GATE A skew symmetric matrix is a square matrix and has property. All Eigen values of skew symmetric matrices are imaginary or zero.

Transcription

1 1. The eigen values of a skew-symmetric matrix are (A) Always zero (B) Always pure imaginary (C) Either zero or pure imaginary (D) Always real Soln.1 A skew symmetric matrix is a square matrix and has property that a ij = -a ji for all A a11 a12... a1n a a... a an1 an2... ann n = Diagonal entries are zero i.e. a ii = o It has the properties (i) (ii) A T = -A where A T is matrix transpose. All Eigen values of skew symmetric matrices are imaginary or zero. Option (C) is the correct 2. The trigonometric Fourier series for the waveform f(t) shown below contains (A) Only cosine terms and zero value for the dc component (B) Only cosine terms and a positive value for the dc component (C) Only cosine terms and a negative value for the dc component (D) Only sine terms and a negative value for the dc component Soln.2 The function f (t) is an even function and is symmetric to y- axis so it contains only cosine terms.

2 Note that the positive part of waveform has amplitude A while the negative part has amplitude 2A. DC value is the average value i.e. the area of the waveform lying above and below this line will be equal. So the line for average value will be shifted to negative value, so DC value is negative. Option (C) is correct 3. A function n(x) satisfied the differential equation d 2 n(x)/dx 2 -n(x)/l 2 = 0 where L is a constant. The boundary conditions are: n(0)=k and n( ) =0. The solution to this equation is (A) n(x) = K exp(x/l) (B) n(x) = K exp(-x/ L ) (C) n(x) = K 2 exp(-x/l) (D) n(x) = K exp(-x/l) Soln.3 Given differential eqn. is = 0 Let = = 0 = 0 = ± So, the solution is = + Applying the boundary conditions n(0)=k i.e. n(0)=c 1 +C 2 =K n( )=0 then n( )=0 C 1 =0 So C 2 =K Thus n(x) =K exp(-x/l) Option (D) is correct.

3 4. For the two-port network shown below, the short-circuit admittance parameter matrix is Soln. (4) 1 I 2 0.5Ω I 2 2 V 1 0.5Ω 0.5Ω V The governing eqn. for the admittance matrix is = = + = + = = = = ǁ I 0.5 = = = 2 V 0.5 I 0.5 = = = 2 V = = = = ǁ

4 5. For parallel RLC circuit, which one of the following statements is NOT correct? (A) The bandwidth of the circuit deceases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value. Soln. (5) Parallel RLC circuit has the following characteristic equation. + + = 0 Where is the bandwidth From the above eqn. we can draw the following conclusions: Bandwidth RLC circuits independent of L At resonance, imaginary part of input impedance is zero, i.e. It is real. At resonance the admittance is minimum thus input impedance is maximum. Thus the option (D) is correct 6. At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n-channel MOSFET is (A) 450 cm 2 /V-s (B) 1350 cm 2 /V-s (C) 1800 cm 2 /V-s (D) 3600 cm 2 /V-s Soln.(6) Option (B) is correct 7. Thin gate oxide in a CMOS process in preferably grown using (A) wet oxidation (C) epitaxial deposition (B) dry oxidation (D) ion implantation Soln.(7) Dry oxidation alone can provide uniform film with good dielectric properties for thin gate oxide in a CMOS Option (B) is correct

5 8. In the silicon BJT circuit shown below, assume that the emitter area of transistor Q 1 is half that of transistor Q 2. The value of current I 0 is approximately (A) 0.5 ma (C) 9.3 ma (B) 2mA (D) 15mA Soln. (8) + i1 - R=9.3KΩ I 0 Q1 + Q2 β 1 = V β 2 = V KVL to circuit with Q = 0 = = = The emitter area of Q 1 is half that of transfer Q 2 So I i = I 0 /2 Therefore I 0 = 2mA So option (B) is correct

