Exam Questions-Section 2

1. If we manage to __________ our natural resources, we would leave a better planet for our children.

A. uphold B. restrain

C. Cherish D. conserveAnswer & Explanation

Answer: Option D

Explanation:

The clue in this sentence is 'If we manage to __________ our natural resources' and 'better planet'.

This implies that the blank should be filled by a word which means 'preserve' or 'keep for long time'.

Therefore the word 'conserve' is the right answer.

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2. (2)3 + (3)4 = (?)5

A. 4

B. 11

C. none of these

D. not possible

Answer & Explanation

Answer: Option C

Explanation:

(2)3 + (3)4 = 2 + 3 = 5 and (10)5 = 51 + 0 = 5

∴ (2)3 + (3)4 = (10)5 .


3. If g(t) = e- t2 then G(0) is __________ (where g(t)⇔ G(f)

A.

B. unity

C.

D. impulse at f = 0


Answer: Option B

Explanation:

The signal g(t) given above is a Gaussian pulse, and it satisfies the relation.

Therefore, its fourier transform is same as the signal itself in frequency domain.

i.e. G(f) = e-f2

Hence G(f)|f = 0 = e-(0) = 1.


4.

The open loop transfer function of a unity feedback system is: The range of K for which the system is stable is

A.

B. 13 > K > 0

C.

D. -6 < K < ∞Answer & Explanation

Answer: Option A

Explanation:

Apply R-H.


5. A discrete time linear shift-invariant system has an impulse response h[n] with h[0] = 1, h[1] = - 1, h[2] - 2, and zero otherwise. The system is given an input sequence x[n] with x[0] - x[2] - 1, and zero otherwise. The number of nonzero samples in the outputsequence y[n], and the value of y[2] are, respectively

A. 5, 2

B. 6, 2

C. 6, 1

D. 5, 3Answer & Explanation

Answer: Option D

Explanation:

Use convolution to get the result.


6. For a series resonant circuit at low frequency circuit impedance is __________ and at high frequency circuit impedance is __________ Fill in the blanks respectively

A. capacitive, inductive

B. inductive, capacitive

C. resistive, inductive

D. capacitive, resistive


Answer: Option A

Explanation:

From this curve; we can select option (a) as correct answer.


7. The network is

A.

First order system and the pole

B.

Second order system of the two poles are

C.

Second order system of the poles is

D. none of the above


Answer: Option B

Explanation:

System T.F. =

Characteristic equation:

So system is second order system of the two poles

are .


8. A signal with ± 10 V ranged and 1 KHz band width is being digitized using a sample and hold circuit and a 10 bit quantizer. The minimum sampling rate is :

A. 62831.9 V/sec

B. 125663.8 V/sec

C. 31415.95 V/sec

D. NoneAnswer & Explanation

Answer: Option A

Explanation:

Maximum sampling rate = 2fmax V.


9. The circuit below represents function X(A, B, C, D) as:

A. Σ(5, 8, 13, 14)

B. ∏(0, 1, 2, 3, 4, 6, 7, 8, 10, 11, 12)

C. Σ(5, 9, 13, 14)

D. ∏(0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 15)


Answer: Option C

Explanation:

X = A B C(0. D + 0.D) + A B C(0.D + 0.D) + A B C(0.D + 1.D) + A B C(0.D + 0.D)

+ A B C(0.D + 0.D) + A B C(0.D + 1.D) + ABC(1.D + 0.D)

= ∑(5, 9, 13, 14).


10. The truth table corresponding to the given input digital gate :

A.

B.

C.

D.


Answer: Option B

Explanation:

The circuit behaves like NAND gate.


11.

The z-transform of a signal converges if and only if

A.

B.

C.

D.


Answer: Option A

Explanation:

For u(n), a right handled sequence,

|z| > , |z| =

∴ |3z| > 1; |2z| > 1

∴ < 1; < 1.


