Basic Electronics

Basic Electronics
EM Theory
Digital Communication
Digital Electronics
Microprocessor & Microcontroller
Electronic Devices
Digital Signal Processing (DSP)
Analog Electronics
Telecommunication
RF and Microwave Engineering
Signal & System
Analog Communication
Circuit Theory
Wireless Communication
VLSI
Antenna Theory
Control System

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1.

Pick out the underneath selection in terms of ascending command of band gap energy

A)

Silicone, diamond, graphite

B)

Diamond, silicone, graphite

C)

Graphite, diamond, silicone

D)

Graphite, silicone, diamond

View Answer

Correct Answer : D

Diamond takes the maximum bandgap energy, nearby 6eV, and this high bandgap that differs the complete valance band from the conduction band and that’s why Diamond acts as an insulator. However Graphite is also a manner of Carbon, such as Diamond, its crystal symmetry is dissimilar and bandgap energy is nearby 1eV, building it a conductor. Bandgap energy of Silicon is nearby 1.21eV that can be developed only through rise in temperature, building it a semiconductor.

2.

What is the value of band gap energy or germanium and silicone?

A)

0.72 ev and 1 ev

B)

0.61 ev and 2 ev

C)

0.53 ev and 1.1 ev

D)

The above options are wrong

View Answer

Correct Answer : A

Underneath is the band gap energy connections through temperature for both Germanium and Silicon.

Egsi (T) = 1.21 – 3.60 x 10-4

Egge (T) = 0.785 – 2.23 x 10-4

Replacing T = 300 in both the overhead expressions, so Band gap energies as specified underneath;

Egsi (300) = 1.1 eV

Egge (300) = 0.72 eV

3.

Electrical features of a semiconductor are contingent on the absorption of holes and free electrons.

A)

True

B)

False

C)

Both option A and option B

D)

The above options are wrong

View Answer

Correct Answer : A

Current Density in some conducting material is specified like:

J = ( nµn + pµp ) eE= σE

Wherever, n = concentration of electrons, p = concentration of holes, E = applied electric field, e = charge and σ = conductivity.

Consequently, it can be realized that the electrical features be determined by on concentration of holes and electrons.

4.

Transference of charge carriers in a semiconductor is attained by: –

A)

Diffusion

B)

Conduction

C)

Both option A and option B

D)

The above options are wrong

View Answer

Correct Answer : C

Two techniques where charge carriers are moved in semiconductors are – conduction and diffusion. Conduction in semiconductors is skillful through the drive of holes and free electrons. When holes are positive charge-carrying particles and electrons are negative charge-carrying particles and most importantly current direction of electron and hole are similar. Diffusion is triggered through the drive of charge carriers from a higher concentration zone to a lower concentration zone and of course, this is not impossible because of the non-uniform concentration in semiconductors.

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1.

Which of the following is zero?

A)

Grad Div A

B)

Div Gradient V

C)

Div Curl A

D)

Curl Curl A

View Answer

Correct Answer : C

The divergence of the curl of a vector field vanishes

2.

Which of the following is zero?

A)

Grad div

B)

Curl grad

C)

Div grad

D)

Curl curl

View Answer

Correct Answer : B

The curl of the gradient of a scalar field vanishes

3.

Convert the point (3, 4, 5) from cartesian to spherical coordinates

A)

$\left(7.07,{45}^{\circ},{53}^{\circ}\right)$

B)

$\left(0.707,{45}^{\circ},{53}^{\circ}\right)$

C)

$\left(7.07,{54}^{\circ},{63}^{\circ}\right)$

D)

$\left(0.707,{54}^{\circ},{63}^{\circ}\right)$

View Answer

Correct Answer : A

r = √(x^{2}+y^{2}+z^{2}) = √50 = 7.07

= cos^{-1}(z/r) = cos^{-1}(5/5√2) = 45⁰

= tan^{-1}(y/x) = tan^{-1}(4/3) = 53⁰.

4.

The curl of a vector gives a

A)

Scaler

B)

Vector

C)

Zero value

D)

Non zero value

View Answer

Correct Answer : B

Curl is always defined for vectors only. The curl of a vector is a vector only. The curl of the resultant vector is also a vector only

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1.

Assuming the signals to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is

A)

16 dB

B)

32 dB

C)

48 dB

D)

64 dB

View Answer

Correct Answer : C

SNR = 1.76 + 6.02 n≈ 6n≈ 6 × 8= 48 dB

2.

