2-01 What problems should the physical layer 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 communication means are different, so that the data link layer will not 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)
The main characteristics of uniquely identifying the physical layer of the data circuit between two adjacent systems : (1) Because before OSI, many physical regulations or protocols have been formulated, and in the field of data communication, these physical regulations It has been adopted by many commercial devices. In addition, the physical layer protocol covers a wide range. So far, no new physical layer protocol has been formulated according to the abstract model of OSI. Instead, the existing physical procedures are used to determine the physical layer. To describe the mechanical, electrical, functional and regulatory characteristics of the interface with the transmission media. (2) Because there are many ways of physical connection, and there are many types of transmission media, the specific physical protocol is quite complicated.
2-02 What is the difference between the classification and the protocol?
Answer: The procedure specifically refers to the physical layer protocol
2-03. Try to give a model of the data communication system and explain the role of its main components.
Answer: Source point: The source point device generates the data to be transmitted. The source point is also called the source station.
Transmitter: Usually the data generated by the source can be transmitted in the transmission system after being encoded by the transmitter. Receiver: Receives the signal from the transmission system and converts 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 destination is also called the destination station transmission system: signal physical channel
2-04. 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 information. Signal: It is the electrical or electromagnetic representation of data. Analog data: An analog signal that carries information. Analog signal: a continuously changing signal. Digital signal: A signal with a limited number of discrete values. Numerical data: data with discrete values. Code: The basic waveform representing different discrete values when the waveform in the time domain (or time domain for short) is used to represent a digital signal.
Simplex communication: that is, there is only one direction of communication and no interaction in the opposite direction. Half-duplex communication: that is, both parties can send information, but not both parties can send at the same time (of course, they can not receive at the same time). This way of communication is that one party sends and the other receives, and then vice versa after a while. Full-duplex communication: That is, both parties in communication can send and receive information at the same time. Baseband signal (ie basic frequency band signal)-the signal from the source. 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 (that is, the channel can only pass through a certain frequency range).
Computer network
2-05 What are the characteristics of the physical layer interface? What does each contain?
Answer: (1) The mechanical characteristics indicate the shape and size of the connector used in the interface, the number and arrangement of leads, fixing and locking devices, etc. (2) The electrical characteristics indicate the range of voltages that appear on each line of the interface cable. (3) The functional characteristics indicate what a certain level of voltage appears on a certain line means. (4) The characteristics of the procedure describe the sequence of occurrence of various possible events for different functions.
2-06 What factors limit the transmission rate of data in the channel? Can the signal-to-noise ratio be increased arbitrarily? What is the significance of Shannon's formula in data communication? What is the difference between "bits/second" and "symbols/second"?
Answer: The symbol transmission rate is limited by the Nyquist criterion, and the information transmission rate is limited by the Shannon formula. The meaning of Shannon’s formula in data communication is: as long as the information transmission rate is lower than the limit transmission rate of the channel, the transmission can be achieved without error. . Bit/s is the unit symbol transmission rate of the information transmission rate, also known as the modulation rate, waveform rate, or symbol rate. One symbol does not necessarily correspond to one bit.
2-07 Assume that the highest symbol rate of a channel restricted by the Nyquist 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 a data rate (b/s) can be obtained?
Answer: C=RLog2(16)=20000b/s4=80000b/s( Computer Network Xie Xiren Seventh Edition)
2-08 Assuming that a 3KHz bandwidth telephone channel is used to transmit 64kb/s data (error-free transmission), how high should this channel have a signal-to-noise ratio (represented by ratio and decibel, respectively)? What problem does this result indicate?)
Answer: C=Wlog2(1+S/N)(b/s)
W=3khz, C=64khz----àS/N=64.2dB is a signal with very high S/N ratio requirements source
2-09 Use Shannon’s formula to calculate. 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%, ask how many times the S/N ratio should be increased ? How many times should the signal-to-noise ratio S/N increase based on the calculation just now? 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%?
Answer: C = W log2(1+S/N) b/s-àSN1=2*(C1/W)-1=2*(35000/3100)-1
SN2=2*(C2/W)-1= 2*(1.6C1/w)-1=2(1.635000/3100)-1
SN2/SN1=100 The signal-to-noise ratio should be increased to about 100 times. C3=Wlong2(1+SN3)=Wlog2(1+10SN2) C3/C2=18.5%
If the S/N ratio is increased to 10 times on this basis, the maximum information communication rate can only increase by 18.5% Left and right
2-10 What are the commonly used transmission media? What are the characteristics of each?
Answer: twisted pair shielded twisted pair STP (Shielded Twisted Pair) unshielded twisted pair UTP (Unshielded Twisted Pair) coaxial cable 50 W coaxial cable 75 W coaxial cable optical cable wireless transmission: shortwave communication/microwave/satellite communication
2-11 Suppose there is a twisted pair cable whose attenuation 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 cable is to be increased to 100 kilometers, how much should the attenuation be reduced?
Solution: The working distance of the link using this twisted pair is =20/0.7=28.6km, the
attenuation should be reduced to 20/100=0.2db
Xie Xiren
2-12 Try to calculate the bandwidth of the light wave operating between 1200nm and 1400nm and between 1400nm and 1600nm. Suppose the propagation speed of light in the fiber is 210e8m/s.
Solution: V=LF-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 channel multiplexing technology? What are the commonly used channel multiplexing technologies?
Answer: In order to maximize channel utilization by sharing channels. Frequency division, time division, code division, wave division.
2-14 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)
SDH (Synchronous Digital Hierarchy) synchronous digital series
STM-1 (Synchronous Transfer Module) Level 1 Synchronous Transfer Module
OC-48 (Optical Carrier) Level 48 Optical Carrier
2-15 Code Division Multiple Access CDMA Why can all users use the same frequency band for communication at the same time? Won't interfere with each other? What are the advantages and disadvantages of this multiplexing method?
Answer: 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 strong anti-interference ability, and its frequency spectrum is similar to white noise, and it is not easy to be found by the enemy. Take up a lot of bandwidth.
2-16 There are 4 stations in total for code division multiple access communication. The chip sequence of the 4 stations is
A: (-1-1-1+1+1+1-1+1+1) B: (-1-1+1+1+1+1+1-1)
C: (-1+1-1+1+1+1+1-1-1) D: (-1+1-1) 1-1-1-1 + 1-1)
Now we have received such a chip sequence S: (-1 + 1-3 + 1-1-3 + 1 + 1). Ask which station sent 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 sends 1
S•B=(＋1-1－3-1－1-3－1－1)/8=－1, B sends 0
S •C=(＋1＋1＋3＋1-1－1-3－1-1)/8=0, C does not send
S•D=(＋1＋1＋3－1＋1＋3＋1-1)/8=1, D sends 1
2-17 Try to compare xDSL, HFC and The advantages and disadvantages of FTTx access technology?
Answer: The xDSL technology is to use digital technology to transform the existing analog telephone subscriber line so that it can carry broadband services. Low cost and easy to implement, but the bandwidth and quality vary greatly. The biggest advantage of the HFC network is that it has a very wide frequency band and can use the cable TV network that has considerable coverage. To transform the existing 450 MHz one-way transmission cable TV network into a 750 MHz two-way transmission HFC network requires considerable capital and time. FTTx (fiber to...) where the letter x can mean different things. It can provide the best bandwidth and quality, but the current line and engineering costs are too large.
2-18 Why in ASDL technology, the transmission rate can be as high as several megabits per second in a bandwidth of less than 1MHz?
Answer: Relying on advanced DMT coding, frequency division multi-carrier parallel transmission, so that one symbol per second is equivalent to multiple bits per second