Solved Papaers - 2007

8/3/2019 Solved Papaers - 2007

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SAIMEDHA-HYD,Blore,TPT,VJA S Academy Page 1

GATE 2007

1. If E denotes expectation, the variance of a random variable X is given by(a)E (b) E (c) E (d)

2. The following plot shows a function y which varies linearly with x. the valueof the integral I = . is

(a)1.0 (b) 2.5 (c) 4.0 (d) 5.03. For || 1, coth(X) can be approximated as

(a)X (b) (c) (d) 4.

lim

is:

(a)0.5 (b) 1 (c) 2 (d) not defined5. Which one of the following functions is strictly bounded?

(a) (b) (c) (d) 6. For the function , the linear approximation around x = 2 is:

(a)3 (b) 1-x (c) 3 22 1 2 (d) 7. An independent voltage source in series with an impedance

delivers a maximum average power to a load impedance when(a)

(b)

(c)

(d)

8. The RC circuit shown in the figure is

(a) A low pass filter (b) A high pass filter(c) A band pass filter (d) A band reject filter

1

1

2

2

3

3x

y

-1

R C

CR

+ +

- -


2/40


9. The electron and hole concentrations in an intrinsic semiconductor are per

at 300 K. Now, if acceptor impurities are introduced with a concentration

of

per

(where

), the electron concentration per

at 300 K will

be:

(a) (b) (c) (d) 10.In a junction diode under reverse bias, the magnitude of the electrical

field is maximum at

(a)The edge of the depletion region on the p-side(b)The edge of the depletion region on the n-side(c)The junction(d)The centre of the depletion region on the n-side

11.The correct full wave rectifier circuit is:

12.In a trans conductance amplifier, it is desirable to have(a)A large input resistance and a large output resistance(b)A large input resistance and a small output resistance(c)

A small input resistance and a large output resistance(d)A small input resistance and a small output resistance

13.X = 01110 and Y = 11001 are two 5-bit binary numbers represented in twoscomplement format. The sum of X & Y represented in twos complement

format using 6-bits is:

(a)100111 (b) 001000 (c) 000111 (d) 10100114.The Boolean function Y = AB + CD is to be realized using only 2-input NAND

gates. The minimum number of gates required is:

(a)2 (b) 3 (c) 4 (d) 5

(b) +

-

+

+

-Output

Input

(a)

+

+

-Output

Input

+

-Output

Input

-

+

+

-Output

Input

(c)

(d)


3/40


15.If the closed-loop transfer function of a control system is given asT(s) =

,then it is

(a) An unstable system (b) An uncontrollable system(c) A minimum phase system (d) A non-minimum phase system

16.If the Laplace transform of a signal y(t) us Y(s) = , then its final value is:(a)-1 (b) 0 (c) 1 (d) unbounded

17.If R() is the autocorrelation function of a real, wide-sense stationary randomprocess, then which of the following is NOT true?

(a)R() = R() (b) |R | R0(c) R() = - R() (d) The mean square value of the process is R(0)

18.If S(f) is the power spectral density of a real, wide-sense stationary randomprocess, then which of the following is ALWAYS true?

(a)S(0) S(f) (b) S(f) 0 (c) S(-f) = -S(f) (d) 0 19.A plane wave of wavelength is travelling in a direction making angle 30

with positive x axis and 90 with positive y-axis. The field of the planewave can be represented as ( is constant)(a) . (b) . (c)

. (d)

.

20.If C is a closed curve enclosing a surface S, then the magnetic field intensity , the current density and the electric flux density are related by(a) . . (b) . . (c) . . (d) . .

21.It is given that , , are M non-zero, orthogonal vectors. Thedimension of the vector space spanned by the 2M vectors, , , , , is:(a)2M (b) M+1(c) M (d) dependent of the choice of, , 22.Consider the function f(x) = 2. The maximum value of f(x) in theclosed interval [-4, 4] is:

(a)18 (b) 10 (c) -2.25 (d) Indeterminate23.An examination consists of two papers, paper 1 and paper 2. The probability

of failing in paper 1 is 0.3 and that in paper 2 is 0.2. Given that a student has

failed in paper 2, the probability of failing in paper 1 is 0.6. The probability of

a student failing in both the papers is:

(a)0.5 (b) 0.18 (c) 0.12 (d) 0.06


4/40


24.The solution of the differential equation under the boundaryconditions (i) y =

at x = 0 and (ii) y =

at x =

, where k,

and

are

constants, is(a)y = .exp (b) y = .exp (c) y = .sinh (d) y = .exp

25.The equation 4 4 0 is to be solved using the Newton Raphsonmethod. If x = 2 is taken as the initial approximation of the solution, then the

next approximation using this method will be:

(a) (b) (c) 1 (d) 26.Three functions f1(t), f2(t) and f3(t), which are zero outside the interval [0, T],

are shown in the figure. Which of the following statements are correct?

(a) and are orthogonal (b) and are orthogonal(c) and are orthogonal (d) and are orthonormal

27.If the semi-circular contour D of radius 2 is shown in the figure, then thevalue of the integral . is :

t

T

2

-2

T/3

2T/3

t

t

12

1

1

-3


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(a) (b) j (c) (d) 28.Two series resonant filters are shown as in Figure. Let the 3-dB bandwidth of

the Filter 1 be B1 and that of Filter 2 be B2. The value of is

(a)4 (b) 1 (c) (d) 29.For the circuit shown in the Figure, the Thevinin voltage and resistance

looking into X-Y are

(a)4/3 V, 2 (b) 4 V, 2/3 (c) 4/3 V, 2/3 (d) 4 V, 2 30.In the circuit shown, is zero volts at t = 0 sec. For t > 0, the capacitor

current , where t is in seconds, is given by

00

2

D

jw

j2

-j2

Filter 1

4 /4

Filter 2

2A

X

Y

i

2i+

-

1

1 2


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(a)0.50 exp(-25t)mA (b) 0.25 exp(-25t)mA(c) 0.50 exp(-12.5t)mA (d) 0.25 exp(-6.25t)mA

31.In the AC network shown in figure, the phasor, voltage

(in Volts) is

(a)0 (b) 530

(c) 12.530 (d) 173032.A junction has built-in potential of 0.8 V. The depletion layer width at a

reverse bias of 1.2 V is 2. For a reverse bias of 7.2 V, the depletion layerwidth will be

(a)4 (b) 4.9 (c) 8 (d) 12 33.Group I lists four types of p-n junction diodes. Match each device in Group I

with one of the options in Group II to indicate the bias condition of the device

in its normal mode of operation.

