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Reply To: MS outpur power in 1800

#54623
RH
Guest

Hi Yan,

Free space path loss=32.44+20Log(f)+20Log(d)

where f=Frequency,
d=distance between MS and BTS.
So,you can see from tha formula that,as f (Frequency) is higher,path loss will be higher.
So,at a same distance,1800 band should have lower coverage then 900 band due to higher path loss.