The minimum sampling frequency (f_{s})min required to avoid slope overload when x(t) = cos(2p800t) and d = 0.1 is

A)

23.12 kHz

B)

50.25 kHz

C)

98.12 KHz

D)

75.67 kHz

View Answer

Correct Answer : B

To avoid slope overload

${f}_{s}\ge 2\pi {f}_{m}\left(\frac{{A}_{m}}{\delta}\right)$,

Given that f_{m}= 800 Hz, A_{m}= 1 and d = 0.1.

Therefore,

$\u3016\left({f}_{s}\right){\u3017}_{min}=2*\pi *800*\left(\frac{1}{0.1}\right)=50.25KHz$

3.

In a BPSK signal detector, the local oscillator has a fixed phase error of 20°. This phase error deteriorates the SNR at the output by a factor of

A)

cos20°

B)

cos220°

C)

cos70°

D)

cos270°

View Answer

Correct Answer : B

In BPSK, if local oscillator in the detector has a fixed phase error f, then the output power would reduce by a factor cos2f. Therefore, the SNR deteriorates by a factor cos2f.

Given that f = 20°. Therefore,

SNR at the output deteriorates by a factor of cos (220).

4.

The code word length of a PCM system is increased from 6 to 8 bits, the signal to improve the noise ratio factor by quantization,

A)

8/6

B)

12

C)

16

D)

8

View Answer

Correct Answer : C

SNRQ α 2*2hα 4n

as n = 2

So SNRQ increases by 16.

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1.

The representation of octal number (532.2)_{8} in decimal is ________

A)

(346.25)_{10}

B)

(532.864)_{10}

C)

(340.67)_{10}

D)

(531.668)_{10}

View Answer

Correct Answer : A

(532.2)_{8} = 5 * 8^{2} + 3 * 8^{1} + 2* 8^{0} + 2 * 8^{-1} = 346.25 = (346.25)_{10}

2.

If the decimal digit is a fraction when its binary equivalent is achieved by ________ the number continuously by 2

A)

Subtracting

B)

Adding

C)

Multiplying

D)

Dividing

View Answer

Correct Answer : C

First of all multiplying the decimal number continually by 2, the binary correspondent can be obtained by a collection of the integer part. But, when it’s an integer, subsequently its binary equivalent is determined by dividing the number by 2 also collecting the remainders.

Example:-

1101.0111 = (1×2^{3}) + (1×2^{2}) + (0×2^{1}) + (1×2^{0}) + (0×2^{-1}) + (1×2^{-2}) + (1×2^{-3}) + (1×2^{-4})

= 8 + 4 + 0 + 1 + 0 + 1/4 + 1/8 + 1/16

= 8 + 4 + 0 + 1 + 0 + 0.25 + 0.125 + 0.0625 = 13.4375_{10}

3.

The decimal equivalent of the binary number (1011.011)_{2} is ________

A)

(11.375)_{10}

B)

(10.123)_{10}

C)

(11.175)_{10}

D)

(9.23)_{10}

View Answer

Correct Answer : A

:(1011.011)_{2} = 1 * 2^{3} + 0 * 2^{2}+ 1 * 2^{1}+1* 2^{0} + 0 * 2^{-1} +1 * 2^{-2}+ 1 * 2^{-3}= 11.375

= (11.375)_{10}

4.

An important drawback of binary system is ________

A)

It requires very large string of 1’s and 0’s to represent a decimal number

B)

It requires sparingly small string of 1’s and 0’s to represent a decimal number

C)

It requires large string of 1’s and small string of 0’s to represent a decimal number

D)

It requires small string of 1’s and large string of 0’s to represent a decimal number

View Answer

Correct Answer : A

An important drawback of a binary system is it requires very large string of 1’s and 0’s to represent a decimal number The main advantage of using binary is that it is a base that is easily represented by electronic devices. Binary System is also used in coding, less computations, and also less computational errors. The main disadvantage of binary number is hard to read and write for humans because of the large number of a binary of an equivalent decimal number.

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1.

A microprocessor is a _______ chip integrating all the functions of a CPU of a computer

A)

Multiple

B)

Single

C)

Double

D)

Triple

View Answer

Correct Answer : B

A CPU, ROM, RAM, and data input/output circuitry into a single IC is called a microprocessor.

2.