6 9. The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and the effect of output resistance r 0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is TRUE? (A) The input resistance R i increases and the magnitude of voltage gain A V decreases (B) The input resistance R i decreases and the magnitude of voltage gain A V decreases (C) Both input resistance R i and the magnitude of voltage gain A V decrease (D) Both input resistance R i and the magnitude of voltage gain A V increase Soln. (9) By removing C E Input impedance increases by (1+ R E ) Voltage gain decreases by the same factor Option (A) is correct 10. Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is (A) -R 2 /R 1 (B) -R 3 /R 1 (C) - (R 2 R 3 )/R 1 (D) - (R 2 +R 3 )/R 1 Soln. (10) Since the OP-AMP is ideal, voltage at point A will be zero. The above circuit can be simplified to

7 R 1 R2 A - V 0 + R 3 DC V i writing the current eqn. = = = The option (C) is correct 11. Match the logic gates in Column A with their equivalents in Column B. (A) P 2, Q-4, R-1, S-3 (B) P-4, Q-2, R-1, S-3 (C) P 2, Q-4, R-3, S-1 (D) P-4, Q-2, R-3, S-1 Soln. (11) P gate = + = So P matches with 4 (P-4) Q gate = = + Matches with (Q-2) R gate Exor ie. Now see gate 3 = + = + = + ie. is EXOR gate R matches with 3 Now gate 1

8 + = + ie. Ex Nor ie. S So S-1 Option D is correct 12. For the output F to be 1 in the logic circuit shown, the input combination should be (A) A = 1, B= 1. C = 0 (B) A = 1, B= 0,C = 0 (C) A = 0, B= 1. C = 0 (D) A = 0, B= 0, C = 1 Soln. (12) Best way is to try the various inputs (A) A=1, B=1,C=0 A=1, B=1 output of gate 1 is 0 output of gate 2 is 1 c=0 output of gate 3 is 0 with these inputs F = 0 similarly try options B & C both give F = 0 only option D ie A = 0, B = 0, C = 1 Gate 1 output 0 Gate 2 output 1 C is 1 F = 1 Thus proved Option (D) is correct 13. In the circuit shown, the device connected to Y 5 can have address in the range

9 (A) FF (C) 2E00 2EFF (B) 2D00 2DFF (D) FD00 FDFF Soln. (13) To connect to Y 5, inputs CBA in 101 A 15 A 14 A 13 A 12 A 11 A 10 A 9 A 8 A 7 A 6 A 5 A 4 A 3 A 2 A 1 A D 0 0 (Min.) (Max.) 2 D F F Option (B) is correct 14. Consider the z-transform X(z) = 5z 2 + 4z ; 0< z <. The inverse z-transform x[n] is (A) 5δ[n + 2] + 3δ[n] + 4δ[n 1] (B) 5δ[n - 2] + 3δ[n] + 4δ[n + 1] (C) 5 u[n + 2] + 3 u[n] + 4 u[n 1] (D) 5 u[n - 2] + 3 u[n] + 4 u[n + 1] Soln. (14) We know that + X(z)= 5z 2 + 4z < < x(n) = Option (A) is correct 15. Two discrete time systems with impulse responses h 1 [n] = δ[n -1] and h 2 [n] = δ[n 2] are connected in cascade. The overall impulse response of the cascaded system is (A) δ[n - 1] + δ[n - 2] (B) δ[n - 4] (C) δ[n - 3] (D) δ[n - 1] δ[n - 2] Soln. (15) We know h = = h = 2 = Overall impulse response of cascaded systems in Z- domain:

10 H(z) = H 1 (z).h 2 (z) = z -1. z -2 = z -3 Overall impulse response in discrete time domain is h = Option (C) is correct 16. For an N-point FFT algorithm with N = 2 m which one of the following statements is TRUE? (A) It is not possible to construct a signal flow graph with both input and output in normal order (B) The number of butterflies in the mth stage is N/m (C) In-place computation requires storage of only 2N node data (D) Computation of a butterfly requires only one complex multiplication Soln. (16) Computation of a Butterfly requires only one complex multiplication for N point Fast Fourier Transform (FFT) with N=2 m. It requires only one complex multiplication to compute a Butterfly Option (D) is correct 17. The transfer function Y(s)/R(s) of the system shown is (A) 0 (B) 1/(s+1) (C) 2/(s+1) (D) 2/(s+3) Soln. (17) P(s) From the above diagram.

