Google News
logo
ECE : Communication Systems - Quiz(MCQ)
A)
A high gain D.C. amplifier
B)
A cathode follower stage
C)
A double-tuned amplifier
D)
None of the above

Correct Answer :   A high gain D.C. amplifier

A)
Space diversity
B)
Directional antenna
C)
Frequency diversity
D)
Broad band antenna

Correct Answer :   Frequency diversity

A)
half duplex arrangement
B)
duplex arrangement
C)
Either (A) or (B)
D)
Neither (A) nor (B)

Correct Answer :   duplex arrangement


Explanation : Separate frequencies for transmission from base and portable units allows two way transmission and is called duplex arrangement.

A)
resistance
B)
inductance
C)
shunt capacitance
D)
series capacitance

Correct Answer :   series capacitance


Explanation : Since Xc for high frequencies is low, the high frequencies are shunted to ground and are not transmitted.

A)
low pass filter
B)
high pass filter
C)
band stop filter
D)
band pass filter

Correct Answer :   low pass filter


Explanation : Woofer is a low frequency loud speaker covering the range 16 Hz to 500 Hz.

A)
Filter method
B)
Phase cancellation method
C)
Good attenuation characteristics
D)
All of the above

Correct Answer :   All of the above

A)
To reduce bandwidth
B)
To increase bandwidth
C)
To prevent overmodulation
D)
To regulate oscillator I/P voltage

Correct Answer :   To prevent overmodulation

A)
Clockwise
B)
Anticlockwise
C)
Mostly anticlockwise but some times clockwise
D)
Clockwise or anticlockwise depending on frequency of data stored

Correct Answer :   Anticlockwise

A)
upto 100 bps
B)
upto 250 bps
C)
upto 400 bps
D)
upto 600 bps

Correct Answer :   upto 600 bps


Explanation : When data rate in bits per second is upto 600, modem is low speed.

A)
it reduces the bandwidth requirement to half
B)
it results in better reception
C)
it avoids phase distortion at low frequencies
D)
None of the above

Correct Answer :   it reduces the bandwidth requirement to half


Explanation : VSB (vestigial side band) transmission transmits one side band fully and the other side band partially thus, reducing the bandwidth requirement.

A)
frequency changing
B)
frequency interleaving
C)
frequency adjustment
D)
frequency amalgamation

Correct Answer :   frequency interleaving


Explanation : In frequency interleaving the space between bundles is used for colour signal.

A)
does not depend on frequency
B)
increases as frequency is increased
C)
decreases as frequency is increased
D)
either (a) or (c) depending on the temperature

Correct Answer :   decreases as frequency is increased


Explanation : Atmospheric noise decreases as frequency is increased

A)
tuning
B)
detection
C)
rectification
D)
None of the above

Correct Answer :   tuning


Explanation : In varactor diode the applied reverse bias controls the width and therefore capacitance of depletion layer.

This capacitance is used for tuning.

A)
1
B)
2
C)
3
D)
4

Correct Answer :   3


Explanation : Shot noise, partition noise and thermal noise.

A)
1000 ohm
B)
700 ohm
C)
500 ohm
D)
300 ohm

Correct Answer :   300 ohm

A)
Is caused by reflection
B)
Is due to transverse nature of waves
C)
Is always vertical in an isotropic medium
D)
Results from longitudinal nature of waves

Correct Answer :   Is due to transverse nature of waves

A)
High frequency
B)
Medium frequency
C)
Low frequency
D)
Very high frequency

Correct Answer :   Medium frequency

A)
512 W
B)
588 W
C)
650 W
D)
750 W

Correct Answer :   512 W


Explanation :

A)
2
B)
4
C)
6
D)
8

Correct Answer :   4


Explanation : 24 = 16

A)
remains the same
B)
increases by 25%
C)
increases by 33.3%
D)
increases by 50%

Correct Answer :   increases by 33.3%


Explanation :

A)
FDM
B)
TDM
C)
PWD
D)
PCM

Correct Answer :   PCM


Explanation : An analog number cannot be converted into an exact digital number.

This is called quantizing error.

A)
CB
B)
CC
C)
CE
D)
CE or CC

Correct Answer :   CE


Explanation : Common emitter connection has high power gain, and good current and voltage gains.

A)
radiation
B)
dielectric heating
C)
conductor heating
D)
All of the above

Correct Answer :   All of the above


Explanation : All the three cause dissipation of energy as heat.

A)
1 V/m
B)
2 mV/m
C)
10 V/m
D)
50 mV/m

Correct Answer :   50 mV/m


Explanation :

A)
high boost to bass and little boost to treble
B)
equal boost to bass and treble
C)
less boost to both bass and treble
D)
less boost to bass and more boost to treble

Correct Answer :   high boost to bass and little boost to treble


Explanation : For depth in sound the low frequency notes, (i.e., bass) need more amplification.

A)
0.1 km
B)
1 km
C)
10 km
D)
50 km

Correct Answer :   50 km


Explanation : Mobile telephone has about 50 km range.

A)
a combination of disk and cone far apart
B)
a combination of disk and cone in close approximity
C)
a combination of disk and cone at λ/2 spacing
D)
a combination of disk and cone at λ/4 spacing

Correct Answer :   a combination of disk and cone in close approximity


Explanation : It is a combination of disc and cone in close approximity.

28 .
In a two tone AM system the two modulating frequencies are 2000 p and 4000 p rad/sec. If carrier frequency is 2p x 106 rad/sec the frequencies of upper sidebands are
A)
1.001 MHz and 1.004 MHz
B)
1.002 MHz and 1.004 MHz
C)
1.001 MHz and 1.002 MHz
D)
1.002 MHz and 1.008 MHz

Correct Answer :   1.001 MHz and 1.002 MHz


Explaination :

29 .
In a system an input resistance of 20 kΩ causes a noise voltage of 15μV. If two such input resistances each of 20 kΩ are used in parallel, the total noise voltage will be about
A)
7.5 μV
B)
10.6 μV
C)
15 μV
D)
30 μV

Correct Answer :   10.6 μV


Explaination :

A)
16 frequency bands
B)
8 frequency bands
C)
4 frequency bands
D)
2 frequency bands

Correct Answer :   4 frequency bands


Explanation : Dolby A system divides the frequency in 4 bands viz.f < 80 Hz, 80 Hz < f < 2999 Hz, f > 3000 Hz, f > 9000 Hz.