In 8086 microprocessor , the address bus is ________ bit wide

A)

12 bit

B)

10 bit

C)

16 bit

D)

20 bit

View Answer

Correct Answer : D

Intel 8086 is a 16 bit integer processor. It has 16-bit data bus and 20-bit address bus.

3.

RST 7.5 interrupt is?

A)

Vectored and maskable

B)

Vectored and non maskable

C)

Direct and maskable

D)

Direct and non-maskable

View Answer

Correct Answer : A

RST 7.5 interrupt is a vectored and maskable interrupt with vector address 3CH.

4.

Memory is an integral part of a ______ system

A)

Supercomputer

B)

Microcomputer

C)

Minicomputer

D)

Mainframe computer

View Answer

Correct Answer : B

A Microcomputer is determined partly by characteristics of components of the system, i.e. primarily the memory, the disk units, the display, the keyboard, the flexibility of the hardware, and the operating system and other software.

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1.

A Self bias configuration contains R_{D}=3.3, R_{s}=1 KΩ, R_{G}=**1M**Ω and g_{m}=1.5mS. Determine A_{v}?

A)

-2

B)

3

C)

-4

D)

5

View Answer

Correct Answer : A

A_{v}=-g_{m} R_{D}/ (1+g_{m} R_{S}) because R_{S} is given and it is not bypassed.

A_{v}=-1.5×3.3KΩ/ (1+1.5×1)

A_{v}=-2.

2.

In germanium semiconductor material at T 400 K the intrinsic concentration is (10^{14} per cc)

A)

26.8

B)

18.4

C)

8.5

D)

3.6

View Answer

Correct Answer : C

n_{i}^{2} = N_{c }N_{v}

Property |
Si |
GaAs |
Ge |

Bandgap Energy |
1.12 |
1.42 |
0.66 |

Dielectric Constant |
11.7 |
13.1 |
16.0 |

Effective density of states in conduction band N |
2.8 *10 |
4.7 * 10 |
1.04*10 |

Effective density of states in valance band N |
1.04*10 |
7.0 * 10 |
6.0 * 10 |

Intrinsic carrier concentration n |
1.5*10 |
1.8*10 |
2.4*10 |

Mobility Electron Hole |
1350 480 |
8500 400 |
3900 1900 |

3.

Two semiconductor material have exactly the same properties except that material A has a bandgap of 1.0eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of material B is

A)

2016

B)

47.5

C)

58.23

D)

1048

View Answer

Correct Answer : B

=

4.

The thermal-equilibrium concentration of hole p0 in silicon at T = 300 K is 1015 cm^{3}. The value of n_{0} is,

A)

3.8 x 10^{8} cm^{3}

B)

4.4 x 10^{4} cm^{3}

C)

2.6 x 10^{4} cm^{3}

D)

4.3 x 10^{8} cm^{3}

View Answer

Correct Answer : B

p_{0 }=N_{v} _{F }– E_{v} = kT ln (

At 300 K, N_{v} = 10*10^{19} cm^{-3}

E_{F} – E_{v} = 0.0259 ln (

n_{0} = N_{c}

At 300 K, N_{c} = 2.8 * 10^{19} cm^{-3}

E_{c} – E_{F} = 1.12 – 0.239 = 0.881 eV

n_{0} = 4.4 * 10^{4} cm^{-3}

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1.

The period of the signal is

A)

1

B)

C)

D)

3

View Answer

Correct Answer : A

Here fundamental frequency

Hence,

The minimum value of n for which M is integer is 3

2.

The system is

A)

Time Invariant and Linear

B)

Time Variant and Non Linear

C)

Time Variant and Linear

D)

Time Invariant and Non Linear

View Answer

Correct Answer : B

If the input is delayed by k units in time and applied to system then it will be

If the output is delayed by k units in time and applied to system then it will be

Here, it’s a time variant system.

The outputs due to the signals and are

Weighted sum of the output is

The output due to weighted sum of inputs is

Thus, the system is non linear

3.

The step response of is

4.

The impulse response is

A)

Causal and Stable

B)

Non causal and Stable

C)

Causal and Unstable

D)

Non causal and Unstable

View Answer

Correct Answer : B

is not zero for less than zero, so the system is non-causal

Now,

Hence, the system is stable.

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1.

