最新網(wǎng)絡(luò)優(yōu)化專業(yè)試題庫(kù)-

1、網(wǎng)絡(luò)優(yōu)化專業(yè)考試復(fù)習(xí)題一、填空題1、在MSC中與周期性位置更新時(shí)間T3212相對(duì)應(yīng)的參數(shù)是。答案:BTDM2、通過(guò)指令可以看到指定小區(qū)當(dāng)前ICMBAND值,從而判斷小區(qū)可能受到上行干擾。答案:RLCRP3、在缺省情況下,與訓(xùn)練序列碼(TSC相等的代碼是。答案:BCC4、使用LAPD信令壓縮,好處是減少了和間的物理鏈路,從而優(yōu)化傳輸方案,節(jié)約傳輸資金投入。答案:BSC 、 BTS5、在手機(jī)通話過(guò)程中,用來(lái)傳遞切換命令消息的邏輯信道是。答案:FACCH6、全速率業(yè)務(wù)信道和半速率業(yè)務(wù)信道傳送數(shù)據(jù)最快速度為和。答案:9.6kbit/s、4.8 kbit/s7、使用GPS配合TEMS測(cè)試,GPS選項(xiàng)設(shè)置

2、為時(shí),才能正常記錄經(jīng)緯度信息。答案:NMEA8、在BSC終端上用指令,可以看到指定硬件是否有告警(Fault Code Class xx以及可能更換的部件(Replace Unit。答案:RXMFP9、同一小區(qū)中,每個(gè)Channel Group中最多可以容納個(gè)頻率。答案:1610、發(fā)生Intra-cell切換的可能原因?yàn)?。答?服務(wù)小區(qū)信號(hào)質(zhì)量差而信號(hào)強(qiáng)度較強(qiáng)11、當(dāng)一個(gè)小區(qū)參數(shù)BCCHTYPE=NCOMB,SDCCH=2時(shí),該小區(qū)SDCCH的定義數(shù)為_(kāi)答案:1612、在BSC終端上提取即時(shí)統(tǒng)計(jì)文件的指令是。答案:SDTDP: rptid=xxx,int=1;13、用來(lái)處理小區(qū)參數(shù)軟件的是_答

3、案:IPOS14、在BSC A中定義BSC B中的小區(qū)作為外部小區(qū)時(shí),該小區(qū)的LAC應(yīng)與相同。答案:BSC B15、頻率多重復(fù)用技術(shù)是增加網(wǎng)絡(luò)容量的常用方法,它的英文縮寫是_答案:MRP16、交換機(jī)上記錄的超過(guò)時(shí)間提前量(TA掉話主要指測(cè)量到TA超過(guò)參數(shù)指定的值時(shí),系統(tǒng)強(qiáng)制結(jié)束通話。答案:MAXTA17、應(yīng)用分集接收技術(shù)可以大大降低移動(dòng)通信接收信號(hào)不穩(wěn)定的概率,分集技術(shù)主要包括_、_、_和_.答案:空間分集,時(shí)間分集,頻率分集,極化分集18、對(duì)_ 和_參數(shù)進(jìn)行修改可以緩解話音信道擁塞:答案:KOFFSET、ACCMIN19、_、_、_可降低移動(dòng)系統(tǒng)中干擾的方法答案:動(dòng)態(tài)功率控制、不連續(xù)發(fā)射、跳

4、頻20、_、_、_和_可能產(chǎn)生隨機(jī)接入失敗答案:上下行功率不平衡、存在同BCCHNO同BSIC情況、覆蓋不好、存在干擾21、在_和_情況下可以推斷小區(qū)上行信號(hào)可能受到干擾。答案:(BSC監(jiān)控終端上發(fā)現(xiàn)該小區(qū)各BPC的ICMBAND值不都為1、(占用開(kāi)啟上下行動(dòng)態(tài)功率控制的小區(qū)時(shí),手機(jī)從待機(jī)狀態(tài)進(jìn)入通話狀態(tài)后RQxlev下降而手機(jī)發(fā)射功率沒(méi)有下降22、用來(lái)處理統(tǒng)計(jì)數(shù)據(jù)的常用軟件包括_、_.答案:TURTLE、 SPOS23、產(chǎn)生信令信道掉話的可能原因有_、_和_.答案:TCH擁塞;上下行鏈路質(zhì)量低于BSC交換屬性定義的BADQUL、BADQDL;信令錯(cuò)誤導(dǎo)致測(cè)量報(bào)告丟失24、在TEMS Inve

