Online Electronics and Communication Engineering Test - Communication Systems Test 3

Instruction:

  • This is a FREE online test. Beware of scammers who ask for money to attend this test.
  • Total number of questions: 20.
  • Time allotted: 30 minutes.
  • Each question carries 1 mark; there are no negative marks.
  • DO NOT refresh the page.
  • All the best!

Marks : 2/20


Total number of questions
20
Number of answered questions
0
Number of unanswered questions
20
Test Review : View answers and explanation for this test.

1.
In CCIR-B system, the time between start of one H syn. pulse and next is
64 μs
6.4 μs
640 μs
0.64 μs
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

The frequency of H pulses is 15625 Hz.

Hence .


2.
In a CD player the speed of CD is
constant at 500 rpm
constant at 200 rpm
varies from 200 to 500 rpm
none of the above
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

Since the circumference of outer tracks is more than that of tracks near the centre, the speed of disc is varied from 200 rpm to 500 rpm.


3.
In a 100% amplitude modulated signal, the power in the upper sideband when carrier power is to be 100 W and modulation system SSBSC, is
100 W
66 W
50 W
25 W
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

Modulation index = 100% 1

Pc = 100 W


4.
Trinitron is a monochrome picture tube.
True
False
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

Frequency band for local programs is fixed.


5.
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
0.1 kHz sinusoidal
10.1 kHz sinusoidal
a linear function
a constant
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

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.


6.
In cellular telephone each cell is designed to handle
45 two way communications
90 two way communications
45 one way communications
180 two way communications
Your Answer: Option
(Not Answered)
Correct Answer: Option
Explanation:

Each cell is designed for 45 two way conversations.


7.
Which is a digital coding technique?
PCM
DM
DPCM
All
Your Answer: Option
(Not Answered)
Correct Answer: Option

8.
The IF frequencies in a TV receiver
38.9 MHz for video and 33.4 MHz for audio
38.9 MHz for both video and audio
33.4 MHz for both video and audio
33.4 MHz for video and 38.9 MHz for audio
Your Answer: Option
(Not Answered)
Correct Answer: Option

9.
In FDM systems used for telephone, which modulation scheme is adopted?
AM
DSB-SC
SSB-SC
FM
Your Answer: Option
(Not Answered)
Correct Answer: Option

10.
Noise can be reduced by increasing sampling rate
true
false
can't say
insufficient data
Your Answer: Option
(Not Answered)
Correct Answer: Option

11.
In cell communication, cell stands for
Geographical Area
Battery
Location
No meaning
Your Answer: Option
(Not Answered)
Correct Answer: Option

12.
If the circumference of a spherical target is between 1 to 10 wavelength in a radar system, the situation is called
resonance region
Rayleigh region
optical region
binomial region
Your Answer: Option
(Not Answered)
Correct Answer: Option

13.
The maximum permissible distance between two samples of a 10 kHz signal
100 μ sec
50 μ sec
10 μ sec
none
Your Answer: Option
(Not Answered)
Correct Answer: Option

14.
Square law modulators utilise
linear range of V-I characteristics of a diode
linear range of V-I characteristics of a triode
non-linear region of V-I dynamic characteristics of a diode
non-linear region of V-I dynamic characteristics of a triode
Your Answer: Option
(Not Answered)
Correct Answer: Option

15.
If F(ω) is Fourier transform of f(t), then Fourier transform of f(t) cos2 ωct is
F(ω) + [F(ω + 2ωc) + F(ω - 2ωc)]
F(ω) + [F(ω + 2ωc) + F(ω - 2ωc)]
F(ω) - [F(ω + 2ωc) + F(ω - 2ωc)]
F(ω) - [F(ω + 2ωc) + F(ω - 2ωc)]
Your Answer: Option
(Not Answered)
Correct Answer: Option

16.
In AM the amplitude of carrier components ... while in FM it
remains constant, does not remain constant
does not remain constant ... remains constant
remains constant... also remains constant
does not remain constant, also does not remain constant
Your Answer: Option
(Not Answered)
Correct Answer: Option

17.
If carrier modulated by a digital bit stream had one of the possible phases of 0°, 90°, 180°, 270°, then the modulation is called.
BPSK
QPSK
QAM
MSK
Your Answer: Option
(Not Answered)
Correct Answer: Option

18.
The currents fed to the two dipoles of turnstile antenna are
in phase
in quadrature
in phase opposition
either (a) or (c)
Your Answer: Option
(Not Answered)
Correct Answer: Option

19.
Consider the following statements as regards BPSK and QPSK:
  1. In BPSK we deal individually with each bit of duration Tb
  2. In QPSK we lump two bits to form a symbol
  3. In both BPSK and QPSK signal changes occur at bit rate
  4. In QPSK the symbol can have any one of four possible values
Which of the above statements are correct?
1, 2, 3, 4
1, 2, 3
1, 2, 4
2, 3, 4
Your Answer: Option
(Not Answered)
Correct Answer: Option

20.
A pre-emphasis circuit provides extra noise immunity by
converting the phase modulation to frequency modulation
pre-amplifying the whole audio band
boosting the base frequencies
amplifying the higher audio frequencies
Your Answer: Option
(Not Answered)
Correct Answer: Option

*** END OF THE TEST ***
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