1. Steele, R, Mobile Radio Communications, Pentech Press, 1992 ; TK6570; [1.2.5,2.7]

Transcription

1 3. The Moble Rado Channels One of the dstngushng features of moble communcatons s the channel behavor, of whch most s unwelcome. We ll look at these :. path loss analogous to nverse square law, but worst, much worse. Inverse fourth power??. Shadowng varaton due to obstacles blockng the path (makes handoff dffcult problem). 3. fadng tens of db and phase reversal n a fracton of a second. Very challengng, very destructve. We wll elaborate on fadng : - two phenomena; fadng n tme, dsperson (delay spread, frequency selectve fadng). Keep them separate n your mnd! - dfferences between moble and base - fadng spectra, fade rate, delay profle. 3. Path Loss References :. Steele, R, Moble Rado Communcatons, Pentech Press, 99 ; TK6570; [..5,.7] 3-

2 . Lee, Wllam C. Y., Moble Communcatons Engneerng : Theory and Applcatons, McGraw Hll, 998, nd Edton; TK6570; [3.-3.4] Free space nverse square law s optmstc. Even n the absence of local scatterers and obstacles, there are large geometry objects wth reflectons. Consder the followng smple propagaton model. Let s(t) be the transmtted complex envelop of the transmtted bandpass sgnal j fct st!() = Re ste () π. The receved bandpass sgnal (gnorng free space path { } loss for the tme beng) can be wrtten as : j π fc ( t x/ c) { ( ) } j( ωct xβ) s( t x c) e rtx!(, ) = Re st x/ ce { } jxβ jωct { s( t x c) e e } = Re / = Re / (plane wave) where x s the dstance from the transmtter to the recever (n the drecton of propagaton), c s the speed of lght, ωc = π fc s the carrer frequency n rad/s, β = π / λ, and λ = c / fc s the wavelength. Let h a and h m be the heght of the transmttng and recevng antennae respectvely, and d the horzontal dstance between them. d h a h m 3-

3 Any phase dfference between the drect and reflected rays leads to partal cancellaton n addton to the / d free space loss. Drect path length : ha hm xd = d + ( ha hm) = d + ( ha hm) / d d + d Reflected path length : ha + hm xr = d + ( ha + hm) d + d Dfferental path length : x = x r x = d hahm / d Sum of arrvals (note nverse square and a reflecton coeffcent of ) : jx dβ sum() t = s ( t xd / c) e s ( t xr / c) e d d jxd β jβ x s( t xd / c) e ( e ) d snce s( t x / c) s( t x / c) d r jx β (narrowband processes, tme scale >> x / c r ) Receved power s proportonal to : E sum( t) = E s( t x / c) ( cos( β x ) ) d Po β x = for β x (far away) " d d 4π hh a m = Po λd 3-3

4 that s the nverse fourth power of the dstance. Example : Consder the case ha = 30 λ, hm = 5 λ, d' = d/ λ. The plot of the normalzed power as a functon of d ' s as shown below. There are hm / λ maxma and the farthest one s at d' = 4( ha / λ)( hm / λ). The nverse fourth power phenomenon kcks n after the last peak. Realty s not a plane earth obstacles, terran varatons, etc ntroduce dffracton, reflectons. However, the exponent s usually between 3 and 4 when averaged over a large ensemble of real confguratons [M. Hata, Emprcal Forms for Propagaton Loss n Land Moble Rado, IEEE Trans. On Vehcular Technology, Aug 980] Consequences of faster than nverse square bad news - need a powerful transmtter for reasonable range - accentuated near-far problem 3-4

5 - n ncrease n transmt power doesn t gve a proportonal ncrease n coverage area. P r = P / r t k Good news Edge of coverage defned by specfc receved power level, so / k k / P r = P / r. Ths means A / A = ( r / r) = ( P / P) k - coverage areas are relatvely sharply defned whch s good for cellular layout 3. Shadowng Hlls, large buldngs cause varatons n the receved power ndcated by path loss. These obstacles prevent the exstence of a drect lne of sght path between the transmtter and the recever. When the recever s n a shadow, the sgnal strength wll be weak. Shadowng can also be caused by folage attenuaton and precptaton. Ths s especally true when transmttng at super hgh frequency (tens of GHz). Shadowng s partcular problematc for nbound (moble to base) sgnals where the transmtted power s low. 3-5

