Channel Modelling ETIM10 Lecture no: 3 Fading Statistical description of the wireless channel Fredrik Tufvesson Department of Electrical and Information Technology Lund University, Sweden Fredrik.Tufvesson@eit.lth.se 016-01- Fredrik Tufvesson - ETIN10 1
Contents Why statistical description Large scale fading Fading margin Small scale fading without dominant component with dominant component Statistical models Measurement example 016-01- Fredrik Tufvesson & Carl Gustafson- ETIN10
Why statistical description Unknown environment Complicated environment Can not describe everything in detail Large fluctuations Need a statistical measure since we can not describe every point everywhere There is a x% probability that the amplitude/power will be above the level y 016-01- Fredrik Tufvesson & Carl Gustafson- ETIN10 3
The WSSUS model Assumptions A very common wide-band channel model is the WSSUS-model. Roughly speaking it means that the statistical properties remain the same over the considered time (or area) Recalling that the channel is composed of a number of different contributions (incoming waves), the following is assumed: The channel is Wide-Sense Stationary (WSS), meaning that the time correlation of the channel is invariant over time. (Contributions with different Doppler frequency are uncorrelated.) The channel is built up by Uncorrelated Scatterers (US), meaning that the frequency correlation of the channels is invariant over frequency. (Contributions with different delays are uncorrelated.) 016-01- Fredrik Tufvesson & Carl Gustafson- ETIN10 4
What is large scale and small scale? 016-01- Fredrik Tufvesson - ETIM10 5
Large-scale fading Basic principle Received power D C d B Position A B C C A 016-01- Fredrik Tufvesson - ETIM10 6
Large-scale fading Log-normal distribution A normal distribution in the db domain. pdf ( L ) db Deterministic mean value of path loss, L 0 db db pdf ( LdB ) 1 = exp ( L ) db L0 db πσfdb σ FdB 016-01- Fredrik Tufvesson - ETIM10 7
Large-scale fading Example What is the probability that the small scale averaged amplitude will be 10 db below the mean if the large scale fading can be described as log-normal with a standard deviation of 5 db? pdf ( L db ) P out 10 = Pr db 0 Q σ db { L L + 10} = Q = ( ) 0. 0 L 0 db What if the standard deviation is 10 db instead? 016-01- Fredrik Tufvesson - ETIM10 8
The Q(.)-function Upper-tail probabilities x Q(x) x Q(x) x Q(x) 4.65 0.00001 4.107 0.0000 4.013 0.00003 3.944 0.00004 3.891 0.00005 3.846 0.00006 3.808 0.00007 3.775 0.00008 3.746 0.00009 3.719 0.00010 3.540 0.0000 3.43 0.00030 3.353 0.00040 3.91 0.00050 3.39 0.00060 3.195 0.00070 3.156 0.00080 3.11 0.00090 3.090 0.00100.878 0.0000.748 0.00300.65 0.00400.576 0.00500.51 0.00600.457 0.00700.409 0.00800.366 0.00900.36 0.01000.054 0.0000 1.881 0.03000 1.751 0.04000 1.645 0.05000 1.555 0.06000 1.476 0.07000 1.405 0.08000 1.341 0.09000 1.8 0.10000 0.84 0.0000 0.54 0.30000 0.53 0.40000 0.000 0.50000 016-01- Fredrik Tufvesson - ETIM10 9
Large-scale fading, why log-normal? Many diffraction points adding extra attenuation to the pathloss. This is, however, only one of several possible explanations. α N α 1 α Total pathloss: Ltot = L d α α K αn L = L d + α + α + K + α ( ) 1 ( ) tot db db 1 db db N db If these are considered random and independent, we should get a normal distribution in the db domain. 016-01- Fredrik Tufvesson - ETIM10 10
Two waves Wave 1 E RX1 (t)=a 1 cos(πf c t-π/λ*d 1 ) k 0 =π/λ E=E 1 exp(-jk o d 1 ) E TX (t)=a cos(πf c t) E RX (t)=a cos(πf c t-π/λ*d ) E E=E exp(-jk o d ) E 1 Wave 016-01- Fredrik Tufvesson - ETIM10 11
