Principles of Computer Networks (Xie Xiren 8th Edition) Chapter 2 After-School Exercises Answers

Chapter two

1. What problems does the physical layer need to solve? What are the main characteristics of the physical layer?
Answer: The main problems to be solved by the physical layer:
(1) The physical layer should shield the physical equipment and transmission media as much as possible, and the different communication means, so that the data link layer cannot feel these differences, and only consider completing the protocol of this layer. and service.
(2) Give its service users (data link layer) the ability to transmit and receive bit streams (usually serially transmitted bit streams) on a physical transmission medium. For this reason, the physical layer should solve the problem of physical connection. Build, maintain, and release issues.
(3) A data circuit is uniquely identified between two adjacent systems.
The main characteristics of the physical layer:
(1) Before OSI, many physical regulations or protocols have been formulated, and in the field of data communication, these physical regulations have been adopted by many commercialized devices. In addition, the physical layer protocol involves Therefore, a new set of physical layer protocols has not been formulated according to the abstract model of OSI, but the existing physical regulations have been used to define the physical layer as the mechanical, electrical, functional and regulatory characteristics that describe the interface with the transmission medium.
(2) Since there are many ways of physical connection and many types of transmission media, the specific physical protocol is quite complicated.
2. What is the difference between a protocol and an agreement?
Answer: In the early days of data communication, various rules used for communication were called "protocols". Later, computer networks with architecture began to use the term "protocol". The previous "rules" were actually "protocols". However, due to habit, the old name "procedure" is sometimes still used for the previously formulated procedures.
3. Give the model of the data communication system and explain the function of its main components.
Answer: Source : The source device generates the data to be transmitted. The source point is also called the source station.
Transmitter : Usually the data generated by the source is encoded by the transmitter before it can be transmitted in the transmission system.
receiver: Receive the signal sent by the transmission system and convert it into information that can be processed by the destination device.
End point : The end point device obtains the transmitted information from the receiver. The end point is also known as the destination station transmission system: signal physical channel
4. Try to explain the following terms: data, signal, analog data, analog signal, baseband signal, bandpass signal, digital data, digital signal, symbol, simplex communication, half Duplex communication, full duplex communication, serial transmission, parallel transmission.
Answer: Data : is the entity that carries the information.
Signal : An electrical or electromagnetic representation of data.
Analog Data : Analog signals that carry information.
Analog Signal : A continuously changing signal.
Baseband Signal : The signal from the source. The data signals that represent various text or image files output by a computer are all baseband signals.
Bandpass signal: After the baseband signal is modulated by the carrier, the frequency range of the signal is moved to a higher frequency band for transmission in the channel.
Numeric data : data whose values are discontinuous values.
Digital signal : A signal that takes a finite number of discrete values.
Symbol : A basic waveform representing different discrete values when a digital signal is represented by a waveform in the time domain.
Simplex communication : that is, there is only one direction of communication and no interaction in the opposite direction.
Half-duplex communication : that is, both communication and both parties can send information, but both parties cannot send at the same time (and certainly cannot receive at the same time). This communication method is that one party sends and the other party receives, and then vice versa after a period of time.
full duplex communication: That is, both sides of the communication can send and receive information at the same time.
Serial transmission : Use a data line to transmit data, 1 bit at a time, and multiple bits need to be transmitted one after the other.
Parallel transfer : Use multiple data lines to transfer multiple bits at a time.
5. What are the characteristics of the interface of the physical layer? What does one contain?
Answer: Mechanical characteristics : the shape and size of the connector used in the interface, the number and arrangement of the leads, the fixing and locking device, etc.
Electrical Characteristics : Indicates the range of voltages that appear on each line of the interface cable.
Functional characteristics : Indicate what the voltage of a certain level on a line means.
Procedural characteristics : Describe the sequence of occurrence of various possible events for different functions.
6. What factors limit the transmission rate of data in the channel? Can the signal-to-noise ratio be arbitrarily increased? What is the significance of Shannon's formula in data communication? What is the difference between "bits per second" and "symbols per second"?
Answer: 1. The transmission rate of data in the channel is limited by factors such as bandwidth and signal-to-noise ratio.
2. The signal-to-noise ratio cannot be increased arbitrarily.
3. The meaning of Shannon's formula: As long as the information transmission rate is lower than the limit information transmission rate of the channel, there must be some way to achieve error-free transmission.
4. Bits/s is the unit symbol transmission rate of the information transmission rate. The transmission rate is also called modulation rate, waveform rate or symbol rate. One symbol does not necessarily correspond to one bit.

