1、频谱:36个频点(带宽7.2MHz);复用系数:Reuse factor=9; 每个基站配有3个扇区(假设每扇区配有1个BCCH,1个SD),每个基站覆盖范围0.4875平方公里。请计算该规划网络的最大容量为每平方公里多少爱尔兰? 每个扇区的载频数为:36/9=4(个)每扇区TCH信道数为: 4*8-2=30(个)每扇区话务根据爱尔兰B表查得为:21.9爱尔兰则每个基站的话务为:3*21.9=65.7爱尔兰则网络容量为:65.7*(1/0.4875)=134.8爱尔兰/平方公里频率规划类计算:某地区网络中有一新建开发区,面积40KM2,预计人口80000人,手机用户渗透率约80,根据无线链路预
2、算的结果该地区平均小区覆盖面积0.5KM2,话务模型预测每用户忙时通话时长90S,可用频谱带宽为15.6MHz,请按GOS=2%,频率复用采用MRP方式(BCCH 3*6=18,TCH 3*5=15)的模式进行该区域的网元数量测算。计算步骤: 1)覆盖需求:满足覆盖需求共需小区数量为:40/0.5=80个小区 2) 容量需求:忙时全网总话务量:80000*0.8*90s/3600=1600erl 每小区载频配置计算:总频点数15.6/0.2=78个,BCCH频点18个(3*6复用),非BCCH频点60个(3*5复用),最大配置可以做到每小区5块载频 60/(3*5)+1=5 每小区可用TCH信
3、道数:5*8-1(BCCH)-2(SDCCH)=37 每小区可承载话务量:28.25erl 满足容量需求共需小区数:1600/28.25=57 综上所述: 取最大值,该地区需要80个小区1. 下图为某地区网络的基站分布和配置,请采用MRP频率复用技术为该网络确定频率分配方案和复用模型,并为基站A、B分配频率。(8分)1. Reuse pattern:TRXcell numberfrequency neededreuse patternfrequency number145213*721(94-74) 23*618(1-18)340193*515(20-34)425123*412(36-47)5
4、1683*39(49-57)61379(59-67)11rest10total20394Frequency plan:BCCHTRX2TRX3TRX4TRX5TRX6TRX7TRX8A120364959guard+rest,total 10 frequencysB1875262C18030445565A2932375060B28626415363C27914315666A39222385161B385927425464C3781532465767A4912339B4842843C4773347A59024B58329C5761734A689B682C67518A788B781C774guard
5、733548586869707172Cell A and cell B:计算SDCCH容量 已知条件:2 TRXs/Cell,服务等级1% blocking probability,1.5 min/call/subs/BH,location updates once in 60 min.,call assignment (7 s/Call including ciphering and authentication),计算每个小区应配置的SDCCH数目;答案:每小时每用户话务量:1.5*60/3600=0.025ERL假设配1条SD时2TRX基站可承载话务量(呼损1%):7.35ERL此时可承
6、载用户数:7.35/(0.025)=294通话产生的SDCCH话务量为:294*7/3600=0.57ERL位置更新产生的SD话务量为:该扇区每小时共产生0.57*2=1.14ERL的SDCCH话务量在呼损1%的情况下,根据爱尔兰B表查询,需要5个SDCCH/8信道,即1条SDCCH时隙LAC容量寻呼能力计算:PAGING GROUP计算:MFR*(CCCH数-AG) CCCH BLOCK在COMBINE为3,NONCOMBINE为9 PAGING MESSAGE以及容纳用户数计算: COMBINE时,CCCH=3,设定AG=2,PCH=1,每BLOCK包含的寻呼消息数为3(IMSI与TMSI
7、平均.2个IMSI,4个TMSI),每个复帧时长:4.615ms*51=235ms每小时传输复帧数为:3600/235*1000=15319个 每小时BLOCK数=复帧数*PCH BLOCK数=15319*1=15319 每小时PAGING MESSAGE数量=15319*3=49575 每用户需要两个消息进行寻呼,每小时可寻呼用户数:总寻呼消息量/2=22978.5 NON-COMBINE时: CCCH=9,AG=2,PCH=7,每BLOCK包含寻呼消息数3,(IMSI与TMSI平均), 每小时传输复帧数为: 每小时BLOCK数=每小时复帧数*PCH BLOCK数=15319*7=10723
8、3 每小时PAGING MESSAGE数量=每小时BLOCK数*3=321702 可寻呼用户数:总寻呼消息量/2=160851在一个LAC区内,所有小区采用uncombined(BCCH/SDCCH)信道配置,并且AG设为2。如果采用TMSI寻呼用户,每个BLOCK最大可以发送4个寻呼消息,采用IMSI寻呼,每个BLOCK最大可以发送2个寻呼消息,平均为3个寻呼消息/BLOCK,对于每个用户,需要用2个寻呼消息。请计算一个小时之内最大可以寻呼多少个用户?(1) 在一个复帧内,UNCOMBINED方式共有9个BLOCK作为CCCH信道,因为AG=2,所以PCH为7个BLCOK;(2) 7个BLO
9、CK可以发送的消息数为7321个;(3) 因为每个复帧的时长为235MS,所以 系统可以发送的消息数量为3600211000235321702个;(4) 由于每个用户需要2个消息进行寻呼,所以系统在一个小时内可以寻呼的用户数量为3217022160851个链路预算类计算:上行:下行:1. 以下为某型号基站的典型无线链路参数值:基站发射功率 43dBm, 合路-双工器损耗4.5 dB, 馈线损耗 3dB, 基站天线增益 15dBi, 手机灵敏度 -102dBm,人体损耗3dB,手机天线增益 -2dB设下行允许最大损耗为y,则: 43-4.5-3+15 -y-3+(-2)= -102 y=147.