The circuit shown represents * *

A)

A band pass filter

B)

A voltage controlled oscillator

C)

An amplitude modulator

D)

A monostable multivibrator

View Answer

Correct Answer : D

The given circuit represents a mono stable multivibrator. To design this monostable multivibrator circuit using an op-amp, the op-amp is used with the positive feedback. The time for which the output goes into the unstable state can be given by the expression:

$T=RC\mathrm{ln}(1+\frac{{R}_{2}}{{R}_{1}})$

Where ${R}_{1}$ and ${R}_{2}$ are the resistors used with the op-amp in the positive feedback configuration.

2.

In the differential amplifier shown in the figure, the magnitudes of the common-mode and differential mode gains are A_{CM} and A_{d}, respectively. If the resistance R_{E}is increased, then

A)

A_{cm} increases

B)

common-mode rejection ratio increases

C)

A_{d }increases

D)

common-mode rejection ratio decreases

View Answer

Correct Answer : B

The magnitude of the common-mode is A_{CM }and differential-mode gain is A_{d}.

${A}_{cm}$ = $\frac{-{R}_{c}}{2{R}_{\epsilon}}$

If ${R}_{\epsilon}$ is increased, then A_{CM }decreases

A_{d} = $\frac{1}{2}{g}_{m}{R}_{c}$

If ${R}_{\epsilon}$ is increased, then A_{d }does not get affected

CMRR = $\frac{{A}_{d}}{{A}_{cm}}$ = $\frac{1}{2}$${g}_{m}$ ${R}_{c}$ $\left(-\frac{2{R}_{\epsilon}}{{R}_{c}}\right)$ = $-{g}_{m}{R}_{c}$

Hence if ${R}_{\epsilon}$is increased, then CMRR (Common mode rejection ratio) increased

3.

In the circuit shown below, what is the output voltage (V_{out}) if a silicon transistor Q and an ideal Op-amp are used?* *

A)

− 15 V

B)

- 0.7 V

C)

+ 0.7 V

D)

+ 15 V

View Answer

Correct Answer : B

As the transistor, Q is a silicon transistor

V_{out} + V_{BE} = 0

V_{out} = − V_{BE}

= −0.7 V.

4.

The circuit below implements a filter between the input current i_{1 }and output voltage

A)

Low pass filter

B)

Band pass filter

C)

Band stop filter

D)

High pass filter

View Answer

Correct Answer : D

Explanation: When * $\omega =0$*

Inductor performs as a SC

${v}_{0}$ $=0$

And when

Inductor performs as an OC

⇒ ${v}_{0}$ = ${i}_{1}{R}_{1}$

So, it acts as a high pass filter

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1.

Switching systems are connected by

A)

Subscriber lines

B)

Trunks

C)

Traffic line

D)

None of these

View Answer

Correct Answer : B

The links that are connecting two switching systems are known as the trunk. It is a component of the telecommunication network.

2.

The instantaneous resistance of a microphone for a simple telephone communication is

A)

${R}_{i}={R}_{0}-\mathrm{sin}\omega t$

B)

R_{i}=0

C)

${R}_{0}={R}_{i}$

D)

None of these

View Answer

Correct Answer : A

When a sound wave hits the diaphragm of the microphone the instantaneous resistance is calculated by the formula r_{i} = r_{0}- rsin (ωt). Where r_{0 }is the quiescent resistance of the microphone with no speech signal. r, is the maximum variation of resistance. And r_{i}= is the instantaneous resistance.

3.

Which of the following is the standard feature in all electronic exchange

A)

SPC

B)

PSC

C)

PSP

D)

SSS

View Answer

Correct Answer : A

SPC. Space division electronic switching system uses electromechanical switching network with stored program control.

4.

In synchronous duplex mode operation

A)

Processors execute the same set of instuctions

B)

Hardware coupling is provided between two processors

C)

Processors execute the same set of instructions

D)

All of these

View Answer

Correct Answer : D

In the synchronous duplex mode of operation, the Processors execute the same set of instructions and compare. Hardware coupling is provided between two processors. Processors execute the same sets of instructions. After comparing if a mismatch occurs the faulty processor is detected and taken out.

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1.

An air-filled rectangular waveguide is with a cross section of 5cm x 3 cm. Find the cutoff frequency (in MHz) of this waveguide in $T{E}_{21}$ mode.

A)

7810 MHz

B)

810 MHz

C)

7000 MHz

D)

7500 MHz

View Answer

Correct Answer : A

Here a

Cut-off frequency ${f}_{c}$ = $\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^{2}+{\left(\frac{n}{b}\right)}^{2}}$ where c = Velocity of the light =3 x ${10}^{8}$ m/s

*=* 7810 MHz

2.

