1) Overhead bits are
a. Framing and synchronizing bits
b. Data due to noise
c. Encoded bits
d. None of the above
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Answer
Explanation
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ANSWER: Framing and synchronizing bits
Explanation:
No explanation is available for this question!
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2) ISI may be removed by using
a. Differential coding
b. Manchester coding
c. Polar NRZ
d. None of the above
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Answer
Explanation
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ANSWER: Differential coding
Explanation:
No explanation is available for this question!
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3) Timing jitter is
a. Change in amplitude
b. Change in frequency
c. Deviation in location of the pulses
d. All of the above
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Answer
Explanation
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ANSWER: Deviation in location of the pulses
Explanation:
No explanation is available for this question!
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4) Probability density function defines
a. Amplitudes of random noise
b. Density of signal
c. Probability of error
d. All of the above
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Answer
Explanation
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ANSWER: Amplitudes of random noise
Explanation:
No explanation is available for this question!
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5) Impulse noise is caused due to
a. Switching transients
b. Lightening strikes
c. Power line load switching
d. All of the above
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Answer
Explanation
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ANSWER: All of the above
Explanation:
No explanation is available for this question!
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6) In coherent detection of signals,
a. Local carrier is generated
b. Carrier of frequency and phase as same as transmitted carrier is generated
c. The carrier is in synchronization with modulated carrier
d. All of the above
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Answer
Explanation
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ANSWER: All of the above
Explanation:
No explanation is available for this question!
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7) Synchronization of signals is done using
a. Pilot clock
b. Extracting timing information from the received signal
c. Transmitter and receiver connected to master timing source
d. All of the above
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Answer
Explanation
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ANSWER: All of the above
Explanation:
No explanation is available for this question!
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8) In DPSK technique, the technique used to encode bits is
a. AMI
b. Differential code
c. Uni polar RZ format
d. Manchester format
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Answer
Explanation
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ANSWER: Differential code
Explanation:
No explanation is available for this question!
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9) The technique that may be used to reduce the side band power is
a. MSK
b. BPSK
c. Gaussian minimum shift keying
d. BFSK
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Answer
Explanation
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ANSWER: Gaussian minimum shift keying
Explanation:
No explanation is available for this question!
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10) In MSK, the difference between the higher and lower frequency is
a. Same as the bit rate
b. Half of the bit rate
c. Twice of the bit rate
d. Four time the bit rate
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Answer
Explanation
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ANSWER: Half of the bit rate
Explanation:
No explanation is available for this question!
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