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2011数学建模(美赛)b题

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2021-02-08 10:25
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2021年2月8日发(作者:floria)


Team #12110











Page 1 of 9


Minimizing the Number of repeaters


Introduction




Very high frequency (VHF) is the radio spectrum



whose frequency band ranges from 30MHz to


300MHz. VHF is always used for radio stations and television broadcasts.


In addition, it is also used


by


signal


transmission


of


sea


navigation


and


aviation.


Because


the


radio


spectrum


of


VHF


is


transmitted through straight lines, a signal is easily influenced by geographical factors easily.


Thus,


signals


become


weak


when


it


is


transmitted


and


some


low- power


users


need


repeaters


to


amplify


them and increase the transmission distance. We consider the situation in which every two repeaters


are too close or the separate frequency is not far enough apart which can interference with each other.


In order to mitigate the interference caused by the nearby repeaters, this paper employs a continuous


tone- coded squelch system (CTCSS). We associate to each repeater a separate subaudible tone



that


is,


the


subaudible


tone


(67Hz-250.3Hz)


is


added


to


VHF.


In


this


way,


repeaters


recognize


signals


attached to the same subaudible tones just like secret keys.


In this system, the nearby repeaters can


share the same frequency pair. When users send the signals at one frequency, different repeaters with


subaudible


tones


can


recognize


signals


from


the


users


the


same


subaudible


tone.


If


the


users


in


a


certain area contact with each other, we should consider the signal



s coverage area of the users and


the repeaters. As long as the users



signals are accepted by repeaters, the signals could be amplified


to


transmit


farther.


At


the


same


time,


the


repeaters


attached


with


the


subaudible


tones


could


only


recognize the users with the same subaudible tones. Hence, we can consider repeaters corresponding


to the number of the users, which leads to the problem


of frequency channel.


When the number of


users in this area increases, we can add repeaters.


If two repeaters have different subaudible tones,


they


would


not


communicate


with


each


other.


Thus,


we


should


consider


the


problem


of


how


the


repeaters


communicate


with


each


other


when


they


have


different


subaudible


tones.


In


the


mobile


communication system



the spectrum is influenced by many factors such as reflex



diffraction and


dispersion.


Therefore, when the radio spectrum transmits in the mountainous area



we should still


consider the factors above.








Repeaters


[4]




Repeaters are a type of equipment which can amplify signals



make up the deamplification signals


and support far distance communication.




CTCSS


[5]


CTCSS(Continuous


Tone


Controlled


Squelch


System


)


is


short


for


subaudible


tones,


whose


frequency


ranges


from


67Hz


to


250.3Hz.


It


is


added


to


the


radio


spectrum


to


make


the


signal


carry


with


a


unique secret key.


Assumption



The users in the area is uniform distributed


The signal of the radio spectrum in the


area can’t be effected by environment



In a certain period of time there are a small number of users removing


All repeaters have the same standard



Analysis and solution of the model to the first problem



The problem is to find a least number of repeaters in an area of radius 40 miles so that the users in this


area can communicate with each other. Considering that the given area is flat, we assume that the signal of


Team #12110











Page 2 of 9


each repeater covers a circular area and the repeater lies in the center of the circle. The following Figure 1


shows the relationship of three adjacent repeaters.


A










B











C






Figure 1.


The relationship of three adjacent repeaters

































For case B of Figure 1, if three circles are tangent to each other, then we find that the center area cannot


be covered by the singles. In order to make the signal cover the triangle area, we have to consider adding a


circle, which means that we need to add an extra repeater. Undoubtedly, it is wasteful.


For case C, if the


intersection of three circles is not null, similar to case B, we also have to add another repeater. Thus, it is


easy to find that case A, comparing with cases B and C, is optimal. Thus, we obtain the largest covering area


and the smallest number of repeaters.



When


more


circles


intersect


together,


we


get


some


closely


linked


hexagons,


as


shown


in


Figure


2.