11 P (s) = R (s) Q (s) And So So = = 0 P(s) = R(s) -0 = R(s) = = therefore = Option (B) is correct 18. A system with transfer function Y(s)/X(s) = s/(s+p) has an output y(t) = cos(2t-π/3) for the input signal x(t) = p cos(2t - π/2) Then, the system parameter p is (A) 3 (B) 2/ 3 (C) 1 (D) 3/2 Soln. (18) From the given input and output signals, we can find the phase difference between the output and input signals i.e. = = = 0 and Given ω=2rad/sec. (from the given equation) = = = 0 Which is also 30 0 = 0 = 0 = = Option (B) is correct 19. For the asymptotic Bode magnitude plot shown below, the system transfer function can be (A) (10s+1)/(0.1s+1) (B) (100s+1)/(0.1s+1)

12 (C) (100s)/(10s+1) (D) (0.1s+1)/(10s+1) Soln. (19) At ω=0.1, slope changes from odb to +20dB Factor + So, system transfer function T(s) = At ω=0.1, magnitude =0dB 20log k =0 thus T(s) = Option (A) is correct = or, k=1 20. Suppose that the modulating signal is m(t) = 2cos (2πf m t) and the carrier signal is x c (t) = A c cos(2πf c t), which one of the following is a conventional AM signal without over-modulation? (A) x(t) = A cm (t) cos(2πf c t) (B) x(t) = A c [1 + m(t)]cos(2πf c t) (C) x(t) = A c cos(2πf c t) + Ac/4 m(t) cos(2πf c t) (D) x(t) = A c cos(2πf m t) cos(2πf c t) + Ac sin(2πf m t) sin(2πf c t) Soln. (20) The conventional AM signal is given as x(t) = A c [1+m(t)] cos 2πf c t = A c cos (2πf c t) + A c m (t) cos 2πf c t Let us check option (B) Modulation index =2/1=2 (since modulating signal has amplitude 2) So, it is AM signal with over modulation. Now check option (c) X(t)=A c cos(2 f c t) +A c /4 m(t)cos(2 f c t) Hence Modulation Index =2.A c /4 =1/2 Thus, it is AM signal without over modulation Option (C) is correct 21. Consider an angle modulated signal x(t) = 6cos[2πx10 6 t+2sin(8000πt) + 4cos(8000πt)] V. The average power of x(t) is : (A) 10W (B) 18W (C) 20W (D) 28W

13 Soln. (21) Angle modulated signal as given is x(t) = 6 cos [2π sin (8000πt)+4 cos (8000πt) V. Here ω c = 2π x 10 6 ω m = 800 π so the wave is x(t) = 6 cos [ω c t + 2 sinω m t+4 cosω m t] Average power = ( )2 =( )2 =36/2=18w Option (B) is correct 22. If the scattering matrix [S] of a two port network is then the network is (A) lossless and reciprocal (C) not lossless but reciprocal (B) lossless but not reciprocal (D) neither lossless nor reciprocal Soln. (22) The given matrix is 02 0 = 0 0 Since, 0 0 =cos0+jsin0= =cos90 0 +jsin90 0 For reciprocal network S ij = S ji i.e. S 12 = S 21 In the present problem S 12 =S 21 = j 0.9 So the network is reciprocal. Now it could be checked for the condition of lossless. For the losses network s-matrix has to be unitary. It gives rise to two conditions = Or =

14 Applying condition (1) to = 0 2 i.e. + i.e. (0.2) 2 + (j 0.9) 2 or, so the network is not lossless Option (C) is correct 23. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω /m. if the line is distortion less, the attenuation constant (in Np/m) is (A) 500 (B) 5 (C) (D) Soln. (23) For distortion less line attenuation constant and velocity of propagation should be independence of frequency. Given Z 0 = 50Ω R = 0.1 Ω/m Since the line is distortion less it should satisfy the following: LG = CR i.e. = Characteristic impedance = = and Attenuation constant = = Or α = For the present case α = =0.002 Np/m Option (D) is correct 24. Consider the pulse shape s(t) as shown. The impulse response h(t) of the filter matched to this pulse is

