At time 9:17, the example of Rho_B/Rho_A = 1.25 --> The Power of PDSCH_B in this cause should be = -2.03 dB + 18.22 right? (in UA-cam is -2.03+15.22). By the way, thank you very much for the great explanation.
Please explain, why Ppdsch_B= 20.88mW? In the scenario when Pb=0, Pa=-3. If Pa = -3, as in first example (Pb= 1, Pa= -3) RS Power is 18.2 (66.67mWatt) not 15.2. it means Ppdsch_B should be 0.6264*66.67=41.47 mW. Right?
Hi, The RS power formula include the value of Pb: Prs=Psingle_port - 10*Log(12*Nrb) + Log(1+Pb). in the second example Pb=0 not 1, so the value of the RS power is 15.2 for Pb=0 and 18.2 for Pb=1. Kind Regards, Smart Telecom Education
Great Explain, Could you pls start series on CA
At time 9:17, the example of Rho_B/Rho_A = 1.25 --> The Power of PDSCH_B in this cause should be = -2.03 dB + 18.22 right? (in UA-cam is -2.03+15.22). By the way, thank you very much for the great explanation.
Hi, Thankyou very nicely explaned. Could you please give the spec ref for Prs=Psingle_port - 10*Log(12*Nrb) + Log(1+Pb).
Thank you!
Can you explain please why you add 10*log10(1+Pb) to Reference Power formula? is it just P_RS = P_singlePort -10*log10(12*Nrb)?
Please explain, why Ppdsch_B= 20.88mW? In the scenario when Pb=0, Pa=-3. If Pa = -3, as in first example (Pb= 1, Pa= -3) RS Power is 18.2 (66.67mWatt) not 15.2. it means Ppdsch_B should be 0.6264*66.67=41.47 mW. Right?
Hi,
The RS power formula include the value of Pb:
Prs=Psingle_port - 10*Log(12*Nrb) + Log(1+Pb).
in the second example Pb=0 not 1, so the value of the RS power is 15.2 for Pb=0 and 18.2 for Pb=1.
Kind Regards,
Smart Telecom Education
can you please share link of that slides?
5g.smarttelecomedu.com/lte-downlink-power-calculation/