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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.

Phone: 0744-2429714 Mobile: 9001297111 , 9829567114, 9829597114, 9001297242 Website: www.vpmclasses.com E-Mail: [email protected] /[email protected] Address: 1-C-8, SFS, TALWANDI, KOTA, RAJASTHAN, 324005

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For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exams

Pattern of questions : MCQs

There are a total of 65 questions carrying 100 marks.

Questions (56-65) belongs to general aptitude (GA).Questions (56-60) will carry 1-mark each, andquestion (61-65) will carry 2-marks each

GATE - ELECTRONICS AND COMMUNICATION

1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714

Web Site www.vpmclasses.com [email protected]

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Total marks : 100

Duration of test : 3 Hours

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MOCK TEST PAPER

Questions (1-25) will carry 1-mark each and questions (26-55) will carry 2-marks each.

For Q.1-25 and Q.56-60 1/3 mark will be deductedfor each wrong answer.For Q.26-51 and Q. 61-65 2/3 mark will be deducted for each wrong answer.The question pairs (Q.52, Q.53) and (Q.54, Q.55) are linked questions.For Q.52 &54 2/3 mark will be deducted. There is no negative marking for Q.53 &Q.55.

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Q.48-51 are common data questions. If first question is attempted wrongly then answer of second question will not be evaluated.

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1 In the matrix equation Px = q, which of the following is a necessary condition for the existence of at

least one solution for the unknown vector x?

(A) Augmented matrix [Pq] must have the same rank as matrix P

(B) Vector q must have only non - zero elements

(C) Matrix P must be singular

(D) Matrix P must be square

2. An arbitrary vector X is an eigen vector of the matrix A =1 0 00 a 00 0 b

, if (a, b) =

(A) (0, 0)

(B) (1, 1)

(C) (0, 1)

(D) (1, 2)

3. The integration of logx.dx has the value

(A) (x log x – 1)

(B) log x – x

(C) x (log x – 1)

(D) None of these

4. If f(x) = | x |, then the interval [–1, 1], f(x) is

(A) Satisfied all the conditions of Rolle’s Theorem

(B) Satisfied all the conditions of Mean Value Theorem

(C) Does not satisfied the conditions of Mean Value Theorem

(D) None of these

5. Differential equation,

2

2d x dx10 25x 0dt dt

Will have a solution of the form

(A) (C1 + C2 t)e–5t

(B) C1 e–2t

(C) C1 e–5t + C2 e5t

(D) C1 e–5t + C2 e2t



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Where C1 and C2 are constants.

6. For the differential equation dy 5y 0dt

with y(0) = 1, then general solution is

(A) e5t

(B) e–5t

(C) 5e–5t

(D) 5te

7. For | z | = 1, where C is the circle, is

(A)

(B)

(C)

(D) None of these

8. If A and B are independent and P (C)= 0, then A, B and C are independent

(A) True

(B) False

(C) Both (a) and (b)

(D) None of these

9. Following are the value of a function

y(x): y(–1) = 5, y(0), y (1) = 8

dydx

at x = 0 as per Newton’s central difference scheme is

(A) 0

(B) 1.5

(C) 2.0

(D) 3.0

10. For any two events A and B

(A) P(B) = P(A B) + P( A B)

(B) P(A B) = P(A) +P(B) – P(A B)

(C) P(A/B) P(A).

(D) All of these



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11. The time constant of the network shown in the figure is

(A) 2 RC (B) 3 RC

(C) RC2

(D) 2 RC3

12. In the network shown in the figure, the effective resistance faced by the voltage source is

V

(A) 4 (B) 3 (C) 2 (D) 1 13. Two coils having equal resistance but different inductances are connected in series. The time

constant of the series combination is the (A) Sum of the time constants of the individual coils (B) Average of the time constants of individual coils (C) Geometric mean of the time constants of the individual coils (D) Product of the time constants of the individual coils

R



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14. When a unit impulse voltage is applied to an inductor of 1 H, the energy supplied by the source is (A) (B) 1 J

(C) 12

J

(D) 0. 15. In the circuit shown in figure, IB = 10 Micro-amperes, the value of resistance Rin is:

(A) 1 kilo-ohm (B) 10 kilo-ohm

(C) 100 kilo-ohm (D) 1 mega-ohm

16. The voltage transfer ratio V2/ V1 for the network shown in the figure is

(A) 113

(B) 213

(C) 513

(D) 413

17. If I(t) = 14

(1 – e–2t) u (t), where u (t) is a unit step voltage, then the complex frequencies associated

with i (t) would include

(A) s = 0 and s = j2

(B) s = j2 and s = –j2

(C) s = –j2 and s = –2

(D) s = 0 and s = –2

12

–12



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18. The circuit shown in Fig-I is replaced by that in Fig-II. If current I remains the same, then R0 will be

