2011國(guó)際數(shù)學(xué)建模競(jìng)賽試卷

1、AbstractThe repeaters can pick up signals,lify them, and retransmitthem. Through repeaters, users can communicate with others farapart. Due to the limit of channel, the number of people everyrepeater canmodate has limit. We can use CTCSS technologyto assign different PL tones to different repeaters

2、to distinguishdifferent repeaters. Then the questiones a coverage problem.We use cellular mto simulate it, and find some limits, then wechalk up the optimal solution.he flat area with 1000 simultaneous users, we amet usersdistribute averagely and the number of people in each area does notexceed theu

3、m use of repeater, then we need only to cover allarea. We use cellular mto calculate, and cellular usually isreplaced by Regular hexagon. We findt the minimum number ofrepeaters nerepeater cansary is 43. Then the number of peoplet amodate is 155. Forthe users dont distributeaveragely, as long as the

4、 number is lessalso applicable. When the populationn this limit, this mises 10000, we changesome parameters. We cut down every area. But it will also reducetheum number of users of each repeater, so it is nesary tocalculate appropriate areas. We use optimization ml to calculate.sary is 93.We findt t

5、he minimum number of repeaters neOnce think about the countercheck of mountains in practice, we canuse directional and nondirectional repeaters to work together. The nondirectional ones can be built on the top, and directional ones on mountainside or valley. The ones on the top can be used as transf

6、ersion betn other repeaters. We can use Omand 8m.Using different kinds of repeaters, we increase the radius largely,and eliminate blind spots.Resement of ProblemRepeaters are used to extend areas the VHF radio spectrum can transmit. But if there are too many repeaters in the same area, they willerfe

7、re with one another. Toe this problem, the CTCSS technology can be used. Its knownt thespectrum available is 145 to 148 MHz, and the transmitter frequency and the receiver frequency must differ with 600 kHz. Besides, there are 54 different PL tones available. We need to solve how many repeaters shou

8、ld be use and how to choose the localitiesto ensure people in 40can communicate simultaneously.ysis of ProblemThe problem ist we should choose the minimum number of repeaterst canmake users far apart communicate with each other. The key ist we should ensureevery two users can contact, and eliminate

9、theerference. Through the CTCSStechnology and the rational allocation of the PL tones, we can meett.Question OneTo make 1000 peoplehe flat area can communicate with each other, we need to calculate he theoretical radius of each repeater. Then we assign the PL tones anddivide the spectrumo some chann

10、els to get theum number of peopletcan use a repeater. Later we compare this number with the actual number in each areato decide the actual radius of each repeater.ast the problemes how to use radius.the minimum number of circles to cover the circular area of radius 40Question TwoWhen the number of p

11、eople increases to 10000, the population density increases.Because the limithe population of each repeater, it is nesary to add repeaters.We can change some parameters of the mto calculate.Question ThreeOnce think about the countercheck of mountains in practice, we must take theincreased account of

12、the attenuation of signal. It reacts on the areat repeater workson. Then we should select a good placement of repeaters, and use directional and nondirectional repeaters to work together.Amption of mThe users usehones.Users use same equipments and the repeaters are the same. The users distribute ave

13、ragely.Besides the attenuation of signal, there is no other reason of signal.t affect the transmitExplanation of symbolsF: Communication frequency (MHz)h1: The height of A poh2: The height of B po(m)(m)d: The distance betPt1: Radio transmisn poA and B (m) er (dBm)PA1: The gain of Walkie-talkie anten

14、na (dB)RA: The gain of repeater antenna (dB)RR: The sensitivity of repeater as a receiver (dBm)Mand SolutionQuestion OneDetermine the scope of each repeater:1. A practical formula of the transmisloss of radio wave:LM(dB)=88.1+20lgF-20lgh1h2+40lgdThe limit of this formula ist the hypsography is about

15、 15m and the communicationdistance is lessn 65 km.Calculate the distance of wireless communication system:We have knownt:a. System operating frequency: 145-148MHzb. RepeatTransmitarameters:er: 20W (43dBm)Receiver sensitivity: -116dBmGain of nondirectional antenna: 9.8dBmThe height of antenna: 30mc.

