Calculation of the total eNodeB radiated power based on the reference signal power

Ask the Expert!

Ask the Expert!
Members can sign in to post a question to our Subject Matter Experts.

Calculation of the total eNodeB radiated power based on the reference signal power

  • 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


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



    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?

  • Hi,

    The maximum power transmitted over the whole bandwidth is typically 43dBm.

    Then, the RS EPRE depends on the configuration but is usually around 15dBm. Thus, I think your first assumption might be wrong.



  • Thank you for your response Marko.

    Yes, I agree in that RS_EPRE=15 dBm is a more "typical" value. However, I was concerned with the fact that the standard supports up until 50 dBm (see PDSCH-config section in 6.3.2 of TS 36.331). I wonder such a high value is supported? Would it make sense to deploy an eNodeB with an RS_EPRE=50 dBm?

  • Even though the standard allows the RS power to go up to 50 dBm, I doubt we'd ever see that in real life. Total cell power requirements aside, a UE could detect that cell and camp on it a hundred kilometers away, and yet be unable to receive any services because it would be uplink limited (a typical UE tops out at around 24 dBM).