Chapter 2 Physical Layer

Chapter 2 Physical Layer

1.What problems should the physical layer solve? What are the main features of the physical layer?

Answer: (1) main problems to be solved at the physical layer:

① The physical layer should be as much as possibleBlock the differences between physical devices, transmission media, and communication methodsSo that the above data link layer does not feel the existence of these differences, and focuses on completing the protocols and services at this layer.

② Provide its service users (data link layer) the ability to transmit and receive bits on a physical transmission media (generally sequential bits. To this end, the physical layer should solve the problem of establishing, maintaining and releasing physical connections.

③ Uniquely identifies a data circuit between two adjacent nodes.

(2) Main features of the physical layer:

① Since many physical procedures or protocols have been developed before OSI, and these physical procedures have been adopted by many commercialized devices in the field of data communication. In addition, physical layer protocols involve a wide range of applications. Therefore, we have not developed a new set of physical layer protocols based on the OSI abstract model, but have followed the existing physical procedures,Determines the physical layer as the mechanical, electrical, functional, and procedural characteristics that describe and transmit media interfaces.

② Because there are many physical connections and many media transmission types, the specific physical protocols are quite complicated.

 

2.What are the features of physical layer interfaces? What content does each contain?

Answer: Physical Layer interfaces include mechanical, electrical, functional, and procedural features.

(1)The mechanical characteristics indicate the shape and size of the connector used for the interface, the number and arrangement of leads, and the fixing and locking devices.

(2) electrical characteristics are described inWhat is the voltage range of the interface cable?. That is, what voltage represents 1 or 0.

(3) FeaturesWhat is the significance of the voltage of an electric plane displayed on an online line?.

(4) procedure featuresSequence of various possible events of different functions.

 

3.What is the significance of the neeness criterion and Shannon formula in data communication?

A: The Nai's guidelines indicate:The transmission rate of the code element is limited and cannot be increased at will. Otherwise, the receiver cannot correctly determine whether the code element is 1 or 0.(Because there is mutual interference between code elements ). The neeshield criterion is derived under ideal conditions. In actual conditions, the maximum data transmission rate is smaller than that obtained under ideal conditions. The task of telecom technicians is to find a better transmission code waveform under actual conditions and convert bits into suitable transmission signals. It should be noted that the Nai's criteria do not limit the information transmission rate (B/S. To increase the information transmission rate, each transmitted code element must be able to represent more bit information. This requires a good coding technology.

The Shannon formula provides the information transmission rate limit.That is, for a certain amount of transmission bandwidth (in Hz) and a certain signal-to-noise ratio, the upper limit of the information transmission rate is determined. This limit cannot be broken through. To increase the transmission rate of information, you must try to increase the transmission lineBandwidthOr you must try to increaseSignal-to-Noise RatioAnd there is no other way. The Shannon formula tells us that to get an infinite amount of information transmission rate, there are only two ways: either using an infinite amount of transmission bandwidth (which is obviously impossible) or making the signal-to-noise ratio infinitely large, that is, the use of non-noise transmission channels or the use of infinite transmission power (of course, these are also impossible ).

 

4.Which types of media are commonly used for transmission? What are their characteristics?

A: Common transmission media are:Twisted Pair wires, coaxial cables, optical fiber cables, and electromagnetic waves.

Twisted pair Cable features: (1) it can defend against certain electromagnetic interference and reduce interference after twisted pair, but cannot be completely eliminated; (2) it can be used for analog transmission and digital transmission; (3) it is inexpensive.

Coaxial Cable features: (1) Coaxial Cable has stronger anti-interference performance than twisted pair cables; (2)Baseband coaxial cable is only used for digital transmission, and broadband coaxial cable can be used for analog and digital signal transmission(3) The price is more expensive than twisted pair wires, but cheaper than optical fiber cables.