5、stigation測(cè)量中,接收?qǐng)鰪?qiáng)(RxLev中出現(xiàn)黑色豎線表示發(fā)生1次切換,這條黑線與第三層信息中的切換信息對(duì)應(yīng),準(zhǔn)確地說(shuō)該豎線的對(duì)應(yīng)信息為_(kāi) 答案:Handover complete25、在開(kāi)啟GPRS的小區(qū)中,使用RLCRP指令查看專用PDCH的狀態(tài)為:答案:始終為Busy26、當(dāng)一個(gè)小區(qū)的參數(shù)T3212值設(shè)為25時(shí),表示周期性位置更新時(shí)長(zhǎng)為。答案:2.5小時(shí)27、愛(ài)立信頻率規(guī)劃軟件TCP是 _的英文縮寫。答案:TEMS CellPlanner28、無(wú)線鏈路超時(shí)參數(shù)RLINKT在指派專用信道后從設(shè)定值啟動(dòng),每個(gè)SACCH單元成功被解碼后該值增加,如果解碼沒(méi)有成功則該值減小。答案:2,12

6、9、整個(gè)切換過(guò)程由、共同完成。答案:MS、BTS、BSC、MSC30、由于陰影效應(yīng)引起的衰落稱為。答案:對(duì)數(shù)正態(tài)衰落31、SDCCH上能承載_ _、_、_、_業(yè)務(wù)。答案:呼叫建立、短信息、位置更新、周期性登記、補(bǔ)充業(yè)務(wù)登記等業(yè)務(wù)二、選擇題1、以下各項(xiàng)中對(duì)手機(jī)接收功率不產(chǎn)生影響的是:A、天饋線損耗B、基站天線增益C、分集接收增益D、路徑損耗答案:(C2、不屬于無(wú)線覆蓋預(yù)測(cè)模型是:A、Algorithm 9999B、Okumura-HataC、PrinthanaD、Walfish-Ikegami 答案:(C3、手機(jī)從服務(wù)小區(qū)A重選到另一位置區(qū)中的小區(qū)B,CRHA =4dB,CRHB=6dB,那么實(shí)

7、際重選中小區(qū)重選的滯后值為 dB。A、2B、4C、6D、10答案:(B4、假設(shè)用戶甲始終在小區(qū)A的覆蓋區(qū)內(nèi),T3212A=20,甲開(kāi)機(jī)后10分鐘接聽(tīng)了1個(gè)3分鐘的電話,此后再?zèng)]有進(jìn)行呼叫,那么甲的手機(jī)將在分鐘后進(jìn)行第一次周期性位置登記。A、7B、120C、113D、107答案:(B5、與一個(gè)載波配置很多的小區(qū)相比,采用蜂窩小區(qū)模式并進(jìn)行頻率復(fù)用的結(jié)果是。A、增加了信道利用率、投資和干擾B、增加了容量和投資,但是減少了干擾C、增加了容量,但是減少了投資和干擾D、增加了容量、投資和干擾答案:(D6、在網(wǎng)絡(luò)規(guī)劃中考慮系統(tǒng)平衡(system balance時(shí),必須假設(shè)。A、手機(jī)和BTS的靈敏度相同B、

8、手機(jī)和BTS的路徑損耗相同C、手機(jī)和BTS的天線損耗相同D、手機(jī)和BTS的合路器損耗相同答案:(B7、TCP/IP協(xié)議通過(guò)(來(lái)區(qū)分不同的連接。A.IP地址B.端口號(hào)C.IP地址+端口號(hào)D.MAC地址答案:C8、E-GSM頻段是指(A.90 - 915MHz;935 - 960MHzB.900 - 915MHz;921 - 960MHzC.880 - 915MHz;925 - 960MHzD.824 - 849MHz;869 - 894MHz答案:C9、當(dāng)進(jìn)行1800M網(wǎng)絡(luò)覆蓋規(guī)劃時(shí)應(yīng)采用的傳播模型為(A.Okumura-HataB.Cost231-HataC.Cost231 Walfish-I