15 Soln. (24) Given pulse shape S(t) is as shown

16 Impulse response of the matched filter is h(t) =s(t-t) s(t+t) is left side shifted version of s(t) by T. s(t-t) is the mirror image of s(t+t) on the Y-axis. Option (C) is correct 25. The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m 2 ) is: (A) 1/30π (B) 1/60π (C) 1/120π (D) 1/240π Soln. (25) Given = V/m ϵ r = 4 Magnitude of the time average power density

17 P av = Where = = ϵ = = = ϵ = = W/m2 Option C is correct 26. If ey = X 1/x, then y has a (A) maximum at x= e (B) minimum at x= e (C) maximum at x= e -1 (D) minimum at x= e -1 Soln. (26) Given: = Taking log, = = = Differentiating = + = + = 0 rr, + = 0 = 0 = i.e. = = = = For = = =

18 Since Hence maximum at = Option (A) is correct 27. A fair coin is tossed independently four times. The probability of the event the number of time heads shown up is more than the number of times tails shown up is : [2 marks] (A) 1/16 (B) 1/8 (C) ¼ (D) 5/16 Soln. (27) There are only five events when the number of times heads show up is more than the number of times tails show up are the following: H H H H H H H T H H T H H T H H T H H H Since the fair coin in tossed independently four times the probability of each event is = All the event are mutually exclusive so the total probability = = Option (D) is correct 28. If A = xy âx+ x2 ây then c A.dl over the path shown in the figure is (A) 0 (B) 2/ 3 (C) 1 (D) 2 3

19 Soln. (28) Given = + h hhh = + = + = + From :- P Q =, = 0 = = 2 = = Q R = = 0 = 2 = = 8 R S = = 0

20 = = 2 = 2 = 2 S P = = 0 = = = 2 = = + = Option C is correct 29. The residues of a complex function X(z) = (1-12z)/[z(z-1)(z-2)] at its poles are (A) 1/2, -1/2 and 1 (B) 1/2, -1/2 and -1 (C) 1/2, -1 and -3/2 (D) 1/2, -1 and 3/2 Soln. (29) Given function = To find residue at its poles Residue at z = a of G(Z)= lim Poles are at 0, 1 & 2 For Pole at 0 Residue: = lim = For Pole at z = 1

21 Residue: = lim = Pole at z =2 Option C is correct = lim = 30. Consider differential equation dy(x)/dx - y(x) = x with the initial condition y(0) = 0. Using Euler s first order method with a step size of 0.1, the value of y (0.3) is (A) 0.01 (B) (C) (D) Given f(t) = L -1 [(3s+1)/(s 3 +4s 2 +(k-3)s)]. If Limx f(t) = 1, then the value of k is : (A) 1 (B) 2 (C) 3 (D) 4 Soln. (31) lim = lim Given that = Since lim = Thus, lim = lim = = = Option (D) is correct = = 32. In the circuit shown, the switch S is open for a long time and is closed at t=0. The current i(t) for t 0+ is

22 (A) i(t)= e -1000t A (C) i(t)= e -1000t A (B) i(t)= e -1000t A (D) i(t)=0.375e -1000t A Soln. (32) At t = 0 -, the circuit is shown below 10Ω 1.5A 10Ω 10Ω S 0.75A 0.75A The current of 1.5A source will divide equally in both10ω resistors At t = Ω S 1.5A 10Ω 15mH 10Ω Current in inductor will not change instantaneously, so the current of 0.75 ma which was flowing through first 10Ω resister will get divided to this branch and the switch branch 0 = 0 2 = 0 = 0 The current will divide into three 10Ω branches (inductor is short circuited) = 0 Where R is equivalent resistor seen across L with current source opened.

23 10Ω 10Ω 10Ω R R = 10 + (10 10) R = = 15 Ω = = 0 02 Option A is the correct 33. The current I in the circuit shown is (A) -j1a (B) J1A (C) 0A (D) 20A Soln. (33) The given circuit is redrawn jωl A ω= 10 3 rad/sec. - 1Ω 1 jωc (L = 20mH, C = 50µH) At node A + + = 0