(A) Zero

(B) R

(C) 2R

(D) 4R

19. Thevenin equivalent of a network is shown in the given figure. For maximum power transfer to the variable and purely resistive load RI, its resistance should be

(A) 60

(B) 80

(C) 100

(D) Infinity

20. The admittance parameter Y12 in the 2-port network in the given figure is

(A) – 0.2 mho

(B) 0.1 mho

(i) (ii)



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(C) – 0.05 mho

(D) 0.05 mho

21. The capacitor in the circuit as shown above is initially charged to 12 V with S1 and S2 open. S1 is

closed at t = 0 while S2 is closed at t = 3.

Waveform of the capacitor correct is represented by

(A)

(B)

(C)

(D)



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22. z-matrix for the network shown in the given figure is

(A) 2s 1 2s

32s 2ss

(B) 2s 1 –2s

3–2s 2ss

(C) 2s 1 2s

3–2s 2ss

(D)

32s –2s2

32s 2ss

23. A sample of silicon at T = 300 K is a doped with boron at a concentration of 2.5 × 1013 cm–3 and with arsenic at a concentration of 1 × 1013 cm–3. The material is

(A) p – type with p0 = 1.5 × 1013 cm–3

(B) p – type with p0 = 1.5 × 107 cm–3

(C) n – type with n0 = 1.5 × 1013 cm–3

(D) n – type with n0 = 1.5 × 107 cm–3

24. The 6V zener diode shown in the figure, has zero zener resistance and a knee current of 5mA. The minimum value of R so that the voltage across it does not fall below 6 V is



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(A) 1.2 k ohms (B) 80 ohms (C) 50 ohms (D) 0 ohms

25. Analog multiplier X of given figure has the characteristics up = v1v2, output of this circuit is

+

10 kW

RX

vss

vs

v0

(A) vsvss

(B) s

ss

v–v

(C) –vsvss

(D) s

ss

vv

26. An n-channel JFET having a pinch-off voltage (VP) of –5V shows a transconductance (gm) of 1 mA/V, when applied gate-to-source voltage (VGS) is –3V. Its maximum transconductance (in mA/V) will be

(A) 1.5

(B) 2.0

(C) 2.5

(D) 3.0

27. For the circuit shown in the given figure, assuming ideal diodes, output waveform V0 will be



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(A)

(B)

(C)

(D)

28. Boolean expression for the output of the logic circuit shown in the figure is

(A) Y = AB + AB + C

(B) Y = A B + AB +C

(C) Y = AB + B A + C

(D) Y = AB +B +C

29. A full-adder can be implemented with half-address and OR gates. A 4-bit parallel full adder without any initial carry requires

(A) 8 half-adders, 4-OR gates (B) 8 half-adders, 3-OR gates (C) 7 half-adders, 4-OR gates (D) 7 half-adders, 3-OR gates



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30. If vector field

F = (xy + z3)ux + (3x2 – z)uy + (3xz2 – y)uz is ir-rotational, then value of , b and is

(A) = 6, = = 1

(B) = = 1, = 0

(C) = 0, = = 1

(D) = = = 0

31. For a quarter wavelength ideal transmission line of characteristic impedance 50 and load impedance 100 , the input impedance will be

(A) 25

(B) 50

(C) 100

(D) 150 32. Find Laplace transform of the function u|t| – u(t – 2)

(A) –2se – 1

s

(B) –2s1– e

s

(C) 2s

(D) –2s

33. For the network shown in figure below, what is the driving point impedance Z(s)?

1r–

Z 12 H



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(A) 2

2

s s 4s 4s

(B) 2

2

2s s 4s 4s

(C) 2

2

2s s 4s 2s

(D) None of these

34. The response of a network is,

i(t) = K t e–at for t 0,

Where is real positive.

The value of t at which the i(t) will become maximum, is

(A) α

(B) 2α

(C) 1

(D) α2

35. If L[f(t)] = 22(s 1) ,

s 2s 5

then f(0+) and f() are given by

(A) 0, 2 respectively

(B) 2, 0 respectively

(C) 0, 1 respectively

(D) 25

,0 respectively

36. The transfer function H(s) of a stable system is



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2

2s 5s – 9H(s)

(s 1)(s – 2s 10)

The impulse response is

(A) – e–t u(t) + (et sin 3t + 2et cos 3t) u(t)

(B) – e–t u(t) – (et sin 3t + 2et cos 3t) u(– t)

(C) – e–t u(t) – (et sin 3t + 2et cos 3t) u(t)