(a)P-1, Q-2, R-1, S-2 (b) P-2, Q-1, R-1, S-2(c) P-1, Q-2, R-1, S-2 (d) P-2, Q-1, R-2, S-2

Group I Group II

P. Zener Diode 1. Forward Bias

Q. Solar Cell 2. Reverse Bias

R. LASER Diode

S. Avalanche Photodiode

5 5

j 3-j 3

A

B

530 A

20 K

20 K10 V 4


7/40


34.The DC current gain () of a BJT is 50. Assuming that the emitter injectionefficiency is 0.995, the base transport factor is

(a)0.980 (b) 0.985 (c) 0.990 (d) 0.99535.Group I lists four different semiconductor device. Match each device in Group

I with its characteristic property in Group II.

(a)P-3, Q-1, R-4, S-2 (b) P-2, Q-1, R-3, S-2(c) P-3, Q-4, R-1, S-2 (d) P-3, Q-2, R-1, S-4

36.For the Op-Amp circuit shown in the figure, is

(a)-2 V (b) -1 V (c) -0.5 V (d) 0.5 V37.For the BJT circuit shown, assume that the of the transistor is very large

and 0.7 . The mode of operation of the BJT is

(a)Cut-off (b) saturation (c) normal active (d) reverse active

Group I Group II

P. BJT 1. Population inversion

Q. MOS Capacitor 2. Pinch-off voltage

R. LASER diode 3. Early Effect

S. JFET 4. Flat-band Voltage

1K

1K

2 K

1K

1V +

-

1 K

10 K

10 V

2 V

+

+-

-


8/40


38.In the Op-Amp circuit shown assume that the diode current follows theequation, I =

exp /

. For

2,

, and for

4,

. The

relationship between the

and

is

(a) 2 (b) (c)

2(d)

2

39.In the CMOS inverter circuit shown, if the transconductance parameters ofthe NMOS transistor are 40/ and theirthreshold voltages are 1, the current I is

(a)0 A (b) 25 A (c) 45 A (d) 90A40.For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the

knee current is negligible and the Zener dynamic resistance is 10 . If the

input voltage ( ) range from 10 to 16 V, the output voltage () range from

(a)7.00 to 7.29 V (b) 7.14 to 7.29 V (c) 7.14 to 7.43 V (d) 7.29 to 7.43 V

2 K

+

-

5 V

2.5 V

I

PMOS

NMOS

200


9/40


41.The Boolean expression Y = can be minimizedto

(a)Y =

(b) Y =

(c) Y = (d) Y = 42.The circuit diagram of a standard TTL NOT gate is shown in the figure.

When Vi = 2.5 V, the modes of operation of the transistor will be

(a): reverse active; : normal active; : saturation; : cut-off(b): reverse active; : saturation; : saturation; : cut-off(c): normal active; : cut-off; : cut-off; : saturation(d): saturation; : saturation; : saturation; : normal active

43.In the following circuit, X is given by

(a)X = (b) X = (c) X = AB+BC+AC (d) X =

4 K 1.4 K 100

1 K

=5V

0 0

1 1

1 1

0

X

0

A B C

4-to-1MUX

4-to-1MUX

YY


10/40


44.The following binary values were applied to the X and Y inputs of the NANDlatch shown in the figure in the sequence indicated below: X = 0, Y = 1; X = 0,

Y = 0; X = 1, Y = 1; The corresponding stable P, Q outputs will be

(a)P = 1, Q = 0; P = 1, Q = 0, P = 1, Q = 0 or P = 0, Q = 1(b)P = 1, Q = 0; P = 0, Q = 1, P = 0, Q = 1 or P = 0, Q = 1(c)P = 1, Q = 0; P = 1, Q = 1, P = 1, Q = 0 or P = 0, Q = 1(d)P = 1, Q = 0; P = 1, Q = 0, P = 1, Q = 1

45.For the circuit shown in, the counter state (, ) follows the sequence

(a)00, 01, 10, 11, 00 (b) 00, 01, 10, 00, 01(c) 00, 01, 11, 00, 01 (d) 00, 10, 11, 00, 10

46.An 8255 chip is interfaced to an 8085 microprocessor system as an I/Omapped I/O as shown in the figure. The address line A0 and A1 of the 8085

are used by the 8255 chip to decode internally its three ports and the control

register. The address lines A3 to A7 as well as the IO/ signal are used foraddress decoding. The range of addresses for which of the 8255 chip would

get selected is

P

Q

X

Y

>>

Clock


11/40


(a)F8H FBH (b) F8H FCH (c) F8H FFH (d) F0H F7H47.The 3 dB bandwidth of the low pass signal ., where u(t) is the unit

step function, is given by,

(a) (b) 2 1 (c) (d) 1 Hz48.A Hilbert transformer is a

(a)Non-linear system (b) Non-causal system(c) Time-varying system (d) Low-pass system