9、kegamiD.Keenan-Motley答案:B10、根據(jù)GSM規(guī)范,如果基站的接收靈敏度為-110dBm,那么為了保證正常工作,其載頻的第一鄰頻干擾電平最大值為:(A.9dB;B.-9dB;C.-101dBm;D.-119dBm答案:C11、基站 EIRP的計(jì)算公式為(A.BTS max. transmit power -BTS combiner and jumper loss -BTS feederand connector loss + BTS antenna gainB.BTS max. transmit power -BTS combiner and jumper lossC.BT

10、S combiner and jumper loss -BTS feeder and connector lossD.以上都不對(duì)。答案:A12、全向站和三小區(qū)定向站在小區(qū)覆蓋半徑相同的情況下,定向站覆蓋面積比全向基站覆蓋面積 ( A.小B.大C.相等D.都有可能答案:A13、移動(dòng)用戶滲透率與下面那個(gè)因素?zé)o關(guān)。(A.當(dāng)?shù)亟?jīng)濟(jì)發(fā)展情況;B.固定電話滲透率;C.人口總數(shù);D.話費(fèi)與移動(dòng)資費(fèi)競(jìng)爭(zhēng)情況答案:C14、某地頻率規(guī)劃中,除主BCCH頻點(diǎn)外,為TCH分配6M頻帶,采用1×3跳頻,為保證同基站內(nèi)不出現(xiàn)同鄰頻碰撞,最大站型配置為(A.S444B.S555C.S666D.S777答案:C15

11、、路徑損耗降低3dB時(shí),郊區(qū)環(huán)境下,小區(qū)覆蓋半徑增加約(A.5%B.10%C.22%D.15%答案:C16、當(dāng)基站采用微波傳輸時(shí),需要保證采用微波的基站之間(A.視通傳播B.電離層傳播C.衍射傳播D.反射傳播答案:A17、手機(jī)在占用下列哪個(gè)信道時(shí)肯定是滿功率發(fā)射(A. TCCHB. RACHC. CCBCCHD. SCDCCH答案:B18、某小區(qū)的NCC為3,BCC為4,其BSIC為:(a:70 b:340 c:120 d:28答案:D19、可以同時(shí)接入GSM語(yǔ)音和GPRS業(yè)務(wù)的GPRS手機(jī)終端類型是:(A. Type AB. Type EC. Type FD. Type G答案:A20、一個(gè)

12、GSM900天線的增益是15dBd, 在計(jì)算系統(tǒng)增益時(shí)數(shù)值應(yīng)該是(;A. 110dB;B. 120dB;C. 150dB;D. 17dB;答案:D21.一個(gè) Combined-BCCH/CCCH Multiframe 結(jié)構(gòu)中有多少 SDCCH , 一個(gè) Non-combined SDCCH Multiframe 結(jié)構(gòu)中有多少 SDCCH? ( A. 9 and 8B. 8 and 3C. 3 and 12D. 4 and 9E. 4 and 8答案:E22.手機(jī)在占用下列哪個(gè)信道時(shí)肯定是滿功率發(fā)射(A. CTCHB. RACHC. CCBCHD. CSDCCH答案:B23.網(wǎng)絡(luò)中出現(xiàn)同頻同BS

13、IC時(shí),下列描述那一個(gè)是正確的(A有可能會(huì)出現(xiàn)切換成功率低,話音接通率低,信令接通率低(B一定會(huì)出現(xiàn)切換成功率低,話音接通率低,信令接通率低(C對(duì)信令接通率無(wú)影響(D對(duì)通話質(zhì)量有影響答案:A24.通過(guò)BCCH載頻的TS0來(lái)傳送的下行邏輯信道有( (ACCCH、SCH、BCCH、RACH、PCH。(BFCCH、SCH、BCCH、AGCH、PCH。(CBCCH、SDCCH、SACCH、AGCH、PCH。(DBCCH、SDCCH、FACCH、AGCH、PCH。答案:B25.BSC和MSC之間的接口為?(AA接口(BAbis接口(CGb接口(DE接口26.“CHANNEL REQUEST”這個(gè)消息是在