12. If the characteristic equation of a closed loop system is s2 + 2s + 2, then the system is :

A. over damped

B. critically damped

C. under damped

D. undamped


Answer: Option C

Explanation:

1 + G(s) H(s) = 0

s2 + 2s + 2 = 0

ξ < 1, under damped.


13. 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


Answer: Option B

Explanation:

The given digits are 2, 2, 3, 3, 3, 4, 4, 4, 4 we have to find the numbers that are greater than 300

∴ The first digit can be 3 or 4 but not 2.

Now, let us fix the first, second and third digits as 3, 2, 2, then the fourth place can be filled in 3 ways.

∴ The number of ways is 3 similarly, we fix first third and fourth place as 3, 2 and 2 respectively (4) so the second place can be filled in 3 ways again,

The number of ways is 3

Now, we fix first, second and fourth, previous cases and we obtain the same result.

∴ The number of ways is 3 so, the total number of ways is 9 similarly this can done by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the a ways each

The number of numbers starting with 3 is 27

Similarly by taking 4 as the first digit we get 27 numbers

∴ The number of numbers that are greater than 3000 is 27 + 27 = 54

But, 3222, 4222, is not possible as there are only two 2's, 3333 is not possible as there are only three 3's

∴ The total number of numbers that are greater than 3000 is 54 - 3 = 51.


14. Consider circuit with 4 : 16 Demux below: Now:

1. f1 = (2, 6, 11, 12)2. f2 = (2, 5, 14)

3. = (2, 3, 4, 5, 9, 13, 14)

A. only 1 and 3 are correct

B. only 1 is correct

C. only 2 is correct

D. 2 and 3 are correct


Answer: Option D

Explanation:

f1 = y2 + y6 + y11 + y12

and f2 = y1 + yl0 + yl3

f1(m, n, o, p) = (3, 4, 9, 13)

f2(m, n, o, p) = (2, 5, 14)

∴ = Π(2, 3, 4, 5, 9, 13, 14).


15.

The value of the integral is

A.

B.

C.

D.


Answer: Option A

Explanation:

Compare this with

where s = 3.


16. The following circuit can be represented as :

A. C

B. f(A, B, C) = (0, 1, 2, 3, 4, 5, 6, 7)

C. (A ⊕ B)C

D. C


Answer: Option D

Explanation:


17. A two port network is described by the relation V1 = 2I1 + 3V2 I2 = - I1 + 2V2 Then Z-parameter of such network is

A.

B.

C.

D.


Answer: Option D

Explanation:


18. Find Y in the circuit below :f1(E, F) = ∑(0, 1, 2, 3)f2(A, B, E, F) = ∑(0, 1, 2, 3, 5, 6, 9)f3(A, B, E, F) = ∏(4, 7, 8, 10, 11, 12, 13, 14, 15)

A. f2

B. f2

C. 0

D. ∑(0, 1, 2, 3, 5, 6, 9)


Answer: Option C

Explanation:


19. Calculate the stability factor and change in Ic from 25°C to 100°C for, β = 50, RB/RE = 250, ΔICO = 19.9 nA for emitter bias configuration

A. 42.553, 0.85 μF

B. 40.91, 0.58 μF

C. 38.53, 0.85 μF

D. 41.10, 0.39 μF


Answer: Option A

Explanation:


20. The star equivalent C1 C2, C3 of the delta network is respectively

A.

B.

C.

D.


Answer: Option A

Explanation:

∴ .


21. In a radar system, the range of R1 is achieved at a frequency f1. Then the range R2 at frequency R2 = 8f1 would be __________ [Neglect effect of 1 on the Beam width.]

A. R2 = R1

B. R2 = 4R1

C. R2 = 2 R1

D.R2 =


Answer: Option C

Explanation:

The range of the radar is directly proportional to the square root of the frequency.

∴ R ∝

∴ R1 ∝

and R2 ∝

∴

∴ .