24 = + + = = + + = = = = Option A is correct 34. In the circuit shown, the power supplied by the voltage source is 100W (A) 0W (B) 5W (C) 10W (D) Soln. (34) The given circuit is redrawn i+ i i+2 Since we have to find the power supplied by the voltage source. So if we can find the current drawn from the source, power supplied can be evaluated. So the loop considered is as shown in the figure. Let the current drawn be I Amp. Current in different branches is as per the figure. Applying KVL to the given loop

25 10 = (i+3)(1+1) + (i+2)(1+1) 10 = 2i i + 4 4i = 0 i = 0 Hence power drawn from the supply P = V. i = = 0 Watts Option (A) is correct 35. In a uniformly doped BJT, assume that N E, N B and N C are the emitter, base and collector dopings in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following conditions is TRUE? (A) N E =N B =N C (B) N E >>N B and >N B >N C (C) N E =N B and N B <N C (D) N E <N B <N C Soln. (35) For proper transistor action the lightly doped base between the heavily doped emitter and moderately doped collector. Thus option B is the most nearest option i.e. N E >>N B (Heavily doped emitter as compared to base) N B >N C (it should be N B < N C ) Option B is correct 36. Compared to a p-n junction with N A =N D =10 14 /cm 3, which one of the following statements is TRUE for a p-n junction with N A =N D =10 20 /cm 3? (A) Reverse breakdown voltage is lower and depletion capacitance is lower (B) Reverse breakdown voltage is higher and depletion capacitance is lower (C) Reverse breakdown voltage is lower and depletion capacitance is higher (D) Reverse breakdown voltage is higher and depletion capacitance is higher Soln. (36) In the given problem the doping concentration has been increased significantly. With increase in doping the breakdown voltage decreases. With more of doping depletion width (d) decreases. Depletion capacitance is given by C = εa/d, so it increases Option C is correct 37. Assuming that flip-flops are in reset condition initially, the count sequence observed at Q A in the circuit shown is

26 (A) (B) (C) (D) Soln. (37) See the given diagram Given all flip flops are in reset condition i.e. the outputs are 0 initially, = = = 0 = = = = 0 Since connected together = = 0 After one clock pulse: = (content of D A transfers to Q A ) = 0 = Now = = = = = = 0 After two clock pulses = = = 0 = = 0 = = = After three clock pulse = 0 = = = = = = 0 = =

27 So the sequence observed at = 0000 Option D is correct 38. The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP- AMP and practical diodes) Soln. (38) The above circuit can be redrawn as follows:

28 R +20V 4R D 2 V 1 R A - + D 1 V0 We choose the input voltage from one of the option say V 1 = -10 V Point A is at the ground potential = + = + 0 = Now at V 1 = -5 V = + or V 0 = -5+5 = 0 V For V 1 > -5 V, both diodes are conducting Thus option (B) is correct So, V 0 = The Boolean function realized by the logic circuit shown is (A) F=Σm(0,1,3,5,9,10,14) (C) F=Σm(1,2,4,5,11,14,15) (B) F=Σm(2,3,5,7,8,12,13) (D) F=Σm(2,3,5,7,8,9,12) Soln. (39) The given logic circuit having 4 X 1 M U X

29 = = = = = = = h = = = = Option (D) is correct 40. For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is (A) 00H (B) 45H (C) 67H (D) E7H Soln. (40) Contents of various registers during the execution of the programs line by line are: 3000 MVI A, 45 H (Move 45 H to Acc) (A) MOV B,A (Move the content of Acc. To B) (B) STC (Set carry flag ) i.e. CS = CMC (complement the carry) CS = RAR (Rotates the content of Acc. To right with carry) CS A 7 A A 1 A i.e

30 3006 XRA B (Content of B is to be Ex-ored with Acc.) (B) (A) A B (67H) Option (C) is correct 41. A continuous time LTI system is described by d 2 y(t)/dt 2 + 4dy(t)/dt + 3y(t) = 2dx(t)/dt + 4x(t) Assuming zero initial conditions, the response y(t) of the above system for the input x(t)=e - 2t u(t) is given by (A) (e t -e 3t )u(t) (C) (e -t +e -3t )u(t) (D) (e t +e 3t )u(t) (B) (e -t -e -3t )u(t) Soln. (41) A CTLTI system is described by Since the system is continuous time LTI system we take Laplace transform on both sides, assuming zero initial condition = 2 + = = Given that input is = = = = = Taking inverse Laplace transform on both sides = Option B is correct

