below is Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 34 GATE EC 2012 www.gatehelp.com (A) Av . 200 (C) Av . 20 SOL 1.44 (B) Av . 100 (D) Av . 10 Option (D) is correct. DC Analysis : Using KVL in input loop, VC 100IB 0.7 = 0 VC = 100IB + 0.7 ...(i) IC - IE = 13.7 VC = ( + 1) IB 12k 13.7 VC = 100I B 12 # 103 ...(ii) Solving equation (i) and (ii), IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 35 GATE EC 2012 www.gatehelp.com This is a shunt-shunt feedback amplifier. Given parameters, r = VT = 25 mV = 2.5 k IB 0.01 mA 100 = 0.04 s gm = = r 2.5 # 1000 Writing KCL at output node v0 + g v + v0 v = 0 m RC RF v 0 : 1 + 1 D + v :gm 1 D = 0 RC RF RF Substituting RC = 12 k , RF = 100 k , gm = 0.04 s v 0 (9.33 # 10 5) + v (0.04) = 0 v 0 = 428.72V Writing KCL at input node vi = v + v + v vo Rs Rs r RF ...(i) vi = v 1 + 1 + 1 v 0 : Rs Rs r RF D RF vi = v (5.1 10 4) v 0 # Rs RF Substituting V from equation (i) vi = 5.1 # 10 4 v v 0 0 428.72 Rs RF MCQ 1.45 vi Rs = 10 k (source resistance) = 1.16 # 10 6 v 0 1 # 10 5 v 0 10 # 103 vi 5 3 = 1.116 # 10 10 # 10 1 Av = v 0 = - 8.96 vi 10 # 103 # 1.116 # 10 5 If VA VB = 6 V then VC VD is (A) 5 V (B) 2 V (C) 3 V (D) 6 V Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 36 SOL 1.45 GATE EC 2012 www.gatehelp.com Option (A) is correct. In the given circuit VA VB = 6 V So current in the branch will be IAB = 6 = 3 A 2 We can see, that the circuit is a one port circuit looking from terminal BD as shown below For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e. IDC = IAB = 3 A The total current in the resistor 1 will be I1 = 2 + IDC = 2+3 = 5A So, VCD = 1 # ( I1) = 5 V MCQ 1.46 The maximum value of f (x) = x3 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41 SOL 1.46 (By writing KCL at node D ) (D) 46 Option (B) is correct. f (x) = x3 9x2 + 24x + 5 df (x) = 3x2 18x + 24 = 0 dx Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 37 GATE EC 2012 & www.gatehelp.com df (x) = x2 6x + 8 = 0 dx x = 4, x = 2 d 2 f (x) = 6x 18 dx 2 d 2 f (x) For x = 2, = 12 18 = 6 < 0 dx2 So at x = 2, f (x) will be maximum f (x) MCQ 1.47 max = (2) 3 9 (2) 2 + 24 (2) + 5 = 8 36 + 48 + 5 = 25 Given that 5 3 10 A=> H and I = >0 1H, the value of A3 is 20 (A) 15A + 12I (C) 17A + 15I SOL 1.47 (B) 19A + 30I (D) 17A + 21I Option (B) is correct. Characteristic equation. A I = 0 5 3 =0 2 5 + 2 + 6 = 0 2 + 5 + 6 = 0 Since characteristic equation satisfies its own matrix, so A2 + 5A + 6 = 0 & A2 = 5A 6I Multiplying with A A3 + 5A2 + 6A = 0 A3 + 5 ( 5A 6I) + 6A = 0 A3 = 19A + 30I Common Data Questions Common Data for Questions 48 and 49 : With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 38 GATE EC 2012 www.gatehelp.com (i) 1 connected at port B draws a current of 3 A (ii) 2.5 connected at port B draws a current of 2 A MCQ 1.48 With 10 V dc connected at port A , the current drawn by 7 connected at port B is (B) 5/7 A (A) 3/7 A (C) 1 A SOL 1.48 (D) 9/7 A Option (C) is correct. When 10 V is connected at port A the network is Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh . IL = VTh,10 V RTh + RL For RL = 1 , IL = 3 A 3= VTh,10 V RTh + 1 ...(i) For RL = 2.5 , IL = 2 A = VTh,10 V RTh + 2.5 ...(ii) Dividing above two Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 39 GATE EC 2012 www.gatehelp.com 3 = RTh + 2.5 2 RTh + 1 3RTh + 3 = 2RTh + 5 RTh = 2 Substituting RTh into equation (i) VTh,10 V = 3 (2 + 1) = 9 V Note that it is a non reciprocal two port network. Thevenin voltage seen at port B depends on the voltage connected at port A . Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin s resistance remains same. Now, the circuit is For RL = 7 IL = MCQ 1.49 VTh,10 V = 9 = 1A 2 + RL 2 + 7 For the same network, with 6 V dc connected at port A , 1 connected at port B draws 7/3 A. If 8 V dc is connected to port A , the open circuit voltage at port B is (B) 7 V (A) 6 V (C) 8 V SOL 1.49 (D) 9 V Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 and IL = 7 A 3 VTh, 6 V = RTh # 7 + 1 # 7 3 3 = 2#7 +7 = 7V 33 This is a linear network, so VTh at port B can be written as Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 40 GATE EC 2012 www.gatehelp.com VTh = V1 + where V1 is the input applied at port A . We have V1 = 10 V , VTh,10 V = 9 V ...(i) ` 9 = 10 + When V1 = 6 V , VTh, 6 V = 9 V ...