14、哪個(gè)信道上發(fā)送的。(ARACH (BAGCCH(CPCCH(DSCDCCH答案:A27.由于阻擋而產(chǎn)生的類似陰影效果的無(wú)線信號(hào)衰落稱為什么?(A快衰落(B慢衰落。(C多徑衰落(D短期衰落。答案:B28.關(guān)于BSC的功能,一下描述正確的是(A分配無(wú)線信道(B測(cè)量無(wú)線信號(hào)的強(qiáng)度(C本地交換(D存儲(chǔ)移動(dòng)臺(tái)位置區(qū)信息答案:A29.TDMA幀中51復(fù)幀的時(shí)間間隔為?( (A1200ms (B2400ms(C235ms (D4800ms答案:C30、以下關(guān)于Idle BA list的敘述,正確的選項(xiàng)有:A、參與小區(qū)選擇和小區(qū)重選過(guò)程B、盡量把網(wǎng)中所有BCCHNO都定義進(jìn)Idle BA list,以加快小區(qū)

15、選擇的速度C、手機(jī)通過(guò)系統(tǒng)信息5來(lái)接收Idle BQA listD、必須與Active BQA list完全一致答案:(A、B31、在同一BSC中,各小區(qū)的等無(wú)線參數(shù)設(shè)置應(yīng)是一致的。A、ATTB、CQROC、T3212D、LAC答案:(A、C、D32、小區(qū)負(fù)荷分擔(dān)功能要求進(jìn)行話務(wù)分擔(dān)的目標(biāo)小區(qū)必須具備以下條件:A、空閑信道大于CLSACC參數(shù)規(guī)定值。B、HOCLSACC參數(shù)設(shè)為ON。C、與服務(wù)小區(qū)屬同一BSC。D、與服務(wù)小區(qū)屬同一層。答案:(A、B、C、D33、關(guān)于天線增益以下說(shuō)法正確的是:A、是天線在其最大輻射方向的增益B、是與基準(zhǔn)天線比較得出的相對(duì)值C、dQBi是天線增益相對(duì)于半波振子的參

16、考值D、dQBd=dBi+2.15答案:(A、B34、當(dāng)一個(gè)小區(qū)的SDCCH參數(shù)設(shè)為2時(shí),該小區(qū)的SDCCH配置數(shù)可能為:A、16C、20D、220答案:(A、C35、下面各項(xiàng)中可能發(fā)生在手機(jī)被叫過(guò)程中的行為有:A、手機(jī)通過(guò)RACH發(fā)起接入請(qǐng)求B、MSC向被叫手機(jī)所處位置區(qū)中所有BSC發(fā)出尋呼報(bào)文C、BSC向被叫手機(jī)所處位置區(qū)中所有小區(qū)發(fā)出尋呼命令D、小區(qū)收到尋呼命令后在相應(yīng)尋呼組所屬尋呼子信道上發(fā)出尋呼消息答案:(A、B、C、D36、不符合我省現(xiàn)行BSIC編碼規(guī)范的選項(xiàng)有:A、14B、68C、530D、49答案:(A、B、D37、是新建基站時(shí)必須考慮的因素。A、頻率規(guī)劃B、天線隔離度C、電源

17、供應(yīng)D、與BSC間的傳輸連接答案:(B、C、D38、路測(cè)前應(yīng)提前準(zhǔn)備好的相關(guān)文件包括:A、小區(qū)TQCH配置信息B、基站經(jīng)緯度信息C、小區(qū)BCCH、BSIC信息D、小區(qū)CGI信息答案:(B、C、D39、以下關(guān)于載波干擾比C/I的說(shuō)法正確的有:A、指接收到的希望信號(hào)電平與非希望信號(hào)電平的比值B、手機(jī)占用非跳頻小區(qū)中的TCH通話時(shí)C/I值保持不變C、載波干擾比通常使用百分比形式D、GSM規(guī)范中要求同頻載波干擾比9dB答案:(A、D40、通過(guò)調(diào)整天線下傾角可以控制小區(qū)的覆蓋范圍,關(guān)于天線傾角以下說(shuō)法正確的有:A、天線的下傾方式可以分為機(jī)械下傾和電子下傾B、天線的下傾角度過(guò)大時(shí)會(huì)導(dǎo)致天線波束嚴(yán)重變形C、