49.The frequency response of a linear, time-invariant system is given byH(f) =

. The step response of the system is(a)51 . (b) 5 1 .(c) 1 . (d) 1

.50.A 5-point sequence x[n] is given as x[-3] = 1, x[-2] = 1, x[-1] = 0, x[0] = 5, x[1]

= 1. Let X() denote the dicrete-time Fourier transform of x[n]. The value of . is(a)5 (b) 10 (c) 16 (d) 5+j10

51.The z-transform X[z] of a sequence x[n] is given by X[z] = .. It is giventhat the region of convergence of X[z] includes the unit circle. the value of x[0]

is

(a)-0.5 (b) 0 (c) 0.25 (d) 0.552.A control system with a PD controller is shown in the figure. If the velocity

error constant Kv = 1000 and the damping ratio = 0.5, then the value of KP

and KD are

8255

10 /

+

-

100 10


12/40


(a)KP = 100, KD = 0.09 (b) KP = 100, KD = 0.9(c) KP = 10, KD = 0.09 (d) KP = 10, KD = 0.9

53.The transfer function of a plant is T(s) = . The second-orderapproximation of T(s) using dominant pole concept is

(a) (b) (c) (d) 54.The open-loop transfer function of a plant is given as G(s) = . If the plant

is operated in a unity feedback configuration, then the lead compensator that

can stabilize this control system is

(a) (b) (c) (d) (b)55.A unity feedback control system has open-loop transfer function G(s) =. The gain K for which is s = -1+j1 will lie on the root locus of this

system is

(a)4 (b) 5.5 (c) 6.5 (d) 1056.The asymptotic Bode plot of a transfer function is as shown in the figure. The

transfer function G(s) corresponding to this Bode plot is

(a) (b) (c) (d) .57.The state space representation of a separately excited DC servo motor

dynamics is given as 1 11 10 010 ; where is the speed of

the motor is, is the armature current and u is the armature voltage. Thetransfer function

of the motor is

101 20 100

0

20

40

60

-20 dB/ decade

-40 dB/ decade

-60 dB/ decade

||


13/40


(a) (b) (c) (d) 58.

In delta modulation, the slope overload distortion can be reduced by(a)Decreasing the step size (b) Decreasing the granual noise(c) Decreasing the sample rate (d) Increasing the step

59.The raised cosine pulse p(t) is used for zero ISI in digital communications.The expression for p(t) with unity roll-off factor is given by p(t) =

The value of p(t) at t =

is(a)-0.5 (b) 0 (c) 0.5 (d)

60.In the following scheme, if the spectrum M(f) of m(t) is as shown, thenspectrum Y(f) of y(t) will be

61.During transmission over a certain binary communication channel, bit errorsoccur independently with probability p. The probability of AT MOST one bit

in error in a block of n bits is given by

(a) (b) 1- (c) np 1 1 (d) 1- 1 62.In a GSM system, 8 channels can co- exist in 200 kHz bandwidth using

TDMA, A GSM based cellular operator is allocated 5 MHz bandwidth.

0 B-B

f

M(f)m(t)

y(t)

Cos(2

Sin (2

Hilbert

Transform

Y(f)

B-B 0

f(a)

-B-2B B 2B0

f

Y(f)

(b)

-B B0f

Y(f)

(c) -B-2B B 2B f

Y(f)

(d) 0


14/40


Assuming a frequency reuse factor of, i.e. a five-cell repeat pattern, the

maximum number of simultaneous channels that can exist in one cell is

(a)200 (b) 40 (c) 25 (d) 563.In a Direct sequence CDMA system the chip rate is 1.2288 x 10 chips for

second. If the processing gain is desired to be AT LEAST 100, the data rate

(a)Must be less than or equal to 12.288 x 10 bits per sec(b)Must be greater 12.288 x 103 bits per sec(c)Must be exactly equal to 12.288 x 103 bits per sec(d)Can take any value less than 122.88 x 103 bits per sec

64.An air-filled rectangular waveguide has inner dimensions of 3 cm x 2 cm. thewave impedance of the TE20 mode of propagation in the waveguide at a

frequency of 30 GHz is (free space impedance

377

(a)308 (b) 355 (c) 400 (d) 461 65.The field (in A/m) of a plane wave propagating in free space is given by cos sin . The time average power flowdensity in Watts is:

(a) (b) (c) 50 (d) 66.The field in a rectangular waveguide of inner dimensions a x b is given by . Where H0 is constant, a and b are the

dimensions along the x-axis and the y axis respectively. The mode ofpropagation in the waveguide is:

(a)TE20 (b) TM11 (c) TM20 (d) TE1067.A load of 50 is connected in shunt in a 2-wire transmission line of Z0 = 50

as shown in the figure. The 2-port scattering parameter matrix (S-matrix) of

the shunt element is

(a) (b) 0 11 0 (c)

(d)

68.The parallel branches of a 2-wire transmission line are terminated in 100 and 200 resistor as shown in the figure. The characteristic impedance of

50

=50

=50


15/40


the line is Z0 = 50 and each section has a length of. The voltage reflection

co-efficient at the input is:

(a) (b) (c) (d) 69.A dipole is kept horizontally at a height of above a perfectly conducting

infinite ground plate. The radiation pattern in the plane of the dipole

( plane) looks approximately as

70.A right circularly polarized (RCP) plane wave is incident at an angle of 600 tothe normal, on an air-dielectric interface. If the reflected wave is linearly

polarized , the relative dielectric constant is:

100

200

50

50

50

44

4

z

y

(a)z

y

(b)

z

y

(c)z

y

(d)


16/40


(a)2

(b)

3(c) 2 (d) 3

Common data for questions 71, 72, 73 :

The figure shows the high-frequency capacitance-voltage (C-V) characteristics

of a Metal / / silicon (MOS) capacitor having an area of 1 x 10-4 cm2,Assume that the permittivities () of silicon and Sio2 are 1 x 10-12 F/cm and3.5 x 10-13 F/cm respectively.