(ii) ` 7 = 6 + Solving (i) and (ii) = 0.5 , = 4 Thus, with any voltage V1 applied at port A , Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V (open circuit voltage) VTh,8 V = 0.5 # 8 + 4 = 8 = Voc 1 Common Data for Question 50 and 51 : In the three dimensional view of a silicon n -channel MOS transistor shown below, = 20 nm . The transistor is of width 1 m . The depletion width formed at every p -n junction is 10 nm . The relative permittivity of Si and SiO 2 , respectively, are 11.7 and 3.9, and 0 = 8.9 # 10 12 F/m . MCQ 1.50 The gate source overlap capacitance is approximately (A) 0.7 fF (B) 0.7 pF (C) 0.35 fF SOL 1.50 (D) 0.24 pF Option (B) is correct. Gate source overlap capacitance. Co = W ox 0 (medium Sio 2 ) tox Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 41 GATE EC 2012 www.gatehelp.com 9 6 12 3 = 20 # 10 # 1 # 10 # 9 .9 # 8.9 # 10 1 # 10 = 0.69 # 10 15 F MCQ 1.51 The source-body junction capacitance is approximately (A) 2 fF (B) 7 fF (C) 2 pF SOL 1.51 (D) 7 pF Option (B) is correct. Source body junction capacitance. Cs = A r 0 d A = (0.2 m + 0.2 m + 0.2 m) # 1 m + 2 (0.2 m # 0.2 m) = 0.68 m2 d = 10 nm (depletion width of all junction) 12 12 # Cs = 0.68 # 10 # 11.7 9 8.9 # 10 10 # 10 15 = 7 # 10 F Linked Answer Questions Statement for Linked Answer Question 52 and 53 : An infinitely long uniform solid wire of radius a carries a uniform dc current of density J MCQ 1.52 The magnetic field at a distance r from the center of the wire is proportional to (A) r for r < a and 1/r 2 for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a SOL 1.52 (D) 0 for r < a and 1/r 2 for r > a Option (C) is correct. For r > a , Ienclosed = ( a2) J # H : dl = Ienclosed H # 2 r = ( a2) J H = Io 2 r H \ 1 , for r > a r Io = ( a2) J For r < a , Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 42 GATE EC 2012 Ienclosed = So, www.gatehelp.com J ( r 2) Jr 2 =2 a a 2 # H : dl = Ienclosed 2 H # 2 r = Jr2 a H = Jr 2 2 a H \ r, MCQ 1.53 for r < a A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below. The magnetic field inside the hole is (A) uniform and depends only on d (B) uniform and depends only on b (C) uniform and depends on both b and d (D) non uniform SOL 1.53 Option (B) is correct. Assuming the cross section of the wire on x -y plane as shown in figure. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 43 GATE EC 2012 www.gatehelp.com Since, the hole is drilled along the length of wire. So, it can be assumed that the drilled portion carriers current density of J . Now, for the wire without hole, magnetic field intensity at point P will be given as H 1 (2 R) = J ( R2) H 1 (2 R) = JR 2 Since, point o is at origin. So, in vector form H1 = J (xax + yay) 2 Again only due to the hole magnetic field intensity will be given as. (H 2) (2 r) = J ( r 2) H 2 = Jr 2 Again, if we take Ol at origin then in vector form H2 = J (xlax + ylay) 2 where xl and yl denotes point P in the new co-ordinate system. Now the relation between two co-ordinate system will be. x = xl + d y = yl So, H2 = J [(x d) ax + yay] 2 So, total magnetic field intensity = H1 + H2 = J dax 2 So, magnetic field inside the hole will depend only on d . Statement for Linked Answer Question 54 and 55 : Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 44 MCQ 1.54 GATE EC 2012 The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a = 3, b = 1 SOL 1.54 www.gatehelp.com (D) a = 3, b = 1 Option (A) is correct. j + a GC (s) = s + a = s+b j + b Phase lead angle = tan 1 a k tan 1 a k a b J N 1 K a bO = tan K 2O K1 + O ab P L (b a) = tan 1 c ab + 2 m For phase-lead compensation > 0 b a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true. MCQ 1.55 The phase of the above lead compensator is maximum at (B) 3 rad/s (A) 2 rad/s (C) SOL 1.55 (D) 1/ 3 rad/s 6 rad/s Option (A) is correct. = tan 1 a k tan 1 a k a b 1/a 1/b d = =0 2 1+ 2 d 1 +a k ab k a 1 + 2 = 1 + 1 2 a ab2 b b a2 1 1 = 2 1 1 ab ab b a b l = ab = 1#2 = 2 rad/ sec General Aptitude (GA) Question (Compulsory) Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 45 GATE EC 2012 www.gatehelp.com Q. 56 - Q. 60 carry one mark each. MCQ 1.56 If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321 (A) 2.23 (B) 4.33 (C) 11.37 (D) 27.64 SOL 1.56 Option (D) is correct option. Let 1.001 = x So in given data : x1259 = 3.52 x2062 = 7.85 Again x3321 = x1259 + 2062 = x1259 x2062 = 3.52 # 7.85 = 27.64 MCQ 1.57 Choose the most appropriate