18、當(dāng)天線的機(jī)械下傾角度增大到一定數(shù)值時(shí)應(yīng)考慮天線前后輻射比D、與沒(méi)有下傾時(shí)相比,天線采用電子下傾后各個(gè)波瓣的功率下降幅度是相同的答案:(A、B 、C、D41、目前我省使用的微蜂窩主要型號(hào)有:A、RBS200B、RBS2302C、BTSITED、MAXITE答案:(B、D42、解決TCH擁塞的辦法主要有:A、開(kāi)啟小區(qū)負(fù)荷分擔(dān)功能B、調(diào)整基站發(fā)射功率C、增加TCH數(shù)量D、調(diào)整切換邊界答案:(A、B、C、D43、關(guān)于不同頻段電磁波傳播特性的描述正確的是:A、GSM1800的衰耗比GSM900多6dBB、1800MHz比900MHz的建筑物貫穿損耗小C、低頻段電磁波的繞射能力D、1800MHz和900M

19、Hz頻段的傳播特征差別很大答案:(A、B44. Which cell plan is designed first and is the basis for future planning?A. FinalB. PQreliminaryC. CoverageD. InitialE. Nominal答案:E45. What is the name of the type of fading which is due to shadowing?A. Atmospheric fadingB. Multi-path fadingC. Rayleigh fadingD. Log Normal fadin

20、gE. Path loss答案:D46. Which unit is used for measuring traffic?A. ErlangB. Bit/secondC. HertzD. Subscribers/kHzE. Subscribers/minute答案:A47. What type of interference is produced by other transmitters on the same frequency?A. Double interferenceB. Co-channel interferenceC. Adjacent channel interferenc

21、eD. Reflected interferenceE. Timing interference答案:B48. What does Grade of Service (GoS mean?A. Number of available subscriber servicesB. Number of unsuccessful call set-ups due to congestionC. Channel utilizationD. Number of successful call set-upsE. Number of dropped calls答案:B49. Which cell patter

22、n is recommended for use with frequency hopping?A. 4/12B. 3/9C. 7/21D. 7/7E. None of these答案:B50. Which of the following is not necessary to consider when surveying a potential site?A. Space for antennasB. Frequency allocationC. Electrical power availabilityD. Contract with ownerE. Availability of t

23、ransmission to the BSC答案:B51. What is used to prevent “the ping-po ng effect”, that is, rapid, consecutive handovers?A. Dynamic power controlB. Frequency hoppingC. HysteresisD. Cell load sharingE. Discontinuous transmission答案:C52. Which of the following will not increase system capacity?A. Increasin

24、g the number of available frequencies.B. Changing the modulation scheme.C. Implementing half rate vocodersD. Going from a 4/12 to a 3/9 reuse pattern.E. Splitting cells.答案:B53. Which of the following is used for planning network dimensioning, frequency planning, and making coverage and interference

25、predictions?A. Channel Event Recording (CERB. TEMS cell plannerC. TEst Mobile System (TEMSD. Cellular Network AdministrationE. Cellular Network Analyzer (CeNA答案:B54. Which cell plan is designed first, and is the basis for future planning?A Base CoverageB FinalC PreliminaryD NominalE Ultimate答案:D55.