A)
25 dB
B)
50 dB
C)
75 dB
D)
100 dB

Correct Answer :   50 dB


Explanation : Hi-Fi system has a S/N ratio at least equal to 50 dB.

A)
25 times each but alternately
B)
50 times each but alternately
C)
25 times each
D)
50 times each

Correct Answer :   25 times each but alternately

A)
photo voltaic
B)
photo emission
C)
Both (A) and (B)
D)
photo conduction

Correct Answer :   photo conduction


Explanation : Photo conduction uses the fact that resistivity of some materials depends on intensity of light falling on their surface.

A)
2.9 kV
B)
4.5 kV
C)
29 kV
D)
45 kV

Correct Answer :   29 kV


Explanation :

A)
CW receivers
B)
TRF receivers
C)
pulsed receivers
D)
super heterodyne receivers

Correct Answer :   super heterodyne receivers


Explanation : Now-a-days only super heterodyne radio receivers are used.

A)
6
B)
1/6 times
C)
increase by 20 kHz
D)
increase by 30 kHz

Correct Answer :   6


Explanation : BW = 2fm(1 + β)
for β >> 1

B.W. a 2fm.

A)
low
B)
infinite
C)
zero
D)
zero or low

Correct Answer :   infinite


Explanation : For zero frequency capacitive reactance is infinite.

38 .
Consider the following test signal patterns

1. Ramp
2. Grill
3. Grey shades
4. Sine pulse
5. Picture line up generating equipment

Which of the above are used in TV studios?
A)
1, 2 and 3 only
B)
1, 2 and 5 only
C)
2, 3, 4 and 5 only
D)
All 1, 2, 3, 4 and 5

Correct Answer :   All 1, 2, 3, 4 and 5


Explaination : All these are standard test signals used in TV studios.

A)
open circuit
B)
short circuit
C)
complex impedance
D)
pure reactance

Correct Answer :   complex impedance


Explanation : If line is OC or SC or terminated in pure reactance, SWR is 1.

A)
to synchronise colours
B)
to maintain colour sequence
C)
for interlacing of horizontal lines
D)
to ensure I and Q phase correctly

Correct Answer :   to ensure I and Q phase correctly


Explanation : Colour burst consists of a short train of about 10 cycles sent to the receiver along with synchronous signals. It ensures I and Q phase correctly.

A)
10
B)
100
C)
0.01
D)
0.1

Correct Answer :   100


Explanation :

A)
14
B)
12
C)
10
D)
8

Correct Answer :   14


Explanation : The 14 pulses are preliminary pulse, start pulse, stop pulse and 11 other pulses.

A)
Saticon
B)
Vidicon
C)
New vicon
D)
Plumbicon

Correct Answer :   New vicon


Explanation : New vicon has very high sensitivity and very wide spectral response which extends well into infrared region.

44 .
A computer monitor uses 180 W in normal operation and 20 W in sleep mode. The computer is on all the 24 hours but is in sleep mode for 18 hours a days. The energy units saved in 30 days are
A)
0.54 units
B)
2.88 units
C)
4.8 units
D)
86.4 units

Correct Answer :   86.4 units


Explaination :

45 .
Consider the following functions

1. It detects the AM video carrier IF so as to produce the video signal
2. It produces and separates out the inter carrier sound IF of 5.5 MHz
3. It amplifies the video signal
4. It amplifies the audio signal

The video detector in a colour TV performs which of the above functions
A)
1 and 2
B)
1, 2 and 3
C)
1, 2 and 4
D)
1, 2, 3 and 4

Correct Answer :   1 and 2


Explaination : Video detector does not amplify any signal.

A)
horn
B)
helical
C)
parabolic
D)
log periodic

Correct Answer :   helical


Explanation : Helical antenna produces a circularly polarized radiation if its two components are at right angles and are equal. It is very commonly used in radio telemetry.

A)
pulse code modulated infrared light
B)
demodulation
C)
pulse code modulated ultra violet light
D)
Either (A) or (B)

Correct Answer :   pulse code modulated infrared light


Explanation : Infrared light is used in TV remote control.

A)
squelch
B)
double conversion
C)
variable selectivity
D)
variable sensitivity

Correct Answer :   variable sensitivity


Explanation : Variable sensitivity prevents overloading.

A)
increases
B)
decreases
C)
first increase, then decreases
D)
first decreases then increases

Correct Answer :   decreases


Explanation : When magnitude of reverse bias is increased, thickness of depletion layer increases.

Capacitance is inversely proportional to thickness.

A)
class A
B)
class B
C)
class C
D)
Class A or Class C

Correct Answer :   class B

52 .
24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM is
A)
1.68 MHz
B)
3.072 MHz
C)
81.6 kHz
D)
240 kHz

Correct Answer :   1.68 MHz


Explaination :

Bandwidth ≈ 24 (ln2 128) 10 kHz = 1.68 MHz.

A)
audio amplifiers
B)
RF amplifiers
C)
Both audio and RF amplifiers
D)
None of the above

Correct Answer :   RF amplifiers


Explanation : RF amplifiers use resonant loads (L and C in parallel).

A)
an inductance
B)
a resistance
C)
a capacitance
D)
a conductance

Correct Answer :   an inductance


Explanation : Mass has inertia and is analogous to inductance.