22.

With a line charge along z-axis the field intensity is E = ; which of the following figure shows an exact field distribution?

A.

B.

C.

D.


Answer: Option D

Explanation:

The given figure fails to show the symmetry with respect to Φ.

The figure shows symmetry with respect to f, also the length of arrow is decreasing away from the charge shows that magnitude E is decreasing away from line charge.

But problem with this figure is longest lines must be shown in most crowded region.

Here we use lines of fixed segments but different thickness. But this attempt also makes the region crowded near origin.

This figure shows compromise. A symmetrical distribution of lines (at every 45°) shows azimuthal symmetry and arrowheads show direction.


23. The logic function f(A, B, C) = m(0, 2, 4, 5, 6) can be represented by :

1.

2.

3.

A. 1 only

B. 1, 2 only

C. 1, 3 only

D. 1, 2 and 3Answer & Explanation

Answer: Option B

Explanation:

Now the Variable Entered Map (VEM) is:


24. In amplitude modulation, carrier signals A cos ωt has its amplitude A modulated in proportion with message bearing (low frequency) signal m(t). The magnitude of m(t) is chosen to be __________ .

A. less than 1

B. less than or equal to 1

C. more than 1

D. none of these


Answer: Option B

Explanation:

For proper recovery of signal |m(t)| ≤ 1.


25. For real values of x, the minimum value of the function f(x) exp (x) + exp (- x) is

A. 2 B. 1

C. 0.5 D. 0


Answer: Option A

Explanation:

f(x) = ex + e-x

f(x) = ex - ex

f(x) = 0 ⇒ ex - e-x = 0

f'(x) = ex + e-x ⇒ +ve for x = 0

Thus minimum.

Minimum f(x) = e0 + e0 = 2.


26. The logic realized by the circuit shown in figure below is

A. F = A - C

B. F = A + C

C. F = B ⊙ C

D. F = B ⊕ C


Answer: Option B

Explanation:

F = A BC + ABC + A B C + ABC

= B(AC + AC) + B(AC + AC)

B(A ⊕ C) + B(A ⊕ C)

(A ⊕ C) (B ⊕ B)

A ⊕ C.


27. An image uses 512 x 512 picture elements. Each of the picture elements can take any of the 8 distinguishable intensity levels. The maximum entropy in the above image will be

A. 2097152 bits

B. 648 bits

C. 786432 bits

D. 144 bitsAnswer & Explanation

Answer: Option C

Explanation:

Each picture element can be represented by 3 bits. Thus total entropy = 512 x 512 x 3 = 786432.


28. A tachometer has a sensitivity of 4 V/1000 rpm. Express the gain constant of the tachometer in the units volts/(rad/sec).

A. 0.0489

B. 0.0381

C. 2

D. none of theseAnswer & Explanation

Answer: Option B

Explanation:

4 V/100 rpm = V/rad/sec

= 0.0381 V/rad/sec.


29.

Consider the signal x(n) = 1 + sin n + 3 cos n + cos . Then Fourier series coefficients are __________ .

A. 1, - j, + j, j, - j

B. 1, + j, - j, - j , + j

C. 1, 3 - j, 3 + j, j, - j

D. 1, 3 + j, 3 - j, - j, + j


Answer: Option A

Explanation:

The signal x(n) = 1 + sin n 3 cos n + cos n + cos

is periodic with period N, and we can expand x[n] directly in terms of complex exponentials to obtain

Collecting terms, we find that

Thus for Fourier series coefficients for this example are

a0 = 1,

a1 = + = - j,

a-1 = - = + j,

a2 = j,

a-2 = - j.


30. A certain system has transfer function is a parameter. Consider the standard negative unity feedback configuration as shown below.

Which of the following statements is true?