6 One remedy s to use auxlary (or satellte) antennas, usually for nbound transmsson. Why not outbound (.e. smulcast)? Smultaneous transmsson on same frequency from dfferent stes gves destructve nterference n some places f exactly the same frequency; beats f not exactly the same; garblng or ISI f envelopes not matched n tme. Shadowng varaton s usually log-normal. That means f you take the db equvalent of the receved sgnal strength, t wll be a Gaussan random varable. A typcal varaton s 6-8 db. 3.3 Slow Fadng Here we examne the fne structure of the receved sgnal for moble channels. Dscusson s lmted to the pdf (tme varatons n a later secton), so t apples to very slow fadng (large λ, slow or motonless moble) so that the fade rate f d " modulaton bandwdth Fadng Mechansm 3-6

7 Receve superposton of several reflectons, each wth own path length from transmtter x and complex reflecton coeffcent a (magntude and phase change). Average path length : x N x N = = Dfferental path length : x = x x Receved sgnal : rtx!(, ) = Re ast ( x / ce ) j( ω t βx ) β { } = Re as( t x / c) e e { } = Re as( t [ x+ x]/ c) e e c j x jω t c jβ x j( ω t βx) c Assume that modulaton s(t) changes slowly enough to be unaffected by dfferental delays x / c,.e. B x / c". Then rtx!(, ) Re ae st ( x/ ce ) j( ωct βx) = Re gst ( x/ ce ) jβ x j( ω t βx) where g s a complex gan, constant at that pont n space. As vehcle moves, set of dfferent path lengths change, get tme varyng g(t). c The last equaton depcts the phenomenon of flat fadng. We examne next the frst order statstcs of g, the complex gan. 3-7

8 3.3. Gaussan Model Flat Fadng Examne the frst order statstcs of g, the complex gan. Apply central lmt theorem to real and magnary part of jβ x g = ae = gi + jgq and t becomes Gaussan : where p g g ( g) = exp πσ σ ( ) g = E g = E gi + E gq σ σ If you plot the pdf pg ( g ) as a functon of g I and g Q, you see crcular symmetry. g Q constant prob. contour g I j g g jg re ψ = + =, where r = g, 0 r In polar co-ordnate, I Q ψ = arctan g / g, π ψ π. Can show that ( Q I ) < and 3-8

9 Consequently r p (, r ψ ) e π σ r /σ r, ψ = r r /σ pr ( r) = e σ r 0 (Raylegh) p ψ ( ψ ) = π Clearly r, ψ are ndependent. The receved power s proportonal to z = r, whch s exponentally dstrbuted. Follows from change of varables n Raylegh, or from r = gi + gq (sum of ndependent squared Gaussan s χ and χ wth degrees of freedom s exponental wth mean σ ). In other word p z e z σ z /σ z ( ) = 0 How good s the Gaussan approxmaton? As shown n the dagram below, even 6 to 0 components s a close approxmaton for r r = σ π /, except for the tals. 3-9

10 The dstrbuton of the resultant of N unt vectors wth random relatve Pr a / N > y where a s phase. The x-component of each pont represents ( ) the magntude of the sum and y s the correspondng value on the y-axs. Moble satellte wth a lne of sght component (LOS) s better modeled by Rce fadng. In ths case, the complex gan g becomes g = g + g s d g = Kσ P = g = Kσ s s s g g d = = E gd σ Pd g s 3-0

11 where g s s the specular (LOS) component and g d s the dffuse component. K represents the power rato n the specular and dffuse components. The total power s P = E g = gs + E gd = σ ( + K). Snce g has a non zero mean of g s, t means z = g s non central χ wth degree of freedom, or the pdf of r = z = g s Rcean : r r r K pr() r = exp K I o σ σ σ and phase [see also Eqn of Proaks] K ( ) cos ( ψ ) K pψ ( ψ ) = e + 4π K cos( ψ) e [ Q( K cos ψ)] π 3-

12 Note that σ tself vares wth the lognormal dstrbuton determned by shadowng. The K factor depends on terran. 3.4 Fadng Autocorrelaton Functon and Power Spectrum We need second order statstcs of the complex gan process f we are to descrbe ts tme varaton. It s a Gaussan process, so that s all we need. If second order statstcs are constant n tme, than t s WSS Power Spectrum and Autocorrelaton at Moble References :. Jakes, Wllam C. Jr., Mcrowave Moble Communcatons, Wley, 974; TK6570; [Chapter ].. Lee [7.] Consder the case we transmt an unmodulated carrer and a movng recever. Consequences :. The receved complex envelop s the fadng gan g. Snce the recever s movng, g changes wth tme,.e. a random process g(t).. When there s a relatve moton between the transmtter and recever, there exsts a Doppler frequency. How do we relate the two? 3-

