# 程序代写 COMP90007 Internet Technologies Week 4 Workshop – cscodehelp代写

COMP90007 Internet Technologies Week 4 Workshop

Semester 2, 2021

Suggested solutions

© University of Melbourne 2021

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Question 1 (Sampling)

■ Consider a telephone signal that is bandwidth limited to 4 kHz.

❑ (a) At what rate should you sample the signal so that you can completely reconstruct the signal?

min. sampling rate = 2 × 4000 = 8 kHz = 8000 samples/s

❑ (b) If each sample of the signal is to be encoded at 256 levels, how many bits are required for each sample?

256 possible values per sample requires log2(256) = 8 bits/sample

❑ (c) What is the minimum bit rate required to transmit this signal? 8 bits/sample × 8000 samples/sec = 64 kbps

Note: This is a direct application of the Sampling Theorem and forms the basics of the application of the theorem, i.e. without considering data rates.

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Question 2 (Sampling)

■ Is the Sampling theorem true for optical fibre or only for copper wire?

• The Sampling theorem is a property of mathematics and has nothing to do with technology.

• The Sampling theorem is independent of the transmission medium. The Sampling theorem states that if you have a function which does not contain any frequency components (sines or cosines) above f, then by sampling at a frequency of 2f, you capture all the information there is.

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Question 3 ( Rate)

■

•

Given a noiseless 4 kHz channel, what is the maximum data rate of the communication channel?

A noiseless channel can carry an arbitrarily large amount of information, e.g. there can be an infinite number of signalling levels, this is because there is no noise. This is a neat observation and the level information is not restricted by the question in any way. Shannon specifies a limit on the information rate based on given noise level.

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Question 4 ( Rate)

■ The bandwidth of a television video stream is 6 MHz. How many bits/sec are sent if four-level digital signals are used? Assume a noiseless channel

The maximum baud rate is 12 M symbols/sec

Four levels of signalling provide: log2 4 = 2 bits/symbol

Hence, the total data rate is: 12 M symbols/sec × 2 bits/symbol = 24 Mbps

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Question 4 ( Rate)

■ The bandwidth of a television video stream is 6 MHz. How many bits/sec are sent if four-level digital signals are used? Now assume a S/N of 20dB (i.e. 100).

Using Shannon’s theorem, we have: B x log(1+S/N)

= 6MHz x log2(1+100) = 6MHz x 6.65 = 39.9Mbps

Using Nyquist’s theorem, we have: 2B x log2 V =2*6MHzxlog2 4=12MHzx2=24Mbps

The bottleneck is therefore the Nyquist limit, giving a maximum channel capacity of 24Mbps.

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Question 5 (Framing)

The following character encoding is used in a data link protocol:

A: 01000111 B: 11100011 FLAG: 01111110 ESC: 11100000

Show the bit sequence transmitted (in binary) for the four-character frame payload A B ESC FLAG, when each of the following framing methods are used:

(a) Character count

(b) Flag bytes with byte stuffing

(c) Starting and ending flag bytes, with bit stuffing

Answer:

1. 00000101 01000111 11100011 11100000 01111110 5 A B ‘ESC’ ‘FLAG’

2. 01111110 01000111 11100011 11100000 11100000 11100000 01111110 01111110 FLAG A B ESC ‘ESC’ ESC ‘FLAG’ FLAG

3. 01111110 01000111 110100011 111000000 011111010 01111110 FLAG A B ‘ESC’ ‘FLAG’ FLAG

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Question 6 (Framing)

The following data fragment occurs in the middle of a data stream for which the byte-stuffing algorithm as described in the lecture is used:

A B ESC C ESC FLAG FLAG D.

What is the output after stuffing?

Answer:

After stuffing we get:

A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.

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