Obviously, it looks like a honeycomb structure. In fact, the honeycomb pattern is one of the most efficient


arrangement for radio spectrum. It transmits by the wireless medium of microwave, satellites and radiation.


The


structure


has


a


feature


of


point- to-point


transmission


or


multicast.


It


is


widely


used


in


UN


Urban


Network, Campus Network and Enterprise Network.



Figure 2. some circles intersecting together form the closely linked hexagons







Now we have a circle with radius of 40 miles. Then we analyze the distances of signals from users and


repeaters covering in


the circle.


Because the differences


for the users and repeaters in


energy


and height,


they have different covering distances. We calculate the distances with the theory of space loss. The formula


[6]


is



L


M


?


88.1< /p>


?


20


lg


F< /p>


?


20


lg


h< /p>


1


h


2


?


40


lg


d


L


M


F


,



the wireless loss




the communication working frequency(MHz)



h


1


h


2



the height of the repeater (m)



the height of the user(m)


d


Team #12110












the distance between the user and repeater(km)



F


?


150


M


H


Z


,


h


1


?


1.5


m


Page 3 of 9


We assume that



and


h


2


?


3


0


m


, under the condition of the cable loss and antenna


gain, we obtain the system gain


SG


?


P


t

< br>i


?


P


A


?


(


R


A


?


C


L


?


R


R


j


)


(


i


?


1,


2

< p>
j


?


1,


2)

< p>
.


The system gain is the allowed decay maximum of the signal from the users to repeaters. If the system


gain


value


is


higher


than


the


wireless


loss,


the


users


could


communicate


with


each


other.


Reversely,


the


users could not communicate. We make the system gain value equals to the wireless loss, thus, we get the


extremity distance between the user and repeater. Then we have


SG


?


L


M



We choose a typical repeater and the user facility. Thus, the parameters


[6]


and data of the repeaters are as


follows


The transmitting power


P


t


1


?


20

< p>
W


(43


dB


m


)


R


R


1


?


?


116


dB

< p>
m



The receiving sensitivity






The antenna gain of the repeaters


R


A


The cable loss


C


L


?


2


dB



The parameters of the interphone



The transmitting power


?


9.8


dB


P


t


2


?


4


W

< p>
(36


dB


m


)



The receiving sensitivity

< p>
R


R


2


?


?


116


dB


m





SG


1


?


144.2


dB


The antenna gain of the interphone

< br>P


A


?


0


dB


The


system


gain


of


the


system


from


users


to


repeaters


distance from the users to repeaters


system


from


the


repeaters


to


users


d


2

?


20.7


km


.


Thus,


we


get


the


sending


d


1


?


13.8


km


. Prove in the same way, we have the system gain of the


,


the


sending


distance


from


the


repeaters


to


the


users


S


G

2


?


1


5


1.2


d


B



?


1


3.8


km


According


to


the


sending


distance


D


1



between


user


and


repeater


as


well


as


the


property


of < /p>


D


2


?


23.0 9


km


regular hexagon, we calculate the distance between two repeaters. We obtain that


described in Figure 3. Because


D


2


, which is



is shorter than


D


2


'


, users in this area cannot communicate with each


other.


Thus,


we


consider


the


sending


distance


D


2


'



between


two


repeaters


firstly.


Then


we


calculate


the


distance between the user and the repeater again shown in Figure 4. Finally, we get that


D


1


'


?


1


2.4


km


.


Team #12110











Page 4 of 9


D2=2 3.09km


D1=13.8km


Figure 3. the calculation distance according to the sending distance from users to repeaters.






D2' =21.45km


D1'=12.4km



Figure 4. the calculation distance according to the sending distance from repeaters to the users.


According to the calculated distance


D


1


'


?


12.4


km

< p>
D


2


'


?


21.45


km


, we know that the given circle has


a


radius


of


40


miles.


We


firstly


consider


the


signals




covered


area


of


the


repeaters.


Thus,


we


get


the


distribution of the repeater stations in this area showed in Figure 5. The number of repeater stations is 37.


However, we need to decide the amount of repeaters distributing in one station.


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