71.The gate oxide thickness in the MOS capacitor is:(a)50 nm (b) 143 nm (c) 350nm (d) 1 m

72.The maximum depletion layer width in silicon is(a)0.143

m (b) 0.857

m (c) 1

m (d) 1.143

m

73.Consider the following statements about the C-V characteristics plot:S1 : The MOS capacitor has n type substrate

S2 : If positive charges are introduced in the oxide, the C-V plot will shift to

the left

Then which of the following is true?

(a)Both S1 and S2 are true (b) S1 is true and S2 is false(c) S1 is false and S2 is true (d) Both S1 and S2 are false

C

0

V

1pF

7 pF

RCP

dielectric

Linearly

Polarized

60 60 1

air


17/40


Common data for Questions 74 & 75:

Two 4-ray signal constellations are shown. It is given that

and

constitute an ortho normal basis for the two constellations. Assume that the

four symbols in both the constellations are equi-probable. Let denote thepower spectral density of white Gaussian noise.

74.The ratio of the average energy of constellation 1 to the average energy ofconstellation 2 is:

(a)4a2 (b) 4 (c) 2 (d) 875.If these constellations are used for digital communications over an AWGN

channel, then which of the following statements is true?

(a)Probability of symbol error for constellation 1 is over(b)Probability of symbol error for constellation 1 is higher(c)Probability of symbol error is equal for both the constellation(d)The value of N0 will determinate which of the two constellations has a

lower probability of symbol error.

Linked Answer Questions : Q.76 to Q 85 Carry Two Marks Each

Statement for Linked Answer Questions 76 & 77:

Consider the Op-Amp circuit shown in the Figure.

R

C

+

-

0

2

2

22

Constellation 1

a

a

-a

-a

0

Constellation 2


18/40


76.The transfer function/

is:

(a) (b) (c) (d) 77.If Vi = V1 sin() and V0 = V2 sin( ), then the minimum and maximum

values of (in radians) are respectively(a)- (b) 0 and (c) and 0 (d) - and 0Statement for Linked Answer Questions 78 & 79:

An 8085 assembly language program is given below:

Line 1 : MVI A, B5H

2 : MVI B, 0EH

3 : XRI 69H4 : ADD B

5 : ANI 9BH

6 : CPI 9FH

7 : STA 3010H

8 : HLT

78.The contents of the accumulator just after execution of the ADD instructionin the line 4 will be

(a)C3H (b) EAH (c) DCH (d) 69H79.After execution of line 7 of the program, the status of the CY and Z flags willbe

(a)CY = 0, Z = 0 (b) CY = 0, Z = 1 (c) CY = 1, Z = 0 (d) CY = 1, Z = 1Statement for Linked Answer Questions 80 & 81:

Consider a linear system whose state space representation is x(t) = Ax(t). If

the initial state vector of the system is x(0) = 12, then the system responseis x(t) = 2. If the initial state vector of the system changes to x(0) =

1

1, then the system response becomes x(t) =

80.The eigen value and eigenvector pairs , for the system are(a)1, 11 2, 12 (b) 2, 11 1, 12(c) 1, 11 2, 12 (d) 2, 11 1, 12

81.The system matrix A is:(a) 0 11 1 (b) 1 1 1 2 (c) 2 1 1 1 (d) 0 1 2 3


19/40



An input to a 6-level quantizer has the probability density function f(x) asshown in the figure. Decision boundaries of the quantizer are chosen so as to

maximize the entropy of the quantizer output. It is given that 3 consecutive

decision boundaries are -1, 0 and 1.

82.The value of a and b are:(a)a = and b = (b) a = and b = (c) a =

and b = (d) a = and b = 83.assuming the reconstruction levels of the quantizer are the mid-points of the

decision boundaries, the ratio of the signal power to quantization noise power

is:

(a) (b) (c) (d) 28


In the Digital-to-Analog converter circuit shown in the figure below,

VR = 10V and R = 10k.

R R R 2R

2R 2R 2R 2R

R

+

-

i

f(x)

a

b

X

51-1-5 0


20/40


84.The current i is:(a)31.25

A (b) 62.5

A (c) 125

A (d) 250

A

85.The voltage V0 is:(a)-0.781 V (b) -1.562 V (c) -3.125 V (d) -6.250 V

ANSWERS

1-a 2-b 3-c 4-a 5-d 6-a 7-d 8-c 9-d 10-c

11-c 12-a 13-c 14-b 15-d 16-d 17-c 18-b 19-a 20-d

21-c 22-a 23-c 24-d 25-b 26-b 27-a 28-d 29-d 30-a31-d 32-a 33-b 34-b 35-c 36-c 37-b 38-d 39-c 40-c

41-d 42-b 43-a 44-c 45-b 46-c 47-a 48-b 49-b 50-b

51-b 52-b 53-d 54-c 55-d 56-d 57-a 58-d 59-c 60-a

61-c 62-b 63-a 64-c 65-d 66-a 67-b 68-d 69-b 70-d

71-a 72-b 73-c 74-b 75-b 76-a 77-c 78-b 79-c 80-a

81-d 82-a 83-d 84-b 85-c

EXPLANATIONS

1. (a) Var (X) = E 2 2. (b) The equation of the straight line : y = x+1

The value of integral I = 1 2.5 3. (c) [2, p 776] coth (x) = ! ! 4. (a) lim lim . 0.5 [Using L Hospital Rule]5. (d) For strictly bounded function f(x), condition is 0 . Only

will satisfy the criterion.


