Network Principle Course Summary 001

Tag: pos value reserved ASD limited receiver bit get mod

1. Physical Layer

Physical Layer

• Protocol: RJ45, CLOCK, IEEE802.3 (Repeater, hub)
• Role: Transmission of bits via media, determination of mechanical and electrical specifications (bit bit)

1.1 Communication Fundamentals

• Data-the entity that transports the message.

• Signal (signal)--the electrical or electromagnetic performance of the data.

• Analog signal (analogous signal)-the value of the parameter representing the message is continuous.

• Digital signal-The value of the parameter representing the message is discrete.

• Code--Represents the basic waveform of different discrete values when a waveform representing a digital signal is used in a time domain (or short domain).

• Channel--generally used to represent a medium that transmits information in one direction.

• Unidirectional Communication (single-work communication)-there can be only one direction of communication without the opposite direction of interaction.

• Bidirectional alternating communication (half-duplex communication)-both parties to a communication can send information, but not both (and certainly not simultaneously).

• Two-way simultaneous communication (full-duplex communication)-both sides of the communication can send and receive information at the same time.

• Baseband signals (i.e. basic band signals)-signals from sources. Data signals such as computer output that represent a variety of text or image files belong to the baseband signal.

• Baseband signals often contain more low-frequency components, even DC components, and many channels do not transmit this low-frequency component or DC component. Therefore, the baseband signal must be modulated (modulation).

1.2 Nyquist theorem

Field of communication:
Bandwidth: The difference between the highest frequency of the signal and the lowest frequency (the range of signal frequencies allowed through the communication line). The unit is Hz.

In the computer network:
Bandwidth: The ability of a network's communication line to transmit data (the "highest data rate" that can be passed between two points in a network over a period of time). Unit is bit/s.

Sampling theorem (Nyquist): When converting an analog signal to a digital signal, assuming that the maximum frequency in the original signal is F, the sampling frequency F sample must be greater than or equal to twice times the maximum frequency F, in order to ensure that the sampled digital signal retains the information of the original analog signal intact.

In any channel, the rate of the code element transmission is capped , or there will be a problem of inter-code crosstalk , so that the receiving end of the code element Judgment (i.e. recognition) becomes impossible.

If the bandwidth of the channel is wider, that is, the more high-frequency components of the signal can be passed, then the code element can be transmitted at a higher rate without the inter-code crosstalk.

Nyquist theorem formula:
Noise-free Maximum transfer rate Cmax = F Sample * log2n = 2f * LOG2N (BIT/S)

Where F represents the bandwidth of an ideal low communication, and N represents the number of discrete levels for each code element.

eg1：对一个无噪声的4KHz的信道进行采样，可达到的最大数据传输速率是 (无限大)。        原因：在4KHz的信道上，采样频率要达到8KHz（每秒进行8K次采样）。        如果每次采样可以取得16bit的数据，那么信道就可以发送128kbit/s的数据。    如果每次采样可以取得1024bit的数据，那么信道就可以发送8Mbit/s的数据。    所以，只要每个码元能够携带更多的比特，最高码元传输速率就可以是无限大的。    用奈奎斯特定理公式解释就是：无噪声最大传输速率Cmax = 8K * log2N，其中N未知，所以最大传输速率可以无限大。（但是香农定理不允许最大传输速率为无限大）

1.3 Shannon's theorem

Snr: The ratio of the average power of the signal to the average power of the noise. (commonly recorded as: S/N, in DB db)
The formula is: signal-to-noise ratio (db) = LOG10 (S/N) (db)

eg2：    当 S/N = 10 时，信噪比为 10 dB    当 S/N = 1000时，信噪比为 30 dB。

Shannon formula:
The channel limit information transmission rate C can be expressed as:
C = W log2 (1+s/n) (bit/s)

Where: W is the channel bandwidth (in Hz);
S is the average power of the signal transmitted within the channel;
N is the Gaussian noise power inside the channel.

Shannon Formula shows:

• The greater the signal-to-noise ratio in the channel's bandwidth or channel, the higher the limit transmission rate of the information.

• as long as the information transmission rate is lower than the channel limit information transmission rate, it is certain to find some way to achieve error-free transmission .

• If the channel bandwidth W or signal-to-noise ratio S/n has no upper limit (of course, the actual channel can not be such), the channel limit information transfer rate C also has no upper limit.

• The transmission rate of information that can be achieved on the actual channel is much lower than that of Shannon's limit transmission rate.

