Computer Network Chapter 2 Review Notes (Physical Layer)

One, the physical layer concept

确定与传输媒体的接口有关的一些特性:
(1)机械特性:描述接线器的形状。引脚数目和排列。线序的描述。
(2)电气特性:指明在接口电缆的个条线上出现的电压的范围。
(3)功能特性:指明某条线上出现的电压的意义。
(4)过程特性:指明对于不同功能的各种可能事件的出现顺序。

2. Basic knowledge of data communication

A data communication system can be divided into three parts:
source system (sender, sender), transmission system (transmission network), destination system (receiver, receiver)
Insert picture description here
Source systemIt generally includes the following two parts:
Source: the source device generates the data to be transmitted.
Transmitter: Usually, the digital bit stream generated by the source can be transmitted in the transmission system after being encoded by the transmitter.

Destination systemIt generally includes the following two parts:
Receiver: Receives the signal from the transmission system and converts it into an information destination that can be used by the destination device
: The end device obtains the digital bit stream from the receiver, and then outputs the information. The destination is also called the destination station or the destination.

According to the different values ​​of the representative information in the signal, the signal can be divided into the following two categories:

(1) Analog signal, or continuous signal: representative signal is continuous
(2) Digital signal, or discrete signal: representative information is discrete

3. Several basic concepts about channels

(1) One-way communication: also known as simplex communication, which can only communicate in one direction without interaction in the opposite direction (wireless broadcasting, TV broadcasting)
(2) Two-way communication: also known as half-duplex communication, that is communication Both parties can send messages. However, both parties cannot send at the same time
(3) Two-way simultaneous communication: also known as full-duplex communication, that is, both parties can communicate at the same time.

Note: The signals from the source are all baseband signals (ie, basic frequency band signals). Data signals that represent various text or image files output by a computer are all baseband signals. Baseband signals often contain more low-frequency components, even DC components, and many channels cannot transmit such low-frequency or DC components. In order to solve this problem, the baseband signal must be modulated.

Modulation can be divided into two categories:

  • 1. It only transforms the waveform of the baseband signal so that it can adapt to the channel characteristics. The converted signal is still a baseband signal, and this type of modulation is called baseband modulation.
    -The operation of converting a digital signal into another form of digital signal is calledcoding

  • 2. It is necessary to use the carrier for modulation, move the frequency range of the baseband signal to a higher frequency band, and convert it into an analog signal, so that it can be well transmitted in the analog channel (band pass modulation)

Common encoding methods

Insert picture description here

  • Non-return-to-zero (NRZ): positive level represents 1, negative level represents 0
  • Zero return system: positive pulse represents 1, negative pulse represents 0
  • Manchester: an upward transition in the center of the bit period represents 0, and a downward transition in the center of the bit period represents 1, but it can also be defined in reverse
  • Differential Manchester: At the center of each bit, there is always a change. A transition at the beginning of the bit represents 0, and no one is 1.

Basic bandpass modulation method

Insert picture description here

1. Amplitude modulation, abbreviated as amplitude modulation, achieves the purpose of transmitting information by changing the amplitude of the output signal. Generally, the amplitude change of the high-frequency signal output at the modulating end has a certain functional relationship with the original signal, and the demodulation end is used to demodulate and output the original signal.

2. Frequency modulation is to load information on the radio to obtain a modulated wave. It is a modulation method that uses the instantaneous frequency change of the carrier to express information, and expresses different information by using different frequencies of the carrier.

3. Phase modulation is a modulation method in which the deviation of the carrier phase from its reference phase changes in proportion to the instantaneous value of the modulation signal. Phase modulation and frequency modulation are closely related. During phase modulation, frequency modulation occurs at the same time; during frequency modulation, phase modulation occurs at the same time, but the law of change between the two is different.