21/40


6. (a) According to Taylor series, f(x) = f(a) + ! ! ,Now putting f(x) =

and a = 2, it can be derived as

f(x) = ! 2 ! 2 1 2 ! 2 3 7. (d) For maximum average power transfer to the load, load impedance must be 8. (c)

Using Laplace transform of the given network, it can be derived as . Now putting RC = 1 (assumption),

. Hence it has two dominant poles with 3-dB Bandwidth

5.

Hence it is Bandpass Filter.

9. (d) [3, p 27] In intrinsic semiconductor, hole concentration p must be equalto the electron concentration n, so that n = p = ni, where ni is called intrinsic

concentration. When acceptor impurity NAis introduced, semiconductor

becomes P type. The total positive charge density becomes NA+ np and total

negative charge density is np, where suffix P shows P-type semiconductor.

Since semiconductor is electrically neutral, the magnitude of the negative

charge density must equal that of the positive charge density or NA+ np = pp.As in P-type semiconductor NA>> ni, pp

NA. Now according to Mass-action

Law, np = nppp = or np = . Hence np = = .10.(c) [3, p 50, 51, 64] According to Poissons equation i.e. the field

intensity curve is proportional to the integral of the charge density curve.

Usually there is transition of polarity in charge density at p+ n (or normal p

n) junction. Hence integration of such curve provides maximum value at the

p+ n junction.

11.(c) [10, p 74] Figure (c) is the correct choice where total path is complete

R C

CR

+ +

- -


22/40


Type of Neg Feedback Input Resistance Output Resistance

Voltage Series

Or,

Series Shunt

Or,

Voltage Amplifier

Increases Decreases

Current Series

Or,

Series Series

Or,

TransconductanceAmplifier

Increases Increases

Current Shunt

Or,

Shunt Series

Or,

Current Amplifier

Decreases Increases

Voltage Shunt

Or,

Shunt Shunt

Or,

TransresistanceAmplifier

Decreases Decreases


23/40


12.through diodes and load impedance both when supply is of positive andnegative polarity.

13.(a) Trans conductance amplifier means Current-Series amplifier. Hence inputand output resistance both must increases. [3, p 420][6, p 676]

14.(c) In 2 complement format to convert N bit to (N+M) bit digit, sign bit of N-bit digit should be extended by M times. Hence 01110 + 11001 = 00111 or

000111. In otherway, (01110 + 11001) is equivalent to (001110 + 111001),

hence sum is 000111.

15.(b) The minimum number of gates required is 3. Applying Demorgans Law, itcan be derived

16.(d) Closed loop transfer function is given by T(s) =

Here all poles are at left side of s-plane. Hence it is stable system.[5, p 604] if the input transfer function of a linear time-invariant system

has pole-zero concellation, the system will be either not state-controllable or

unobservable, depending on how state variables are defined. If the transfer

function does not have pole-zero concellation, the system can always be

represented completely controllable and observable state model. Thus here

the system with given transfer function is controllable system.

Here one zero s = 5 is at right half of the s-plane. Hence it is non-minimum

phase system

Definition of Non-minimum phase system is as follows: [5, p 349]

A transfer function which has one or more zeros in the right half s-plane is

known as non-minimum phase transfer function.

17.(d) Final value Theorem states that when no poles are on imaginary axis oron the right hand side of the s-plane, lim lim Hence in the given Y(s) =

is having one pole is on the right half s-plane18.(c) [9, p 36] The auto-correlation function is an even function of i.e. .

AB

CD

Y

is equivalent to

=.


24/40


19.(b) [9, p 47] The power spectral density of a stationary process is alwaysnonnegative i.e.

0

20.(a) (a) As plane wave is travelling in a direction making an angle 30 withpositive x-axis and 60 with positive z-axis and 90 with positive y axis, the field equation is given by . ...... which canbe simplified as . .. .. .

21.(d) One of the Maxwell Laws (Amperes Law) states that whereH being magnetic flux intensity, J being current density and D being electric

flux density. Applying Stokes theorem [14, p 246], . . over this Maxwells equation, it becomes

. . . Hence the

solution.

22.(c) There are M non-zero orthogonal vectors in the dimension of the vectorspace spanned by the 2M vectors as and , and , and so on are notorthogonal. Hence there are M dimension required to represent them.

23.(a) Given function f(x) = x2 x 2. To find out maximum /minimum point inthe closed interval, it is necessary to find out where its derivable is zero and

its double derivative is negative /positive respectively. Hence 0 2

1 0.Hence x = 0.5. Now

x = 0.5 = 2 > 0 = positive. Hence there is one

minima in the interval [-4, 4], which implies f(x) is maxima at x = -4 or x =

+4. Hence|x=-4 = 18 or |x=4 = 1024.(c) [14, p 601] P(A) = Probability of failing in Paper 1 =0.3 and P(B) =

Probability of failing in Paper 1 = 0.2. It is given probability of failing in

paper 1 given that a student has failed in Paper 2 is P(A|B)=0.6. Hence usingtheorem, P(A | 0.2 0.6 0.12

25.(d) To solve these type of problem, back-calculate using the options provided.The boundary conditions (i) y = y1 at x = 0 and (ii) y = y2 at x =

, where k, y1

and y2 are constants, are satisfied by the two options i.e. (a) and (d). Now outof these two, options (d) y = (y1 y2).exp(-x / k) + y2 satisfies the equation .26.(b) Using Newton Raphson Method [14, p 650], . Hence with

x0 = 2 and f(x) = x3 x2 + 4x 4 = 0, it can be derived x1 =

27.(b) Two functions are orthogonal if 0 and two functions areorthonormal if

1

.