A)
rectification property of diode
B)
linear portion of static characteristics of diode
C)
linear portion of dynamic characteristic of diode
D)
square law portion of dynamic characteristics of diode

Correct Answer :   rectification property of diode


Explanation : Detection (also called demodulation) is due to rectification property of diode.

A)
555 kHz
B)
505 kHz
C)
500 kHz
D)
9 kHz

Correct Answer :   9 kHz


Explanation :

A)
Cable Antenna Television
B)
Channel Antenna Television
C)
Community Antenna Television
D)
Common Antenna Television

Correct Answer :   Community Antenna Television


Explanation : In USA the first cable service was called community antenna television.

A)
separate motor for each operation
B)
only one motor for all operations
C)
one motor for capstan and another for take up reel
D)
a total of 3 motors

Correct Answer :   only one motor for all operations


Explanation : For reasons of economy only one motor is used.

A)
is located at a height of 358, 00 km
B)
is motionless in space but keeps spinning
C)
appears stationary over earth's magnetic pole
D)
orbits the earth within a 24 hour period

Correct Answer :   orbits the earth within a 24 hour period


Explanation : It orbits the earth in a 24 hour period. Thus it appears to be stationed over one spot of the globe.

A)
38.5 W
B)
400 W
C)
512.5 W
D)
615.5 W

Correct Answer :   512.5 W


Explanation :

Sideband frequencies will be fc ± fm.

61 .
If sound intensity is 8 W/m2, the dB level of this sound is about
A)
5 dB
B)
50 dB
C)
129 dB
D)
1290 dB

Correct Answer :   129 dB

A)
1000 km
B)
10000 km
C)
36000 km
D)
55000 km

Correct Answer :   36000 km

A)
n
B)
2n
C)
Less than n
D)
greater than 2n

Correct Answer :   greater than 2n

A)
3 to 30 MHz
B)
30 to 300 MHz
C)
300 to 3000 MHz
D)
3000 to 30000 MHz

Correct Answer :   300 to 3000 MHz

A)
Carbon
B)
Condenser
C)
Ribbon
D)
Moving coil

Correct Answer :   Condenser

A)
shot noise
B)
flicker noise
C)
Johnson noise
D)
transit-time noise

Correct Answer :   transit-time noise

A)
has a poor efficiency
B)
has a higher efficiency
C)
requires higher audio power
D)
uses class C amplifier for amplification

Correct Answer :   has a poor efficiency

A)
DM
B)
PAM
C)
AM
D)
PCM

Correct Answer :   AM

A)
ASK
B)
FSK
C)
PSK
D)
All of the above

Correct Answer :   All of the above

71 .
A discrete zero memory information source has 40 symbols and each symbol is equally likely. The minimum number of bits required to code the source with uniform length code and entropy of the source are
A)
5 and 5.03
B)
5 and 6.64
C)
6 and 5.83
D)
6 and 5.32

Correct Answer :   6 and 5.32

A)
LC oscillator
B)
crystal oscillator
C)
RC oscillator
D)
Either (A) or (B)

Correct Answer :   crystal oscillator

A)
10.7 MHz
B)
4.5 MHz
C)
75 kHz
D)
455.5 kHz

Correct Answer :   10.7 MHz

A)
one electron gun
B)
three electron guns
C)
one electron gun with three separate in line cathodes
D)
None of the above

Correct Answer :   one electron gun with three separate in line cathodes

A)
1"
B)
1/2"
C)
1/4"
D)
1/8"

Correct Answer :   1/4"


Explanation : Home tape recorders use 1/4 in (i.e., 6.3 mm) tape having two tracks of 2.5 mm width each with an in between separation of 1.3 mm.

A)
1 W
B)
0.1 W
C)
10⁻⁵ W
D)
10⁻¹² W

Correct Answer :   10⁻¹² W


Explanation : It can detect very low power signals.

A)
25 W
B)
50 W
C)
66 W
D)
100 W

Correct Answer :   25 W


Explanation :

A)
0.5 kV
B)
12 kV
C)
24 kV
D)
32 kV

Correct Answer :   24 kV


Explanation : Peak RF voltage delivered to load = 2 (8 + 4) = 24 kV.

81 .
Consider the following

1. Generation of SSB signals
2. Design of minimum phase type filters
3. Representation of band pass signals

Which of the above applications is Hilbert transform used?
A)
1 only
B)
1, 2 only
C)
1 and 3
D)
1, 2 and 3

Correct Answer :   1, 2 and 3


Explaination : Hilbert transform is used in all the applications.

A)
Vidicon
B)
Plumbicon
C)
Saticon
D)
GCD imager

Correct Answer :   Plumbicon


Explanation : Dark current is the small signal current even when there is no illumination on the face plate of camera tube.

A)
0.66 m
B)
6.6 m
C)
66 m
D)
0.066 m

Correct Answer :   6.6 m

A)
1 kHz
B)
3 kHz
C)
10 kHz
D)
50 kHz

Correct Answer :   3 kHz

A)
A fundamental sine wave and harmonics
B)
Fundamental and subharmonic sine waves
C)
A fundamental sine wave and even harmonics
D)
A fundamental sine wave and odd harmonics

Correct Answer :   A fundamental sine wave and odd harmonics

A)
35 kHz
B)
70 kHz
C)
1280 kHz
D)
1520 kHz

Correct Answer :   35 kHz

A)
Reactance FET modulator
B)
Varactor diode modulator
C)
Armstrong modulator
D)
Reactance bipolar transistor

Correct Answer :   Armstrong modulator

A)
88 MHz
B)
100 kHz
C)
108 MHz
D)
150 kHz

Correct Answer :   150 kHz


Explanation : Frequency Modulation (FM) is a method of transmitting information (such as audio signals) by varying the frequency of the carrier wave in accordance with the modulating signal. In commercial FM radio broadcasting, the audio signal (voice or music) is used to modulate the carrier frequency, and this modulated signal is transmitted over the airwaves to be received by FM radios.