31 42. The transfer function of a discrete time LTI system is given by = Consider the following statements: S1: The system is stable and causal for ROC: z >½ S2: The system is stable but not causal for ROC: z <¼ S3: The system is neither stable nor causal for ROC: ¼< z <½ Which one of the following statements is valid? (A) Both S1 and S2 are true (C) Both S1 and S3 are true (B) Both S2 and S3 are true (D) S1, S2 and S3 are all true Soln. (42) = = = 2^ + ^ ROC due to first term. > > 2 nd term given rise to > > Intersection of these ROCs < < For ROC > the system is stable and causal for ROC (composite ROC) i.e. < < In this case ROC is not exterior of > so not causal ROC does not include the unit circle so not stable. Option (C) is correct

32 43. The Nyquist sampling rate for the signal s(t) = sin(500πt)/πt x sin(700πt)/πt is given by (A) 400 Hz (C) 1200Hz (B) 600 Hz (D) 1400 Hz Soln. (43) Given Using the intently so, = 2sinA sinb = cos(a-b) cos(a+b) cos200 cos200 here = 200 = = 00 So, = 2 = 2 00 = 200 Option C is correct 44. A unity negative feedback closed loop system has a plant with the transfer function G(s) = 1/(s 2 +2s+2) and a controller Gc(s) in the feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is (A) Gc(s) = (s+1)/(s+2) (C) Gc(s) = [(s+1)(s+4)]/[(s+2)(s+3)] (B) Gc(s) = (s+2)/(s+1) (D) Gc(s) = 1 + 2/s + 3s Soln. (44) Block diagram of the system is as follows = =

33 Since input is unity Then = Given = im = = = 0 Steady state error = lim = = lim = lim = For e ss to be minimum G c (s) should be maximum. In option (D) = + + (since = ) Option (D) is correct 45. X(t) is a stationary process with the power spectral density Sx(f)>0 for all f. The process is passed through a system shown below. Let Sy(f) be the power spectral density of Y(t). Which one of the following statements is correct? (A) Sy(f)>0 for all f (B) Sy(f)=0 for f >1kHz (C) Sy(f)=0 for f=nfo, fo=2khz, n any integer (D) Sy(f)=0 for f=(2n+1)fo=1khz, n any integer Soln. (45) From the given system diagram = + Taking Fourier on either side = +

34 = = + ^ = ^ ^2 = ^ = 2 ^2 + 2 = = = 0 = = 0 = 2 + where f 0 = 1 KHz and n an integer Option D is correct 46. A plane wave having the electric field component Ei = 24cos(3x10 8 t-βy)âz V/m and traveling in free space is incident normally on a lossless medium with µ=µo and ε=9ε 0 which occupies the region y 0. The reflected magnetic field component is given by (A) [cos(3x10 8 t+y)â x ] /10π A/m (C) -[cos(3x10 8 t+y)â x ] /20π A/m (B) [cos(3x10 8 t+y)â x ] /20π A/m (D) -[cos(3x10 8 t+y)â x ] /10π A/m Soln. (46) Given = 2 cos 0 Above given Electric field component is in Z direction and the wave is travelling in y-direction. We can write the magnetic field component which is normal to y i.e. = = ^8 _ = cos 0 = = = = = = 2 4 = 1 2 Thus, = cos 0 + This indicates that the reflected wave is travelling in y direction Option A is correct

35 47. In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration (VSWR) on the 60W line is (A) 1.00 (B) 1.64 (C) 2.50 (D) 3.00 Soln. (47) The circuit is redrawn short z 0 =30Ω λ/8 z 1 z 0 =60Ω z 0 =302 z L =30Ω z 2 λ/4 The expression for the input impedance of the line is = Input impedance due to shorted λ/8 stub is = 0 = 0 Input impedance due to Load = 0 = 02 = 02