13 Assume the moble s travelng through a set of scatterers at a velocty of V. φ (Top Vew) The scatterer at an angle of φ to the drecton of moton ntroduces a Doppler shft of V f = cos( φ) = fd cos( φ) λ to the unmodulated carrer, where fd = V / λ s the maxmum Doppler frequency. The Doppler shft s the largest (.e. f d ) when the moble s travelng towards the scatterer and the smallest (.e. fd ) when away from the scatterer Conclude : - the transmsson of an unmodulated tone produces multple tones wth dfferent frequences because of the scatterers and the relatve moton. 3-3

14 - The power dstrbuton, as a functon of the Doppler frequency shft, s the power spectral densty of the receved complex envelop, whch as mentoned earler, s the channel complex gan process g(t). - The nverse Fourer transform of the PSD gves the autocorrelaton functon. Note :. Both φ and φ contrbute to the same Doppler shft.. Snce f = f cos( φ), d df = fd ( f / fd) dφ 3. Let p( φ ) be the (raw) receved power densty comng from the drecton φ. For sotropc scatterng, ths s a constant. 4. G( α ) be the antenna gan pattern; assume the antenna s pontng at an angle of γ wth respect to the drecton of moton. For a vertcal polarzed antenna, G( α ) s a constant. γ φ V φ 3-4

15 Based on the above, we can conclude that the power arrvng n dφ at φ s proportonal to G ( φ γ) p( φ) dφ. So power n df at f s { φ γ φ + φ γ φ } S ( f) df G ( ) p( ) G ( ) p( ) g f ( f / f ) d df d For sotropc scatterng and vertcal polarzed antenna, σ f < f d Sg ( f) = π fd f 0 otherwse Ths s known as the Jakes spectrum n the lterature. Shape of the fadng spectrum (sold lne) wth the maxmum Doppler frequency normalzed to. From the power spectrum, we can obtan the autocorrelaton functon : 3-5

16 * Rg( τ) = E g( t) g ( t τ) = IFT Sg( f) V = σ Jo( π fdτ) = σ Jo π τ λ x = σ J o π λ A plot of the autocorrelaton functon versus the normalzed (to the wavelength) dstance of travel From the autocorrelaton functon plot, we see that ponts separated by 0.4λ are uncorrelated; however oscllatons de slowly, so, n prncple, correlaton lasts for many wavelengths Exercse: Wrte a computer program to smulate Raylegh fadng wth a Jakes spectrum. As expected from the autocorrelaton functon, sgnal magntude s quasperodc n space, wth a dp every λ / to λ or so. From the complex envelope trajectory, we see that fades are assocated wth rapd phase change and rapd fractonal ampltude change. 3-6

17 g(x) n db x/lambda Magntude of g(x) n db scale. The horzontal lne denotes the root-mean value n db Imag[g(x)] Real[g(x)] Complex envelop trajectory. The pont near (0,0) corresponds to the deep fade at x = 6λ n the ampltude plot. Separaton n space between successve crcles s λ /0. 3-7

18 In summary, the receved sgnal n a flat fadng channel s r( t) = g( t) s( t) where gt ( ) s a zero mean complex Gaussan process wth an autocorrelaton functon of R ( τ ) = σ J ( π f τ). g o d If the transmtted complex envelop s s() t exp( jπ fot) complex envelop s r t g t ( j π f t) =, the receved () = ()exp o. In other word, gt ( ) s the frequency response of the channel at tme t and frequency f o. Snce gt ( ) s actually ndependent of f o, the nstantaneous frequency response at any tme t s flat, hence the term flat fadng; see the fgure below. 3-8

19 3.4. Autocorrelaton at the Base The channel s recprocal n an electromagnetc sense. Ths means the autocorrelaton functon of the complex channel gan s the same at the base as t s at the moble. The spatal correlaton at the base, however, s consderably dfferent - far less decorrelaton when base s moved than when moble moves. Base staton antennas need much wder spacng for dversty (at moble, only half a wavelength or so s enough for decorrelaton). Source of dfference s geometry : moble s surrounded by scatterers, base has sgnal from narrow spatal angle. Jakes derves a detaled model based on a rng of scatterers about the moble, none near the base. It agrees wth experment only n a general sense, and t s mmedately thrown off by scatterers near the base. Use of antenna arrays at the base s currently a very actve research topc, and better models for correlaton among the antennas are needed. 3-9