16、Radio parametersTransmiter: 4W (36dBm)Receiver Sensitivity: -116dBmAntenna gain: 0dBmHeight: 1.5m2. The system gain of repeaters andhones:um attenuation of the signalThe system gain is so calledt thethones transmit to repeaters. On conditiont the antenna has no gain:System Gain (dB) = Transmiter (dB

17、m) - Receiver Sensitivity (dBm)Consider the conditions of antenna gain, the system gain isSG(dB)=Pt1+PA1-(RA+RR)=36+0-(9.8-116)=142.2(dB)t:Pt1: Radio transmiser (dBm)PA1: The gain of Walkie-talkie antenna (dB)RA: The gain of repeater antenna (dB)RR: The sensitivity of repeater as a receiver (dBm)3.I

18、f the system gain is equal to the transmisloss of radio wave, then radiocommunication has reached the limit of energy. If the system gain is lessn thetransmisloss of radio wave, then communications may not be established.Put system gaino the formula of the transmisloss of radio wave:142.2=88.1+20lg1

19、45-20lg1.5*30+40lgdThen we find: d=12.5m4.The above only calculate the distancet uplink signal ( repeaters), and never calculate the distance downlink signalhones). Usually, due to repeaters have larger transmithones transmit to (repeaters transmit to er, the distance ofdownlink signal is moren upli

20、nk signal. Considert the system communicationis two-way, so we often calculate the distance based on the uplink signal.The effective transmisdistance is about 30km.modate:The number of peoplet a repeater canThe communication ofa repeater include two sides: betn repeaters and users,betn two repeaters

21、. All communications use PL tones. Every repeater has its ownPL tone. So, when users transmit massages to repeaters, the only thing it needs to dois putting the repeaters PL toneo signal. Then, the aim repeater can receive, andothers cant. This make the unidirectional communication from users to rep

22、eaters come true. Otherwise, when repeaters transmit massages to users, the only thing is toput users PL tone to the signal. In summary, the number of peoplet a repeater canmodate is assoted with the number of channels and PL tones available.Communications in repeaters and repeaters can make all rep

23、eaters coverageascircle, using Hexagonal mcan get betn each repeaters and the surrounding ofthe six repeaters coverage have overlap. This way, just assigned to all the repeaterswith the same channel by six different PL tones, and to ensuret each repeaters andthe surrounding six of the repeaters have

24、 different PL, which repeaters can be achieved with repeaters by the communication.In accordance witht the transmitter frequency and the receiver frequency differ in600 kHz, there are six frequencies available: 145.0MHz, 145.6MHz, 146.2MHz, 146.8MHz, 147.4MHz, 148.0MHz. Because there are same PL ton

25、es in one channel,to avoiderfering each other, these frequencies can be dividedo three pairs:145.0MHz and 145.6MHz, 146.2MHz and 146.8MHz, 147.4Mhz and 148.0MHz.t is, there are three channels.Considert every repeater communicate will others, so we can assign some PL his channel and PL tones in other

26、tones in one channel to them, and othL toneschannels can be assigned to users. Then the number of people modate is: 543 - n.t a repeater canThen the questionest covering a circle round with the least number of smallcircle. We amet the number of people in a circle is notthe number of, as is to use re

27、gularpeoplet a repeater canmodate. We use Cellular mhexagon replace circle to cover. As the following picture shows:From this, we can summarize Now every repeater covers thet: n=43.um number is 38, but the number of peopleta repeater canmodate is 119. So the ame is right.So the minimum number of rep

28、eaters nesary is 43.For the users dont distribute averagely, as long as the number is lessn 119, thismis also applicable. Then migration has no effect on the communication.Question twoWhen the number of peoplees 10000, if we also take the radius 12.5km, thennumber of peoplet a repeater covers isit c

29、anmodate. So we reducethe radius. The limits are:（54*3-m）*n=10000Nr=64 r=12.5m=1+6* 1+2+( 30-r )/2r+1 n=1+6* 1+2+ (64-r )/2r+1 We can summarizet: n=91Question Three1. OmFor users on Mountain peak, we can place repeatershe two sides of foot of a mountain and the peak. Due to the circle recovery of the signal transmitting andreceiving, the signal from the peak can be received by the repeater otherwise.2.8mhe foot. So isFor usershe valley, we can place the every two end a repeater. Due to the circlerecovery of the repeater, the signal surround betn the