36 = 0 Now load impedance is = + = Magnitude of reflection coefficient = = = 1 = = = 17 = = Option B is correct = COMMON DATA QUESTIONS: 48 & 49 Consider the common emitter amplifier shown below with the following circuit parameters: β=100, g m = A/V, r 0 =, r p =259 W, R S =1kW, R B =93KW, R C =250 W, R L =1kW, C 1 = and C 2 =4.7mF. 48. The resistance seen by the source Vs is (A) 258 Ω (C) 93 KΩ (B) 1258 Ω (D) Soln. (48) Equivalent model for the given circuit is

37 h = = = 2 Resistance seen by the source V s = + h = = = 28 Option B is correct. 49. The lower cut-off frequency due to C 2 is (A) 33.9 Hz (C) 13.6 Hz (B) 27.1 Hz (D) 16.9 Hz Soln. (49) Lower cutoff frequency due to C 2 = = = 2 Option B is correct

38 COMMON DATA QUESTIONS: 50 & 51 The signal flow graph of a system is shown below. 50. The state variable representation of the system can be Soln. (50) Soln. (50) From signal flow graph = 2 = = + 0 = 0 + = 00 = + = + 2 = Thus, Option (D) is correct

39 51. The transfer function of the system is (A) (s+1)/(s 2 +1) (B) (s-1)/(s 2 +1) (C) (s+1)/(s 2 +s+1) (D) (s-1)/(s 2 +s+1) Soln.(51) = = Where P K is the gain of the forward path between U(s) and Y(s) is the determinant of the graph = + h hh+ = + + = hhh h = 2 0 = = 2 0 = = = = = =1 2 =2 = = Option (C) is correct

40 Linked Answer Question: Q. 52 to Q. 53 Carry Two Marks Each Statement of Linked Answer Questions: 52 & 53 The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T=300K, electronic charge=1.6x10-19c, thermal voltage=26mv and electron mobility = 1350cm2/V-s 52. The magnitude of the electric field at x=0.5 µm is (A) 1kV/cm (B) 5kV/cm (C) 10 kv/cm (D) 26kV/cm Soln. (52) For the given silicon sample the electric field will be given = 1 = = 10 6 V/m 10 6 = 0 V/cm This electric field will remain the same throughout the given sample. Option C is correct. 53. The magnitude of the electron drift current density at x=0.5 µm is (A) 2.16x10 4 A/cm 2 (B) 1.08x10 4 A/cm 2 (C) 4.32x10 3 A/cm 2 (D) 6.48x10 2 A/cm 2 Soln. (53) Electronic drift current in given by = = = A/cm 2 Option A is correct

41 Statement of Linked Answer Questions: 54 & 55 Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk). Yk=Nk if transmitted bit bk=0 Yk=a+Nk if transmitted bit bk=1 Where Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5αe-α n (This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6v. 54. The value of the parameter α (in V-1) is (A) (B) 10 7 (C) x10-10 (D) 2x10-20 Soln. (54) Given: Power spectral density In time domain = = 0 W/H z = Power = Variance = Which is equal to power since mean is zero = 0 Option (B) is correct

42 55. The probability of bit error is (A) 0.5xe -3.5 (B) 0.5xe -5 (C) 0.5xe -7 (D) 0.5xe -10 Soln. (55) Probability of Bit error = = 0 = 0 Option (D) is correct 0 Question No One Mark Each 56. Which of the following options is the closest in meaning to the world below: Circuitous (A) Cyclic (C) Confusing (B) Indirect (D) Crooked Soln. (56) Circuitous means deviating from a straight course Indirect (A) Cyclic : Recurring in cycle (B) Indirect : Not leading by a straight line (C) Confusing : Lacking Clarity (D) Crooked : Irregular shape Option (B) is correct 57. The question below consists of a pair of related of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed: Worker (A) fallow : land (C) wit : jester (B) unaware : sleeper (D) renovated : house Soln. (57) Unemployed worker Here one is opposite to other (A) Fallow : land Undeveloped Land Appropriate choice. (B) Unaware : steeper, both mean the same