alternate from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he..................the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken SOL 1.57 Option (C) is correct. MCQ 1.58 Choose the most appropriate word from the options given below to complete the following sentence : Give the seriousness of the situation that he had to face, his........was impressive. (A) beggary (B) nomenclature (C) jealousy (D) nonchalance SOL 1.58 Option (D) is correct. MCQ 1.59 Which one of the following options is the closest in meaning to the word given below ? Latitude (A) Eligibility (B) Freedom (C) Coercion (D) Meticulousness SOL 1.59 Option (B) is correct. MCQ 1.60 One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (A) requested that (B) should be given Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 46 GATE EC 2012 (C) the driving test SOL 1.60 www.gatehelp.com (D) instead of tomorrow Option (B) is correct. Q. 61 - Q. 65 carry two marks each. MCQ 1.61 One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above passage ? (A) Through regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances. (B) The legions were treated inhumanly as if the men were animals (C) Disciplines was the armies inheritance from their seniors (D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them. SOL 1.61 Option (A) is correct. MCQ 1.62 Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5 (B) 6 (C) 9 (D) 10 SOL 1.62 Option (A) is correct. Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y . Then from the given data. x + y = 14 20x + 10y = 230 Solving the above two equations we get x = 9, y = 5 So, the no. of notes of Rs. 10 is 5. MCQ 1.63 There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2 (B) 3 (C) 4 SOL 1.63 (D) 8 Option (A) is correct. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 47 GATE EC 2012 www.gatehelp.com We will categorize the 8 bags in three groups as : (i) A1 A2 A 3 , (ii) B1 B2 B 3 , (iii) C1 C2 Weighting will be done as bellow : 1st weighting " A1 A2 A 3 will be on one side of balance and B1 B2 B 3 on the other. It may have three results as described in the following cases. Case 1 : A1 A 2 A 3 = B1 B 2 B 3 This results out that either C1 or C2 will heavier for which we will have to perform weighting again. 2 nd weighting " C1 is kept on the one side and C2 on the other. if C1 > C2 then C1 is heavier. then C2 is heavier. C1 < C 2 Case 2 : A1 A 2 A 3 > B1 B 2 B 3 it means one of the A1 A2 A 3 will be heavier So we will perform next weighting as: 2 nd weighting " A1 is kept on one side of the balance and A2 on the other. it means A 3 will be heavier if A1 = A2 then A1 will be heavier A1 > A 2 then A2 will be heavier A1 < A 2 Case 3 : A1 A 2 A 3 < B1 B 2 B 3 This time one of the B1 B2 B 3 will be heavier, So again as the above case weighting will be done. 2 nd weighting " B1 is kept one side and B2 on the other if B1 = B2 B 3 will be heavier B1 > B 2 B1 will be heavier B1 < B 2 B2 will be heavier So, as described above, in all the three cases weighting is done only two times to give out the result so minimum no. of weighting required = 2. MCQ 1.64 The data given in the following table summarizes the monthly budget of an average household. Category Amount (Rs.) Food 4000 Clothing 1200 Rent 2000 Savings 1500 Other expenses 1800 The approximate percentages of the monthly budget NOT spent on savings is (A) 10% (C) 81% SOL 1.64 (B) 14% (D) 86% Option (D) is correct. Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in Page 48 GATE EC 2012 www.gatehelp.com Total budget = 4000 + 1200 + 2000 + 1500 + 1800 = 10, 500 The amount spent on saving = 1500 So, the amount not spent on saving = 10, 500 1500 = 9000 So, percentage of the amount = 9000 # 100% 10500 = 86% MCQ 1.65 A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a conditions that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that days is (A) 1/4 (B) 1/16 (C) 7/16 SOL 1.65 (D) 9/16 Option (C) is correct. The graphical representation of their arriving time so that they met is given as below in the figure by shaded region. So, the area of shaded region is given by Area of 4PQRS (Area of TEFQ + Area of TGSH ) = 60 # 60 2 b 1 # 45 # 45 l 2 = 1575 So, the required probability = 1575 = 7 3600 16 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in

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