20 3.5 Multpath Spread and Correlaton Bandwdth References : Lee [.5,.6], Steele [Chapter ]. Returnng to Secton 3.3, we now examne the case of large dfferental delays amongst the dfferent paths. By large, we mean the dfferental delays are large wth respect to sgnal varatons n tme. Rewrte the receved sgnal as rt!() = Re ast ( x / ce ) j( ω t βx ) β { τ } = Re as( t ) e e = Re hs ( t τ ) e c c j x jω t where τ = x / c s the delay assocated wth the -th path, and jβ x jωτ c h = ae = ae. jω t c The complex envelop of the receved sgnal s rt ( ) = hst ( τ ) mplyng a (baseband equvalent) channel wth an mpulse response of and a frequency response of ht ( ) = hδ( t τ ) 3-0

21 H f jπ fτ ( ) he = Snce we assume the dfferental delays are large wth respect to sgnal varatons n tme, ths means H( f ) has sgnfcant varaton across the sgnal band (frequency selectve). Alternatve descrpton for the channel : h( τ ) = hδτ ( τ) s defned as the response of the channel at the present moment due to an mpulse appled τ second earler. Ths means when the nput to the channel s s( t ), the output s rt ( ) = h( τ ) st ( τ) dτ = hst ( τ ) Multpath spread (or delay spread), τ d, s the range of τ over whch there s sgnfcant energy, that s sgnfcant values for h. Correlaton bandwdth (coherence bandwdth) s the frequency separaton over whch there s a sgnfcant change n H( f ). It s the wdth of * * H( f) H ( f f) df or E H( f) H ( f f) (loose defnton). A rough approxmaton s: coherence bandwdth / τ d, snce fne structure wdth n one doman (frequency) s roughly the recprocal of the wdth n the other (tme). Lttle pont n beng more precse. 3-

22 Example : Two path model : h( τ ) = hδτ ( ) + hδτ ( τd ). The transfer jπ fτd functon of the channel s H( f) = h+ he. Assume h and h are drawn from an ensemble, the frequency correlaton functon becomes S f E H f H f f * h( ) = ( ) ( ) σ = σ + j f d e π τ σ Cauton : S ( f ) s NOT the transform of autocorrelaton of gan. h Now a general model s and jπ fτ H( f) h( τ ) e d = j π fτ * j π( f f ) s ( τ τ)( ) Sh( f) = E h( ) e d h ( s) e ds = = * jπ fτ j π( f f ) s E h τ h s e e dτds ( ) ( ) τ * jπ fτ j π( f f ) s Ehτ h s e e dτds [ ( ) ( )] If we assume uncorrelated scatterng gans at dfferent delays are uncorrelated (wde-sense statonary uncorrelated scatterers WSSUS), = * E h( τ) h ( s) G( τ) δ( τ s) where G( τ ) s the delay power profle. Then jπ fτ Sh( f) G( τ ) e d = τ : frequency correlaton functon 3-

23 Root mean square (RMS) delay spread s a common measure ( τ τ) G( τ) dτ τ rms = ; τ = τg( τ) dτ G( τ) dτ If the moble moves, then the path length, hence phases change, so coeffcents h n mpulse response are tme varyng. Now t s h( τ,) t = h() t δτ ( τ) It descrbes the response, seen at tme t, to an mpulse τ seconds earler. The tap-delay lne model for the channel s shown below. s( t) τ τ τ τ τ 3 h( t) h ( t) h ( t) h ( t) 3 r ( t ) rt ( ) = h( τ, tst ) ( τ) dτ = h( tst ) ( τ ) Fnally, lets take the transform n the nput/output relaton 3-3

24 where ( ) jπ ft jπ fτ r() t = h( τ,) t S( f) e df e d = S( f) H( f, t) e jπ ft df jπ fτ H( f, t) h( τ, t) e d = τ τ s called the tme varant transfer functon of the channel. j fot Example : If S( f) δ ( f f o ) rt () = H fo, t e π. So H( fo, t ) must be the frequency response of the channel at tme t and frequency f o. A sample H( f,) t s shown below. =, then ( ) 3-4