(D) – e–t u(t) + (et sin 3t + 2et cos 3t) u(– t)

37. In the given figure, spring constant is K, viscous friction coefficient is B, mass is M and the system output motions is x(t) corresponding to input force F (t). Which of the following parameters related to the above system is

1. Time constant = 1M

2. Damping coefficient = B2 KM

3. Natural frequency of oscillation = KM

Select the correct answer using the codes given below: (A) 1, 2 and 3 (B) 1 and 2 (C) 2 and 3 (D) 1 and 3 38. The response c(t) of a system to an input r(t) is given by the following differential equation :

2

2d c(t) dc(t)3 5c(t) 5 r(t).

dtdt

The transfer function of the system is given by

(A) G(s) = 25

s 3s 5

(B) G(s) = 21

s 3s 5

(C) G(s) = 23s

s 3s 5

(D) G(s) = 2s 2

s 3s 5



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39. A control system whose step response is + 0.5 (1+e–2t) is cascaded to another control block whose

impulse response is e–t. The transfer function of the cascaded combination is

(A)

1s 1 s 2

(B)

1s s 1

(C)

1s s 2

(D)

0.5s 1 s 2

40. The mean square value of the shot noise current (A) Varies inversely as average current (B) is independent of average current

(C) Varies as average current

(D) Varies directly as average current 41. The pdf of a random variable x is given by

xK 0 x 1

f (x)0, otherwise

Than value of the k will be- (A) 0 (B) + 1 (C) –1 (D) None of these

42. Power spectral density of stationary noise process N(f) has auto correlation for Rnm () = ke–3 | |

(A) 23k

3

(B) 23k

3 –



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(C) 26k

9

(D) 26k

9 –

43. When output z(t) is related to input y(t) as y = z and probability density function of the input

p(y) = e–y 0 < y <

p(y) = 0 – < y < 0 Probability function will be

(A) ze

2 z

(B) – ze

2 z

(C) e–z (D) ez 44. When spectral density of

fa ; f W

m(t)0; o otherwise

Output SNR at receiver will be

(A) c

0

a A2N

(B) c

0

a A3N

(C) 2

c

0

aA4N

(D) 2

c

0

aA3N



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45. White noise with power density on2

is applied to a low pass network for which | H(0) | = 2. It has a

noise bandwidth of 2 MHz. If average output noise power is 8.1 W in a 1W resistor, then n0 is

(A) 6.25 × 108 W/Hz

(B) 6.25 × 10–8 W/Hz

(C) 1.25 × 108 W/Hz

(D) 1.25 × 10–8 W/Hz

46. A line of length has characteristic impedance Z0. The line is cut into half. Value of impedance is:

(A) 0Z2

(B) 0Z4

(C) 2Z0

(D) Z0

47 Consider a low pass random process with a white noise power spectral density Sx() = N2

as shown

in the figure.

N2

–2 Bp 2 Bpw

The auto correlation function RX() is (A) 2NBsin(2) (B) NBsin(2) (C) NBsin(2) (D) None of these



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Common data Q.48 – 49 Given symmetrical lattice network shown in the figure below.

48. Matrix (Z) will be

(A)

1 2 2 1

2 1 1 2

(Z Z ) (Z – Z )2 2

(Z – Z ) (Z Z )2 2

(B)

1 2 2 1

2 1 1 2

(Z – Z ) (Z – Z )2 2

(Z – Z ) (Z Z )2 2

(C)

1 2 2 1

2 1 1 2

(Z Z ) (Z Z )2 2

(Z – Z ) (Z Z )2 2

(D) None of these

49. Matrix (A) will be

(A)

1 2 1 2

2 1 1 2

1 2

2 1 2 1

(Z Z ) 2Z Z(Z – Z ) (Z – Z )

(Z Z )2(Z – Z ) (Z – Z )

(B)

1 2 1 2

2 1 2 1

1 2

2 1 2 1

(Z Z ) Z Z(Z – Z ) (Z – Z )

(Z Z )2(Z – Z ) (Z – Z )



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(C)

1 2 1 2

2 1 2 1

1 2

2 1 2 1

(Z Z ) Z Z(Z – Z ) (Z – Z )

(Z Z )1(Z – Z ) (Z – Z )

(D) None of these

Common data Q. 50-51

Consider a p – channel enhancement mode MOSFET with k’p = 40 A/V2. The device has following observations:

ID = 0.225 mA at VSG = VSD = 3 V

ID = 1.4 mA at VSG = VSD = 4 V

50. Value of the threshold voltage VTP is

(A) 2.33 V

(B) – 2.33 V

(C) 3.29 V

(D) – 3.29 V

51. The ratio W/L is

(A) 48

(B) 96

(C) 25

(D) 50

Linked Answer Q. 52-53

Given the figure below



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52. Power delivered by sources (1) will be

(A) 2.5 W

(B) 1 W

(C) 5 W

(D) 10 W

53. Power dissipated by source (2) will be

(A) 0

(B) 5W

(C) –5W

(D) 10 W

Linked Answer Q.54-55

A non magnetic medium has an intrinsic impedance 360° 30°.

