Exercise 5.5: Large-scale log-normal fading

Transcription

1 Exercise 5.5: Lrge-scle log-norml fding Since the system is designed to hndle propgtion loss of 135 db, outge will hppen when the propgtion loss is 8 db higher thn the deterministic loss of 17 db = 8. With σ F = 7 db log-norml fding, the outge proility, P out, ecomes P out = Q M db LS σ F = 1 M db erfc LS /σ F = 1 8/7 erfc = 1.7% 1 i. Increse TX power: YES, this will work, since it increses the fding mrgin. Spectrum regul ons my, however, prevent us from doing so. ii. Decrese the deterministic pth loss: YES, this my e possile. It cn e done y mking BS tller, for instnce. iii. Chnge the ntenns: YES, this will lso work.if the signl cn e focused etter the totl loss will ecome smller. The focusing of energy will, however, mke the system more sensitive to the direction in which the TX/RX is in. Spectrum regultions, limiting EIRP my lso prevent us from doing this. iv. Lower the σ F : NO, this will not work in generl. We would need to re-design the rdio propg on environment, which is not exctly relistic. v. Build etter RX : YES, this will work, in principle. The system cn tolerte lower received power if e er receiver is designed. It my, however, not e prcticl solution, since etter receivers tend to e more expensive from severl points of view. Exercise 5.6: Smll-scle Ryleigh fding RX Sensitivity is r min signl mplitude which leds to P out = Pr{r r min } = cdfr min = r min 0 pdfrdr = 1 exp r min rrms 1

2 Inserting the reltion etween the signl mplitude r nd the instntneous received power C C = Kr, K > 0 3 yields the expression P out = 1 exp r min rrms = 1 exp C min/k = 1 exp C min C/K C 4 Using the definition of the fding mrgin eq in the ook, M = C/C min, we get P out = 1 exp 1 5 M which gives us the required fding mrgin, in db, s M = 10 log 10 ln 1 P out 6 Exercise 5.7: Approximtion of Ryleigh fding mrgin For smll vlues of r min, the outge proility, P out, cn e pproximted s P out = Pr{r r min } = cdfr min = 1 exp r min σ r min σ 7 which leds to P out r min σ = r min r rms = 1 M db 8 M db = 10 log 10 P out 9 Define the error ɛ s ɛ = M db M db = 10 log 10 ln 1 P out 10 log 10 P out, P out > 0 10 And the lrgest error, 0.11 db, occurs t the edge of the intervl, nmely where P out = 0.05 = 5%. Given how lrge uncertinties we normlly hve when designing rdio systems, this error is insignificnt.

3 Exercise 5.9: Level crossing rte nd verge durtion of fdes The level crossing rte of Ryleigh fding under isotropic scttering conditions Jkes fding, expressed in terms of the received mplitude r threshold eq in the ook, N R r = r πν mx e r Ω 0 11 Ω0 where Ω 0 is the RMS vlue of the received signl mplitude. Hence, r /Ω 0 = C/ C = 1/M nd the level crossing rte cn e written N R M = πν mx 1 M e 1 M 1 When expressed in terms of mplitude, the product N R radf r = cdfr of the Ryleigh distriution. Now, when expressed in terms of power insted of mplitude, the distriution is exponentil insted. Hence, the verge durtion of fdes, ADF ecomes ADF M = 1 e 1/M N R M = M e 1/M 1 πνmx 13 Exercise 5.10: Lrge- nd smll-scle fding Distnce mx etween ntenns: d mx = 5 km Bse-sttion height: h TX = 0 m Moile-sttion height: h RX = 1.5 m Crrier frequency: f = 450 MHz Lrge scle log-norml fding: σ F = 5 db RX Sensitivity: Cmin RX = 1 dbm Antenn gins: G RX = G TX =.15 db The propgtion loss, L, is given y L = L ground + L Egli s = 0 log 10 d h TX h RX f + 10 log MHz 10 = db The pthloss model is only vlid if the distnce is smller thn the rdio horizon, d h d h 4100 htx + h RX = m > 5 km 15 so we re good to go. 3

4 Figure 1: Link udget With the prmeters given in the prolem sttement, the link udget ecomes s shown in Figure 1. The fding mrgin required for the smll scle fding, reltive to the verge received power, with P out = 5% is M smll = 10 log 10 ln 1 P out = 1.9 db 16 Since the propgtion loss in the prolem gives medin vlue, we need to compenste for the difference etween medin nd verge power of the Ryleigh fding signl. r σ M smll,medin = M smll 10 log 10 = log σ = = 11.3 db r c To otin 95% oundry coverge, we need fding mrgin M lrge ginst σ F = 5 db log-norml fulfilling Mlrge Q = 5%

5 which gives M lrge = = 8. db. 19 d The fding mrgin to e inserted in the link udget, given tht we dd the respective fding mrgins, is M = M smll,medin + M lrge = = 19.5 db 0 With this vlue inserted in the link udget we cn clculte the required trnsmit power. P TX = = 3.6 dbm. 1 Exercise 6.7: CDMA Signls See Section 6... on pges in the ook. T s = 1/W = s Inverse of the ndwidth. Nrrow-nd if 1/W >> τ mx, else wide-nd. With mximum excess dely, τ mx, of 1.3 µs nd chip durtion, T s, of 0.6 µs, the multipth components fll in τ mx /T s = 5 dely ins. This mens tht we experience lekge of energy etween chips nd the chnnel is therefore wide-nd. If the excess dely insted is 100 ns, then the chnnel is much shorter thn chip nd we do not experience ny significnt lekge of energy etween chips. The chnnel is therefore considered s nrrow-nd. Exercise 7.: Okumur nd Okumur-Ht chnnel models Medium-size city h BS = 40 m h MS = m f = 900 MHz d = km 5

6 : Okumur Red vlues of figures in Appendix 7 in Molisch, nd/or look in the lecture slides. L Oku = L free + Corr loss + Corr BS + Corr MS 4π 10 3 = 0 log / = 0 log 10 4π = 134 db 4 : Okumur-Ht h MS = = 1.1 log log A = log log B = log C = 0 Medium-size city 8 These vlues inserted to the Okumur-Ht model yields L Oku-Ht = A + B log 10 d + C = log = db 9 c The two models gree well, which is quite nturl since the second is prmeteriztion of the first. The rnge of vlidity for the Okumur-Ht model is, however, it smller. Exercise 9.1: User influence on coverge Let s ssume tht the MS ntenn is ptch ntenn, since helix ntenns re very rre on terminls these dys. Figure 9.1 then shows tht the medin user User 1 decreses the rdition efficiency y out 3.5 db. Since the propgtion exponent is 3., 3.5 db reduction in the link udget gives Solving for d the new coverge distnce ecomes 10 log 10 d = 3.5 db. 30 d = 9 km. 31 6