ELEC-E7120 Wireless Systems Weekly Exercise Problems 5 Problem 1: (Range and rate in Wi-Fi) When a wireless station (STA) moves away from the Access Point (AP), the received signal strength decreases and the wireless link adjusts modulation and coding accordingly. Answer the following: a) Determine the minimum received signal power (S min) for the different 802.11g modulation and Forward Error Correction (FEC) coding schemes shown in the following table: Rate Modulation Rate FEC S min R indoor R outdoor 6 Mbps BPSK 1/2 24 Mbps 16-QAM 1/2 54 Mbps 64-QAM 3/4 Proceed as follows: - Determine the thermal noise power P N within an 802.11g WLAN-channel of 22 MHz bandwidth. Remember that PP NN = 174 + 10log 10 (ff[hhhh]) in [dbm] for T=300 K. - Assume that the WLAN-receiver has a noise figure (F) of 7 db. - Consider that for a target Symbol Error Probability of 1x10-3 in an AWGN channel without coding, BPSK requires a minimum SNR of 7 db, 16-QAM a minimum SNR of 12 db and 64-QAM a minimum SNR of 16 db. - Assume that coding gain is 7 db for FEC Rate = 1/2 and 3 db for FEC Rate = 3/4 - Include a Rayleigh fading margin of 20 db into the receiver sensitivity Smin The thermal noise power: PP NN = 174 + 10log 10 (22 10 6 ) = 100.5 dbm For 6 Mbps: SSSSSS mmmmmm = 7 db 7 db = 0 db SS mmmmmm = PP NN + NN + SSSSSS mmmmmm + mmmmmmmmmmmm = 100.5 + 7 + 0 + 20 = 73.5 dbm For 24 Mbps: SSSSSS mmmmmm = 12 db 7 db = 5 db SS mmmmmm = PP NN + NN + SSSSSS mmmmmm + mmmmmmmmmmmm = 100.5 + 7 + 5 + 20 = 68.5 dbm For 54 Mbps: SSSSSS mmmmmm = 16 db 3 db = 13 db SS mmmmmm = PP NN + NN + SSSSSS mmmmmm + mmmmmmmmmmmm = 100.5 + 7 + 13 + 20 = 60.5 dbm
b) Estimate the maximum radio range for indoor coverage (Rindoor) assuming a Non Line-of- Sight (NLoS) propagation, and the maximum radio range for outdoor coverage (Routdoor) with Line-of-Sight (LoS) propagation. After that, complete the table above. Proceed as follows: - Assume an Equivalent Isotropically Radiated Power (EIRP) of 20 dbm at the AP, and an isotropic receiver antenna at the wireless STA (0 dbi gain). Remember that the maximum path loss PL = EIRP - Smin. - Consider the following single-slope path loss formulas: Outdoor scenario (LoS): PL(d) = 16.9 log 10 (d) + 40.4 db, d in meters Indoor scenario (NLoS): PL(d) = 43.3 log 10 (d) + 19.1 db, d in meters We can solve the path loss formulas for d: Outdoor scenario (LoS): d = 10 PPPP 40.4 16.9 Indoor scenario (NLoS): d = 10 PPPP 19.1 43.3 For 6 Mbps: For 24 Mbps: For 54 Mbps: Outdoor scenario (LoS): d = 10 (20+73.5) 40.4 16.9 = 1386.8 m Indoor scenario (NLoS): d = 10 (20+73.5) 19.1 43.3 = 52.27 m Outdoor scenario (LoS): d = 10 (20+68.5) 40.4 16.9 = 701.7 m Indoor scenario (NLoS): d = 10 (20+68.5) 19.1 43.3 = 40 m Outdoor scenario (LoS): d = 10 (20+60.5) 40.4 16.9 = 235.9 m Indoor scenario (NLoS): d = 10 (20+60.5) 19.1 43.3 = 26.2 m We can finally complete the table. Rate Modulation Rate FEC S min R indoor R outdoor 6 Mbps BPSK 1/2-73.5 dbm 52.27 m 1386.8 m 24 Mbps 16-QAM 1/2-68.5 dbm 40 m 701.7 m 54 Mbps 64-QAM 3/4-60.5 dbm 26.2 m 235.9 m
Problem 2: (Sectorization principle and directive antennas) Consider the wireless communication system presented in Fig. 1, composed by 7 cell-sites of radius R = 1 km. We are interested in determining the received signal quality that the Mobile Stations (MS) (marked by a red 'X' in the figure) observes from each of the Base Station (BS) (i.e., BS 1, BS 2 BS 7) under different system configurations. Assume that the path loss attenuation (in db scale) is given by L(d) = 137. 4 + 35. 2 log 10 (d), d > 0, (1) where d is the distance between each transmitter-receiver pair in kilometres. Shadow fading and fast fading are not taken into account. For sake of simplicity, we consider that all BSs apply the same transmit power, and that each cell site uses a 3 sector configuration with identical antenna gain patterns: 15 [dddd] 60 θ < 60 G (θ) = 8[dddd] ooooheeeeeeeeeeee (2) Assume that the MS is connected to BS number 1, and provide the final results in db. (a) Determine the SIR that the MS perceives when the cluster size is 3 (i.e., each sector uses a different set of frequencies). (b) Find the new SIR for the MS when the cluster size is reduced to 1 (i.e., all sectors use the same set of frequencies). (c) Assume that the cluster size is 1 (i.e., reuse factor 1/1), but that the MS is now able to apply advanced signal processing techniques to cancel the two different interfering signals that are received with highest power. Determine the value for the SIR in this new situation. (d) Calculate the throughput that the terminal is able to support in the previous three cases using the following modified version of Shannon formula: R = W*A log 2(1 + B*SIR) (3) where A = 0.88 is a parameter that accounts the bandwidth efficiency, B = 0.8 is a parameter that models the SIR efficiency. The bandwidth per sector is 1. W = 20 MHz when reuse 1/1 is applied 2. W = 20/3 MHz when reuse 1/3 is used.
Compare the different strategies presented in (a)-(c) and provide a conclusion. Figure 1: Wireless Communication system composed by 7 cell sites with 120 sectors.