25 At ths pont, want to fnd out f h( τ,) t s stll Gaussan for any t and τ. Look at the complex envelop of the receved sgnal agan. (any gven pont n space) jϖτ c rt ( ) = hst ( τ ) = ae st ( τ ) If the tme varaton of s(t) s comparable to or less than the ndvdual delays (.e very wdeband), then a collecton of separable paths, no opportunty for central lmt theorem. Wth lower resolvng power, can T = τ : k τ τ( k+ ) τ and rewrte r(t) as : aggregate paths nto bns { } k where rt ( ) = hst ( k τ ) h k k = τ T k k j c ae ωτ are stll more or less Gaussan, but wth less plausblty than flat fadng. So channel becomes less Gaussan as the resolvng power (bandwdth) of the sgnal ncreases. In general, the gan h k s tme dependent because of moton of moble, just lke n the flat fadng case. Gven that hk ( t ) are plausbly Gaussan for any t, the tme varant transfer functon H( f, t) = hk ( t) e k jπkf τ can be vewed as a Gaussan process n f, wth autocorrelaton functon 3-5

26 * Sh ( f) = E H( f, t) H ( f f, t) E h ( t) h ( t) e * j π τ( kf $ f + $ f ) = k $ k $ It s reasonable to assume that complex gans n paths wth resolvable delays are ndependent,.e. wth E h t h t * k () $ () Gk k = $ = 0 otherwse G( τ ) = G δτ ( k τ) k k beng the delay power profle of the channel. Consequently S ( f) G e h = k k j π( k τ) f s smply the Fourer transform of the delay power profle, evaluated at f A more general correlaton functon s : S f t = E H f t H f f t t * h(, ) (, ) (, ) = E hk ( t) h ( t t) e k = G φ ( t) e k k G = φ ( ts ) ( f) G $ h * j π τ( kf $ f + $ f ) $ j π( k τ)( f ) where 3-6

27 φ ( t) = G E h t h t E * k( ) k( t) hk ( t) s the normalzed autocorrelaton functon for any delay. Concluson : uncorrelated scatterng s a convenent model snce the jont tme-frequency correlaton functon s separable. 3.6 More on Flat Fadng As shown earler, n the absence of a LOS component, fadng n a moble rado channel can be modeled as a zero mean complex Gaussan process gt ( ) wth an autocorrelaton of R ( τ ) = σ J ( π f τ) (the correspondng fadng spectrum s Sg( f) σ ( π fd f ) characterstcs of ths fadng model. g o d =, f fd ). We now examne the 3.6. Some mportant probablty densty functons References :. S.O. Rce Mathematcal Analyss of Random Nose, BSTJ Vol. 3, pp. 8-33, July 44, and BSTJ Vol. 4, pp , Jan 45.. S.O. Rce, Statstcal Propertes of s Sne Wave Plus Random Nose, BSTJ, Vol. 7, pp , Jan W.B. Davenport, W.L. Root, An Introducton to the Theory of Random Sgnals and Nose, McGraw Hll,

28 The probablty denstes functons gven below are general n the sense that they are applcable to any fadng spectrum Sg ( f ) wth even symmetry. jπ fot Exercse : Consder the fadng process h( t) = e g( t), where g(t) s a fadng process wth a Jakes spectrum. Show that the PSD of h(t) s not symmetrcal about f = 0. Note : the above represents the scenaro where there s a frequency offset between the oscllators n the transmtter and recever, n addton to fadng. Frst order gt () g() t jg () t ate () jψ t ( ) = I + Q = : g pg ( g) = exp πσ σ a p a e π σ a /σ a, ψ (, ψ ) = Dervatve fadng process d gt () = gt () = g() t + jg () t = bte () jθ t dt I Q ( ) : where g p g ( g) = exp πγ γ ( f ) γ = σ π rms (varance of dervatve process) f rms f Sg ( f) df = S ( f ) df g (mean square Doppler spread) 3-8

