Hi,
Could you support to know the reel Rs power using the below config:
MaxPower=46
Pb=1
Pa=-3
BW=10M
Thanks a lot
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Hi,
Could you support to know the reel Rs power using the below config:
MaxPower=46
Pb=1
Pa=-3
BW=10M
Thanks a lot
Ok That a lot,
Actually my MaxPower=20W and i want to increase the power from 18.2 to 21.2, is there any issue if put Pa=-6/Pb=3?
Just Change Rs Power to 212 and keep Pb=1 and Pa=-3, Max Power will increase to 40 W (46 dBm)
if you change Pb=3 and Pa=-6 with Rs=212, Max Power will remain 20W (43 dBm), coverage will increase but less power allocation for PDSCH type A
btw please don't forget to consider single / dual carrier in your power calculation
Hello,
the idea of LTE is to have same power for each RE So the RE power is 80 watts/(Number of RB *12) which in you case will be 21.25 dBm
pa is the power addition and reduction on CRS so you will have 21.25-3=18.25 dBm
pb is the power reduction.addition to Type-B RE (symbols which are including RS)
this is not a good configuration which causes lower RS power and consequently lowercoverage and SINR
if you are using MIMO, it is better consider +3dB power addition in RS power to extend the coverage and better SINR
cheers
Huawei CRS
1.PDSCHCfg.ReferenceSignalPwr 18.2 dBm for 20 watt===>Pb
2.PDSCH Other PDSCHCfg.Pb CellDlpcPdschPa.Pa PcOff 1 dB to -3 dB===>Pa
E******* CRS
1.calculated from[SectorEquipmentFunction]configuredOutputPower[EUtranCellFDD] crsGain 40000 to 300 (3 dB)==>Pa
2.[EUtranCellFDD] pdschTypeBgain 1 ===>(Pb)
Nokia
1.CRS power calculated from [LNCEL] pMax[LNCEL] dlRsBoost 430 (20 watt) 1000 (0 dB)===>Pa