Calculation formula:

信噪比(dB)=10log10(S/N)(dB)       //S为信号的平均功率，N为噪声的平均功率。
香农公式：C=Wlog2(1+S/N)(bit/s)    //C为信道的极限信息传输速率，W是信道带宽
M进制的码元，码元宽度为T秒：C=1/T×log2(M)
无噪:带宽为BHz。  C=2Blog2(M)
有噪:带宽为BHz。  C=Blog2(1+S/N)
最高码元速率R，码元振幅等级n。  C=Rlog2(n)
频率＝传播速率/波长
频带宽度=最高频率-最低频率

7. Assume that the maximum symbol rate of a channel limited by the Nye's criterion is 20000 symbols/sec. If amplitude modulation is used, and the amplitude of the symbol is divided into 16 different levels for transmission, how high the data rate (b/s) can be obtained? (You can refer to the formula above in question 7)
Solution: C=RLog2(16 )=20000b/s×4=80000b/s
8. Assuming that 64kb/s data (error-free transmission) is to be transmitted by a telephone channel with a bandwidth of 3KHz, how high the signal-to-noise ratio should this channel be (using ratio and decibel respectively) to represent? What does this result indicate?) (You can refer to the formula above in question 7)
Solution: C=Wlog2(1+S/N)
64000b/s=3000Hz×log2(1+S/N)
S/N= 64.2dB,
which means that this is a channel with high signal-to-noise ratio requirements.
9. Calculate with Shannon's formula, assuming that the channel bandwidth is 3100Hz and the maximum channel transmission rate is 35Kb/s, then if you want to increase the maximum channel transmission rate by 60%, how many times should the signal-to-noise ratio S/N be increased? If the signal-to-noise ratio S/N is increased by ten times on the basis of the calculation just now, can the maximum information rate be increased by another 20%? (You can refer to the formula above in question 7)
Solution: C=Wlog2(1+S/N)
35000b/s=3100log2(1+S/N)
S/N= $2^{35000/3100}$ -1
means the channel transmission rate becomes 35Kb/s(1+0.6)=56Kb/s
S/N1= $2^{56000/3100}$ -1
At this time, S/N is expanded to 100 times of the original
C1=Wlog2(1+10S/N1)
C1/56Kb/s=118.5%
At this time, the information rate can only be increased by 18.5%
10. What kinds of transmission media are commonly used? What are the characteristics of each?
Answer:Twisted
pair is divided into shielded twisted pair and unshielded twisted pair. It can transmit analog signals as well as digital signals, and the effective bandwidth is up to 250kHz. Generally used as a telephone line to transmit sound signals. Twisted pair cables are easily interfered by external high-frequency electromagnetic waves and have a high bit error rate.
Coaxial
cable is divided into baseband coaxial cable and broadband coaxial cable, due to its high bandwidth (up to 300~400Hz), low bit error rate, and high cost performance.
Optical
fiber The optical fiber uses the optical fiber carrier to propagate the optical signal by using the principle of full inversion of light. Its advantages are small diameter, light weight, large transmission frequency band, large communication capacity, good anti-lightning and electromagnetic interference performance, five crosstalk interference, good confidentiality, and low bit error rate.
Radio microwave communication
Radio microwave communication is divided into ground microwave relay communication and satellite communication. Its main advantages are high frequency, wide frequency band, and large capacity of communication channels; less industrial interference for signals, high transmission quality, and relatively stable communication; not affected by geographical environment, less investment in construction, and quick results. The disadvantage is that the terrestrial microwave relay communication propagates in a straight line in space, and the transmission distance is limited, generally only 50km, and the concealment and confidentiality are poor; although the communication distance of satellite communication is long and the communication cost has nothing to do with the communication distance, the propagation delay is large.
11. Assuming that the attenuation of 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? If the working distance of the twisted pair is increased to 100 kilometers, how much should the attenuation be reduced?
Solution: The working distance of the twisted pair is S=20dB/0.7dB/Km=28.6Km 100Km
=20dB/attenuation
attenuation=0.2dB/Km
The frequency bandwidth of light waves. Assume that the propagation speed of light in the fiber is $2 × 10^8$ m/s. (You can refer to the formula above in question 7
Solution: 1200nm to 1400nm:
frequency = propagation rate/wavelength
Maximum frequency = $2 × 10^8m/s/1200nm=1.67×10^{14}Hz$