The channel bandwidth refers to the range of frequencies that the FM signal occupies in the radio frequency spectrum. In FM broadcasting, the signal is allocated a specific frequency band, and this band needs to be wide enough to accommodate the modulated signal.

The channel bandwidth is influenced by various factors, including the audio quality, the data rate required for the transmission, and the interference considerations to ensure proper reception. In commercial FM broadcasting, a bandwidth of approximately 150 kHz is chosen as a trade-off between audio quality and spectrum efficiency.

Here's why :

1. Audio Quality: A wider bandwidth allows for higher fidelity audio transmission. The wider the bandwidth, the more accurately the original audio signal can be reconstructed at the receiving end. A 150 kHz bandwidth is considered sufficient to provide good audio quality for commercial FM radio broadcasting.

2. Spectrum Efficiency: On the other hand, allocating too much bandwidth for each FM channel would be inefficient in terms of spectrum usage. It would limit the number of radio stations that could coexist within the available radio frequency spectrum, leading to congestion and reduced choices for listeners.

3. Interference: FM channels are typically spaced apart in the radio frequency spectrum to avoid interference between adjacent channels. Allocating a wider bandwidth for each FM channel would require more significant spacing between channels, further reducing the available spectrum for broadcasting.

For these reasons, a practical compromise is made by selecting a channel bandwidth of around 150 kHz in commercial FM systems. This allows for good audio quality while optimizing spectrum utilization and providing enough spacing between channels to minimize interference.

A)
ASK
B)
FSK
C)
PSK
D)
None of the above

Correct Answer :   PSK

A)
very sensitive receiver
B)
fully steerable antenna
C)
high transmitting power
D)
All of the above

Correct Answer :   All of the above

A)
Pulse Repetition Frequency
B)
Pulse Return Factor
C)
Power Return Factor
D)
Pulse Response Factor

Correct Answer :   Pulse Repetition Frequency


Explanation : PRF in radar systems means pulse repetition frequency.

A)
x(t) = cos t + 0.5
B)
x(t) = 2 cos t + 3 cos 3t
C)
x(t) = 2 cos pt + 7 cos t
D)
x(t) = 2 cos 1.5pt + sin 3.5 pt

Correct Answer :   x(t) = cos t + 0.5


Explanation : x(t) = cos t + 0.5

not satisfies the Dirichlet condition.

The integration of constant term is ∞.

A)
Reactance FM modulator
B)
Armstrong modulator
C)
Varactor diode modulator
D)
Reactance bipolar transistor modulator

Correct Answer :   Armstrong modulator


Explanation : It generates FM through phase modulation.

A)
0.1
B)
0.81
C)
0.9
D)
10/9

Correct Answer :   10/9


Explanation : Calling rate is the number of calls per subscriber.

A)
10 times
B)
20 times
C)
50 times
D)
100 times

Correct Answer :   50 times


Explanation : Each frame is divided into odd and even fields. Each field is scanned 25 times so that each frame is scanned 50 times.

A)
0.33 μs
B)
3.3 μs
C)
33 μs
D)
330 μs

Correct Answer :   3.3 μs


Explanation : Since velocity of em waves is 300 m/ms, each kilometer means time delay of 3.3 μs.

A)
33.3%
B)
16.65%
C)
50%
D)
100%

Correct Answer :   33.3%


Explanation : Since signal power = 0.5 PC and total power is 1.5 PC efficiency is 33.3%.

A)
to search submarines
B)
aim a gun at an aircraft
C)
to direct guided missiles
D)
All of the above

Correct Answer :   All of the above


Explanation : Radar is used for all the three purposes.

A)
End fire array
B)
Folded dipole
C)
Rhombic antenna
D)
Broad side array

Correct Answer :   Rhombic antenna


Explanation : Rhombic antenna is a non-resonant antenna capable of operating over 3 to 30 MHz range

A)
71.2 kHz to 71.20 kHz
B)
692 kHz to 732 kHz
C)
702 kHz to 722 kHz
D)
711.9 kHz to 712.1 kHz

Correct Answer :   702 kHz to 722 kHz


Explanation :

Sideband frequencies will be fc ± fm.

A)
About 71%
B)
About 83%
C)
About 92%
D)
100%

Correct Answer :   100%

A)
2 : 1
B)
4 : 1
C)
8 : 1
D)
16 : 1

Correct Answer :   4 : 1


Explanation : Balun has turn ratio of 2 : 1. Therefore, impedance transformation is 4 : 1.

104 .
If γ = (S0/N0)/(Si/Nm) where S0 and S1 are signal power output and signal power input and N0 and Nm are noise power output and noise power input, the value of γ for SSB-SC and DSB-SC amplitude modulated system respectively are
A)
1 and 1
B)
1 and 2
C)
2 and 1
D)
0.5 and 0.5

Correct Answer :   1 and 1


Explaination : SSB - SC and SSB - DB have γ = 1 and are similar from noise point.

A)
Helical
B)
Yagi-Uda
C)
Parabolic reflector
D)
Small circular loop

Correct Answer :   Helical


Explanation : The helical antenna is circularly polarized, i.e., its polarization is evenly divided between vertical and horizontal components.

A)
5 W
B)
25 W
C)
100 W
D)
500 W

Correct Answer :   500 W


Explanation :

A)
50
B)
66.6
C)
75
D)
100

Correct Answer :   66.6


Explanation :

A)
less than 20 km
B)
less than 100 km
C)
less than 500 km
D)
less than 1000 km

Correct Answer :   less than 100 km


Explanation : Television transmission is governed by line of sight distance which is about 100 km or less.