Insert picture description here
Insert picture description here

Mediation: Turn digital into analog signal
Coding: Turn data into digital signal

Transmission media under the physical layer

1. Guided transmission media

Twisted pair : the most commonly used transmission medium. Both analog transmission and digital transmission can use twisted pair, and the communication distance is generally several to ten kilometers. Shielded twisted pair STP (Shielded Twisted Pair) with metal shielding layer, UTP (Unshielded Twisted Pair).
Insert picture description here

Coaxial cable
refers to a cable with two concentric conductors, and the conductor and shielding layer share the same axis. The most common coaxial cable consists of copper conductors separated by insulating materials.
Insert picture description here

Fiber optic
Insert picture description here
Insert picture description here
light wave propagating in the core of
Insert picture description here
the optical fiber advantages:

(1) The transmission loss is small, the relay distance is long, and it is particularly economical for long-distance transmission.
(2) The performance of anti-lightning and electromagnetic anti-interference is good.
(3) No crosstalk interference, good confidentiality, and it is not easy to steal or intercept data
(4) ) Small size and light weight

2. Non-guided transmission media (free space)

Three types of guided transmission media were introduced above. However, if the communication line passes through some mountains or islands, IRS21844S is sometimes difficult to construct. Even in cities, it is not easy to dig up roads and lay cables. When the communication distance is long, laying cables is expensive and time-consuming. However, the propagation of radio waves in free space can achieve a variety of communications quickly. Because this communication method does not use the various guided transmission media introduced in the previous section, free space is called "unguided transmission media".

Microwave relay communication:

Can transmit phone calls, telegrams, graphics, data, etc.

Features: The microwave band frequency is very high, and its frequency range is also very wide, so the capacity of its communication channel is large, the
microwave transmission quality is high, and the anti-interference is strong.
Disadvantages:

(1) Neighboring stations must look straight (often called LOS Line Of
Sight), and there must be no obstacles. Sometimes the signal emitted by an antenna is divided into several slightly different paths to reach the receiving antenna, causing distortion.
(2) The propagation of microwaves is sometimes affected by bad weather.
(3) Compared with the cable communication system, the concealment and confidentiality of microwave communication are poor.
(4) The use and maintenance of a large number of relay stations consumes more manpower and material resources.

The commonly used satellite communication method is a kind of microwave relay communication using artificial synchronous earth satellites located at an altitude of about 36,000 kilometers as repeaters between earth stations. The geostationary communication satellite is an unmanned microwave communication relay station in space. It can be seen that the main advantages and disadvantages of satellite communications should be roughly the same as terrestrial microwave communications.

Satellite communication
Characteristics of satellite communication
The communication distance is long, and the communication cost has nothing to do with the communication distance. The electromagnetic wave energy emitted by the geostationary satellite radiates to the communication coverage area on the earth with a span of more than 18,000 kilometers, accounting for about one-third of the world. As long as three satellites are placed in a synchronous orbit over the earth's equator at an equal distance of 120 degrees, global communication can be basically realized.

Similar to microwave relay communication,The frequency band of satellite communication is very wide, the communication capacity is very large, the signal interference is also small, and the communication is relatively stable.. In order to avoid interference, if the distance between satellites is not less than 2 degrees, only 180 synchronous satellites can be placed over the entire equator. Fortunately, people figured out that different frequency bands could be used on satellites for communication. Therefore, the total communication capacity resources are still very large.
Another feature of satellite communications is that it hasLarger propagation delaySince the antenna elevation angles of the earth stations are not the same, regardless of the ground distance between the two earth stations (one street apart or tens of thousands kilometers apart), the propagation delay from one earth station to another earth station via satellite Between 250-300 ms. Generally it can be 270 ms. This is quite different from other communications (please note: this has nothing to do with the distance between two earth stations). In contrast, the propagation delay of a terrestrial microwave relay communication link is generally taken as 3.3 us/km.