Optical fiber characteristics: (1) high transmission rate, low loss, long relay distance, especially economic for long-distance transmission; (2) good lightning and electromagnetic interference resistance; (3) no crosstalk interference, confidentiality is good, and data is not easily eavesdropped or intercepted; (4) small size, light weight.

Electromagnetic waves, advantages: (1) microwave band frequency is very high, its frequency range is also very wide, so itsLarge communication channel capacity(2) High microwave transmission quality; (3) high reliability of microwave relays; (4) Comparison of microwave relays with cable carriers of the same capacity and length, low investment in construction, quick results. Of course, microwave relay communication also has the following Disadvantages: (1)The adjacent stations must be directly connected without obstacles.. (2) microwave transmission is sometimes affected by bad weather. (3) compared with Cable Communication Systems, microwave communication is concealed andPoor confidentiality(4) It takes some manpower and material resources to use and maintain a large number of trunk stations.

 

5.What is Manchester encoding and differential Manchester encoding? What are their features?

A: Manchester encoding divides each code element into two equal intervals.The first interval is the high level, and the last interval is the low level. Code element 0 is the opposite, from low to high. The advantage of this encoding is that it can ensure a level conversion between the center of each code element, which is very advantageous for the receiver to extract the bit synchronization signal.The disadvantage is that the bandwidth it occupies doubles the original baseband signal..

The rule for differential Manchester encoding isIf the source code is 1, the level of the first half of the source code is the same as that of the second half of the source code. However, if the source code is 0, the level of the first half of the yuan is opposite to that of the last half of the yuan.. No matter the code element is 1 or 0, there must be a level conversion in the middle of each code element. Differential Manchester encoding requires more complex technologies, but it can obtain better anti-interference performance. (Difference, skip to 0, skip to 1)

 

6.What are the physical meanings of transmission latency and transmission latency?

A:Propagation latency refers to the time required for electromagnetic wave transmission in a channel.. It depends on the transmission rate and propagation distance of the electromagnetic wave on the channel.The sending latency is the time required to send data.. It depends on the length of the data block and the data transmission rate on the channel.

 

7.What are the main features of a simulated transmission system and a digital transmission system?

A: analog signal transmission is used to simulate transmission systems. During transmission, noise is added in. After signal attenuationAmplifier AmplificationBecause both the signal and noise are amplified by the amplifier, the final signal will be distorted.

The digital transmission system uses digital signal transmission. noise is also added during transmission, and the pulse signal waveform is deteriorated.Repeater AmplificationThe relay shapers perform Pulse Shaping and regeneration, and the noise is removed. After long-distance transmission, the noise will not be accumulated, so the final signal error is small.

 

8. EIA RS-232.AndRS-449Where are interface standards used?

A: EIA RS-232 is a serial physical interface standard developed by EIA, which is usually used for standard telephone lines (a phone) between DTE-DCE physical interface, the mechanical characteristics, functional characteristics, process characteristics and electrical characteristics between DTE and DCE are specified, and the transmission rate is relatively low.

While EIA RS-449 is generally used for broadband circuit (generally lease circuit), also specifies the mechanical characteristics, functional characteristics, process characteristics and electrical characteristics between DTE and DCE. RS-449 has three standard composition, RS-449 stipulated the mechanical characteristics, functional characteristics and Process Characteristics of DTE-DCE interface, RS-423-A stipulated the electrical characteristics of non-balanced transmission, RS-422-A specifies the electrical characteristics when balanced transmission is adopted. The transmission rate is higher than that of EIA RS-232.

 

9.What are the characteristics of baseband and broadband signal transmission?

Answer: (1) the baseband signal is used to sort the data.After encoding, it is directly sent to the line for transmission., According to the original signalInherent band transmission. OnlyAll the waySignal.

(2) A broadband signal uses a baseband data signal to modulated a carrier at a certain frequency, and its spectrum is moved to a higher frequency. Because the spectrum of each baseband signal is movedDifferent frequency bandsTherefore, they do not interfere with each other. In this way, you canTransmit multiple signals at the same timeThus, the line utilization is improved.

