Chapter 3 Problem Solutions

Transcription

1 Chate Poblem Solutions Poblem A Equation (5) gives P P G G log h log h L 4 log d t t t sys Substituting gives P log 6 log log 4 log 875 m B The wavelength is given by 8 The fee-sace ath loss is then 9 5 m 4d 4 L ath log log 5 The eceived owe in fee sace is given by the ange equation: P P G G L L t t sys ath log 6 7 m C Equation (6) gives hh t A Adi sin d The agument of the sine function is hh t 754 d 5 Then sin That is, the agument is about % lage than the sine Poblem Fom the geomety of Figue, Chate,

2 ht tan i x tan, x which gives x 5 m Similaly, h tan d x tan, d x which gives dx 7 m Adding gives Now Equation (6) gives Equation (7) gives d m hh t A Adi sin Adi sin 8 A di d 5 4 hh t 4 A Adi Adi 9 A di d 5 Relative to the value obtained fom Equation (6), the aoximation of 9A di Equation (7) is log 99 8Adi Poblem We stat by calculating the themal noise floo The noise figue is F 9 o F 794 The noise owe efeed to the eceive inut is P kt B kt B F n ant In decibels this is W P m Then fo a SNR of 7 we need a n signal owe of P 7 m A Fom Equation (5) we have m log 6 log log 6 4 log 844 m 4log d Solving gives d 484 km d Chate,

3 B In fee sace the ange equation gives m log 6 L ath This gives a maximum ath loss of L 54 Now ath 4 d L ath log At 85 MHz the wavelength is 5 m 6 Substituting gives d 4 m Poblem 4 Since the noise and intefeence owes add, we will have to wok with owe athe than with decibels Fom Poblem we know that the 8 noise floo is at 957 W The signal owe is 7 highe, o P n 5 48 W P The signal owe exceeds the sum of intefeence and noise by 6, o by a facto of 98 That is, 5 P P P 957 P Solving gives the intefeence owe as n i i 8 P i 48 W That is, P 6 m Now fo tansmission ove a flat eath we have i 6 m log 6 log log 6 4 log 844 m 4log d, i d whee d i is the distance fom base station B to the hand-held eceive Solving gives d 8 km The seaation between base stations is then i d d d km AB i i Poblem 5 A Equation (4) gives a h e log 75h 497 fo f 4 MHz e = log Then Equation () gives the median ath loss as Chate,

4 log 8 log te loghte logd a he log 8 log log log 5 L uban f h 5 67 B The eceived signal owe is given by P 5 log m m C Using Equation (5), P 5 log 7 log log log 88 m The flat Eath model gives an ovely otimistic estimate of the eceived signal owe Poblem 6 We stat by calculating the themal noise floo The noise figue is F 6 o F 98 The noise owe efeed to the eceive inut is Pn kt BF In decibels this is W P n m Then fo a SNR of 8 we need a signal owe of P 8 5 m A Using the aametes fom at B of Poblem 5 we have 5 5 m log m 7 4 L5 uban db The maximum ath loss is gives L5 uban 5 The Hata model f hte log hte log d ahe log log d log 8 log log 8 log Chate, 4

5 Solving gives d 45 km B Using Equation (5), 5 5 m log 7 log log6 4 4 log d Solving gives d 4 km The flat Eath model oveestimates the maximum ange by an ode of magnitude Poblem 7 Equation (5) gives a h log f 7 h 56log f 8 e e log log 8 45 Then Equation () gives the median ath loss in uban aeas as log 8 log te loghte logd a he log 8 log log log 45 L uban f h 5 67 In sububan aeas we have fom Equation (6) In ual aeas Equation (7) gives L sububan L uban log f log L ual L uban 478 log f 8log f log 8log Poblem 8 In the Lee model, distances ae measued in feet and miles Conveting the aametes given in Poblem 5 gives h 984 ft, h 55 ft, and d 746 mi Now Equation (5) gives t Chate, 5

6 c log log log 85 5 log Table gives P mile 77 m and 48 fo New Yok Equation (49), modified to include system losses, gives 746 P,5 77 m 48log m Poblem 9 In the solution to Poblem 6 we showed that the minimum eceived signal owe is P 5 m Using the numbes fom Poblem 8, we have d 5 m 77 m 48log Solving gives d 98 mi, o 8 km This is lowe than the 45 km obtained in Poblem 6 using the Hata model Pat of the diffeence may be a consequence of the lage value of ath-loss exonent that the Lee model uses fo New Yok City Poblem Fom Equation (64) we have P f m PQ P P Q Q, ath ath whee the fade magin f m is given by f m P Poblem The eceive sensitivity is m 5 95 m Fom the solution to Poblem we have fm PQ 85 Q 7 Chate, 6