3. Channel multiplexing technology

Synchronous communication : Adjust the clock frequency of the sender and receiver to the same, and continuously receive and send the bit stream.
Asynchronous communication : When sending characters, the time interval between the sent characters can be arbitrary, and the receiver is ready to receive at any time

Frequency division multiplexing : divide the broadband into multiple sub-channels, each channel transmits one signal (all users occupy different bandwidth resources at the same time)
Time division multiplexing : send different signals in different time periods to achieve multiple channels the effect of the transmission (the same for all users account for bandwidth at different times)
WDM : light is frequency division multiplexing, a carrier signal transmitting optical fiber
code division multiplexing : a set of mutually orthogonal comprising The code group of the code word is used to transmit multiple signals.

Equipment:
Repeater: Amplify and forward the signal
Hub: Multi-port repeater, similar to repeater

Bandwidth access technology

ADSL technology is an asymmetric digital subscriber line technology to achieve broadband access to the Internet. As a transmission layer technology, ADSL makes full use of existing copper wire resources to provide an uplink bandwidth of 640kbps and a downlink 8Mbps on a pair of twisted pairs , Thereby overcoming the "bottleneck" of traditional users in the "last mile" and realizing broadband connection in the true sense.
Features:

The main feature of ADSL technology is that it can make full use of the existing copper cable network (telephone line network) [1], and install ADSL equipment at both ends of the line to provide users with high broadband services. Another advantage of ADSL is that it can coexist with ordinary telephones on the same telephone line, and ADSL transmission can be carried out while receiving and making calls on an ordinary telephone line without affecting each other. Users can access broadband multimedia information network and WWW through ADSL, and can watch movies and TV programs, hold a video conference, and download data files at a high rate. That's not all. You can also use the phone on the same phone line. Without affecting the other activities mentioned above. Installing ADSL is also extremely convenient and quick. Install ADSL on the existing telephone line, except that the ADSL communication terminal is installed on the user side, without any modification to the existing line. Using ADSL (Asymmetric Digital Subscriber Line, asymmetric digital subscriber line) technology, through a telephone line, at a hundred times faster than ordinary MODEM, browse the Internet, through the network learning, entertainment, shopping, and enjoy advanced data services such as video conferencing The fun of video on demand, online music, online TV, and online MTV has become a reality.
Insert picture description here

Insert picture description here

example

1. As shown in the figure, the data transmission rate of all links in the switching network that adopts the "store-and-forward" mode of packet is 100Mbit/s, the packet size is 1000B, and the packet header size is 20B. If host H1 sends a file with a size of 980000B to host H2, regardless of the packet disassembly time and propagation delay, the time required from H1 sending to H2 receiving is at least ()
Insert picture description here

Correct answer: D
answer analysis: The known packet size is 1000B, and the packet header size is 20B. It can be concluded that the data part of each packet is 980B, so the file with the size of 980000B should be divided into 1000 packets for transmission, each packet 1000B (including the header 20B), so a total of 1000000B of information needs to be transmitted, and the data transmission rate of the link is 100Mbits, which is 12.5MB/s, so the time required for the host H1 to transmit all data is 1MB/12.5MB/s= 80ms. At this time, it happens that the last packet goes out from the host H1 and has not been received by the host H2, and a packet needs to be stored and forwarded twice from the host HI to reach the host
H2 (without considering the propagation delay), and the required time is 2x1000B/12.5MB/ s=0.16ms. In summary, the total time taken is 80ms+0.16ms=80.16ms.

2. In the absence of noise, if the bandwidth of a communication link is 3kHz, 4 phases are used, and each phase has a QAM modulation technique with 4 amplitudes, the maximum data transmission rate of the communication link is ()

Tip: The formula of Nyquist criterion is C=2 W log2V, where W is the bandwidth and V is the type of symbol or how many states

A. 24kbit/s
B. 12kbit/s
C. 96kbit/s
D. 48kbit/s

Correct answer: A. Answer analysis: The maximum frequency is 3kHz
(in analog signals, the bandwidth can be regarded as the maximum frequency), so a sampling frequency of 2 W=6kHz is required ; 4 phases are known, and each Each phase has 4 kinds of amplitudes, which means that 16 states can be represented. The maximum data transmission rate of the communication link is Cmax=2 W log2V=2 3*log216=24kbit/s.

