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發(fā)表于 2013-04-13 13:28:38
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NOKIA設(shè)備切換算法分析
例下面的測(cè)試報(bào)告從BTS 提交給 BSC ,包括服務(wù)小區(qū)和相鄰小區(qū)的測(cè)量結(jié)果:
²
服務(wù)小區(qū)測(cè)量報(bào)告:
Measurment Report
| 1st
| 2nd
| 3rd
| 4th
| 5th
| 6th
| 7th
| 8th
| 9th
| 10th
| 11th
| 12th
| 13th
| 14th
| UL RxLev
| 35
| 36
| 34
| 32
| 25
| 20
| 21
| 18
| 16
| 14
| 12
| 10
| 10
| 9
| DTX Used
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| HoThresholdLevUL = -90dBm
WindowSize = 3, Weighting = 2
Px = 2, Nx = 3
btsMeasAver (BMA) = 1 (no pre-processing in BTS)
²
鄰區(qū)測(cè)量報(bào)告:
| 1st
| 2nd
| 3rd
| 4th
| 5th
| 6th
| 7th
| 8th
| 9th
| Adj1
| -65
| -67
| -71
| -69
| -72
| -70
| -73
| -71
| -75
| Adj2
| -73
| -75
| -74
| -75
| -76
| -77
| -75
| -77
| -74
| Adj3
| -77
| 0
| -80
| -79
| -81
| -79
| 0
| -80
| -79
| Adj4
| -85
| -83
| -87
| -88
| -84
| 0
| -86
| -87
| -86
| Adj5
| -96
| -94
| -97
| -95
| -95
| -96
| -96
| -97
| -94
| Adj6
| -97
| -99
| -98
| -99
| 0
| -97
| 0
| 0
| -96
| allAdjacentCellsAveraged=NO
numberOfZeroResults=2
averagingWindowSizeAdjCell=8
²
服務(wù)小區(qū)對(duì)各鄰區(qū)參數(shù)設(shè)置如下:
| SL
| LMRG
| PRI(Priority)
| OF(hoLoadFactor)
| BLT(btsLoadThreshold)
| BTS load
| Adj1
| -95
| 3
| 3
| 1
| 80%
| 85%
| Adj2
| -95
| 3
| 4
| 2
| 85%
| 90%
| Adj3
| -95
| 3
| 3
| 1
| 85%
| 70%
| Adj4
| -95
| 6
| 4
| 2
| 80%
| 70%
| Adj5
| -95
| 3
| 3
| 1
| 75%
| 70%
| Adj6
| -95
| 3
| 3
| 1
| 80%
| 70%
|
請(qǐng)根據(jù)上述已知條件判斷:
1)、請(qǐng)根據(jù)上述已知條件判斷:什么時(shí)候MS將觸發(fā)切換?
2)、候選目標(biāo)小區(qū)的電平、優(yōu)先級(jí)分別如何?
3)、最終的目標(biāo)小區(qū)是哪個(gè)鄰區(qū)?
答案:
1)、請(qǐng)根據(jù)上述已知條件判斷:什么時(shí)候,MS將觸發(fā)切換
Serving cell:
P1=(35*2+36*2+34*2)/(2+2+2)=35>20
P2=(36*2+34*2+32*1)/(2+2+1) =34.4> 20
P3=(34*2+32*1+25*1)/(2+1+1) =31.25>20
P4=(32*1+25*1+20*1)/(1+1+1) =25.7>20
P5=(25*1+20*1+21*2)/ (1+1+2) =21.75>20
P6=(20*1+21*2+18*2)/(2+2+1)=19.6<20
P7=(21*2+18*2+16*1)/ (2+2+1)=18.8<20
So MS will handover to the best adjacent cell while received the 9th sample, its level is -110+16=-94
2)、候選目標(biāo)小區(qū)的電平、優(yōu)先級(jí)分別如何?
Adjacent cell level:
Adj1=(-65-67-71-69-72-70-73-71)/8=-69.8
Adj2=(-73-75-74-75-76-77-75-77)/8=-75.3
Adj3=(-77-80-79-81-79-80)/6=-79.4
Adj4=(-85-83-87-88-84-86-87)/7=-85.8
Adj5=(-96-94-97-95-95-96-96-97)/8=-95.75
Adj6由于存在3個(gè)0值,因此adjc6此組數(shù)據(jù)不用于計(jì)算鄰區(qū)電平
Adjacent cell priority:
Adj1 and Adj2 are overloaded
Adj1=3-1=2
Adj2=4-2=2
Adj3=3
Adj4=4
Adj5=3
so these 6 adjacent cell’s priority is:
P4> p3> p5>
p1> p2
3)、最終的目標(biāo)小區(qū)是哪個(gè)鄰區(qū)?
Because SL=-95
The RX level of Adj1-Adj4 are all fit for the SL condition, except Adj5.
AV_RXLEV_NCELL(n) > AV_RXLEV_DL_HO + HoMarginLev/Qual(n)
AV_RXLEV_DL_HO=-94
ADJC1 MRG=3:
-69.8>-94+3
滿(mǎn)足切換門(mén)限!
ADJC2 MRG=3:
-75.3>-94+3
滿(mǎn)足切換門(mén)限!
ADJC3 MRG=3:
-79.4>-94+3
滿(mǎn)足切換門(mén)限!
ADJC4:LMRG=6:
-85.8>-94+6
滿(mǎn)足切換門(mén)限!
Accord to the priority above, the best target cell is Adj4
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