Hi all,

I am trying to calculate how much power an eNodeB would radiate if the maximum values of referenceSignalPower (50 dBm according to 6.3.2 in TS 36.331) and p_a (3 dB) were used. However, I am running into weird results and I would appreciate your advice. These are my assumptions:

1) One antenna port

2) Normal cyclic prefix

Let us also configure the PDSCH-ConfigCommon information element as follows:

referenceSignalPower = **50**

p-b=0

Finally, let us assume that the cell under consideration only has one UE and let us configure the following PDSCH-ConfigDedicated for this UE:

p-a=dB3

Next, I calculate the EPRE_PDSCH. Since p-b=0 and only one antenna port is used, then from Table 5.2-1 in TS 36.213, it results that rho_B=rho_A. Then, also from section 5.2 in TS 36.213:

rho_A=delta_power_offset + p-a

Since delta_power_offset=0, then:

rho_A=0+3= 3 dB

Finally, the EPRE_PDSCH is simply:

EPRE_PDSCH = EPRE_cell_specificRS + rho_A = referenceSignalPower + rho_A = 50 + 3 = **53 dBm**

Let us also assume that the cell has a system bandwidth of 20 MHz (N_RB=100 resource blocks). Then, in the first OFDM symbol of the second slot of a subframe in which the PBCH is not transmitted (e.g. the second subframe), there are a total of:

N_RS= N_RB*2 = 200 reference symbols (because each RB carries only 2 RS REs in this symbol)

N_PDSCH_REs = 100*(12-2)=1000 REs carrying the PDSCH

Hence, if each PDSCH RE has a transmit power of EPRE_PDSCH=53 dBm and each RS RE has a transmit power of EPRE_cell_specific_RS=50 dBm, the total power PT radiated by the eNodeB in this symbol is:

PT=N_RS*10^(50/10) + N_PDSCH_REs*10^(53/10) --> **83.4 dBm**.

There are two things that strike me here:

-The EPRE of the reference symbols is lower than the EPRE of the PDSC REs due to p-a=dB3. I always thought that the RSs were the most powerful REs! Is this possible?

-A radiated power of 83.4 dBm sounds like a lot to me. Is this indeed foreseen in LTE deployments?