A)
bandwidth of frequencies having output with in ± 1 dB of the output at 500 Hz
B)
bandwidth of frequencies having output with in ± 3 dB of the output at 500 Hz
C)
bandwidth of frequencies having output with in ± 1 dB of the output at 1000 Hz
D)
bandwidth of frequencies having output with in ± 3 dB of the output at 1000 Hz

Correct Answer :   bandwidth of frequencies having output with in ± 1 dB of the output at 1000 Hz


Explanation : 100 Hz is taken as standard for specifying the output.

A)
1 band
B)
3 or 4 bands
C)
more than 8 bands
D)
large number of bands

Correct Answer :   3 or 4 bands


Explanation : Stereo systems have 3 or 4 bands at 300 Hz, 1000 Hz, 10000 Hz etc. in graphic equaliser system.

111 .
Some sources of TV programs are

1. TV camera
2. Telecine
3. VTR
4. External signal

Which of the above are actually present in a TV studio?
A)
1 and 3
B)
1, 2, 3
C)
1 2 and 4
D)
1, 2, 3 and 4

Correct Answer :   1, 2, 3 and 4

112 .
Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is
A)
a linear function
B)
0.1 kHz sinusoidal
C)
10.1 kHz sinusoidal
D)
None of the above

Correct Answer :  

Y(t) = A cos (2p x 104 t) + A cos (2p x 10.1 x 103 t)

when passed through high pass filter,

Output will be A cos (2p x 10.1 x 103 t) only.

A)
is more in darkness and less in brightness
B)
is more in brightness but less in darkness and brightness
C)
may be more or less in darkness and brightness
D)
None of the above

Correct Answer :   is more in darkness and less in brightness


Explanation : Light dependent resistor (LDR) has high resistance in darkness and low resistance in brightness.

A)
1°
B)
6°
C)
10°
D)
20°

Correct Answer :   6°


Explanation : Tilting of heads is done to reduce cross-talk.

A)
x(t - 2)
B)
(t + 2)
C)
x(t - 12)
D)
x(t + 12)

Correct Answer :   x(t - 2)


Explanation :

A)
Ferric oxide
B)
Ferrous oxide
C)
Chromium di oxide
D)
Either (A) or (B)

Correct Answer :   Chromium di oxide


Explanation : Chromium di oxide has very good retentivity.

117 .
An angle modulated signal is given by s(t) = cos 2p (2 x 106t + 30 sin 150t + 40 cos 150t). The maximum frequency and phase deviations of s(t) are
A)
6 kHz and 80π rad
B)
7.5 kHz, 100π rad
C)
10.5 kHz, 100π rad
D)
10.5 kHz, 140π rad

Correct Answer :   7.5 kHz, 100π rad


Explaination :

A)
third method
B)
filter method
C)
phase shift method
D)
Both (A) and (B)

Correct Answer :   filter method


Explanation : Unwanted frequency is removed by filtering.

A)
series inductance
B)
shunt inductance
C)
shunt capacitance
D)
series capacitance

Correct Answer :   series capacitance


Explanation : Series capacitance has high reactance for low frequencies.

A)
2
B)
4
C)
4.5
D)
7.3

Correct Answer :   2

A)
16 kHz and 20 kHz
B)
20 kHz and 32 kHz
C)
20 kHz and 40 kHz
D)
32 kHz and 40 kHz

Correct Answer :   32 kHz and 40 kHz


Explanation :

A)
- 0.7 R + 0.59 G + 0.11 B
B)
1.3 R - 0.59 G - 0.11 B
C)
0.7 R - 0.59 G - 0.11 B
D)
None of the above

Correct Answer :   0.7 R - 0.59 G - 0.11 B


Explanation : R - Y = R - 0.3R - 0.59C - 0.11B = 0.7R - 0.59G - 0.11B.

A)
Low Q, LC resonant circuit
B)
High Q, LC resonant circuit
C)
Low Q, RLC resonant circuit
D)
High Q, RLC resonant circuit

Correct Answer :   High Q, LC resonant circuit


Explanation : High Q LC resonant circuit provides high selectivity.

A)
998.0 kHz
B)
999.2 kHz
C)
1000 kHz
D)
1002 kHz

Correct Answer :   1000 kHz


Explanation :

Frequency present in the sidebands is equal to = fc ± fm, fc ± 2fm, fc ± 3fm .

A)
0.002
B)
0.02
C)
0.001
D)
0.01

Correct Answer :   0.002


Explanation : Duty Cycle = 2k/1000k = 0.002

A)
0.5 W
B)
5W
C)
50 W
D)
500 W

Correct Answer :   50 W


Explanation :

25 x 103 x 2000 x 10-6 = 50 W.

127 .
Some advantages of optical fibre cables are

1. small diameter
2. Immunity to cross talk and EM interference
3. Laser and LED modulation methods lend themselves ideally to digital operation

Which of the above are correct?
A)
1 only
B)
1 and 2
C)
2 and 3
D)
1, 2, 3

Correct Answer :   1, 2, 3


Explaination : Fibre optic cables have all these advantages.

A)
SAW filter
B)
video detector
C)
chroma amplifier
D)
luminance amplifier

Correct Answer :   luminance amplifier


Explanation : Luminance amplifier changes the luminance levels and provides contrast control.

A)
outside the medium band
B)
within the medium band
C)
depend upon modulation index
D)
depends on Modulating frequency

Correct Answer :   within the medium band


Explanation : Image signal frequency = Received signal frequency - 2 x Intermediate frequency.

= 1 MHz - 2 x 456kHz

=> 1 MHz - 912 kHz => 88 kHz

which lies in medium bond.