54. Loss tangent is

(A) 0.866

(B) 0.5

(C) 1.732

(D) 0.577

55. Dielectric constant is

(A) 1.634

(B) 1.234

(C) 0.936

(D) 0.548

56. No doubt, it was our own government but it was being run on borrowed ideas, using _________solutions.

(A) Worn out (B) Second hand (C) Impractical (D) Appropriate The question below consists of pair of related words followed by four pairs of words. Select the pair that Best expresses the relation in the original pair:



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57. Ratify: Approval: (A) Mutate: change (B) Pacify: conquest (C) Duel: combat (D) Appeal: authority

58. A car goes 35 km in 1 hour, next 270 km in 3 hrs. and next 80 km in 122

hrs. Find the average speed

of the car (A) 59.23 km/h. (B) 61.5 km/h (C) 80 km/h (D) None of these 59. Some critics believe that Satyajit Ray never quite came back to the great beginning he made in this

path breaking film Pather Panchali. ______have endured decades of well-traveled bad prints to

become a signpost in cinematic history.

(A) The bizarre history of its misty origins

(B) Its haunting images

(C) Its compelling munificence

(D) The breathtaking awe it inspires

Choose the most appropriate word from the options given below that is most nearly opposite in

meaning to the given word:

60. Valedictory

(A) sad

(B) Collegiate

(C) Derivative

(D) Generosity

Each of the 11 letters A, H, I M, O,T, U V, W, X and Z appears same when looked at in mirror. They

are called symmetric letters. Other letters in the alphabet are asymmetric letters.



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61. If the area of a given square ABCD is 3 find the total area of the entire figure?

(A) 452

(B) 45

(C) 48

(D) 31

62. The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the amino

terminal portions of both the H and L chains.

(A) H chain

(B) L chain

(C) Both (a) and (b)

(D) None of these

63. In a certain code Language

134 means good and tasty

478 means see good picture

729 means picture are faint

Which number has been used here for faint?

(A) 9

(B) 2

(C) Data are inadequate

(D) 253

64. A bag contains an equal number of one rupee, 50 paisa and 25 paisa coins. If the value of money in

the bag is Rs. 35, find the total number of coins of each type?

(A) 7

(B) 40

(C) 30

(D) 20



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Profit to sale-Table for three companies A, B and C for 1996-97

Companies 1996 1997

Total units 300000 400000

Shares A 5% 25%

B 60% 40%

C 35% 35%

Price A 10% 8%

(per unit) B 7% 14%

(in rupees) C 9% 10%

Profit A 3% 1%

(per unit) B 0.5 5%

(in rupees) C 2% 2.5 65. What is the increase in the total profits of company B in 1997? (A) 800% (B) 900% (C) 750% (D) 789%

ANSWER KEY

Que Ans. Que Ans. Que Ans. Que Ans. Que Ans.

1 A 16 B 31 A 46 D 61 B2 B 17 D 32 B 47 C 62 C3 C 18 D 33 B 48 A 63 C4 C 19 C 34 C 49 A 64 D5 A 20 C 35 B 50 B 65 D6 B 21 A 36 D 51 C7 A 22 A 37 C 52 A8 A 23 A 38 A 53 D9 B 24 B 39 C 54 C10 D 25 B 40 C 55 D11 D 26 C 41 B 56 B12 B 27 D 42 C 57 B13 B 28 B 43 B 58 A14 C 29 D 44 C 59 B15 C 30 A 45 D 60 D



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HINTS AND SOLUTION

1(A) According to Rouche’s theorem, the system is consistent if and only if the coefficient matrix and the

augmented matrix K are of the same rank, otherwise the system is inconsistent.

2.(B) Since the matrix is triangular, the eigen values are , a, b.

If (X1, X2, X3) is an arbitrary eigen vector, say corresponding to 1, then

1 1

2 2

3 3

1 0 0 x x0 a 0 x 1 x0 0 b x x

X2, X3 being not zero, we have, X1 = X1 ; a X2 = X2 which gives

a = 1

and bX = X3 which gives b = 1

(a, b) = (1, 1).