29 Proof : The dervatve process gt ( ) can be obtaned by passng g(t) through a lnear flter wth frequency response jπ f. The output process s Gaussan and has a PSD of S ( f) = ( π f ) S ( f) g g Exercse : Show that f = f / for the Jakes spectrum. rms d Jont pdf of g(t) and gt ( ) : p g g ( g, g) = exp + ( πγ )( πσ ) σ γ gg, Proof : The processes g(t) and gt ( ) can be wrtten as : are jontly Gaussan. Ther cross-correlaton * * * g ( t τ) g ( t τ t) R ( ) gg τ = E g( t) g ( t τ) = lm E g( t) t 0 t * * g( t) g ( t τ) g( t) g ( t τ t) = lm E t 0 t Rg( τ) Rg( τ + t) = lm = Rg ( τ ) t 0 t E g( t) g( t) = R g (0). For Jakes spectrum, the dervatve of the autocorrelaton functon at τ = 0 s 0. Therefore [ ] Conclude : g(t) and gt ( ) are ndependent when evaluated at the same tme nstants. Therefore pgg, ( g, g ) = pg( g) pg ( g ), whch s bascally what the equatons says. 3-9

30 E g( t) g( t ) = 0 Exercse : Show that [ ] symmetry. s true for any Sg ( f ) wth even Jont PDF, polar co-ordnate : jψ jθ jψ jψ Snce g = ae, g = be = ae + jψ ae, we have g = b = a + ψ a Ths nformaton can be used (see [Rce]) to show that p a a a ψ + a ( a, a, ψψ, ) = exp + 4πσγ σ γ aa,, ψψ, where a range from 0 to, ψ range from π to π, and both a and ψ range from to. Because of the form of ths jont pdf, the margnal pdfs can be easly obtaned. Why are these probablty densty functons of mportance? Well, they are drectly related to performance. Wll examne two here frst :. random FM. fade rate and fade duraton More to come later Random FM 3-30

31 Refer to [Lee 7.3] and [Jakes, Chapter ]. In dgtal FM system, a popular demodulator s the lmter/dscrmnator. s( t) LPF Lmter/ Dscrmnator gt ( ) n( t) j ( t) Let s() t = e φ ( ) and gt () = ate () jψ t. Then n the absence of nose, the [ () ()] sgnal at the nput of the LPF s rt () = stgt () () = ate () j ψ t + φ t. The output of the lmter/dscrmnator s approxmately φ( t) + ψ( t) where the frst term s the sgnal component and the second term s called random FM. From the jont pdf p,,, ( a, a, ψψ, ), we can obtan the margnal pdf for the random FM ψ : aaψψ p ( ψ) = p ( a, a, ψ, ψ) dψ da da ψ aa,, ψ, ψ a= 0 a= ψ= π π σ = + γ γ σ ψ 3/ ψ = + 4π f rms π frms 3/ 3-3

32 3 Ths s a long-taled pdf where p ψ ( ψ ) π f rms / ψ for large ψ. So even though ψ ( t) s bounded, ts dervatve s not. It s qute non-gaussan Example: A R b = 4800 bps FSK modem, modulaton ndex h=/, detected wth dscrmnator and no post-flter, transmt at 800 MHz, vehcle at 00 km/hr, what s the rreducble error rate? - A modulaton ndex of ½ means s() t exp( j f t) f = hr b / = Doppler frequency : f = v/ λ = vf / c= 74 Hz. d c = ± where π - Make an error f random FM exceeds sgnal. If the negatve frequency was sent, then the rreducble error probablty s [ ] ψ π f P = Pr ψ > π f = p ( ψ) dψ eo = ( frms / f ) + = for Jakes spectrum ( fd / f ) + When fd / f <<, ( ) Peo fd f 8. Varaton as f d s a characterstc of the error floor. You wll see more error floors n Chapter

33 3.6.3 Frequency and Duraton of Fades The bg questons n relatng fadng to system desgns : - what type of burst error correctng code? - and s burst error correcton useful, anyway? - length of forward error correctng code (FEC)? - depth of nterleavng to break up bursts? - n antenna swtched dversty, how much delay s tolerable? Answers to some can be provded by analyss. Others through measurements and/or smulatons. Prelmnary observatons - Let = gt. As shown n Secton 3.3., zt ( ) ( ) [ ] zo Pr z z = e o z / σ = σ. Thus p ( z) e ( ) z /( σ ) - fades are of varyng depths, and nstantaneous system performance (lke bt error probablty, output sgnal-to-nose rato) may be relatvely soft functon of depth. - But we wll thnk n terms of threshold, snce useful for coded systems, antenna swtchng etc., and some applcablty to uncoded systems - Defne threshold level wth respect to the mean (rms) level. Then at lower levels :. less tme spent n fade state. fewer fades 3. shorter fades 3-33