25/40


Case (a) : 2 1 0 1 2 2 2 0 4 0 Case (b) : 2 1 0 3 2 1 2 0 2 0 Case (c) : 1 1 1 3 2 1 1 3 2 0 Case (d) : 2 1 0 1 2 2 2 0 4 1

28.(a) According to [14, p 509], Cauchys integral formula : If f(s) is analyticwithin and on a closed curve D and if a is any point within D, then f(a) = . Here given and (1, 0) is inside the semi-circle D but not (-1, 0). Hence value of a can be a = 1. Thus

1 .29.(d) for series resonance B = . Hence .30.(d) The Thevenins theorem in a network with dependent source is as follows:

.For calculation of VTH :

Here taking two loops as shown above, it can be derived as 2i i1 (i1 i2)=0,

i = i1 i2 and (i1 i2) 2(i2 + 2) = 0 from which it is found as i2 = 0 and i1 = 4.

Hence VTH = VOPEN = 2 x 2 = 4V.

2A

X

Y

i

2i+

-

1

1 2

2A

X

Y

i

2i+

-

1

1 2


26/40


Form the network shown above, it can be stated that 2A current will pass

through only XY shorted terminal as i current through 1 ohm is zero,

dependent voltage source is also zero hence no effect of 2i voltage source.Hence iSHORT = 2A. Thus 2.

31.(a)

Using Thevenin Principle, which can be drawn as following:

The expression of current through capacitor is as

.

.

Hence 0.5 .32.(d) 530 5 35 3 530 3.40 1730 33.(a) For a p+ n junction, relation between Depletion Capacitance CT and

Depletion Width W and Depletion area A is given by CT = . If Vj is the

junction or barrier potential i.e. Vj = V0 Vd (where V0 contact potential and

Vd is the negative number for an applied reverse bias) then Depletion Width

W

hence

. Putting all the values,

.. .. 0.5, thus W2 = 4 .34. (b)

Group I Group II

P. Zener Diode 2. Reverse Bias

Q. Solar cell 1. Forward Bias

R. LASER Diode 1. Forward Bias

S. Avalanche Photodiode 2. Reverse Bias

20 K

20 K10/s 4

1/sC

10 K

5/s 4

1/sC


27/40


35.(b) [19, p 326] From the theory of Bipolar Junction Transistor, where

is the current transfer ratio,

is the DC current ratio and

is the

emitter injection efficiency. Here = 0.995. Hence B = 0.9853.36.(c)

37.(c) The Op-amp equivalent simplified diagram can be drawn as follows. Thetwo equations are (from the diagram):

Va = 1 0. Solving thersetwo equation, it can bederived 0.5 0.5 .

38.(b) From the figure, 2 0.7 . Hence 1.3 . Thus . 1.3 . Hence 10 10 10 10 1.3 3 .Thus in npn transistor and . Hence both the junction CB and BE are in forward biased. Somode of operation of the BJT is in Saturation.

39.(d) The simplified Op-Amp circuit is given as follows:

Group I Group II

P. BJT 3. Early effect

Q. MOS Capacitor 4. Flat-band voltage

R. LASER Diode 1. Population Inversion

S. JFET 2. pinch-off voltage

1K

1K

2 K

1K

1V

1 K

10 K

10 V

2 V

+

+-

-

3 1.3

1.3 2 n

np


28/40


In general, it can be derived as I = and 0-V = V0.When Vi = 2 V: I = 1 mA = IS exp When Vi = 4 V: I = 2 mA = IS exp Taking ratio of the derived two equations,

2.

40.(c) [6, p 426, 430, 429, 435] From the voltage transfer curve of CMOS, if 2.5 , 2.5 . It is given that 1 and 1

From figure, for NMOS, i.e. NMOS is in saturation and forPMOS, i.e. PMOS is in saturation. Hence both are insaturation. Hence, , , 2 . 5 1 , 2 . 5 1 , , 45 .

41.(c) The zener diode with its dynamic resistance can be equivalently deawn asfollows.

2 K

+ -

V

I 0

5 V

2.5 V I

PMOS

NMOS

D

D

S

S

G

G

2.5 V

200

7 V

10 V to

16 V

10


29/40


It can be derived as 7 10 . Hence when changes from 10 V to16 V,

changes from

7 10 7.143

to

7 10 7.429 42.(d) For solving this type of problem, it is better to use KMAP method.

43.(b) [20, p 201] : Reverse Active; : Saturation; : Saturation; : Cut-off44.(a) For any 4 : 1 MUX, Output Y can be expressed as

1

1

1

1

00

00

01

01

11

11

10

10

ABCD

Y =D+BC+AD+AB

1

1

1

1

00

00

01

01

11

11

10

10

ABCD

Y =D+B+AD

1

1

1

1

1

00

00

01

01

11

11

10

10

ABCD

Y =D+BC+AD

1

1

1

00

00

01

01

11

11

10

10

ABCD

Y =++AD1

1

1

00

00

01

01

11

11

10

10

ABCD

Y =++AB

1


30/40


Y = .From the figure shown above, it can be derived as Z = which can be simplified as X =

45.(c) In the NAND latch given, if one of inputs to NAND gate is 0 thencorresponding output will be 1 otherwise 0.

Hence when X = 0, Y = 0:

Hence when X = 1, Y = 1: As both the external inputs are 1, hence outputstate 0 or 2 depends on feedback path, initially Present P and Q state will be

forced to both 0. Assuming there is propagation delay mismatch in the

feedback path or NAND gate, which will force P = 1 and Q = 0 or P = 0 and Q

= 1.

0 0

1 1

1 1

0

X

0

A B C

4-to-1

MUX

4-to-1

MUXYY

Z

P=1

Q=0

X=0

Y=1

P=1

Q=1

X=0

Y=0

P=1

Q=0

X=1

Y=1


31/40


Or

46.(b) The characteristic equation of the counter can be written as follows:, , , , , .If starting value of counter is 00, then sequence will be 01, 10, 00, 01..