3.(C) dlogx.dx logx.x x. (logx)dxdx

= x log x – 1. dx

= x log x – x

= x (log x – 1)

4.(C) Since f(x) = | x | is continuous is [–1, 1] be it is not differentiable at x = 0 (–1, 1)

5.(A) 2

2d x dx10 25x ]0dt dt

(D2 + 10D + 25) X = 0

(D + 5)2 = 0

D = -5, – 5

Hence solution is, (C1 + C2 t)e–5t



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6.(B) Given: dy 5y 0dt

dy 5 dty

Integrating, we get

loge y = c – 5t

When t = 0, y = 1.

loge1 = c – 5 × 0

c = 0

logey = –5t

y = e–5t

7.(A) Poles of f(z) = 2z 3

z 2z 5

are given by

z2 + 2z + 5 = 0

z = 2 4i 1 2i2

Since, both poles lie outside the circle | z | = 1, therefore f(z) is analytic inside the circle

2z 3 dz 0

z 2z 5

8. (A) P(C) = 0

C =

P(A B C) = P (A B )

= P() = 0

P(A) × P(B) × P(C) = 0 .....Since P(C) = 0

P(A B C) = P (A) – P (B) – P (C)

Hence A, B, C are independent.



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9. (B) 2 1

2 1at x 0

dy y ydx x x

= y(1) y( 1) 8 5 1.51 ( 1) 2

10.(D) (a) B = (A B) ( A B)

P(B) = P(A B) + P( A B)

A B and A B are mutually exclusive.

P( A B) = P(B) – P (A B).

(B) P(A B) = P[A ( A B)]

= P(A) + P( A B)

= P(A) + P(B) – P(A B)

(C) P(A/B) = P(A) P (B/A)

P(A) as P(B/A) 1.

11.(D)

(R) (2R) 2Req RR 2R) 3

R

C

Time constant = (Req) (c)

= 2RC3

12.(B) Current through the 4-ohm resistor = i 3ii – .4 4

Therefore, the voltage drop across it

= 3i 4 3i4

which must equal V.

Thus effective resistance faced by the voltage source is 3



Page 26

13.(B) 1 2 1 2 1 2

1 2

L L L L L L1TR R 2R 2 R R

Thus the time constant is the average of the time constant of individual coils.

14.(C) Current that flows is given by

v|s| = sL (s) v|t| = | t|v|s| = 1

1 = s(1) .|s|

|s| = 1 v | t | 1 Amp.s

Energy supplied = 21 1 1L (1) (1) J2 2 2

15.(C) Let the current B1, flows through resistance Rin when –12 V battery is removed

B1in

12470 1000 R

Similarly current B2 flows in resistance Rin when +12V battery is removed

B2in

121000 1000 R

Since the two voltages are in opposite directions, their currents will subtract. Hence

B1 – B2 = 10 × 10–6

–6

in in

12 12– 10 10(470 1000 R ) (1000 1000 R )

12[(1000 × 1000 + Rin) – (470 × 1000 + Rin)] = 10 × 10–6 (470 × 1000 + Rin). (1000 × 1000 +

Rin)

12 × 53 × 104 = 10–5[470 × 109 + 47 × 104 Rin + 100 × 104 Rin + 2inR ]

636 × 104 × 105 = 2 4 9in inR 147 10 R 470 10

2inR + 1470 × 103 Rin – 16.6 × 108 = 0

Solving for Rin

3 3 2 8

in1470 10 (1470 10 ) 66.4 10

R –2

= 100,000 ohms(rejecting the negative value)

= 100 k.ohms



Page 27

16.(B) Assume current flowing as shown

1 amp

V1

V2

V2

i4

i3 i1 i2

VA

V2 = 4 × 1 = 4 V

21

V 4 2 Amp.2 2

= 2 + 1 = 3 A.

VA = V2 + 2 × 3

VA = 4 + 6 = 10 V

310 5 A2

4 = 3 + 2 = 8A

V1 = VA + 8 × 2 = 10 + 16 = 26 V.

2

1

V 4 2V 26 13

17.(D) i|t| = 2t1 (1 – e ) u | t |4

1 1 1 2| s | –4 s s 2 4s.(s 2)

So complex frequencies associated with i(t) would include s = 0 and s = –2

18.(D) For both the circuits remains the same.

e q

VR

V is also same as given only V. so for same current, Req should be same for circuit (i), all the

three resistors are in parallel,

Req = 32 R || 16 R || 8 R.