7 This is equivalent to f m 5 Q 7 Solving gives f 76 m Fade magin is defined by f m P Substituting gives P f 95 m m m Poblem Equation (5) gives c 5 8 log log log Table gives P mile 7 m and 68 fo Philadelhia Equation (49), modified to include system losses, gives P,5 7 m 68log 6 5 m A A eceived signal owe above the median owe is -5 m Fom Equation (64) we have P P P Q ath 5 m 5 m P P 5 m Q 8 Q 5 6 B At 5 mi the median eceived signal owe is Then, P,5 5 5 mi 7 m 68log 6 4 m 5 m 4 m P P 5 mi 5 m Q 8 Q Chate, 7

8 Poblem The eceive s themal noise floo is given by 5 Pn ktbf W, o P 5 m The eceive sensitivity is 5 above the noise n floo, o 5 m 5 m Now Solving gives, P m P P P 9 Q Q ath 8 P 897 m Using the Lee model fom Poblem d 897 m 7 m 68log 6 Solving gives the maximum oeating ange as d mi o 78 km Poblem 4 A Fom Poblem we have The equied fade magin is f m P P Q ath f 987 m fm 95 Q 6 B The eceive s noise figue is F 4 o F 5 We find the eceive s noise floo to be o Pn kt BF W, P 5 m The eceive sensitivity is above this noise n floo, o m Since the fade magin is given by f P, we have the equied median eceived signal level m P f 987 m m m C The tansmitted owe is 5 W o Pt 454 m The ange equation gives m 454 m 6 L ath, so the maximum ath loss is L 54 ath Chate, 8

9 Fo a medium-sized city Equation (5) gives e log 7 e 56log 8 a h f h f log log 8 96 Then Equation () gives log 8 log te loghte logd a he log 8 loghte loghte log log h L uban f h 5 Setting h h 64 m te log te, gives the equied antenna height as te Poblem 5 Equation (8) gives A A A A A cos f t t The amlitude A is at its eak when f t t f t t n Now Then and cos, o f d c t d c t t, t Substituting and solving fo f gives f nf f, as equied Similaly a the amlitude A is at a null when cos f t t, o f t t n Chate, 9

10 Substituting as above and solving fo f gives f nf f, as equied The diffeence between two successive fequency eaks is f f t t In Figue 6 we ae given t t 5 μs This imlies f MHz The figue shows eaks at 85 MHz, 85 MHz, and 854 MHz, which is consistent with the ediction Poblem 6 Using Equation () with the aveage owe P, P P P P e d P P P e e P 68 Poblem 7 Using Equation () with the aveage owe P,min,min P P,,min P SNR SNRmin P P P,min Pn P n e e P P d 95 Solving gives Conveting to decibels gives,min P n 5 P 95,min P 95 P P,min SNR 95 SNR n min Chate,

11 SNR 9 SNR min Poblem 8 Conveting out of decibels gives the following owes and coesonding delays: Powe Delay Data Relative Powe Delay (μs ) Then Equation () gives k t k 486 μs, Equation () gives t 54 s, Equation () gives 6 μs, and Equation (5) gives B coh 48 khz The maximum system data ate deends on the ulse shae and sectal efficiency used Fo examle, if binay hase-shift keying is used with ectangula ulses, then a 48 khz bandwidth can suot a 74 kbits/s data ate d Poblem 9 8 At a fequency of 9 GHz the wavelength is 58 m 9 9 The eceive s velocity is 6 mh, o v 68 m/s The maximum v 68 Dole shift is then f d 7 Hz Signals aive ove the two 58 aths with Dole shifts of f f cos 5 69 Hz f d d d f cos 85 d 48 Hz The Dole sead is then fdf d 54 Hz The eiod of a fading cycle is given by Equation (4) as t nn 648 ms 54 At a bit ate of 88 Mbits/s, the duation of one bit is T 84 ns Then since T T, the fading is slow Altenatively we can calculate sig nn sig Chate,

12 the coheence time using Equation (47) to be T Since Tsig T coh the fading is slow To esimate the velocity needed fo fast fading, set T T sig coh coh 9 6 f 84 v d 5 ms Solving gives v 48 km/s, o 77,8 mh Poblem Fist, 9 u ath 9 4 Using Figue with an aea coveage of 4 P R Next F gives a equied bounday coveage of 75 Equation (64) gives f m 75 P R P P sens R sens Q ath Solving gives the fade magin f 67 m sens Chate,