 

10.Yes600 mb(MB) data needs to be transmitted from Nanjing to Beijing. One way is to write data to the disk and then take the train to pick up the disks. Another method is to use a computer over a long-distance telephone line (the speed at which information is transmitted is2.4kb/s. Try to compare the advantages and disadvantages of the two methods. If the transfer rate is33.6kb/sWhat is the result?

A: continuous transmission without errors. If 600 kb/s is used, it takes 600 days to transmit 1048576 MB (= 24.3 × 8 = 5033164800bit. If the data is transmitted at a rate of 33.6kb/s, the time is 1.73 days. It is slower and more expensive to take a train than a passenger.

 

11.Yes4Sites for code division and multiplexingCDMACommunication.4Station slice sequence:

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, Station A, Station B, Station C, and Station D send a bit at the same time. If Station E receives the data from the air:

(-1,+1,-3,+1,-1,-3,+1,+1).

Which websites have sent data? What are the sent data bits?

A: You only need to calculate the received sequence and the sequence of the four station codes.Normalized Inner Product:

(-1, + 1,-3, + 1,-1,-3, + 1, + 1) ∙ (-1,-1,-1, + 1, + 1,-1, + 1, + 1)/8 = 1

(-1, + 1,-3, + 1,-1,-3, + 1, + 1) ∙ (-1,-1, + 1,-1, + 1, + 1, + 1,-1)/8 =-1 (-1 indicates sending bits 0)

(-1, + 1,-3, + 1,-1,-3, + 1, + 1) ∙ (-1, + 1,-1, + 1, + 1, + 1,-1,-1)/8 = 0 (0 represents orthogonal .. no relationship)

(-1, + 1,-3, + 1,-1,-3, + 1, + 1) ∙ (-1, + 1,-1,-1, -1,-1, + 1,-1)/8 = 1

The result is that A and D send bits 1,B sends bits 0C does not send data.

 

12. If the signal-to-noise ratio of a channel with a bandwidth of 4 kHz is 30 dB, what is the maximum data rate limited by Shannon theorem?
A: signal to noise ratio (SNR)10 * log10 s/n= 30

Required log10 S/N = 3

∴ S/N = 1000
According to the Shannon theorem, the maximum data rate of the channel with noise is limited.
C =H * log2 (1 + S/N)= 4000 * log2 1001 ≈ 40000 B/S

13. for a low-pass channel with a bandwidth of 3 kHz, if the transmitted code element has eight different physical states to represent the data, the data is represented Based on the nequest sampling theorem, what is the fastest data transmission rate on the non-noise channel?
A: Listen low-pass channel bandwidth H = 3 kHz,

Code element status n = 8, that is, each code element carries log2 8 = 3 binary information
Based on the nequest sampling theorem, this non-noise low-level communication channelFastest code element transmission rateIs (it has no relationship with the hexadecimal)

B = 2 h= 2 × 3000 = 6000 Potter(Because the sampling frequency is twice the highest when the sampling frequency is higher than the bandwidth)

CorrespondingData transmission rateIs
S = 2 H * log2 N = 2x3000 x log2 8 = 18000 B/S

 

 14. PSK with four phases: 0, π/2, π, and 3π/2 (Phase Adjustment), Two different amplitude ask (AM), That is, hybrid modulation technology is used.QAM (Orthogonal amplitude keying)If the baud rate is 4800 Baud, what is the data transmission rate of the modem?
Answer: The number of pending modulation signal states = 4 × 2 = 8
∴ The data transmission rate of the modem is:
4800 × log2 8 = 4800 × 3 = 14400 B/S