Page 28

eq32R R7

…

(1)

For circuit (ii)

Req = (4R|| 2R|| R) + R0

{ R0 is in series with parallel combination of 4R, 2R and R}

eq 04 2R 11R R4 2

eq 0

4 13R R4 13

eq 04R R R7

…(2)

Now (1) and (2) is equal

so 04 32R R R7 7

028R R 4.R7

19.(C) For maximum power transfer to the variable lead resistance RL, RL should have a value equal to the

magnitude of impedance of rest of the circuit

i.e. 2 2= (60) (80)

3600 6400

10000

= 100

20.(C) The equivalent Y – parameter circuit of given circuit is



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YA YC

YB

E = V1 1

Y11 = YA + YB

Y12 = –YB

Y21 = –YB

Y22 = YB + YC

Here A1Y 0.25

B1Y 0.05

20

C1Y 0.1

10

Y12 = –YB = –0.05

21.(A) At t = 3, since 2 is shorted, C jumps to a new value and then decreases with reduced time constant.

22(A) From the circuit

V1 = 1 + (1 + 2) 2s

= (2s + 1)1 + 2s.2

…(1)

And 22 1 2

3V ( )2s.s

2 1 23V 2s. 2s5

…(2)

Then,

2s 1 2sZ matrix 32s 2s

5



Page 30

23.(A) Since Na > Nd thus material is p – type

p0 = Na – Nd

p0 = 2.5 × 1013 – 1 × 1013

p0 = 1.5 × 1013 cm–3

24.(B) Output voltage is regulated to the zener voltage 6 V.

10.6 80 mA50

Zener diode current, zk = 5mA

This is the minimum current drawn by zener hence load current L will be maximum.

Lmax = 80 – 5 = 75 mA

–3

6R 80 ohms75 10

25.(B) v+ = 0 = v–

Let output of analog multiplier be vp

ps vvR R

vs = –vp

vp = vssv0

vs = –vssv0

s0

ss

vv –v

26.(C) Vp = –5V

Vas = –3V

gm = 1ma/V

gm0 = ?



Page 31

Transconductance for JFET is given as –

am m0

p

Vg g 1 –V

–1

am0 m

p

Vg g 1 –

V

–1

m0–3g 1 1 ––5

gm = 1(0.4)–1 = 10.4

mg 2.5 mA / V

27.(D) Voltage at OP – AMP output

V1 = –1–5 sin t

At 0t , V – 6 volts2

At 03t , V 4 volts2

During ‘–‘ve cycle of the waveform, when V1 < – 2V, diode D2 conducts and VD = 2V.

During ‘+’ve cycle, when V1 > 2V, D1 conducts and V0 = –2V

28.(B) X =

X AB AB

And Y X C (AB AB) C

Y (AB AB) C

Y AB AB C

X



Page 32

29.(D) The circuit of a 4-bit full adder using half adders and OR gates is shown in the figure.

HA

A3 B3

C D3

OR OR

HA HA HA

A2 A1 A0B2 B1 B0

C C CD2 D1 D0

HA HA HA

S4 S3 S2 S1 S0 From the figure, it is apparent that 4–bit full odder required seven half adders and 3 OR gates

30.(A)

x y z

3 2 2

v v v

× Fx y z

( xy z ) (3x yz) 3xz y

= (–1 + y)vx – (3z2 – 3z2)vy + (6x – x)vz.

If F is irrotational × F = 0.

–1 + y = 0 y = 1

3z2 – 3z2 = 0 = 1.

6x – x = 0 = 1 = 6

= 6, = y = 1

31.(A) Input impedance,

0in 0

0

Z jR tan( d)Z RR jZ tan( d)

and d2

4 2



Page 33

then, 2 20

inR 50Z 25Z 100

32.(B) f|t| – u|t| – u|t – 2|

Taking laplace transform

–2s 2s

2s1 1 e 1 – eF | s | – e L[v | t |] –s s s s

33.(B) Z(s) =

s21 2ss 22

= 2

2

1 2s 2s s 4s 4 s s 4s

34(C) i(t) = K.t e-at t 0.

at atdi(t) K[e – t . ae ]dt

atdi(t) Ke [1 at]dt

Now, di(t) 0dt

Ke–at(1 – at) = 0

1 – at = 0

1ta

Now

2

at at 2 atd i(t) K –ae – {ae – a t e }dt

= K ae-at[–1 –1 + at]

= K ae–at(at – 2)

Put t = 1a

getting



Page 34

12 aa

2

d i(t) 1K.ae a . – 2adt

= K.a.e–1[–1]

2

2

d i(t) ka–edt

2

2

d i(t)dt

is negative for t = 1a

i(t) will become maximum for t = 1a

35.(B) +

t 0 s 0f(0 ) = lim f | t | lim s.f(s)

2

+2s s 2

2

1s 2 12s.(s 1) sf(0 ) = lim lim

2 5s 2 5 s 1s s

f(0+) = 2.

so

t s 0

f( ) lim f | t | lim s.f(s)

2s 0

2s.(s 1)f( ) lim 0s 2s 5

f(0+)() = 2, 0

36.(D) Given

2

2

s 5s – 9H(s) =(s 1) (s – 2s 10)

2 2 2 2

–1 2(s 1) 3H(s) =(s 1) (s 1) 3 (s 1) 3

System is stable.

h|t| = –e–tv|t| + (2et cos 3t + et sin 3t) u(-t)



Page 35

37.(C) 2

2

d x(t) dx(t)F|t| = m B kx | t |dtdt

The characteristic equation is

ms2 + Bs + K = 0

or s2 + 2ns + 2n = 0

nk B,m 2 km

Thus statement 2 and 3 are correct.