34 g(x) n db x / l a m b d a Fade Rate : - Refer to [Lee.7, 6.4], [Rce 948] - Ths s a level crossng rate (LCR). For a gven ampltude threshold A, the number of upward crossngs per second s n( A) = a p ( A, a) da o aa, where p aa, ( a, a) s the jont pdf of the fadng envelop ( a = g = z) and ts dervatve. The proof below s due to [Rce] : Ampltude of fadng gan a + adt A a t t + dt Tme 3-34

35 - In any tme range t, t+dt, there s an upward crossng f A a a dt and a > 0 Equvalently, we need A a dt a A Therefore A Pr upward crossng n dt da p ( a, a) da [ ] =, a= 0 a= A adt = dt da a p ( A, a) Ths means the expected number of crossngs n any dt, hence the crossng rate s aa, a= 0 n( A) = a p ( A, a) da o Note that the lower lmt of ntegraton s zero because we are nterested n postve crossngs. aa, aa - The jont pdf p, ( a, a) s aa p ( a, a) = p ( a, a, ψψ, ) dψdψ aa, aa,, ψψ, ψ= ψ= π π a a a = exp + πγ σ σ γ Substtuton gves 3-35

36 γ A A na ( ) = exp σ πσ σ = R π fre for Jakes spectrum d where R = A σ s the crossng threshold normalzed to the RMS envelop level. - The fgure below shows the normalzed crossng rate (wth respect to the Doppler frequency) as a functon of the normalzed threshold level R n(a)/fd *log0(R) Normalzed zero crossng (fade) rate versus normalzed threshold level. 3-36

37 It s observed that a. a 6 db decrease n level (R cut n half) means cuttng the fade (zero crossng) rate by half, b. fade rate proportonal to f d, c. maxmum rate at R = 3 db Fade Duraton - Refer to [Lee 6.5], [Jakes.3.5] - Easy, now that we have the other relatons Hence = Pr a / σ R T( R) n( R) d R R e e R T( R) = = R π fre π fr π f d d - Plot of normalzed fade duraton ( ft d ( R) ) vs R s shown n the plot below.. fade duraton s proportonal to ampltude level for R equal 5 db and below. a 6 db decrease n level means fades are half as long (and half as frequent, as we saw earler) 3. fade duraton nversely proportonal to fd 3-37

38 T(R)*fd *log0(R) Normalzed fade duraton versus normalzed threshold - It s shown n [Arnold and Boltman, Interfade Interval Statstcs of a Raylegh dstrbuted wave, IEEE Trans. Communcatons, Sept. 83] that ntervals between fades are approxmately exponentally dstrbuted for R < 5 db or so. The same two authors n [Boltman and Arnold, Fade dstrbuton statstcs of Raylegh dstrbuted wave, IEEE Trans. Communcatons, March 8] show that the fade duraton s not exponental Example : Consder a moble data system. What does a 3 db ncrease n transmt power buy n terms of error burst statstcs? Solutons : 3-38

39 Snce R= A/ σ, ncrease σ by 3 db (.e doublng t) effectvely reduces R to 0.7 tmes ts old value. Snce for small R, the fade duraton s approxmately proportonal to R, so bursts are now 7 as long. Example : Have a moble data system operatng at GHz, vehcle speed 08 km/hr, data rate 9600 bps usng FSK wth dscrmnator detecton, 3 average error probablty (BEP) s 0. Estmate (roughly) burst duraton and frequency, both n bts. Use the followng expressons for the nstantaneous and the average BEPs : P( a) e a / 4 = e (nstantaneous BEP) e = e( ) a( ) = σ a= 0 + (average BEP) P P a p a da where σ s the varance of the underlyng complex fadng process. For ths queston, t also respresents the average SNR n the system. Solutons : - From the BEP expressons, we can deduce that σ 000 or 30 db. - Major approxmaton : bursts defned by nstantaneous BEP > 0.. Ths means A =.5373 s the threshold. - The normalzed threshold s db. R A σ = / = or approxmately

40 - From equatons (or graphs), the normalzed fade rate and normal fade duraton are approxmately 0.4 and 0.0 respectvely. - Snce f = V / λ = 00 Hz, we have na= ( ) 4 deep fades/s and T ( A) = 4µ s d - Bt duraton s T = / 9600 = 04µ s. So average fade duraton s bts - Inter-fade nterval s 9600/n(A) = 686 bts. Exercse : Repeat last exercse but wth a velocty of 50 km/hr. 3-40