26、What is the difference between GSM and CME 20/CMS 40?A CME 20/CMS 40 are GSM-systems with Ericssons mobile stations.B CME 20/CMS 40 are Ericsson implementations of GSM.C CME 20/CMS 40 are new revisions of GSM.D GSM is the French version of CME 20/CMS 40.E CME 20/CMS 40 are GSM systems with, at least

27、, half of theequipment coming from Ericsson.答案:B56. How many ARFCNs can an operator use if allocated 15 MHz uplink and 15 MHz downlink (minus one for guardband?A 14 ARFCNsB 15 ARFCNsC 29 ARFCNsD 75 ARFCNsE 74 ARFCNs答案:E57. What is the name of the type of fading which is due to shadowing?A MultipathB

28、 Qtmospheric fadingC Rayleigh fadingD Log-normal fadingE Path loss答案:D58. A cell has 3 TRUs with an SDCCH/8 configuration. What is themaximum number of telephone calls it can support simultaneously with ahalf-rate vocoder?A 22B 44C 8D 3E 46答案:B59. Which type of interference is produced by other tran

29、smitters on the same frequency?A Double interferenceB Co-channel interferenceC Adjacent channel interferenceD Reflected interferenceE Timing interference答案:B60. Which logical channel must be dimensioned for each cell whenconsidering the call set-up frequency?A SACCHB SDCCHC FCCHD SCHE BCCH答案:B61. Wh

30、ich unit is used for measuring traffic?A ErlangB Bit/secondC HertzD Subscribers/kHzE Subscribers/minute答案:A62. What is the difference between 1 milliwatt and 1 watt?A They cannot be comparedB 30 dBC There is no differenceD 20 log dBm = dBWE 10 dBmW答案:B63. Which of the following problems exists in di

31、gital radio systems, but not in analog radio systems?A Rayleigh fadingB Qtmospheric attenuationC CongestionD Log-normal fadingE Intersymbol interference答案:E64. What cell property is limited by timing advance?A Number of simultaneous callsB RadiusC Frequency of call setupsD Antenna lengthE Color答案:B6

32、5. Which channel is used to send the mobile's measurement report?A SDCCHB FQACCHC MRCHD SACCH (Slow Associated Control CHannelE TCH答案: D66. What does GoS (Grade of Service mean?A Number of available subscriber servicesB Percentage of successful call set-upsC Channel utilizationD Percentage of un

33、successful call set-ups due to congestionE Percentage of dropped calls答案:D67. A particular cell is expected to support 6 Erlangs of traffic at 1% Grade of Service. How many traffic channels are required?A 6B 13C 3D 15E 22答案:B68. What is system balance?A Each cell can handle the same amount of subscr

34、ibersB The network is dimensioned for all mobile classesC All cells are of the same sizeD All BTSs use the same output powerE Uplink and downlink coverage is the same答案:E69. Which of these assumptions is necessary to use in system balance?A The sensitivity is the same for the MS and the BTS.B The pa

35、th loss is the same for the MS and the BTS.C The feeder loss is the same for the MS and the BTS.D The antenna gain is the same for the MS and the BTS.E The combiner loss is the same for the MS and the BTS.答案:B70. Which of these is not a prediction model for radio coverage?A Algorithm 9999B Cost-231C

36、 Okumura-HataD PrinthanaE Walfish-Ikegami答案:D71. Which cell pattern is recommended to use when using frequencyhopping?A 4/12B 3/9C 7/21D 7/7E None of these答案:B72. Reusing frequencies in a cellular pattern (compared to “one big cell”A Increases channel utilization, cost ,and interference.B Increases

37、capacity and cost, but decreases interference.C Increases capacity, but decreases cost and interference.D Increases capacity, cost, and interference.答案:D73. A subscriber makes a two-minute call and a four-minute call during the busy hour. How much traffic does he generate for that hour?A 33 millierl

38、angsB 66 millierlangsC 100 millierlangsD 10 millierlangsE 25 millierlangs答案:C73、所謂公共控制信道是指面向小區(qū)內(nèi)所有手機(jī)的控制信道,以下控制信道中不屬于公共控制信道的有:A、AQGCHB、RQACHC、FACCHD、SDCCH答案:(C、D74、影響信道可用率的原因可能有:A、頻點(diǎn)定義數(shù)與TRX硬件數(shù)不符;B、硬件故障;C、傳輸閃斷;D、鄰小區(qū)使用了相鄰頻點(diǎn);答案:(A、B、C75、手機(jī)能夠正常進(jìn)行小區(qū)選擇,并成功駐留于該小區(qū)的前提是:A、小區(qū)未被禁止接入;B、小區(qū)選擇參數(shù)C1值大于0;C、不屬于“國(guó)內(nèi)漫游禁止位置區(qū)