The modes in a rectangular in a rectangular waveguide are denoted by $T{E}_{mn}$* /* $T{M}_{mn}$

A)

The $T{M}_{10}$ mode of the waveguide does not exist

B)

The $T{E}_{10}$ mode of the waveguide does not exist.

C)

The $T{M}_{10}$ and the $T{E}_{10}$ modes both exist and have the same cutoff frequencies.

D)

The $T{M}_{10}$ and the $T{M}_{01}$ modes both exist and have the same cutoff frequencies.

View Answer

Correct Answer : A

For the

3.

The Phase velocity of an electromagnetic wave propagation in a hollow metallic rectangular wave-guide in the $T{E}_{10}$ mode is

A)

Equal to it’s group velocity

B)

Less than the velocity of light in free space.

C)

Equal to the velocity of light in free space.

D)

Greater than the velocity of light in free space.

View Answer

Correct Answer : D

Inside the waveguide phase, velocity is greater and group velocity is smaller than the velocity of the light.

4.

The phase velocity for the $T{E}_{10}$ mode in an air-filled rectangular waveguide is

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1.

A discrete time signal x[n] = sin (${\pi}^{2}$n), n being an integer, is _________

A)

Periodic with period

B)

Periodic with period ${\pi}^{2}$

C)

Periodic with period $\frac{\pi}{2}$

D)

Not periodic

View Answer

Correct Answer : D

x[n] = sin

N = $\frac{2\pi}{\omega}$ * k

N = $\frac{2\pi}{{\pi}^{2}}$ * k

= $\frac{2k}{\pi}$

Where k is minimum integer, so that N is a natural number.

No integer is possible.

2.

The Dirac delta function * δ(t)* is defined as

A)

B)

C)

= 1

D)

= 1

View Answer

Correct Answer : D

* δ* (

3.

Which one of the following is an Eigen function of the class of all continuous time, linear, time invariant systems (u(t) denotes the unit step function)?

A)

${e}^{j{\omega}_{0}t}u\left(t\right)$

B)

cos$\left({\omega}_{0}t\right)$

C)

${e}^{j{\omega}_{0}t}$

D)

sin(${\omega}_{0}t$)

View Answer

Correct Answer : C

^{$t{e}^{j{\omega}_{0}t}$ }

4.

Let * δ(t)* denote the delta function. The value of the integral ${\int}_{-\infty}^{\infty}\delta \left(t\right)\mathrm{cos}\left(\frac{3t}{2}\right)dt$ is,

A)

$1$

B)

$-1$

C)

$0$

D)

$\frac{\pi}{2}$

View Answer

Correct Answer : A

As for function f(t)

So,

= 1

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1.

FM means

A)

Frequency Modulation

B)

Frequency Modulator

C)

Frequent Frequent Multiplier

D)

Frequency Mixer

View Answer

Correct Answer : A

FM is for the modulation of frequencies. It is the encryption by changing its frequency with regard

to the immediate amplitude of a message signal of the carrier wave. The contents utilized during frequency

modulation are the remaining other choices.

2.

What is a sinusoidal analog signal

A)

It passes in a positive direction and a negative one.

B)

For a half cycle it is positive

C)

For one-half cycle it is negative

D)

Has an endless number of amplitudes within the independent variable set of values

View Answer

Correct Answer : D

Explanation: An analog signal is a sinusoidal wave. A signal with an infinite number of amplitudes within the

range of independent variable values is an analog electrical signal. In a continuous interval, analog signals can

take on any value.

3.

The minimum aerial height required by is

A)

$\raisebox{1ex}{$3\lambda $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

B)

$\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

C)

$2\lambda $

D)

$\lambda $

View Answer

Correct Answer : B

The aerial height should be one-forth higher for successful communication

￼$\lambda $. (H = $\lambda $/4 where = c⁄f). The explanation why the aerial height is held to $\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ was because the impedance is balanced. This means the aerial is 1/4th above ground level at best.

4.

What do you understand with the term analog contact?

A)

A system where one of the characteristics of the carrier signal differs according to the instantaneous modulation signal value

B)

A way to communicate data and computer

C)

Contact numerical coded

D)

An effective long-range contact method;

View Answer

Correct Answer : A

Analog communication means that data is transmitted as a continuous signal via the modulation process. The remaining options are applicable for digital communication when coding is implemented.

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1.