43 (C) Wit : Jester to make jokes and jester is a joker (D) Renovated : Homes House can be renovated. Option (A) is correct 58. Choose the most appropriate word from the options given below to complete the following sentence: If we manage to our natural resources, we would leave a better planet for our children. (A) uphold (C) Cherish (B) restrain (D) conserve Soln. (58) (A) Uphold : cause to remain not appropriate (B) Restrain : Keep under control not appropriate (C) Cherish : Be fond of not related (D) Conserve : Keep in safety and project from harm decay, Loss or destruction : Most appropriate Option (D) is correct 59. Choose the most appropriate word from the options given below to complete the following sentence: His rather casual remarks on politics his lack of seriousness about the subject. (A) masked (C) betrayed (B) belied (D) suppressed Soln.(59) (A) Masked : Hide under a false appearance opposite. (B) Belied : Be in contradiction with Not appropriation (C) Betrayed : Reveal unintentionally most appropriate (D) Suppressed: To put down by force or authority irrelevant Option (C) is correct

44 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is: (A) 2 (B) 17 (C) 13 (D) 3 Soln. (60) Using set theory. n(a) : Number of people who play hockey =15 n(b) : Number of people who play football =17 n(a B) : persons who play both hockey and football = 10 n(a B) = persons who play either hockey or football or both. Formula = + = + 0 = 22 Thus the people who play neither hockey nor football = = 3 Other method Hence, neither hockey nor football = 3 players. Option (D) is correct Question No Carry Two Mark Each 61. Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause. which of the following statements best sums up the meaning of the above passage: (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in warfare would be undesirable. (D) People in military establishments like to use chemical agents in war. Soln. (61) (A) Modern warfare has resulted in civil strife : Not appropriate (B) chemical agents are useful in modern war fare : Not appropriate

45 (C) (D) Use of chemical agents as war fare would be undesirable : Not appropriate or wrong. people in military establishments like to use chemical agents in war : correct choice. Option (D) is correct 62. If =435 how much is ? (A) 534 (B) 1403 (C) 1623 (D) 1513 Soln. (62) Given =435 From the problem it is obvious that the addition is not in base 10, since the numbers add up to different total in base 10 (which we use in daily life) Intution: Consider units place and consider the total of 7+6 (base 10), which is 13 (base 10). Since we have 5 in units place, the intuition would be that this would be base 8. Lets check the sum for base 8 and the full sum matches., i.e (137) 8 +(276) 8 = (435) 8 Now solve the addition in base 8 (731) 8 +(672) 8 = (1623) = + = 2h + = h So option (C) is correct skilled workers can build a wall in 20 days; 8 semi-skilled worker can build a wall in 25days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semiskilled and 5 unskilled workers, how long will it take to build the wall? (A) 20 days (C) 16 days (B) 18 days (D) 15 days Soln. (63) Per day work (Rate of 5 skilled workers) = 1 20 So, Per day work of one skilled worker

46 = Per day work of rate of 8 semi-skilled workers = 1 25 Per day work of rate of one semi-skilled worker 1 = = Per-day work of 10 skilled workers = 1 30 Per-day work of one semi-skilled worker 1 = = Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers = = = = 1 15 Thus time to complete the work is 15 days Option (D) is correct 64. Given digits 2,2,3,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000 can be formed? (A) 50 (B) 51 (C) 52 (D) 54 Soln. (64) Given digits are 2, 2, 3, 3, 4, 4, 4, 4 Numbers greater than 3000, will have either 3 or 4 in the thousands place. We can break this into two parts, i.e. either thousands digit is 3 or it is 4.

47 Ist case First digit = 3 Rest 3 digits can be combination of 234, or 3! = 6 223, or 3!/2! = 3 224, or 3!/2! = 3 332, or 3!/2! = or 3!/2! = or 3!/2! = or 3!/2! = or 1 = 1 So total combination = 25 II Case First digit = 4 Rest 3 digit may be combination of 234 or or or or or or or or or 1 Total combination = 26 Required answer = 25+26=51 Option (B) is correct

48 65. Hari (H), Gita (G), Irfan (I) and Saira (S) are sibiligs (i.e. brothers and sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than3 years. Given the following facts: i. Hair s age + Gita s age > Irfan s age + Saira s age. ii. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. iii. There are not twins. In what order were they born (Oldest first)? (A) HSIG (C) IGSH (B) SGHI (D) IHSG Soln. (65) As given H+G> I +S I H< G S (1) And G S = 1 (2) From equations. (1) and (2), we get I H < 1 H > I So order is SGHI Option B is correct