39、”列表中的位置區(qū);D、小區(qū)屬于允許的PLMN;答案:(A、B、C、D76、控制不連續(xù)發(fā)射是否開(kāi)啟的參數(shù)為:A、DQTXFULB、DTXUC、DTXDD、DQTXB答案:(B、C77、網(wǎng)絡(luò)通過(guò)無(wú)線接口向手機(jī)廣播系統(tǒng)消息,這些系統(tǒng)消息通過(guò)信道進(jìn)行傳送。A、BCCHB、SQCHC、SQDCCHD、SACCH答案:(A、D78、下面可以真實(shí)反映網(wǎng)絡(luò)運(yùn)行質(zhì)量的選項(xiàng)包括:A、用戶投訴B、STS話務(wù)統(tǒng)計(jì)C、CQT撥打測(cè)試D、DT撥打測(cè)試答案:(A、B、C、D79、無(wú)線信號(hào)受到建筑物或地形阻擋后,通過(guò)直射、反射、散射等傳播到達(dá)接收端,這些信號(hào)相互迭加產(chǎn)生的矢量和就會(huì)開(kāi)成一個(gè)嚴(yán)重的衰落谷點(diǎn),這就是我們常說(shuō)的。A

40、、陰影衰落B、多徑衰落C、對(duì)數(shù)正態(tài)衰落D、瑞利衰落答案:(B、D80、以下指定基站實(shí)際發(fā)射功率的參數(shù)是:A、BQSPWRB、BSPWRBC、BSPWRTD、BQSTXPWR答案:(B、C81、Locating算法為切換提供依據(jù),以為輸入,輸出是一個(gè)經(jīng)過(guò)該算法判斷可以切換上去的候選小區(qū)列表。A、手機(jī)測(cè)量到的下行鏈路信號(hào)強(qiáng)度B、手機(jī)測(cè)量到的下行鏈路信號(hào)質(zhì)量C、基站測(cè)量到的上行鏈路信號(hào)強(qiáng)度D、基站測(cè)量到的上行鏈路信號(hào)質(zhì)量答案:(A、B、C、D82、話音信道掉話的可能原因有:A、SQCHO參數(shù)設(shè)為ONB、漏定鄰小區(qū)關(guān)系C、鄰小區(qū)與服務(wù)小區(qū)使用鄰4頻D、測(cè)量到TA值超限答案:(B、D83、跳頻參數(shù)HOP

41、的可能取值包括,不同的設(shè)置決定了跳頻功能是否開(kāi)啟以及參與跳頻的邏輯信道。A、ONB、SQDCCHC、TCHD、OFF答案:(A、C、D84、由交換機(jī)自動(dòng)上傳到OSS系統(tǒng)中特定目錄下的話務(wù)統(tǒng)計(jì)文件SDMREL1-JU101000,1034222533,從文件名可以看出哪些信息:A、該文件產(chǎn)生日期為6月10日B、該文件產(chǎn)生日期為8月10日C、該文件統(tǒng)計(jì)時(shí)段為上午9:00-10:00D、該文件統(tǒng)計(jì)時(shí)段為上午13:00-15:00答案:(A、C85、以下關(guān)于Idle BA list的敘述,正確的選項(xiàng)有:A、參與小區(qū)選擇和小區(qū)重選過(guò)程B、盡量把網(wǎng)中所有BCCHNO都定義進(jìn)Idle BA list,以加快

42、小區(qū)選擇的速度C、手機(jī)通過(guò)系統(tǒng)信息5來(lái)接收Idle BQA listD、必須與Active BQA list完全一致答案:(A、B86、在同一BSC中,各小區(qū)的等無(wú)線參數(shù)設(shè)置應(yīng)是一致的。A、ATTB、CQROC、T3212D、LAC答案:(A、C、D87、小區(qū)負(fù)荷分擔(dān)功能要求進(jìn)行話務(wù)分擔(dān)的目標(biāo)小區(qū)必須具備以下條件:A、空閑信道大于CLSACC參數(shù)規(guī)定值。B、HOCLSACC參數(shù)設(shè)為ON。C、與服務(wù)小區(qū)屬同一BSC。D、與服務(wù)小區(qū)屬同一層。答案:(A、B、C、D88、關(guān)于天線增益以下說(shuō)法正確的是:A、是天線在其最大輻射方向的增益B、是與基準(zhǔn)天線比較得出的相對(duì)值C、dQBi是天線增益相對(duì)于半波振子

