Question

:Consider a cellular system operating at 2.4GHz where propagation follows simplified path loss with γ = 3, d0 = 1 and K = −30dB, and with variations about this pathloss due to log-normal shadowing with σ = 4dB. Suppose that for acceptable voice quality an SNR ratio of 40dBm is required at the mobile. Assume the base station transmits at 0dB and the receiver noise in the bandwidth of interest is −50dB.

a) Find the maximum cell size such that a mobile on the cell boundary will have acceptable voice quality 95% of the time.

b) Assume that the cell size is set as two times what you found in part (a). How should you adjust the transmit power to keep the service quality at the same level?

Answer 1

f = 2.4 GHz r=3 do = 1 K = - 30dB. 4dB. SNR= 40 dBm. Pt = 0dB Pran - sodB. Pr[dB m)– Pt (dBm) + K(dB)- lor logo ( 4.) + Y dB .

fc = 2.4 GHz + 1 = 3x108 2.48109, .m 10 OvdB 4 SNR reed 40dBm Pt odB = 1 W. = 10 log o 103 = 300km, Pnoise 50dB. = 20 dBm - 50 = 0 to logio Pn los Pn. 102 x 103 = ln Pa = 102 mW. 10 log 10-2 =- 20 dBm . Preed 20dBm (d/do u= Ptt 10 log K- lor log ,o 3% + (-30) 10x3 log, (d/ ). - - 30 log (d).

pl Preceived - 0.95 > dodan) P ( Preceived M 20dBm - M solutions) -0.95 54dB 0.05 = pl Preceive - M 20-M 6 Yon 20-M 1.65. 4 20 - (-30 log, od) - - 6.6 30logiod 26.6. d = logio -0.8867 id= 0.1298 m b) d'= 2d = 0.2596 m. pe no. M = Pt - 30 - 30 log 0. 1298) = Pt 30 + 26.6 Pt - 3.4. M =

20-M -1.65 4 M = - 26.6 Pt-3.4 = -26.6 Pt = 3.4-26.6 - 23,2 dBm.