Data/Digital Communications

Data/Digital Communications

University

40 Qs

quiz-placeholder

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Data/Digital Communications

Data/Digital Communications

Assessment

Quiz

Other

University

Medium

Created by

Mac Aban

Used 3+ times

FREE Resource

40 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

With four bits of encoding, how many distinct symbols can be represented?

8

16

32

64

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Also referred to as the symbol rate.

Any of these

Bit rate

Information capacity

Baud rate

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

For a given bandwidth, what is the highest theoretical bit rate with binary encoding?

Half the bandwidth

Twice the bandwidth

M times the bandwidth

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

What is the theoretical information capacity that may be propagated at a 10kHz bandwidth and if

the SNR is 30dB?

100 kbps

99 kbps

50 kbps

49 kbps

Answer explanation

SOLUTION: I = 10k x log2 (1+10^3) = 99.7kbps (use 99kbps)

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

What is the theoretical information capacity that may be propagated at a 10kHz bandwidth if the

system uses 64 levels of encoding?

120 kbps

60 kbps

90 kbps

240 kbps

Answer explanation

SOLUTION: I = 2(10k) x log2(64) = 120kpbs

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A telephone line has a bandwidth of 3.2 kHz and a signal-to-noise ratio of 45 dB. A signal is

transmitted down this line using a four-level code. Calculate the maximum theoretical data rate.

12.8 kbps

42.52 kbps

47.84 kbps

6.4 kbps

Answer explanation

SOLUTION: I = 3200log2(1+10^4.5)=47.84kbps; Also I=2*3200*log2(4) = 12.8kbps.

Therefore max I = 12.8kbps

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

What is the minimum bandwidth to secure 4Mbps transmission speed using 8-bit modulation at

27dB SNR?

500 kHz

450 kHz

250 kHz

550 kHz

Answer explanation

SOLUTION: Using Shannon’s Limit: 4Mbps = B log2(1 + 10^2.7); B = 446kHz

Using Shannon-Hartley: 4Mbps = 2B*8; B = 250kHz

Therefore: Min B = 450kHz

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