A. The closed loop system in never stable for any value of a

B. For some positive values of , the closed loop system is stable, but not for all positive values

C. For all positive values of , the closed loop system is stable

D. The closed loop system is stable for all values of , both positive and negative


Answer: Option C

Explanation:

Closed loop system transfer function

Use Routh criteria

Thus for all positive value of '' this will be stable.


31. If a signal f(t) has energy E, the energy of the signal f(2t) is equal to __________ .

A. E B. E/2

C. E/4 D. 2E


Answer: Option B

Explanation:

Energy content of a signal x(t), E = |f(t)|2 dt

Now, E' = |f(2t)|2 dz for signal f(2t)

Putting 2t = z, we get

E' = |f(t)|2 dz = .


32. For a npn BJT transistor fβ is 1.64 x 108 Hz. Cμ = 10-14 F; C = 4 x 10-13 F and DC current gain is 90. Find fT and gm (fβ = cut off frequency, Cμ = capacitance, C = parasitic capacitance, gm = transconductance, fT = gain BW product)

A. fT = 1.47 x 1010 Hz ; gm = 38 milli mho

B. fT = 1.64 x 108 Hz ; gm = 30 milli mho

C. fT = 1.47 x 109 Hz ; gm = 38 mho

D. fT = 1.33 x 1012 Hz; gm = 0.37 m-mhoAnswer & Explanation

Answer: Option A

Explanation:

∴ fT = 90 x 1.64 x 108 = 1.47 x 1010 Hz

gm = 2fT(Cμ + C) = 2 x x 1.47 x 1010

= (10-14 + 4 x 10-13)

gm= 38 mμ.


33. The impulse response h(t) of a linear time-invariant continuous time system is described by h(t) = exp (t) u(t) + exp (βt) u (- t), where u(t) denotes the unit stepfunction, and and β are constants. This system is stable if

A. is positive and β is positive

B. is negative and β is negative

C. is positive and β is negative

D. is negative and β is positive


Answer: Option D

Explanation:

h(t) = e+tu(t) + eβtu(- t)

For h(t) to be stable h(t) dt < ∞

It will happen when is negative and β is positive.


34. The probability density function of a random variable x is as shown.

The value of A is:

A.

B.

C.

D.


Answer: Option C

Explanation:

= 1

= 1

Solve this to get value of A = 1/5.


35.

ωL =

A. 108 rad/sec

B. 104 rad/sec

C. 102 rad/sec

D. 10 rad/secAnswer & Explanation

Answer: Option D

Explanation:

R1 = (10 + 10) = 20 kΩ

.


36. An 8 level encoding scheme is used in a PCM system of 10 kHz channel BW. The channel capacity is

A. 80 kbps

B. 60 kbps

C. 30 kbps

D. 18 kbps


Answer: Option B

Explanation:

C = 2B log2 M;

M = 8 = 23, log2 M = 3 bits

= 2 x 10 x 103 x 3 = 60 kbps.


37. For the power amplifier circuit shown below, the maximum power dissipated by both output transistor is

A. 9.66 W

B. 30.11 W

C. 31.66 W

D. 33.66 W


Answer: Option C

Explanation:

The maximum power dissipated by both output transistor is maximum


38. For static electric and magnetic fields in an homogenous source-free medium, which of the following represents the correct form of Maxwell's equations?

A. Δ.E = 0

Δ x B = 0

B. Δ.E = 0Δ.B = 0

C. Δ x E = 0Δ x B = 0

D. Δ x E = 0Δ.B = 0


Answer: Option D

Explanation:

Curl of electric field is zero. Divergence of magnetic field is zero.


39. A periodic rectangular signal, x(t) has the waveform shown in the given figure. Frequency of the fifth harmonic of its spectrum is __________

A. 200 Hz

B. 250 Hz

C. 400 Hz

D. 1250 Hz


Answer: Option D

Explanation:

The periodic time = 4 ms = 4 x 10-3 s

∴ The fundamental frequency

∴ Frequency of the 5th harmonic

= 250 x 5 = 1250 Hz.