47.(c) A7 to A4 should be 0 for chip selection and A2 is dont care and A1 & A0have all possibilities.

A7 A6 A5 A4 A3 A2 A1 A0

1 1 1 1 1 0 0 0 =F8 H

1 1 1 1 1 0 0 1 =F9 H

1 1 1 1 1 0 1 0 =FA H

1 1 1 1 1 1 1 1 =FF H

48.(a) s(t) = for 3-dBBandwidth, . Hence 1 rad/s or .49.(b) A Hilbert transformer is a Linear, Non-Causal, Time-Invariant and All-

Pass System. The Hilbert transformer is non-causal as its impulse response

h(t) = 0 t < 0.

50.(b) The impulse frequency response of a LTI system is given by H(f) =

.The step response of the system

will be as Y(s) = H(s) X(s) = . 5 5 1 51.(b) [8, p 285] From the expression of DTFT of x[n] i.e.

X . and x[n] = . . Hence at n = 0,x[n=0]=

. or . 2. 0 10.

P=0

Q=1

X=1

Y=1


32/40


52.(b) [2, p 200] Taking inverse z-transform of X[z] = . with ROC || |2|which includes unit circle, it can be derived as x[n] = 0.5 x

2

1.

Hence x[0] = 0.5 x u[-1] = 0.

53.(b) It is given that KV = lim 1000 andlim lim 10 . Hence 10 x 1000 or 100. Now the characteristic equation 1 + G(s) = 0 or 1 + = 0or 10 100 10 0. Now comparing with standardcharacteristic equation i.e. 2 0, it can be derived as 100 2 10 100 or putting 100 & 0.5, 2 0.5 100 10 100 0.9.

54.(d) [5, p 213] [15, p 795] Using Dominant Pair of Complex-pole orDominant Pole Concept i.e. Poles whose real part is much less than the otherpoles or in other words, the poles which are located near to the imaginary

axis are more dominant in nature. As poles moves away from the imaginary

axis in the left half of s-plane, it becomes less and less significant. This

dominant pole tends to extend their effect on the transient response. Hence

the second order approximation of T(s) using dominant pole concept is as

follows: Putting s = 0, in the expression (s+5) i.e. T(s) = or

55.(c) For stability, no poles of the closed loop transfer should be on the righthalf of s-plane or no repeated roots should be on the imaginary axis.

Now the transfer function in option (a) and (b) is not Lead-Compensator asphase-angle is lagging type due to its negative values.

Now out of two options (c) & (d), option (c) is having closed loop transfer poles

in the left-hand side of the s-plane, which can be derived as follows:

In unity feedback transfer function, closed loop transfer function

T(s) = , Hence the

characteristic equation 10 9 10 0 is having roots s = -9.13, -0.433+0.95j, -0.433-0.95j. Hence the closed loop system is stable.

56.(d) [15, p 486] Using Magnitude Condition of Root-Locus, i.e.||| 1. Hence for unity feedback control system ,|| s=-1+j1 = ;Thus 1 1057.(d) From the Bode-Plot, the transfer function G(s) can be written as


33/40


G(s) = where K is the gain at = 0.1 rad/s. Hence

20

. . . 60

. 1000 2000.Hence transfer

function is G(s) = .58.(a) The Laplace transformation of the state-space representation of

separately excited DC servo motor dynamics 1 11 10 010 is

as follows:

s ; 10 10 .Which can besimplified as

10

10 10 1 10 or 59.(d) In delta modulation, the slope overload distortion can be reduced by either

increasing sampling rate or step size.

60.(c) Given expression p(t) = . The value of p(t) at t = can beobtained using LHospital Rule in Mathematics

i.e.lim lim lim

which can be simplified as

lim lim lim 0.561.(a) The output y(t) can be expressed as y(t) = m(t)cos(2 + sin(2

where is the Hilbert transform of m(t). Hence y(t) is LSB of Single-Side-Bnad Modulation with carrier frequency B Hz. Hence correct output is

62.(c) Probability of AT MOST one bit error in a block of n bits= Probability of one bit error + Probability of zero bit error

= n 1 n 1 1 1

B-B 0

f

Y(f)


34/40


63.(b) [C Y Lee] Total Band Width given is 5 MHz. 5 cell repeat pattern isfollowed and hence Band Width per cell =

= 1 MHz. Now using TDM, one

set of 8 channels can co-exist in 200 KHz. Thus in 1 MHz 5 sets of 8channels i.e. 40 simultaneus channels can co-exist.64.(a) [23, p 726] Processing Gain in CDMA equals to . Here it

is given 1.2288 10 / and it is required 100,hence

. / 100 or data rate 12.288 X 10 bits / sec.65.(c) The wave impedance of TEmn mode of propagation in the waveguide of

dimension [a X b] is where

here

10 and =

399.8666.(d) It is known , . Now it is given

cos or cos . Thus using ,

The expression for E Field can be derived as 5cos 53 . So the average Power Flow will be or

Y

Z

X

-Y

90

5

3

53 5


35/40


5 53 = 5. 53.

67.(a) [13, p 555] The x and y component of E field in a rectangle waveguide isas follows: and . Here as only y component exist andin y only sin part exists, then m = 1, n = 0. Hence the mode is TE20.

68.(b) [12, p 169] The 2-port scattering parameter matrix is given as

where b is the reflected / transmitted normalized power wave

and a is the incident normalized power wave and ; ; ; Hence here 0 as there is no reflection due to impedance matchedcondition and

1as there is full transmission.

69.(d) The individual sections are sections. Hence for section, .

Hence here 100 ohms impedance AB is 25 at EF and 200Ohms impedance CD is 12.5 at EF . Thus at EF effectiveimpedance is 25 12.5 . Now impedance in EF is

100

200

50 50

50

44

4

G

H

E

F

A

B

C

D


36/40


equivalently impedance 300 . Hence reflection co-efficient isgiven by =

70.(b) [17, p 370] For vertical antennas, the non-directional radiators would beconsidered to have the same phase, whereas for horizontal antennas the non

directional radiators would have opposite phase.