10、5dbm手机发射功率 33dBm,手机天线增益 -2dB,基站天线增益 15dBi,分集接收增益 5 dB,基站馈线损耗 3dB, 基站灵敏度 -110dBm。设上行允许的最大损耗为x,则:33+(-2)-3-X+15+5-3=-110 X=155dbm请计算:上、下行最大允许无线传播损耗分别是多少?无线链路是上行受限还是下行受限?实际最大允许无线传播损耗是多少?解答: 上行: 下行:越大越不受限:比如上行140db,下行130db则下行受限从计算的结果可以看出,下行受限实际最大允许无线传播损耗是147.5dbm小区选择和重选C1/C2计算:C1,C2公式: C1 = (A - Max(B,0
11、) A = Received Level Average - p1 B = p2 - Maximum RF Output Power of the Mobile Station p1 = rxLevelAccessMin Min. received level at the MS required for access to the system p2 = msTxPowerMaxCCH(一般设为33dBm) Max. Tx power level an MS may use when accessing the system C1=当前平均电平-最小接入电平-MAX(MSTxPowerMAX
12、CCH-手机最大发射功率) 当PET640s时, 未达到PET时间时,C2=C1+REO-TEO 达到PET时间时,C2=C1+REO 当PET=640s时, C2=C1-REO REO:小区重选偏执,TEO:临时偏执,PET:判决门限1) Please calculate C1 for both cells, with cell will the class 2MS clamp on?C1=当前平均电平-最小接入电平C1= Av_RxLevRxLevAccessMinMax(MSTxPwrMaxCCHMaxoutput power of MS)C1 (CELL1) = -80 +98 -0
13、= 18C1 (CELL2) = -83 + 102 -0 = 19Thus, MS camps on CELL2The MS is currently camp on Macrocell, lets assume that C1 ofboth serving cell and the neighbor cell has been measured as 32.2) During time from 0 to 19s, which cell will the MS camp on当PET1-19秒C2=C1+REO-TEOC2 =32+20 -30*1=22 C2 C2 C1, now tar
14、get cell is very attractive and the idlemode MS will camp on the microcell.4) Target cell evaluation for a MS connecting to Cell A. Threeadjacent cells in the same BSC are reported: Cell B=-85 dBm, CellC=-95 dBm, Cell D = -80. All adjacent cells have the same settingsas following,btsLoadThreshold=80
15、%hoLevPriority =3hoLoadFactor=2rxlevmincell (n)=-92 dBm.(Cell A - Cell B, Cell A - Cell C,Cell A - Cell D)The TCH load in Cell B=90%, Cell C=70%, Cell D=75%.Please list the priority list of the candidate cells?先判定电平,再判定负荷门限(超负荷会降低优先级)Cell D (3), Cell B(3-2=1), Cell C2.下表为BTS传给BSC的测量报告,请确定在什么时间会触发切换?
16、 5分35-110MR1st 2nd3rd4th5th6th7th8th9th10thUL RxLevDTX USEDHoThresholdLevUL = -85dBm WindowSize = 3, Weighting = 2Px = 2, Nx = 3btsMeasAver (BMA) = 1 (no pre-processing in BTS) (35*2+30+31*2)/5=32.4 -77.6 (30+31*2+32*2)/5=31.2 -78.8 -77.6 -78.8 -86 -83 -86 Answer : 在收到第8个测量报告后会触发切换。3、一个服务小区有4个邻区(a,b
17、,c,d),服务小区对每个邻区的SL都设为95dBm,请根据下面的数据对4个邻区进行优先级排队(列出计算方法,并简单说明原因)。5分 由于SL=-95,而d小区的接收电平为96,其它小区的接收电平均大于SL,所以d小区将排在最后; a小区的优先级为:422(由于a小区过载),b小区的优先级为:3,c小区的优先级为: c, b小区的优先级相同,但是b的接收电平较c高顺序为b c a d切换和平均算法:Calculate averaged RxLev_DL without considering “Fast HandoverAveraging” method, and determine the
18、handover trigger point.Parameter ValueQ3 name: hoAveragingLevDL level downlink window size (LDWS), weighting (LDW)21) What is the averaging window size?A. Window size=5B. Window size=4C. Window size=3D. Window size=222) What is the concept of weighting scheme?A. To put more weights for the measureme
19、nts with DTXB. To put more weights for the measurements without DTXC. To put more weights for the measurements regardless of DTXstatus23) What is first averaging value?A. -86 dBmB. -89.8 dBmC. -95.0 dBmD. -90.2 dBm计算:第1次平均值=(-86*2)+(-89*2)+(-93)+(-96)/(2+2+1+1)=-89.8 dBm 第2次平均值=(-89*2)+(-93)+(-96)+(
20、-95*2)/(2+1+1+2)=-92.8 dBm 第3次平均值=(-93)+(-96)+(-95*2)+(-96*2)/(1+1+2+2)=-95.2 dBm 第4次平均值=(-96)+(-95*2)+(-96*2)+(-94*2)/(1+2+2+2)=-95.1dBm24) What is last averaging value?=(-96*2)+(-94*2)+(-92)+(-98)/(2+2+1+1)=-95.0 dBm25) What is first px/nx value?A. 0/3B. 1/3C. 2/3D. 3/326) What is second px/nx value?27) If the enaFastAveHO= yes, what is the first averaging value?
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