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Cell power

Created: Aug 7, 2019 16:15:37Latest reply: Oct 17, 2020 14:06:01 1932 6 0 0 0
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How to set cell power to specific value (i.e. 40 W)?

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Neil_Tong
Created Aug 8, 2019 06:17:11

hello,there is a simple computational formulas about cell power.
For exampple, if the bandwidth is 20M, so the corresponding RE number is 1200, and the maxinum transmit power of the RRU is 40W(40000mW), and the PA is -3, so you can get the maxinum reference signal power of the cell as following:

reference signal power(dB)= 10lg(RRU transmit power)-10lg(RE number)-PA=10lg(40000)-10lg(1200)+3=46.0206-30.7918+3=18.2

Because the unit of the reference signal power is 0.1dBm, so  the maxinum value of the reference signal power that we can set through MOD PDSCHCFG is 182.
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Hi,

Reference Signal Power Boosting in LTE (MOD PDSCHCFG)-2698161-1


MML COMMAND : MOD PDSCHCFG   


Each cell in an LTE radio network sends a cell-specific reference signal (RS) from its transmit antennas. The transmit power of a resource element (RE) carrying such reference signal can be set to be the same as, greater than, or less than the transmit power of an RE carrying Physical Downlink Shared Channel (PDSCH). Let’s take a quick look at the reference signal power boosting, where the RS RE uses more power than the PDSCH RE. The RS power boosting may or may not be desirable from the perspective of the RF performance.

The relative transmit power levels of the RS and the PDSCH have implications on the downlink channel estimation, the amount of downlink interference, and the interpretation and the use of the Channel Quality Indicator (CQI) by the eNodeB. For example, if the RS power is increased, the UE could potentially make the RS measurements (e.g., RSRP and RSRQ) more easily and potentially quantify the downlink channel conditions more reliably. However, the overall interference on the RS RE for a given cell would increase due to multiple neighboring cells transmitting more power on their own RS REs. If the signal-to-interference-plus-noise ratio (SINR) estimated for the PDSCH degrades by a significant amount, the CQI being reported by the UE would be lower. If the reported CQIs are relatively lower, the eNodeB would aim for a lower target throughput by taking actions such as the increased amount of Turbo coding in the PDSCH transmissions. The user-experienced throughput could thus be somewhat lower when RS power is boosted. However, if the enhanced channel estimation and increased reliability of the PDSCH reception lead to fewer HARQ retransmissions, throughput could actually increase in case of the RS power boosting. In summary, the theoretical impact of the RS power boosting on the RF performance is not definitive. Field testing with varying levels of RS power boosting and varying levels of traffic loading is recommended to determine the suitability of the RS power boosting.

The eNodeB broadcasts the transmit power levels of the RS and the PDSCH in SIB 2 using the parameters referenceSignalPower, PA, and PB. The transmit power of an RE carrying the RS (in dBm) is specified as referenceSignalPower. PA influences a parameter called ρA, which is the ratio of the transmit power of the PDSCH RE and the transmit power of the RS RE. ρA is applicable to the OFDM symbols that do not carry RS. PB establishes the relationship between ρA and ρB, where ρB is the ratio of the transmit power of the PDSCH RE and the transmit power of the RS RE in the OFDM symbols that carry RS. PA ranges from 0 to 7 and corresponds to the range from -6 dB to +3 dB for ρA. PB ranges from 0 to 3 and corresponds to the range from 5/4 to 1/2 for (ρB /ρA)[1].

Let’s take two numerical examples. Assume that a 30 W power amplifier is used for a transmit antenna of an eNodeB and that 10 MHz downlink bandwidth is deployed in a cell. The nominal transmit power per subcarrier is (30 W/600)= 50 mW. During an OFDM symbol where no RS is present, each subcarrier of the PDSCH is allocated 50 mW.


Example 1: PA =2 and PB =1 (with 3 dB power boosting for RS)


PA =2 implies ρA = 0.5 or -3 dB. Hence, (Power on PDSCH RE/power on RS RE)= 0.5. Since the transmit power allocated to PDSCH RE is the nominal power level of 50 mW, the transmit power allocated to RS RE is (Power on PDSCH RE/0.5)= (50 mW/0.5)= 100 mW. During an OFDM symbol carrying the RS, the number of REs carrying the RS from one transmit antenna is (50 Physical Resource Blocks * 2 REs/Physical Resource Block)= 100 REs. Furthermore, a given antenna does not transmit any power on a set of 100 REs, because such set is used by a different transmit antenna. Hence, out of 600 REs in an OFDM symbol carrying the RS, 100 REs are subject to RS power boosting, 100 REs have no transmit power, and remaining 400 REs have nominal power levels. The total transmit power during the RS-carrying OFDM symbol would be (100 subcarriers * 100 mW per subcarrier for power-boosted RS REs) + (100 subcarriers* 0 mW for null REs) + (400 subcarriers * 50 mW per subcarrier for non-RS REs)= 30 W. Hence, when PA =2 and PB =1, each RS RE is allocated 100 mW, while a non-RS RE (in any OFDM symbol) is allocated 0 mW or 50 mW. referenceSignalPower is set to 10*log10(100 mW)= 20 dBm.


Example 2: PA =4 and PB =1 (with NO power boosting for RS)


PA =4 implies ρA = 1 or 0 dB. Hence, (Power on PDSCH RE/power on RS RE)= 1. Since the transmit power allocated to PDSCH RE is the nominal power level of 50 mW, the transmit power allocated to RS RE is (Power on PDSCH RE/1)= (50 mW/1)= 50 mW. Hence, out of 600 REs in an OFDM symbol carrying the RS, 100 REs are subject to RS power level, 100 REs have no transmit power, and remaining 400 REs have nominal power levels. The total transmit power during the RS-carrying OFDM symbol would be (100 subcarriers * 50 mW per subcarrier for non-power-boosted RS REs) + (100 subcarriers* 0 mW for null REs) + (400 subcarriers * 50 mW per subcarrier for non-RS REs)= 25 W. Hence, when PA =4 and PB =1, each RS RE is allocated 50 mW, while a non-RS RE (in any OFDM symbol) is allocated 0 mW or 50 mW. referenceSignalPower will be set to 10*log10(50 mW)= 17 dB


Reference : post prepared by eugenm

The link : https://forum.huawei.com/enterprise/en/reference-signal-power-boosting-in-lte-mod-pdschcfg/thread/458849-100305

Thanks

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hello,there is a simple computational formulas about cell power.
For exampple, if the bandwidth is 20M, so the corresponding RE number is 1200, and the maxinum transmit power of the RRU is 40W(40000mW), and the PA is -3, so you can get the maxinum reference signal power of the cell as following:

reference signal power(dB)= 10lg(RRU transmit power)-10lg(RE number)-PA=10lg(40000)-10lg(1200)+3=46.0206-30.7918+3=18.2

Because the unit of the reference signal power is 0.1dBm, so  the maxinum value of the reference signal power that we can set through MOD PDSCHCFG is 182.
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how to calculate RE number? Sir
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BetterMing
BetterMing Created Aug 15, 2020 10:47:24 (0) (0)
Occupied bandwidth = Subcarrier width x Number of subcarriers per RB x Number of RBs
The width of each subcarrier is 15 K, and the number of subcarriers per RB is 12.
Therefore, a 20 MHz bandwidth corresponds to 100 RBs (1200 subcarriers).  
I've been searching for this equation thank you very much.
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I've been searching for this equation thank you very much.

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