Question

In a WLAN, the path-loss is given as PL [dB] - 37 +36 log10 (d), where d is the distance between the access point (AP) and a user The AP transmits at the same power level to all users. If the distance between the AP and a user i is 8 times the distance between the AP and a userſ, how many dB less power does user i receive in comparison to user j?

Answer 1

For distance between \(\mathrm{AP}\) and a user \(\mathrm{j}\), let us take the distance as \(\mathrm{d}\)

then for distance between AP and user i, distance is \(8 \mathrm{~d}\)

Power loss for user 'j':

\(P L_{j}=37+36 \log (d)\)

Power loss for user 'i':

\(P L_{i}=37+36 \log (8 d)=37+36 \log (d)+36 \log (8)\)

because \(\log (\) ab) \(=\log a+\log b\)

Thus,\(P L_{i}-P L_{j}=36 \log (8)=32.5 d B\)