A)
precise frequency value
B)
low frequency drift
C)
Both (A) and (B)
D)
Neither (A) nor (B)

Correct Answer :   Both (A) and (B)

A)
2
B)
4
C)
8
D)
16

Correct Answer :   4

A)
dc resistance
B)
ac resistance
C)
Either (A) or (B)
D)
None of the above

Correct Answer :   ac resistance

A)
in phase component
B)
the envelope
C)
quadrature component
D)
None of the above

Correct Answer :   in phase component

A)
PCM
B)
PFM
C)
PWM
D)
PAM

Correct Answer :   PAM

A)
similar phase spectrum
B)
both have similar magnitude spectrum
C)
no relation
D)
both have similar magnitude and Phase spectrum

Correct Answer :   both have similar magnitude spectrum

A)
the type of wires cables used for the system
B)
the quality of sound reception at the receiver telephone
C)
the probability that a called for connection is established
D)
None of the above

Correct Answer :   the probability that a called for connection is established

A)
before the detector stage
B)
after the detector stage
C)
either before or after the detector stage
D)
None of the above

Correct Answer :   before the detector stage

A)
small carrier power
B)
very small carrier power
C)
large carrier power
D)
very large carrier power

Correct Answer :   very large carrier power

A)
twisted nematic crystal
B)
nematic crystals
C)
sematic crystals
D)
cholesteric crystals

Correct Answer :   twisted nematic crystal

A)
decrease, the modulation index increases
B)
remains constant, the modulation index increases
C)
increases, the modulation index increases
D)
remains constant, the modulation index decreases

Correct Answer :   remains constant, the modulation index increases

A)
to amplify the received signal
B)
to provide gain to the transmitted power
C)
to permit the use of same antenna for transmission and reception
D)
Both (A) and (B)

Correct Answer :   to permit the use of same antenna for transmission and reception

A)
after detection
B)
after modulation
C)
before detection
D)
before modulation

Correct Answer :   before modulation

A)
carrier frequency
B)
message frequency
C)
channel frequency
D)
modulation frequency

Correct Answer :   carrier frequency

A)
low frequencies
B)
high frequencies
C)
very high frequencies
D)
intermediate frequencies

Correct Answer :   intermediate frequencies

A)
AM
B)
FM
C)
AM-SC
D)
Phase modulation

Correct Answer :   AM-SC

A)
Product detector
B)
Balanced modulator
C)
Diode balanced modulator
D)
Complete phase shift generator

Correct Answer :   Balanced modulator

A)
unity
B)
nearly 0.67
C)
less than unity
D)
greater than unity

Correct Answer :   greater than unity

A)
prior to modulation
B)
after demodulation
C)
for low frequency components of the signal
D)
None of the above

Correct Answer :   prior to modulation

A)
Multiplexed Telemetry Index
B)
Moving Target Indicator
C)
Maximum Transmission Index
D)
Microprocessor Tracking Index

Correct Answer :   Moving Target Indicator

A)
Crystal oscillator
B)
Multivibrator
C)
Clapp oscillator
D)
Phase shift circuit

Correct Answer :   Crystal oscillator

A)
38.9 MHz for both video and audio
B)
33.4 MHz for both video and audio
C)
38.9 MHz for video and 33.4 MHz for audio
D)
33.4 MHz for video and 38.9 MHz for audio

Correct Answer :   38.9 MHz for video and 33.4 MHz for audio

A)
FM
B)
PCM
C)
SSB suppressed carrier
D)
pulse position modulation

Correct Answer :   PCM

A)
PAM
B)
PPM
C)
PCM
D)
PWM

Correct Answer :   PCM

A)
Shot noise
B)
Impulse noise
C)
Random noise
D)
Transit time noise

Correct Answer :   Transit time noise


Explanation : At high frequencies time period of wave and transit time become comparable.

A)
1
B)
2
C)
4
D)
6

Correct Answer :   2


Explanation : Two IF amplifier stages are very commonly used.

A)
microwave links
B)
open wire link
C)
coaxial cable link
D)
Either (A) or (B)

Correct Answer :   microwave links

A)
2
B)
3 erlang
C)
3.333 erlang
D)
0.333 erlang

Correct Answer :   3.333 erlang


Explanation :

A)
zero directive gain
B)
minimum directive gain
C)
Both (A) and (B)
D)
maximum directive gain

Correct Answer :   maximum directive gain


Explanation : Directivity means maximum directive gain.

A)
dry cells
B)
AC mains
C)
pencil cells
D)
rechargeable cells

Correct Answer :   rechargeable cells

A)
100%
B)
94.4%
C)
83.3%
D)
50%

Correct Answer :   94.4%

A)
tree configuration
B)
star configuration
C)
loop/mesh configuration
D)
All of the above

Correct Answer :   All of the above

A)
Mixer
B)
IF amplifier
C)
Detector
D)
Local oscillator

Correct Answer :   Detector


Explanation : Detection (also called demodulation) is an essential function.

A)
double the highest frequency in audio signal
B)
half the highest frequency in audio signal
C)
equal to highest frequency in audio signal
D)
four times the highest frequency in audio signal

Correct Answer :   double the highest frequency in audio signal


Explanation : For proper sampling and faithful reproduction sampling frequency must be at least twice the highest frequency in the signal.

A)
Better linearity
B)
Improved efficiency
C)
High power output per transistor
D)
The lower modulating power requirement

Correct Answer :   The lower modulating power requirement

A)
Increase the bandwidth
B)
Prevent excessive grid current
C)
Prevent over modulation
D)
Prevent tuned circuit damping

Correct Answer :   Prevent excessive grid current

A)
Costly
B)
Cheaper
C)
Almost equally costly
D)
None of the above

Correct Answer :   Costly

A)
1.07 MHz
B)
10.7 MHz
C)
106 kHz
D)
455 kHz

Correct Answer :   455 kHz

A)
Large bandwidth is required
B)
Quantising noise can be overcome by companding
C)
Suffers from the disadvantage of its incompatibly with TDM
D)
Quantising noise can be reduced by decreasing the number of standard levels

Correct Answer :   Large bandwidth is required

171 .
An audio signal (say from 50 Hz to 10000 Hz) is frequency translated by a carrier having a frequency of 106 Hz. The values of initial (without frequency translation) and final (after frequency translation) fractional change in frequency from one band edge to the other are
A)
200 and 200
B)
200 and 100.1
C)
200 and 1.01
D)
200 and 10.01