15. Briefly describe the differences between asynchronous transmission and synchronous transmission.
A: During asynchronous transmission, data is transmitted in characters and the sending time of the character is asynchronous, that is, the sending time of the last character is irrelevant to the sending time of the previous character, A random period of time can be separated between them, and only one character can be synchronized during transmission, the receiver can seize the opportunity of re-synchronization at the beginning of each new character (that is, to monitor whether there is a "Start bit" logic level on the line to degrade, which indicates the arrival of a new character, starts synchronization of a new character ). During asynchronous transmission, the generator and receiver clock can be generated independently of each other because it does not require precise synchronization between the generator and the receiver clock and does not require high hardware of the clock circuit, as long as their clock error does not exceed 7%, the receiver can use its own receiving clock to correctly sample the data bit of a character sent.
During synchronous transmission, data is transmitted through data lines in units of BIT data blocks (frames). Generally, a frame has several hundred to several thousand characters in length, during such a long data transmission period, the synchronization between the receiving and sending sides should be maintained. At this time, the generator and the receiver clock must be precisely synchronized, and the hardware clock circuit of both sides must be very high, therefore, generally, the sender sends its clock signal to the receiver as follows: (1) "internal synchronization". At this time, only one data line is used to transmit the synchronous signal and embed it into the data signal, use the encoding waveform transmission that can contain synchronous clock signals, such as Manchester, differential Manchester, or dual-pole type zero-encoding. After being transferred to the receiver, the synchronous clock is separated from the receiver, or (2) in the "out-of-sync" method, another clock line is used to transmit the synchronous clock pulse. In this way, the clock line and the data line are separated, and the generator and the receiver are transmitted using two lines. No matter whether it is used for external synchronization or internal synchronization, after the receiver obtains the synchronous clock pulse, it can use the synchronous clock to sample the data line to get one bit of data transmitted.
Asynchronous transmission requires at least 20% overhead because "Start bit", "parity bit", and "Stop bit" must be added before and after each character data bit. ,Synchronous transmission requires only one frame with a few frame headers and frame tail fields, so the overhead and efficiency are much higher than that of asynchronous transmission. The HDLC overhead of common synchronization transmission link control protocols is only 0.6%. (High asynchronous transmission overhead and low synchronization transmission overhead)

 

16. What is baseband transmission? What is the difference between it and band transmission (broadband transmission?
A:Signals generated by computers or terminalsThe frequency of the waveform spectrum starts from zero and the signal does not need to be modulated to a higher frequency band. The occupied frequency range is called the basic frequency band (baseband for short. This signal is called a baseband signal. When transmitting data, it sends the baseband signal to the line in the intact form, which is called the baseband transmission. No modem is required for baseband transmission, and the device cost is small. It is suitable for short-distance data transmission. The local area network uses baseband coaxial cables and twisted pair wires as transmission media for data transmission.
In band transmission (broadband transmission), the baseband signal needs to be modulated by a modem, that is, some parameters (amplitude, frequency, and phase) of the higher frequency carrier waveform using the baseband signal) control, so that these parameters change with the baseband signal, after modulation, the signal becomes a high-frequency tuned signal, and then sent to the transmission medium for transmission. The tuned signal is transmitted to the receiving end through a line and restored to the original baseband signal through demodulation. This type of band transmission not only overcomes the disadvantages that many long-distance telephone lines cannot directly transmit baseband signals, but also transfers signals from different channels to different high-frequency segments, and then transmits the signals on one line after mixing them,To achieve multiplexingTo improveHigh utilization of communication lines. When band transmission,Both the sender and receiver require a modem.

 

In addition, the differences between bandwidth data transmission rate and channel capacity

Bandwidth: the frequency range in which the channel can transmit signals without losing sight. The transmission media designed for different applications has different channel quality and supports different bandwidths.
Channel Capacity: The maximum semaphore that a channel can transmit per unit of time, indicating the channel's transmission capability. The channel capacity is also sometimes expressed as the number of BITs that can be transmitted per unit of time (the channel's data transmission rate and bit rate), expressed in bit/s (B/S) format, note: bps.
Data transmission rate: the maximum number of BITs that a channel can transmit per unit of time. The channel capacity is proportional to the channel bandwidth. The larger the bandwidth, the larger the capacity.