Exercise: Communication Systems

Questions for: Communication Systems

In the FM wave described by equation v = 15 sin (4 x 108 t + 3 sin 1100 t), the maximum frequency deviation is
A:
175 Hz
B:
525 Hz
C:
3 Hz
D:
58.33 Hz
Answer: B

.

Maximum frequency deviation = 3 x 175 = 525 Hz.

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Assertion (A): Electromagnetic waves in free space spread uniformly in all directions from a point source

Reason (R): Free space has no interference or obstacles.

A:
Both A and R are correct and R is correct explanation of A
B:
Both A and R are correct but R is not correct explanation of A
C:
A is correct but R is wrong
D:
A is wrong but R is correct
Answer: A

Since free space has no interference or obstacles, EM waves spread uniformly.

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For an ideal 3000 Hz channel, Nyquist rate is
A:
3000 bps
B:
6000 bps
C:
9000 bps
D:
12000 bps
Answer: B

For 3000 Hz.

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

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The received signal frequency at any time of a super-heterodyne receiver having IF = 456 kHz is 1 MHz. The corresponding image signal is
A:
within the medium band
B:
outside the medium band
C:
depend upon modulation index
D:
depends on Modulating frequency
Answer: A

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.

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If N bits are lumped together to get an N-bit symbol, the possible number of symbols is
A:
2N
B:
2N - 1
C:
2N - 2
D:
2N - 4
Answer: A

Each bit can have two levels. Hence 2N.

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