Correct Answer :   200 and 1.01

172 .
A)
Video head for recording and laser beam for playback
B)
Laser beam for recording and video head for playback
C)
Laser beam for both recording and playback
D)
None of the above

Correct Answer :   Laser beam for both recording and playback

A)
Zero capacity
B)
Infinite capacity
C)
Small capacity
D)
None of the above

Correct Answer :   Infinite capacity

A)
Is less than that of dipole
B)
Is more than that of dipole
C)
Is the same as that of dipole
D)
May be equal, more or less than that of dipole

Correct Answer :   Is more than that of dipole

A)
Connect a booster
B)
Twist the transmission line
C)
Use a longer transmission line
D)
Change the antenna orientation of location

Correct Answer :   Change the antenna orientation of location

A)
Is useful in reducing depolarization effect on received wave
B)
Is useful in discrimination between reception of adjacent beams
C)
Involves critical alignment of transmitting and receiving antenna
D)
None of the above

Correct Answer :   Is useful in reducing depolarization effect on received wave

A)
increases
B)
decreases
C)
remains the same
D)
Either (A) or (B)

Correct Answer :   increases


Explanation : Because of this reason ground waves can be used for frequencies upto about 1600 kHz only.

A)
varies directly as temperature
B)
is constant at all temperatures
C)
varies inversely as absolute temperature
D)
varies directly as square root of absolute temperature

Correct Answer :   varies directly as square root of absolute temperature

A)
10 m
B)
100 m
C)
500 m
D)
1000 m

Correct Answer :   100 m


Explanation : The range depends on quality of instrument and is about 100 m.

A)
Adding magnetism in mask and mounting frame
B)
Increasing resistance of coils to decrease time constant
C)
Removing residual magnetism in mask, shield etc...
D)
Decreasing resistance of coils to increase time constant.

Correct Answer :   Removing residual magnetism in mask, shield etc...


Explanation : Since gauss is a unit of magnetic flux, degaussing means removal of magnetism.

A)
bandwidth
B)
carrier frequency
C)
transmission loss
D)
transmitted power

Correct Answer :   bandwidth


Explanation : Rate of information depends on bandwidth.

A)
0.1
B)
0.5
C)
0.7
D)
0.9

Correct Answer :   0.5


Explanation :

A)
series traps
B)
shunt traps
C)
absorption traps
D)
All of the above

Correct Answer :   All of the above

A)
voltage and current nodes coincide
B)
voltage and current antinodes coincide
C)
voltage nodes and current antinodes as well as current nodes and voltage antinodes coincide
D)
Both (A) and (B)

Correct Answer :   voltage nodes and current antinodes as well as current nodes and voltage antinodes coincide


Explanation : When standing wave is set up voltage maxima and current minima and voltage minima and current maxima coincide.

A)
telegraphy
B)
telephony
C)
radio transmission
D)
satellite communication

Correct Answer :   telegraphy


Explanation : Frequency shift keying (FSK) is a system of frequency modulation used in telegraphy.

A)
8
B)
12
C)
16
D)
8/6

Correct Answer :   16


Explanation :

Given n1 = 6, n2 = 8,

then (SNR)Q = 22(n2 - n1)  =>16.

A)
increased by 6 dB
B)
increased by 12 dB
C)
decreased by 6 dB
D)
decreased by 12 dB

Correct Answer :   increased by 12 dB

A)
directly windings
B)
through a transformer having two rotating windings
C)
through a transformer having two stationary windings
D)
through a transformer having one stationary and one rotatory winding

Correct Answer :   through a transformer having one stationary and one rotatory winding


Explanation : Rotatory transformer, in VCR, has one stationary and one rotating winding.

A)
0.5 μV
B)
1 μV
C)
100 μV
D)
1000 μV

Correct Answer :   1000 μV


Explanation : It is about 1 mV.

A)
1.299 MHz
B)
1.54 MHz
C)
2 MHz
D)
192 kHz

Correct Answer :   1.54 MHz


Explanation :

Given, n = 24, fm = 3.4 kHz

M = 128,

2N = 128 => n = 7

But fs = 2fm, here we will consider fs (sampling) frequency

instead at 2fm => 2 x 3.4 kHz => 6.8 KHz.

B.W. = [24(7 + 1)] 8 kHz = 1.54 MHz.

A)
two stationary video heads
B)
four stationary video heads
C)
two rotating video heads
D)
four rotating video heads

Correct Answer :   two rotating video heads


Explanation : VCR has two rotating leads about 180° apart.

A)
64
B)
256
C)
512
D)
1024

Correct Answer :   1024


Explanation : 210 = 1024.

A)
0.7
B)
3
C)
5
D)
7

Correct Answer :   0.7


Explanation : Kell factor indicates reduction in vertical resolution and varies from 0.65 to 0.85.

A)
1 μV/m
B)
2 μV/m
C)
5 μV/m
D)
10 μV/m

Correct Answer :   1 μV/m

A)
0.1
B)
0.2
C)
0.4
D)
0.7

Correct Answer :   0.2


Explanation :

A)
a capacitance in series and a resistance in parallel
B)
a resistance in series and a capacitance in parallel
C)
a capacitance in series and an inductance in parallel
D)
an inductance in series and a capacitance in parallel

Correct Answer :   a capacitance in series and an inductance in parallel


Explanation : Series capacitance and shunt inductance is high pass circuit.

200 .
A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is as
A)
1500 bits/sec
B)
3000 bits/sec
C)
4500 bits/sec
D)
6000 bits/sec

Correct Answer :   4500 bits/sec


Explaination :

A)
frequency modulation
B)
amplitude modulation
C)
either amplitude or frequency modulation
D)
neither amplitude nor frequency modulation

Correct Answer :   amplitude modulation

A)
3 W
B)
30 W
C)
300 W
D)
500 W

Correct Answer :   30 W


Explanation :

P = 20 x 103 x 1500 x 10-6 = 30 W.