38.(A) Given

2

2

d c(t) 3dc(t) 5c(t) 5r(t)dtdt

Taking laplace transform

s2c(s) + 3s c(s) + 5 c(s) = 5.R(s)

(s2 + 3s + 5) c(s) = 5 R(s)

2

c(s) 5 G | s |R(s) s 3s 5

39.(C) Given

h1(t) = +0.5(1 + e–2t)

H1(s) = + 1 10.5s s 2

1

2s 1H (s) 0.5

s(s 2)

and h2(t)= e–t

21H (s)

s 1

The transfer function of the cascaded combination is

H(s) = H1(s) . H2(s)



Page 36

2(s 1) 1H(s) 0.5s(s 2) (s 1)

1H(s)

s(s 2)

40.(C) The current in FET or BJT under dc conditions is a constant at every instant. However current from

the emitter to the collector consists of a stream of individual electrons or holes and it is only the time

averaged flow measured as constant current.

The fluctuation observed in the number of covers called that noise is given as

2n 2q dcB

Where, q = 1.6 × 10–19c

B = Band width

41.(B) we know that property of Pdf

x–

f (x)dx 1

1

0

K dx 1

10K x 1

K1

42.(C) jnm hm

–

s ( ) R ( )e d

0

(3 j ) (3 j )

– 0

k e d k e d

k k3 j 3 j

2

6k9



Page 37

43.(B) y z dy 1p(y) e e , –dx 2 z

zdy ep(z) p(y) , 0 z

dz 2 z

44.(C) Mean value of signal power

m m–

p s (f )df

r

f2 a df

= a

2 2 2c m c c

00 0 0

A p A a A a(SNR)4 N 4 N 4N

45.(D) 2

0 nyy

n | H[0] |P 0.12

–7

0 6

0.1(2 ) 10n84(4 ) (10 )

n0 = 1.25 × 10–8 W/Hz

46.(D) When the line is cut into half, no change in the impedance, hence impedance remains same as Z0

47.(C) xNS | | rect2 4 B

We know

RX() SX()

sin( t) rect2

Here, W = 2B

x2 B NR ( ) sin(2 B )

2

Rx() = NB sin(2B)



Page 38

48.(A) The network is rearranged as

I2

2+

–V2

2'

Z1 Z2

Z2Z1

1 1

Z– Parameter equations are

V1 = Z111 + Z12.2

V2 = Z211 + Z22.2

2

111 1 2

1 0

V 1Z (Z Z )2

2

221 2 1

1 0

V 1Z (Z – Z )2

1

112 2 1

2 0

V 1Z (Z – Z )2

1

222 1 2

2 0

V 1Z (Z Z )2

1 2 2 1

2 1 1 2

Z Z Z – Z2 2[Z]

Z Z Z Z2 2

49.(A) The a parameter equations are as below

V1 = AV2 + B2

1 = CV2 + D2



Page 39

2

1 21 1 2

2 2 102 1

1(Z Z )V (Z Z )2A1V (Z Z )(Z Z )2

2

1 1

12 2 10 2 1

2CV (Z Z )Z Z

2

2

1 21

1 21 1 2

2 1 22 1V 01

1 2

2Z . ZZ ZV 2Z ZB

Z ZZ ZZ Z

Since the network is symmetrical,

D = A

2 1 1 2

2 1 1 2

2 1

2 1 2 1

(Z Z ) 2Z Z(Z Z ) (Z – Z )

[A](Z Z )2

(Z Z ) (Z Z )

50.(B) D = kn[VSG + VTP]2

1

2

2Sa TPD1

2D2 Sa TP

(V V )(V V )

2

TP2

TP

(3 V )0.225 0.1611.4 (4 V )

VTP = –2.33 V

51.(C) "

2n oxD(Sat) SG TA

C W V V2 L

20.040 W0.225 (3 2.33)2 L

W 25L



Page 40

52.(A) and 53(D). Referring to primary side, the equivalent circuit becomes

5 0°+

–

+

–5 –60

Applying super-position theorem

15i 0.5 0A

10

210i 60 = 1 –60A10

I = –i1 + i2

= –0.5 + j0 + 0.5 – j0.868

= –j 0.868

–10 –60°

I2

10W

Active component of I through R = 0

Power dissipated in each resistor = 0

Now, power delivered by

Source (1) = 5 × 0.5 = 2.5 W

Source (2) = 10 × 1 = 10 W

–5 0°

I1

10W



Page 41

54.(C) ntan2 tan 60 1.732

55.(D) 12 4

1

rr1

2 4

120

360 0.548(1 1.732 )

56.(B) No doubt, it was our own government but it was being run on borrowed ideas, using second hand solutions.