The Nyquist theorem indicates that the rate of transmission of the code element is limited;
Shannon formula gives the limit of information transmission rate.

eg3：电话系统的典型参数是信道带宽为3000Hz，信噪比为30dB，则该系统的最大数据传输率为 （30kbit/s）    原因：信噪比公式：信噪比（dB）= 10 log<sub>10</sub>（S/N） （dB）    10log<sub>10</sub>(S/N) = 30，即：log<sub>10</sub>(S/N) = 3。所以S/N = 1000    香农公式：C = W log<sub>2</sub>(1+S/N)   (bit/s)     所以有C= 3000 log<sub>2</sub>（1001） ,即C=30000bit/s = 30kbit/s

1.4 Transmission Media

Transmission media is also referred to as a transmission medium or transmission medium.
The transmission medium is not a physical layer, and the transmission medium is the layer below the physical layer.

Transmission media can be divided into guided transmission media (along solid media) and non-guided transmission media (free space).
Guided transmission media divided into: twisted pair, coaxial cable, optical fiber
Non-guided transmission media is divided into: short wave, microwave, electromagnetic wave

1.5 Physical Layer interface characteristics (mechanical, electrical, functional, process)

• Mechanical properties: Specifies the shape and size of the connector used in the interface, number of leads and arrangement, fixation and locking device, etc.

• Electrical characteristics: Indicates the range of voltages that occur on each line of the interface cable (the characteristic is used to indicate which line of the interface cable the voltage should be in the range, i.e. what voltage represents 1 or 0).

• Feature: Indicates the meaning of the voltage at a certain level that appears on a line.

• Process characteristics: Indicates the sequence of occurrences of various possible events for different functions.

2. Computer network Principle chapter II Assignment 2.1 blank question

Problem number	Topics
1.	In order to improve the transmission rate of information, it is necessary to use the multi-input method to try to make each code element carry more bits.
2.	Assuming that the 333 kb/s data (Error-free transmission) is to be transmitted using a telephone channel with a bandwidth of up to khz, the channel should have a signal-to-noise ratio of db (the result is retained to bits).
3.	In the channel Multiplexing technology, the technology of all users occupying different bandwidth resources at the same time is called Frequency division multiplexing, while the technology of all users occupying the same band width at different times is called time division multiplexing.
4.	Shielded twisted pair is on the outside of the twisted pair and a wire braided shielding layer, the main purpose is to improve the resistance of the twisted pair of electromagnetic interference ability.
5.	The purpose of communication is to transmit messages, such as text, images, voice, etc. and need to be transported through the data entity.
6.	The unit of data transmitted on the physical layer is bit, and the English notation is bit (derived from binary digit).
7.	Commonly used three types of guided transmission media are twisted pair, coaxial cable, as well as optical fiber/fiber.
8.	In the way of mutual information interaction, communication has the following three basic methods: Simplex communication, full-duplex communication and half-duplex (bidirectional alternating communication) communication.
9.	Assuming that a twisted pair has an attenuation rate of 0.6 db/km (at 1 khz), the operating distance of the link using the twisted pair is 50 km, if the attenuation is allowed to be in the range of DB. If you want to increase the working distance of this twisted pair to 150 km and allow the attenuation of a single dB, the attenuation rate should be reduced to 0.3 db/km.

2.2 Selection Questions

2. (C) characteristics indicate the meaning of the voltage at a certain level on a line.

   A. 机械B. 电气C. 功能D. 过程

   Parsing: "What is the meaning", that is, functional characteristics.

4. (B) The characteristic is used to describe the line of the interface cable where the voltage should be the range, that is, what the voltage represents 1 or 0.

   A. 机械B. 电气C. 功能D. 过程

   Analysis: the "range of voltage", that is, electrical characteristics.

6. The physical layer's (A) feature is used to describe the shape and size of the connector used in the interface, the number of pins and arrangement, fixation and locking devices, and so on.

   A. 机械B. 电气C. 功能D. 过程

   Analysis: The appearance of "shape and size", that is, mechanical properties.

8. It is assumed that the maximum code rate for a channel that is limited by the Nyquist criterion is 30000 yards per second. If amplitude modulation is used to divide the amplitude of the code element into 8 different classes to transmit, the highest data rate available is (D)

   A. 60000 b/sB. 70000 b/sC. 80000 b/sD. 90000 b/s

   Explanation: Bit rate (data transfer rate) = Baud rate (code element transfer rate) * N, each change can represent 8 different signals, so each change can represent a number of bits n= log28 = 3. That is, the bit rate is 90000bit/s.

10. The characteristics of the following options, which are not part of the computer network physical layer, are (D).

    A. 机械特性B. 电气特性 C. 功能特性 D. 规程特性

    Analysis: Physical layer interface features include: mechanical properties, electrical characteristics, functional characteristics, process characteristics.