minimum frequency = $2$ $10^8m/s/1400nm=1.43 × 10^{14}Hz$
Bandwidth = highest frequency - lowest frequency = 23.8THz $1400nm$
to 1600nm:
highest frequency = $2 × 10^8m/s/ 1400nm=1.43×10^14Hz$
lowest frequency = $2$ $10^8m/s/1600nm=1.25 × 10^14Hz$
frequency bandwidth = highest frequency - lowest frequency = 17.86THz $13.$
Why use channel multiplexing technology? What are the commonly used channel multiplexing techniques?
A: In order to maximize the channel utilization by sharing the channel.
Frequency division, time division, code division, wavelength division.
14. Try to write the full text of the following English abbreviations and give a brief explanation. FDM,TDM,STDM,WDM,DWDM,CDMA,SONET,SDH,STM-1,OC-48.
Answer: FDM(frequency division multiplexing)
TDM(Time Division Multiplexing)
STDM(Statistic Time Division Multiplexing)
WDM(Wave Division Multiplexing)
DWDM (Dense Wave Division Multiplexing)
CDMA (Code Wave Division Multiplexing)
SONET (Synchronous Optical Network) Synchronous Optical Network
SDH (Synchronous Digital Hierarchy) Synchronous Digital Series
STM-1 (Synchronous Transfer Module) Level 1 Synchronous Transfer Module
OC-48 (Optical Carrier) Class 48 optical carrier
15. Why can code division multiple access (CDMA) enable all users to communicate using the same frequency band at the same time without interfering with each other? What are the advantages and disadvantages of this reuse method?
A: Each user uses different code patterns that are specially selected to be orthogonal to each other, so they will not interfere with each other. The signal sent by this system has a strong anti-jamming capability, and its spectrum is similar to white noise, which is not easy to be detected by the enemy. Occupies a large bandwidth.
16. A total of 4 stations carry out code division multiple access communication. The chip sequence of 4 stations is 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) The following chip sequence S is now received: (-1+1-3+1-1-3+1+1). Which station sent the data? Did the station sending the data send a 0 or a 1?
Solution: S•A=(+1-1+3+1-1+3+1+1)/8=1, A sends 1
S•B=(+1-1-3-1-1-3+1-1)/8=-1, B sends 0
S •C=(+1+1+3+1-1-3-1-1)/8=0, C does not send
S•D=(+1+1+3-1+1+3+1-1)/8=1, D sends 1
17. Try comparing xDSL, HFC and FTTx Advantages and disadvantages of access technology?
Answer: xDSL technology is to transform the existing analog telephone subscriber line with digital technology so that it can carry broadband services. Low cost and easy to implement, but large differences in bandwidth and quality. The biggest advantage of the HFC network is that it has a wide frequency band and can utilize the cable TV network that already has a considerable coverage. To transform the existing 450 MHz one-way transmission cable TV network to 750 MHz two-way transmission HFC network requires considerable capital and time. FTTx (Fiber To...) Here the letter x can mean different things. It can provide the best bandwidth and quality, but the current line and engineering costs are too high.
18. Why in ASDL technology can transfer rates as high as several megabits per second in less than 1MHz bandwidth?
A: Relying on advanced DMT coding, frequency division and multi-carrier parallel transmission, so that transmitting one symbol per second is equivalent to transmitting multiple bits per second
. 19. What are EPON and GPON?
Answer: EPON (Ethernet Passive Optical Network) is a new type of optical fiber access network technology. It adopts point-to-multipoint structure and passive optical fiber transmission to provide various services on Ethernet. It adopts PON technology in the physical layer, uses the Ethernet protocol in the link layer, and realizes the access of Ethernet by using the topology structure of PON. Therefore, it combines the advantages of PON technology and Ethernet technology: low cost; high bandwidth; strong scalability, flexible and fast service reorganization; compatibility with existing Ethernet; convenient management and so on.
GPON (Gigabit-Capable PON) technology is the latest generation broadband passive optical integrated access standard based on the ITU-TG.984.x standard. It has many advantages such as high bandwidth, high efficiency, large coverage, and rich user interfaces. Most operators regard it as an ideal technology to realize broadband and comprehensive transformation of access network services.