The impedance of an ideal current source should be

A)

0

B)

Infinite

C)

Greater than 0 but lesser than infinity

D)

None of these

View Answer

Correct Answer : B

For ideal current source lossless, internal current=0, so internal resistance=$\infty $

2.

In an R-L circuit, the phase angle between voltage and current is

A)

$30\xb0$

B)

$90\xb0$

C)

$180\xb0$

D)

Greater than $0\xb0$ but lesser than $90\xb0$

View Answer

Correct Answer : D

For the R-L circuit, the phase angle is between $0\xb0$and $90\xb0$

3.

Laplace Transform analysis gives

A)

Time-domain response only

B)

Frequency domain response only

C)

Both (a) & (b)

D)

The real response only.

View Answer

Correct Answer : B

By definition of Laplace Transformation: $F\left(s\right)={\int}_{0}^{\infty}{e}^{-st}f\left(t\right)d\left(t\right)$ where $s=\sigma +j\omega $

4.

The numbers of links for a graph having ‘n’ nodes & ‘b’ branches are

A)

b-n+1

B)

n-b+1

C)

b+n-1

D)

b+n

View Answer

Correct Answer : A

Link is a branch of a graph that is not present in a tree that is required to make a loop. A tree is a part of a graph that has no closed part.

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1.

Consider the path loss in a certain situation is given by $\overline{PL\left(d\right)}=37+38{\mathrm{log}}_{10}\left(f\right)where\overline{PL\left(d\right)}$d,f is in dB, Km, and Hz respectively. Assume transmit power ${P}_{tx}=10watt$ and frequency of operation is 1GHz, What is the received power at a distance 10Km?

A)

-199.7 dBm

B)

-199.7 dB

C)

200dBm

D)

200 dB

View Answer

Correct Answer : A

$\overline{)PL\left(d\right)}=37+38{\mathrm{log}}_{10}\left(d\right)+18.3{\mathrm{log}}_{10}\left(f\right)\phantom{\rule{0ex}{0ex}}\overline{)PL\left(d\right)}=37+38{\mathrm{log}}_{10}\left(10\right)+18.3{\mathrm{log}}_{10}\left({10}^{9}\right)=239.7dB\phantom{\rule{0ex}{0ex}}ReceivedPower=10-229.7dB=-199.7dBm$

2.

The speed of a train is 100 Km/hr. The angle of arrival of the received signal relative to the direction of motion is 30°. The communication between the station and the train takes place at a frequency of 900MHz.What’s the received frequency when the train moves towards the station?

A)

900.000722 MHz

B)

900 MHz

C)

800 MHz

D)

700 MHz

View Answer

Correct Answer : A

$V=100\frac{Km}{hr}=\frac{100x1000m}{3600s}=\frac{250}{9}m/s\phantom{\rule{0ex}{0ex}}{f}_{d}=V*\frac{\mathrm{cos}\theta}{\lambda}=\frac{V{f}_{c}}{c}*\mathrm{cos}\theta =\frac{250x900x{10}^{6}x\mathrm{cos}(30\xb0)}{9x3x{10}^{8}}\phantom{\rule{0ex}{0ex}}Frequency=900x{10}^{6}+72.168=900.000722Mhz$

3.

The speed of a train is 100 Km/hr. The angle of arrival of the received signal relative to the direction of motion is 30°. The communication between the station and the train takes place at frequency of 900MHz.What’s the received frequency when the train moves towards the station?

A)

900.000722 MHz

B)

899.999927 MHz

C)

800 MHz

D)

700 MHz

View Answer

Correct Answer : B

$V=100\frac{Km}{hr}=\frac{100x1000m}{3600s}=\frac{250}{9}m/s\phantom{\rule{0ex}{0ex}}{f}_{d}=V*\frac{\mathrm{cos}\theta}{\lambda}=\frac{V{f}_{c}}{c}*Cos\left(\theta \right)=\frac{250x9x{10}^{6}xCos(30\xb0)}{9x3x{10}^{8}}\phantom{\rule{0ex}{0ex}}Frequencyis900x{10}^{6}-72.168=899.999927MHz$

4.

A channel with Rayleigh fading and average received power 20 dBm, What’s the outage probability that the received power is below 5 dBm.