12. The typical transmission rate for (C) is the highest in the following transport media.

    A. 双绞线B. 同轴电缆C. 光缆D. 无线介质

14. In the following transmission medium, C is the best anti-electromagnetic interference.

    A. 双绞线B. 同轴电缆C. 光缆D. 无线介质

    Analysis: Guided transmission media (twisted pair, coaxial cable, optical fiber) and non-guided transmission medium (mechanical wave, electromagnetic wave and other wireless media), optical fiber communication capacity is the largest, transmission loss is the smallest, the longest relay distance, the best anti-EMI ability, the highest transmission rate.

3. Computer network Shehiren Seventh Edition exercise Answer (chapter 2nd)

2-01 What problems should the physical layer solve? What are the main features of the physical layer?

For:
Physical layer to solve the main problem:
(1) The physical layer should be as far as possible to shield off the physical equipment and transmission media, communication means of different, so that the data link layer does not feel these differences, only consider the completion of this layer of protocols and services.
(2) The ability of the service user (the data link layer) to transmit and receive bit streams (typically serial sequentially transmitted bitstream) on a physical transmission medium, so that the physical layer should solve the problem of establishing, maintaining, and releasing physical connections.
(3) uniquely identifies a data circuit between two adjacent systems

The main features of the physical layer are:
(1) Since many physical protocols or protocols have been developed prior to the OSI, and in the field of data communication, these physical procedures have been adopted by many commercialized devices, and the physical layer protocol covers a wide range, so far there is no new set of physical layer protocols according to the OSI abstract model, Instead, the physical layer is defined as the mechanical, electrical, functional, and procedural characteristics that describe the interface with the transmitting media, with existing physical procedures.
(2) Due to the many ways of physical connection, there are many kinds of transmission media, so the specific physical protocol is quite complex.

2-02 What is the difference between a layer and a protocol?

A: The procedure refers specifically to the physical layer protocol

2-03 try to give the model of data communication system and explain the function of its main composition construction.

For:
SOURCE point: The source point device produces the data to be transferred. The source point is also known as the source station.
Transmitter: The data generated by the source point is usually encoded by the transmitter before it can be transmitted in the transmission system.
Receiver: Receives the signal transmitted by the transmission system and converts it into information that can be processed by the destination device.
End point: The endpoint device obtains the transmitted information from the receiver. The end point is also called the destination station
Transmission system: signal physical Channel

2-04 try to explain the following nouns: data, signal, analog data, analog signal, baseband signal, band-pass signal, digital data, digital signal, code element, single-work communication, half-duplex communication, full-duplex communication, serial transmission, parallel transmission.

For:
Data: Is the entity that transports information.
Signal: It is the electrical or electromagnetic performance of the data.
Analog data: An analog signal for transporting information.
Analog signal: a continuously changing signal.
Digital signal: A signal that takes a finite number of discrete values.
Numeric data: Data with values of discontinuous values.
Code: Represents a basic waveform of different discrete values when a waveform representing a digital signal is used in a time domain (or short domain).
Simplex communication: That is, there is only one direction of communication without the opposite direction of interaction.
Half-duplex communication: Communication and both can send information, but cannot both send simultaneously (of course, can not simultaneously receive). This means of communication is sent by the other party to receive, over a period of time again in turn.
Full-duplex communication: That is, both sides of the communication can send and receive information at the same time.
Baseband signals (i.e. basic band signals)-signals from sources. Data signals such as computer output that represent a variety of text or image files belong to the baseband signal.
With communication number-after the baseband signal is modulated by the carrier, the frequency range of the signal is moved to a higher band for transmission in the channel (that is, only within a certain frequency range can pass the channel).

What are the features of the 2-05 physical layer interface? What does it contain?

For:
(1) Mechanical properties Interface shapes and sizes, number of leads and arrangement, fixation and locking devices, etc.
(2) Electrical characteristics indicate the range of voltages that appear on each line of the interface cable.
(3) The function characteristic indicates the meaning of the voltage of a certain level appearing on a line.
(4) The procedure characteristic describes the order of occurrence of various possible events for different functions.

2-06 what are the factors that limit the transmission rate of data in the channel? Can signal-to-noise ratio be increased arbitrarily? What is the meaning of Shannon formula in data communication? What is the difference between "bits per second" and "yards per second"?

For:
code element Transfer rate is limited by Nyquist criterion, and information transmission rate is limited by Shannon Formula.
The meaning of Shannon formula in data communication is that, as long as the information transmission rate is lower than the limit messaging rate of the channel, the transmission can be achieved without error.
Bit/s is the unit of information transfer rate
The code-element transfer rate is also known as modulation rate, waveform rate, or symbol rate. A code element does not necessarily correspond to a bit.