The total E field can be derived as 180or = 180 2 s in 180 Hence 1 ;= |1 2 sin 180 2 sin 180|= 1 2 sin 180 2 sin 180=

1 2sin 2sin 2

1 cos 2 sin now

0 0, 30 2, 90 0which shows (c) is the

correct plot.

71.(d) [17, p 147] Polarizing angle / Brewster angle: angle at which, there is noreflected wave when the incident wave is parallel (or vertically polarized). If

the incident wave is not entirely parallel polarized, there will be some

reflection, but the reflected wave will be entirely of perpendicular (or

horizontal polarization0. [ 13, p 455] The Brewster angle is also known as

the polarizing angle because an arbitrarily polarized incident wave will be

reflected with only the component of E perpendicular to the plane of

incidence.

for parallel polarization, where

is the angle

of incidence. Here tan 60 3 372.(a) [19, p 270] From figure, 7 7 10 and from data given, 3.5 10/ and MOS capacitor area A = 10 . Hence

d = . 0.5 1050

Real

Image

Ground 180

2 7 0

0

9 0


37/40


73.(b) [19, p 270] The maximum depletion layer which is given by where

can ne derived from

.Here it is given,

1 ,

semiconductor permitivitty 10/ and A = 10. Now puttingthe given value, it can be derived as and hence 0.857 .74.(c) Case S1 : MOS capacitor has an p-type substrate as shown in Fig 6-16 in

[19]

Case S2 : If positive charges are introduced in the oxide, they increase the

depletion layer and decrease the capacitance, so C-V curve will shift to the

left direction, because now small voltage is needed to decrease the

capacitance.

75.(b) The constellation diagram shows the amplitude and phase of themodulated signal. Hence the ratio of the energy of the two constellation

scheme is calculated as follows:

Ratio =

4

76.(b) The constellation 2 is having more orthogonal relations between thepoints. Hence probability of symbol error in Constellation 2 is less.

77.(a) The Laplace-transformed equivalent simplified Op-Amp circuit

0

2

2

22

Constellation 1

2a

a

-a

-a

0

Constellation 2


38/40


. From these two equations, it canbe derived as

78.(c) As , , 2 2 Thus | = or 0 and = 0 or .Hence minimum and maximum values of are respectively [0 and ] or [- 0].

79.(b) The 8085 program is as followsLine 1: MVI A, B5 H : A = B5 H

2: MVI B, 0E H : B = 0E H

3: XRI 69 H : A = A69H = B5H69H = DCH4: ADD B : A = A + B = DCH + 0EH = EAH

Hence answer is EAH.

80.(c) Continued from Line 4:Line 5: ANI 9B H : A = A9BH = EAH9BH = 8AH6: CPI 9F H : compare immediate with Accumulator and

as 9F H is greater that 8A H which is stored

in Accumulator, carry Flag is Set and Zero

Flag is reset.

7: STA 3010 H : Store Accumulator Direct (Copy content of

Accumulator to a memory location specified by the operated 3010 H). Here no

flags are affected.

Hence Carry Flag is Set and Zero Flag is reset.

81.(a) [5, p 588] The unforced response of system x(t) using state transitionmatrix (t) can be written as x(t) = 0 0. Usinggiven values of x(t) and x(0), (t) can be found out as 2 2 2 2 and hence

= 1

2 3

|

|

12 3

0 or = -1, -2

R

1/sC

(s)

(s)

(s)


39/40


Now eigen vector corresponding to = -1 is 1 12 1 3 0 or and eigen vector corresponding to = -2 is 2 12 1 3 0 or .Hence correct answer is 1 11 2, 12

82.(d) From the expression above | | 12 3 00 0 12 3 1 00 1 0 1 2 3. Hence A = 0 1 2 383.(a) From property of Probability Density Function, 1 which is

nothing but area under the PDF curve. Here area under the PDF curve is

4b+2a+4b which is equal to 1 hence 4b + a =

.

From symmetry and data given, the decision boundaries are at [-3, -1, 0, +1,

+3]. Hence output of the quantizer will be at [-4, -2, -0.2, +0.5, +2, +4]. Hence

entropy of the 6 level quantizer output will be H = 4 2 0.5 . 0.5 . 2 4 Now P(-4) = . 2 , P(-2) = . 2 P(-0.5) =

.

, P(0.5) =

.

,

P(2) = . 2 , P(+4) = . 2 . HenceH = 2 2 2 2 = -2 (4b2 + )Case (a) : , 2 4 0.7781Case (b) : , 2 4 0.7739Case (c) :

, 24 0.7526

Case (d) : , 24 0.6778Thus case (a) is having maximum entropy and hence (a) is the solution.

N = .. ..


40/40

84.(d) The noise power is given by = 4b | 2 |.. b aThe signal power is given by = N =

.

. .

= 2b | | b aHence signal-to-noise ratio is given by , 28

85.(b) Here Node Voltage Analysis is performed to find out I as well as potentialVAand VB is required to find out iAand iB for Q85. Given VR = 10V and R =

10K.

0 5 2 0 20; 0 2 5 2 0;

0 0 2 4 0

Solving simultaneous equations (Use Calculator fx-991 MS to solve theequations), it can be derived that VA= 5, VB = 2.5 and VC = 1.25.

Hence i = . 62.5

86.(c) From equivalent simplified Figure of Op-Amp based DAC, it can bederived as . 3.125

R R R2R

2R2R2R

2R

V = 0 V = 0

i

2R 2R R

V = 0

5 1.25

Solved Papaers - 2007

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