A)

0.3

B)

0.2

C)

0.0311

D)

0.1

View Answer

Correct Answer : C

${{p}_{a}}^{2}=\frac{1}{{\Omega}_{p}}{e}^{-\frac{x}{{\Omega}_{p}}}\phantom{\rule{0ex}{0ex}}{\Omega}_{p}=20dBm=-10dB\phantom{\rule{0ex}{0ex}}or,{\Omega}_{p}={10}^{-1}\phantom{\rule{0ex}{0ex}}{P}_{o}={10}^{-2.5}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{{10}^{-2.5}}\frac{1}{{10}^{-1}}{e}^{-\frac{x}{{10}^{-1}}}dx=0.0311\phantom{\rule{0ex}{0ex}}$

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1.

For the operation of enhancement only N-channel MOSFET, value of Gate voltage has to be

A)

Positive

B)

Negative

C)

Zero

D)

Complex

View Answer

Correct Answer : A

The gate voltage must be positive for the formation of the channel in an N-channel MOSFET.

2.

MOS transistor

A)

Has only one p-junction

B)

Has only two electrodes

C)

Gate electrode is in direct contact with silicon

D)

Conducts when sufficient voltage is applied

View Answer

Correct Answer : D

The channel is formed and the current will flow when a threshold voltage is applied.

3.

The threshold voltage of a MOSFET can be lowered by

1) Using thinner gate oxide

2) Reducing the substrate concentration

3) Increasing the substrate concentration

A)

3 alone is correct

B)

1 and 2 are correct

C)

1 and 3 are correct

D)

2 alone is correct

View Answer

Correct Answer : C

Threshold voltage can be controlled by the gate oxide of the MOSFET and by substrate concentration

4.

The polarity of the inversion layer in a MOSFET is the same

A)

Charge on gate electrode

B)

Minority carriers in the drain

C)

Majority carriers in substrate

D)

Majority carrier. In source

View Answer

Correct Answer : D

The channel formed is of same type as source type

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1.

In the case of an antenna radiating in free space, the electric field at a distance of 1 km is found to be 12mV/m. Given that intrinsic impedance of the free space is 120 pW, the magnitude of average power density due to this antenna at a distance of 2 km from the antenna (in nW/${m}^{2}$ ) is____.

A)

15.2

B)

47.7

C)

24.7

D)

34.3

View Answer

Correct Answer : B

E at a distance of 1 km = 12 mv/m

E at distance of 2 km, = 6 mv/m

${P}_{avg}=\frac{1}{2}\frac{{E}^{2}}{\eta}=\frac{1}{2}*\frac{{(6*{10}^{-3})}^{2}}{377}=47.7nW/{m}^{2}$

2.

The radiation pattern of an antenna in spherical coordinates is given by

F(θ) = cos4θ, 0 ≤ θ ≤ π/2

The directivity of the antenna is

A)

10 dB

B)

12.6 dB

C)

11.5 dB

D)

18 dB

View Answer

Correct Answer : A

$D=\frac{\left(4\pi {U}_{max}\right)}{{\pi}_{rad}}\phantom{\rule{0ex}{0ex}}f\left(\theta \right)=co{s}^{4}\theta ,0\le \pi /2\phantom{\rule{0ex}{0ex}}{\pi}_{rad}={\int}_{0}^{2\pi}{\int}_{0}^{(\pi /2)}f\left(\theta \right)sin\theta d\phi d\theta \phantom{\rule{0ex}{0ex}}{\pi}_{rad}={\int}_{0}^{2\pi}{\int}_{0}^{(\pi /2)}co{s}^{4}\theta sin\theta d\phi d\theta \phantom{\rule{0ex}{0ex}}{\pi}_{rad}=\frac{2\pi}{5}\phantom{\rule{0ex}{0ex}}D=\frac{\left(4\pi {U}_{max}\right)}{\left({\displaystyle \frac{2\pi}{5}}\right)}=10{U}_{max}\phantom{\rule{0ex}{0ex}}D=10{\left[f\right(\theta \left)\right]}_{max}=10\phantom{\rule{0ex}{0ex}}D\left(dB\right)=10{{\mathrm{log}}_{10}}^{\left(10\right)}=10dB$

3.

For a Hertz dipole antenna, the Half Power Beam Width (HPBW) in the E-plane is

4.

A transmission line is feeding 1 Watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated

A)

10 Watts

B)

1 Watt

C)

0.1 Watt

D)

0.01 Watt

View Answer

Correct Answer : B

Gain of amplifier = 10 dB

But the gain of antenna = Directive gain

So, radiated power will be 1 watt.

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