A)
degrees
B)
metres
C)
ratio of two angles
D)
ratio of two powers

Correct Answer :   degrees


Explanation : Beam width is the angular separation between the two half power points on the power density radiation pattern. It is expressed in degrees.

A)
0.32 μs
B)
3.2 μs
C)
32 μs
D)
320 μs

Correct Answer :   32 μs

A)
grid
B)
any anode
C)
Both (A) and (B)
D)
cathode

Correct Answer :   cathode


Explanation : Video signal is applied to cathode.

A)
magnetic recording and retrieval
B)
optical recording and retrieval
C)
optical recording but magnetic retrieval
D)
magnetic recording but optical retrieval

Correct Answer :   optical recording and retrieval


Explanation : CD uses optical recording and retrieval.

207 .
A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is
A)
21 kHz
B)
22 kHz
C)
42 kHz
D)
44 kHz

Correct Answer :   44 kHz


Explaination :

A)
after detector
B)
before mixer
C)
before detector
D)
after power amplifier

Correct Answer :   after detector


Explanation : Squelch, i.e., muting is added after detector to suppress noise when no carrier is present at input.

A)
1000.3 kHz
B)
999.7 kHz
C)
998 kHz
D)
700 kHz

Correct Answer :   700 kHz


Explanation : 1000 ± 0.3 and 1000 ± 2 kHz will be present.

A)
linear from edge to centre
B)
nonlinear from centre to edge
C)
linear from centre towards edge
D)
nonlinear from edge to center

Correct Answer :   linear from centre towards edge


Explanation : It is linear from centre to edge.

A)
used in UHF range
B)
used in radar
C)
circularly polarized
D)
used as a direction finding antenna

Correct Answer :   used in UHF range


Explanation : Discone antenna is a combination of disc and cone. It is omnidirectional and is used in VHF and UHF range especially at airports.

A)
5 kHz
B)
35 kHz
C)
70 kHz
D)
5 MHz

Correct Answer :   35 kHz

A)
53 kV
B)
40 kV
C)
20 kV
D)
10 W

Correct Answer :   20 kV


Explanation :

214 .
A 1000 kHz carrier wave modulated 40% at 40000 Hz is applied to a resonant circuit tuned to a carrier frequency and having Q = 140. What is the degree of modulator after transmission through the circuit?
A)
40%
B)
20%
C)
0.54
D)
0.27

Correct Answer :   0.27


Explaination :

A)
about 1 kV
B)
about 15 kV
C)
about 100 kV
D)
about 1000 kV

Correct Answer :   about 15 kV


Explanation : EHT voltage is 1 kV for each diagonal inch of CRT screen.

A)
45 two way communications
B)
45 one way communications
C)
90 two way communications
D)
180 two way communications

Correct Answer :   45 two way communications


Explanation : Each cell is designed for 45 two way conversations.

A)
8 kHz
B)
21 kHz
C)
27 kHz
D)
72 kHz

Correct Answer :   72 kHz


Explanation : Since SSB modulation is used, bandwidth = 24 x 3 = 72 kHz.

A)
100 W
B)
1000 W
C)
10000 W
D)
100000 W

Correct Answer :   10000 W


Explanation :

A)
balun
B)
slotted line
C)
directional coupler
D)
quarter wave transformer

Correct Answer :   balun


Explanation : Balun gives 4 : 1 impedance transformation and is suited for coupling a coaxial line to two wire line.

A)
3000 bps
B)
6000 bps
C)
9000 bps
D)
12000 bps

Correct Answer :   6000 bps


Explanation : For 3000 Hz.

Nyquist rate is twice, i.e., 6000 bps.

A)
1000 bps
B)
5000 bps
C)
20000 bps
D)
50000 bps

Correct Answer :   5000 bps


Explanation : When data rate is in the range of 2400 to about 10800 bits per second, modem is designated as high speed.

A)
preliminary pulse
B)
start pulse
C)
erase pulse
D)
either start pulse or preliminary pulse

Correct Answer :   preliminary pulse


Explanation : The first pulse is preliminary pulse followed by start pulse.

A)
poor AF response
B)
diagonal clipping
C)
poor AGC operation
D)
negative peak clipping

Correct Answer :   negative peak clipping

A)
low pass filter
B)
high pass filter
C)
band pass filter
D)
band stop filter

Correct Answer :   high pass filter


Explanation : Tweeter is high frequency loudspeaker covering the range above 5000 Hz.

A)
3000 bps
B)
6000 bps
C)
12000 bps
D)
None of the above

Correct Answer :   12000 bps


Explanation :

Noiseless channel: Nyquist's Theorem - If the signal has V discrete levels over a transmission medium of bandwidth H, the maximum data rate = 2H log2 V bits/sec.

A noiseless 3-kHz channel cannot transmit binary signals at a rate exceeding 6000 bps (= 2 x 3000 log2 2).

A)
25%
B)
50%
C)
66.6%
D)
83.3%

Correct Answer :   83.3%

A)
only erase head coil
B)
only record head coil
C)
coils of both erase head and record head
D)
None of the above

Correct Answer :   only erase head coil

A)
1.3 mm
B)
130 mm
C)
13 mm
D)
None of the above

Correct Answer :   13 mm

A)
SIOs
B)
Fading
C)
Faradays rotation
D)
Atmospheric storms

Correct Answer :   Fading


Explanation : Tropospheric scatter propagation is subject to two types of fading called Rayleigh fading (due to multipath propagation) and fading due to variations in atmospheric conditions along the propagation path.

A)
right to left
B)
left to right
C)
left to right for odd fields and right to left for even field
D)
left to right for even fields and right to left for odd field

Correct Answer :   left to right