57.(B) Ratify: Approval:: duel: combat

58.(A) Total distance covered

= 35 + 270 + 80 = 385 km.

= 1 + 3 + 122

hrs.

= 13/2 hrs

Average speed = D 385 2T 13

= 59.23 km/hr.

59.(B) some critics believe that Satyajit Ray never quite came back to the great beginning he made in this

path breaking film Pather Panchali. Its haunting images have endured decades of well-travelled bad

prints to become a signpost in cinematic history.

60(D) Generosity is nearly opposite to Valedictory

61.(B) Count the number of squares in the figure and multiply it by 3 i.e. 45.



Page 42

62(C) The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the Amino

terminal portions of both the H and L chains.

63.(C) 4 = good 7 = picture and 2 and 9 = are and faint respectively

64.(D) X X X 7X35 or1 2 4 4

= 35 or 7X = 35 × 4 or X = 20 65(D) The increase in the total profits of company B in 1997 is 789%.

Please find below in Table – the details of the Course for GATE 2013 Electronics & Communication Engineering offered by VPM CLASSES:

TABLE - GATE 2013 ENGINEERING GATE 2013

CORRESPONDENCE COURSE (with GATE Aptitude study material) (without GATE Aptitude study material)

Course A Course B Rs.7,100/- Rs.6,300/-

6 volumes of theory 6 volumes of theory 1 volume of theory covering various aspects of

GATE General Aptitude 36 Topic-wise Unit test papers (UTPs)

covering the GATE syllabus 12 Topic-wise Unit test papers covering the

GATE General Aptitude syllabus 6 Volume Test Papers (VTPs) for better

practice and revision of your syllabus 36 Topic-wise Unit test papers (UTPs)

covering the GATE syllabus 12 Full length test papers (on GATE pattern)



Page 43

6 Volume Test Papers (VTPs) for better practice and revision of your syllabus

Previous 2 year solved question bank (2011-2012)

12 Full length test papers (on GATE pattern) Hints and solutions with all test papers Previous 2 year solved question bank (2011-

2012)

Hints and solutions with all test papers

NOTE: All prices include service tax.

HOW TO APPLY

Mode of Payment of fee: Option 1 - Demand Draft

For the course you want to get enrolled with us, please prepare a Demand Draft (D.D.) of

an amount equal to the fee mentioned against your course of interest. Please prepare the

D.D. in favor of "VPM Classes" payable at any bank at Kota, Rajasthan.

Items to send:

1) Duly filled application form.

2) D.D. for the course you wish to get enrolled in

3) Xerox copy of 10th, Bachelor degree & Master Degree (if applicable) mark sheets

4) 2 additional passport sized photographs

5) Your application / roll nos. of the various exams (IIT JAM / NET / GATE / TIFR / IISc etc.)

you are appearing for (when applicable).

Please send the above items to:

VPM Classes

1-C-8, Sheela Chowdhary Road, SFS, Talwandi, Kota, Rajasthan PIN - 324005.



Page 44

(Tel No.: 0744-2429714 or 09001297111 | E-mail: [email protected]

/[email protected]/ [email protected]

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Please deposit the fee by cash/cheque in any of the 2 VPM Classes a/c nos. mentioned

below:

Money can be transferred from ANY bank in India to the above account numbers using the

following information:

Details Punjab National Bank

(PNB) HDFC

VPM CLASSES account no.

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1) The Deposit slip in the name of VPM Classes received from the bank, bearing amount, date, bank details (bank name, branch, and bank seal). 2) The completely filled VPM Classes Application form.

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you are appearing this year (when applicable).

Your 1st dispatch will be sent within 3 working days of us receiving this e-mail with all

the details. In case of Online Test Series, the login information will be provided within 12

hours of enrollment.



Page 45

(Note: In case of payment by cheque, the 1st dispatch will be subject to realization of the cheque). Then please send the following items by speed post as well:

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Please send the above items to:

VPM Classes

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(Tel No.: 0744-2429714 or 09001297111 | E-mail: [email protected] /

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