43、的參考值D、dQBd=dBi+2.15答案:(A、B89. The following events all require an SDCCH. Which event has normally the longest holding time?A Normal updatingB IQMI attach/detachC RegistrationD Call set-upE SMS point-to-point答案:E90. Use the enclosed Erlang table1 to work out which is the optimum SDCCH configuration u

44、sing four TRXs, if the SDCCH/TCH traffic ratio is 30% and the required GoS on the TCH is 2%. (Assume no cell broadcast, no immediate assignment on the TCH and a GOS on the SDCCH of 0,7%.A SDCCH/8B SQCCH/8 + SDCCH/4C 2*SDCCH/8D SDCCH/4E 10*SDCCH/8答案:C91. How many paging requests can, at the most, fit

45、 into one paging block?A 10B 20C 4D 60E 80答案:C92. AGBLK must be set to a non-zero value (that is = 1 in Ericssons GSM system, if:A Cell broadcast is used in a cell with a combined configuration.B Cell broadcast is used in a cell with a non-combined configuration.C System information, type 4, is sent

46、 on the BCCH.D System information, type 5, is sent on the BCCH.E MFRMS=2 or MFRMS=9答案:B93. Which is the most important factor for determining the paging capacity in the BTS?A MFRMSB AQBLKC Paging strategyD Combined or non-combined mappingE The use of cell broadcast答案:D94. To reduce co-channel interf

47、erence by means of antenna tilt, some prerequisites must be met. Which of the below is not a prerequisite?A Short site-to-site distanceB Sector antennasC High gain antennasD High mounted antennasE Narrow vertical beam答案:B95. Null fill-in is obtained by:A Slightly adjusting the phase of the power fed

48、 to the different antenna elements.B Slightly adjusting the power fed to the different antenna elements.C Slightly adjusting both the power and the phase fed to the different antenna elements.D Slightly tilting the antenna elements with respect to each other.E Continuously moving the antenna element

49、s slightly with respect to each other. 答案:B96. If the duplex distance is 45 MHz and the bandwidth is 14 MHz, which order of intermodulation products is the first to become a problem?A 7B 30C 35D 36E 37答案:A97. A repeater can be used to:A Increase the capacity locally (in a traffic hot spot.B Increase

50、 the capacity along highways (Questions.C Get better coverage on water.D Get better coverage where the area to be covered is blocked by obstacles.E For none of these purposes.答案:D98. A Tower Mounted Amplifier (TMA can be beneficial in some configurations due to its ability to:A Improve the downlink

51、sensitivity and compensate for the downlink feeder loss.B Improve the uplink sensitivity and compensate for the uplink feeder QUE loss.C Improve both uplink and downlink sensitivities.D Improve the compensate for the downlink feeder lossE Improve compensate for the uplink feeder loss.答案:A99. In a fo

52、ur-way power splitter, the /4-transformer changes the impedance of the input from 50 . to:A 1 .B 12.5 .C 50 .D 25 .E 125答案: B100. The wave propagation prediction algorithm 9999 takes a lot into account, but not:A Land usageB Terrain QUE profileC Curvature of the earthD Diffraction loss due to obstru

53、cting buildings答案:D101. Which of the following statements about the urban model is not true?A Permanent screens, used in the half-screen part of the urban mode.B Temporary QUE screens, used in the half-screen part of the urban model.C The recursive microcell part of the urban model uses something ca

54、lled the illusory distance.D The resulting path loss from the urban model is obtained by taking the maximum value of the two contributions, calculated by the half-screen and recursive microcell algorithms.答案:D102. What is the minimum required isolation between two TX antennas at the same site?A 2dBB 30 dBC 4 dBD 5 dBE 6 dB答案: B103. What is the datum