2-07 It is assumed that a channel is subject to a maximum code rate of $20000 per second that is limited by the Nyquist criterion. If amplitude modulation is used to divide the amplitude of the code element into 16 different classes to transmit, how much higher data rate (b/s) can be obtained?

Answer: C=rLog2 (=20000b/s)4=80000b/s

2-08 assuming that the 64KB/S data (error-free transmission) is to be transmitted using a 3KHz-bandwidth telephone channel, how high is the signal-to-noise ratio (expressed in ratios and decibels, respectively)? What does this result indicate? ）

For:
C=WLOG2 (1+s/n) (b/s)
w=3khz,c=64khz----ÀS/N=64.2DB is a very high signal-to-noise source.

2-09 with the Shannon formula calculation, assuming that the channel bandwidth is 3100Hz, the maximum channel transmission rate is 35kb/s, then if you want to increase the maximum channel transmission rate of 60%, the Q/S ratio of S/n should be increased to how many times? How many times should the signal-to-noise ratio s/n be increased on the basis of the calculation just now? If the signal-to-noise ratio s/n is increased to 10 times times on the basis of the calculation just now, can the maximum information rate be increased by 20%?

For:
C = W log2 (1+s/n), so there is s/n=2c/w-1;
s/n1=2c1/w-1=2 (35000/3100) -1=2503, s/n2=2c2/w-1=2 (1.635000/3100) -1=273276;
So the sn2/sn1=108 signal-to-noise ratio should be increased to about 100 times times.
C3=wlong2 (1+S/N3) =wlog2 (1+10s/n2) =66291 b/S
Rate increase = (C3-C2)/c2=18.5%

2-11 assuming that a twisted pair is 0.7db/km (at 1 khz), if 20dB attenuation is allowed, how long is the working distance of the link using this twisted pair of wires? If the working distance of the twisted pair increases to 100 km, what should the attenuation be reduced to?

Solution:
The working distance of the link using this twisted pair is =20/0.7=28.6km
Attenuation should be reduced to 20/100=0.2db

The 2-12 test calculates the bandwidth of the wavelength between 1200nm and 1400nm and the light wave working between 1400nm and 1600nm. It is assumed that the propagation rate of light in the fiber is 2*10e/s.

Solution:
V=l*f-àf=v/l--àb=f2-f1=v/l1-v/l2
1200nm to 1400nm: bandwidth =23.8thz
1400nm to 1600nm: bandwidth =17.86thz

2-13 Why use a channel multiplexing technology? What are the commonly used channel multiplexing technologies?

A: To maximize channel utilization by sharing channels. Frequency Division, Time division, code points, wave points.

2-15-point Multiple access CDMA Why can all users communicate at the same frequency band at the same time without interfering with each other? What are the pros and cons of this multiplexing approach?

A: Each user uses a specially selected cross-orthogonal different pattern, so that each other does not cause interference.
This system sends the signal to have the very strong anti-jamming ability, its spectrum resembles the white noise, is not easy to be discovered by the enemy. Occupy a larger bandwidth.

2-16 a total of 4 stations for Code Division multiple Access communication. 4 Stations of the code sequence listed as
A: ( -1-1-1+1+1-1+1+1) B: ( -1-1+1-1+1+1+1-1)
C: ( -1+1-1+1+1+1-1-1) D: ( -1+1-1-1-1-1+1-1)
Now receive such a chip sequence S: ( -1+1-3+1-1-3+1+1). Ask which station to send the data? Does the station sending the data send 0 or 1?

Solution:
S. A= (+1-1+3+1-1+3+1+1)/8= 1, a Send 1
S. b= (+1-1-3-1-1-3+1-1)/8=-1, b send 0
S. C= (+1+1+3+1-1-3-1-1)/8= 0, c no send
S. D= (+1+1+3-1+1+3+1-1)/8= 1, D send 1

2-17 try to compare the advantages and disadvantages of xDSL, HFC and FTTX access technology.

For:
XDSL technology is to use digital technology to transform the existing analog telephone subscriber line, so that it can carry the broadband service. Low cost, easy to achieve, but the bandwidth and quality of the large difference.
The biggest advantage of HFC networks is its wide frequency band and the ability to take advantage of cable networks that already have considerable coverage. To transform an existing cable television network with one-MHz unidirectional transmission into a two-way, dual-transmission HFC network requires considerable funding and time.
FTTX (fiber optics to ...) Here the letter X can represent different meanings. Provides the best bandwidth and quality, but at this stage the line and engineering costs are too high.

2-18 Why in the ASDL technology, in less than 1MHz of bandwidth can transmit the rate up to a few megabytes per second?

For:
By advanced DMT Coding, frequency division Multicarrier parallel transmission, so that the transmission of one code per second is equivalent to transmitting multiple bits per second

Network Principle Course Summary 001