40. An antisymmetric filter having odd number of coefficient may have a performance as

A. low pass

B. high pass

C. band pass

D. all passAnswer & Explanation

Answer: Option D

Explanation:

Consider an anti-symmetric filter response with order 5.

h(n) = [h(2), h(1), h(0), - h(1), - h(2)]

We know, H(0) = addition of all component values.

Hence H(0) = h(0)

Similarly H() = addition of all component values with alternate negative sign.

= h(2) - h(1) + h(0) + h(1) - h(2) = h(0)

Hence we get all pass filter.


41. A material has conductivity of 105 mho/m and permeability of 4 x 10-7 H/m. The skin depth at 9 GHz is

A. 1.678 μm

B. 26 μm

C. 17 μm

D. 32.32 μm


Answer: Option C

Explanation:


42. 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

A.

B.

C.

D.


Answer: Option D

Explanation:

4 heads and 0 tail 3 head and 1 tail

.


43. The steady state error of a stable type 0 unity feedback system for a unit step function is :

A. 0

B.

C. ∞

D.


Answer: Option B

Explanation:

.


44. Find RTH across the terminals A and B

A. 52.28 Ω

B. 58.28 Ω

C. 52.82 Ω

D. 58.82 Ω


Answer: Option D

Explanation:

There are not independent source in the circuit.

∴ The Thevenin and Norton equivalent will have 0 A and 0 V sources.

To find RTH, a 1 A source is connected as

Writing a nodal equations at n,

0.01Vx + 0.003Vx + 0.0067Vx = 1

0.017Vx = 1

∴ Vx = 58.82 V


45. Find Y- parameters

A.

B.

C.

D. None of these


Answer: Option A

Explanation:

Apply Nodal Analysis

At node V

...(i)

At node V2

...(ii)

At node V3

∴

∴

Put V3 in equation (i)

Put V3 in equation (ii)

.


46. For a second-order system with the closed-loop transfer function The settling time for 2-percent band, in seconds, is :

A. 1.5 B. 2.0

C. 3.0 D. 4.0


Answer: Option B

Explanation:

.


47.

An open loop transfer function is given by has

A. one zero at ∞

B. two zeros at ∞

C. three zeros at ∞

D. four zeros at ∞


Answer: Option C

Explanation:

The root locus is given below.

From root locus, we can see that there are 3 zeros at ∞ .


48. The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are

A. 41 A, 410 W respectively

B. 35 A, 350 W respectively

C. 5 A, 250 W respectively

D. 11 A, 1210 W respectivelyAnswer & Explanation

Answer: Option C

Explanation:

Power = I2R = 25 X 10 = 250 Watts.


49. The cut off voltage for JFET is 5 V. The pinch off voltage is __________ .

A. (5.0)l/2 V

B. 2.5 V

C.V

D. 5 VAnswer & Explanation

Answer: Option D

Explanation:

For JFET

ID = IDSS (1 - VGS/Vp)2

0 = (1 - VGS/Vp)2

∴ VGS = VP

∴ Vp = 5 V.


50. Assertion (A): Conductors do not permit propagation of waves more than a short distance into the conductor at microwave frequencies.

Reason (R): The relaxation time constant for conductors is much small than the period of centimetric EM wave.

A. Both (A) and (R) are true and (R) is the correct explanation of (A)

B. Both (A) and (R) are true but (R) is not the correct explanation of (A)

C. (A) is true but (R) is false

D. (A) is false but (R) is true


Answer: Option B

Explanation:

Depth of penetration = , where = relaxation time.

δ would be small, if σ is large, which itself depends upon relaxation time.

for conductors is of the order of 10-14 s, for λ = 3 x 10-2 cm

∴ period of centimetric EM waves

=

Therefore, even though is much smaller than period of centimetric waves but it is not the correct reason for the assertion given.


Exam Questions-Section 2

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