3. If the limit data transmission rate of the channel without noise is not less than the limit data transmission rate under the condition of a signal-to-noise ratio of 30dB, the number of signal states is at least ()

Note: The conversion method of db and S/N is db=10log10 (S/N)

A. 4
B. 16
C. 32
D. 8

Correct answer: C
answer analysis: The Nyquist sampling theorem can be used to calculate the limit data transmission rate in the absence of noise, and the Shannon second theorem can be used to calculate the limit data transmission rate in a noisy channel. 2Wlog2V≥Wlog(1+S/N), W is the channel bandwidth, N is the number of signal states, and S/N is the signal-to-noise ratio. Substitute the data into the calculation to get N≥32, choose D. Decibel number=10log10S/N

4. The code without synchronization information is (non-return-to-zero code)

5. The basic signal unit for carrying information is Code element

6. (Single-choice question, 2.0 points) In a computer network, the indicator that characterizes the effectiveness of data transmission is (B)
A. Bit error rate
B. Transmission rate
C. Frequency band utilization
D. Channel capacity

7. The equipment belonging to the physical layer is (ABD)
A. Ethernet switch
B. Bridge
C. Repeater
D. Gateway

8. When a certain port of the hub receives data, its action is (B)
A. Forward it from the appropriate port according to the destination address
B. Broadcast it
from all ports except the input port C. Broadcast it from all ports
D . Randomly select a port to forward

9. Regarding the statements about repeaters and hubs, what is incorrect is ()
A. Both can amplify and reshape the signal
. B. The number of network segments interconnected by repeaters or hubs is not limited
. C. Repeaters usually There are only 2 ports, and the hub usually has 4 or more ports
D. All work in the physical layer

Correct answer: B
answer analysis: The number of network segments interconnected through repeaters or hubs is limited, and the 5-4-3 principle needs to be followed. The so-called 5-4-3 rule means that no more than 5 between any two computers Segment lines (including hub-to-hub connection cables, as well as hub-to-computer connection cables), 4 hubs, of which only 3 hubs can be directly connected to computers or network devices. This is the maximum topology allowed by the 10Base-T network and the number of hub layers that can be cascaded. Among them, the hub installed in the middle is the only hub in the network that cannot be directly connected to the computer. After the computer sends data, if it does not get a response within a certain period of time, then it will be considered that the data transmission failed.

10. The waveform diagrams of Manchester encoding and differential Manchester encoding are shown in the figure below. The actual bit string transmitted is (C)
Insert picture description here
A. 1 0 0 1 0 1 1 0 0
B. 1 0 0 0 0 1 1 0 1
C. 0 1 1 0 1 0 0 1 1
D. 0 1 1 1 1 0 0 1 0

11. The signal power of an analog channel that transmits digital signals is 0.62W, the noise power is 0.02W, and the frequency range is 3.5~3.9MHz. The highest data transmission rate of this channel is ()
A. 8Mbit/s
B. 2Mbit/s
C. 4Mbit/s
D. 1Mbit/s

Correct answer: B
answer analysis: calculate the signal-to-noise ratio S/N=0.62/0.02=31, bandwidth W=3.9MHZ-3.5MHz=0.4MHz, according to Shannon’s formula, the highest data transmission rate C=Wlog2(1+S/ N)=0.4MHz*log2(1+31)=2Mbit/s.

12. The network connected by the hub is topologically () type structure
Insert picture description here

1. (
true or
false , 1.0 points) In time division multiplexing, each sub-channel can use the full bandwidth in its own time slice () A. True B. False


Insert picture description here

2. According to Shannon's law, given any analog channel with a determined signal-to-noise ratio, its capacity has no upper limit (×)
A.
True B. False

Insert picture description here

3. Frequency division multiplexing must ensure that the signals of each channel do not overlap each other (√)

Insert picture description here

note:
The value of the digital signal is a certain range of discrete values ​​The value of the
analog data is a certain range of continuous values

Guess you like

Origin blog.csdn.net/weixin_44763595/article/details/111705774