Two waves Wave 1 Wave 1 + Wave Wave 016-01- Fredrik Tufvesson - ETIM10 1
Doppler shifts v r θ Receiving antenna moves with speed v r at an angle θ relative to the propagation direction of the incoming wave, which has frequency f 0. c Frequency of received signal: f = f0 + ν where the doppler shift is ν = v r f0 cos( θ) c The maximal Doppler shift is ν = f max 0 v c 016-01- Fredrik Tufvesson - ETIM10 13
Doppler shifts How large is the maximum Doppler frequency at pedestrian speeds for 5. GHz WLAN and at highway speeds using GSM 900? ν = f max 0 v c f 0 =5. 10 9 Hz, v=5 km/h, (1.4 m/s) 4 Hz f 0 =900 10 6 Hz, v=110 km/h, (30.6 m/s) 9 Hz 016-01- Fredrik Tufvesson - ETIM10 14
Two waves with Doppler Wave 1 v E = E1exp( jk d f cos( θ ) t) 0 1 c 1 c0 E TX (t)=a cos(πf c t) v Wave v E = Eexp( jk d f cos( θ ) t) 0 c c0 E E1 The two components have different Doppler shifts! The Doppler shifts will cause a random frequency modulation 016-01- Fredrik Tufvesson - ETIM10 15
Many incoming waves Many incoming waves with independent amplitudes and phases Add them up as phasors r1, φ1 r, φ r 3 φ 3 φ r φ r 1 φ 1 r3, φ3 r4, φ4 r, φ φ 4 r 4 r ( φ) = ( φ ) + ( φ ) + ( φ ) + ( φ ) rexp j r exp j r exp j r exp j r exp j 1 1 3 3 4 4 016-01- Fredrik Tufvesson - ETIM10 16
Many incoming waves Re and Im components are sums of many independent equally distributed components Re( r) N(0, σ ) Re(r) and Im(r) are independent r 3 φ 3 φ r φ r 1 φ 1 r The phase of r has a uniform distribution φ 4 r 4 016-01- Fredrik Tufvesson - ETIM10 17
Rayleigh fading No dominant component (no line-of-sight) TX X RX Tap distribution D Gaussian (zero mean) Amplitude distribution 0.8 Rayleigh 0.6 Im( a ) Re( a) r = a 0.4 0. 0 0 1 3 No line-of-sight component ( ) pdf r r r = exp σ σ 016-01- Fredrik Tufvesson - ETIM10 18
Rayleigh fading Rayleigh distribution ( ) pdf r r r = exp σ σ Pr r min 0 r r rms = σ Probability that the amplitude is below some threshold r min : r< r = pdf r dr = rmin r min min 1 exp 0 rrms ( ) ( ) 016-01- Fredrik Tufvesson - ETIM10 19
Rayleigh fading outage probability What is the probability that we will receive an amplitude 0 db below the r rms? r min Pr ( r < rmin ) = 1 exp 1 exp( 0.01) 0.01 = rrms What is the probability that we will receive an amplitude below r rms? r min Pr ( r < rmin ) = 1 exp 1 exp( 1) 0.63 = rrms 016-01- Fredrik Tufvesson - ETIM10 0
Rayleigh fading fading margin To Ensure that we in most cases receive enough power we transmit extra power fading margin r rms r min rrms db 10log10 rmin M = M = 0 r min r r = σ rms 016-01- Fredrik Tufvesson - ETIM10 1
Rayleigh fading fading margin How many db fading margin, against Rayleigh fading, do we need to obtain an outage probability of 1%? Pr < = ( r r ) Some manipulation gives r 1 0.01= exp r min min rms r = ln ( 0.99) = 0.01 r rms rmin min 1 exp rrms 016-01- Fredrik Tufvesson - ETIM10 r ln ( 0.99) = r r rms r min = 1% = 0.01 min rms M = = 1/0.01= 100 M db = 0
Rayleigh fading signal and interference Both the desired signal and the interference undergo fading For a single user inteferer and Rayleigh fading: pdf cdf SIR SIR () r = () r = 1 σ where σ = is the mean signal σ to interference 1radio σ r ( σ + r ) σ r ( σ + r ) pdf 0.5 0.45 0.4 0.35 0.3 0.5 0. 0.15 0.1 0.05 0-0 -15-10 -5 0 5 10 15 0 SIR(dB) pdf for 10 db mean signal to interference ratio 016-01- Fredrik Tufvesson - ETIM10 3
Rayleigh fading signal and interference What is the probability that the instantaneous SIR will be below 0 db if the mean SIR is 10 db when both the desired signal and the interferer experience Rayleigh fading? σ rmin 10 Pr ( r < rmin ) = 1 = 1 0.09 ( σ + r (10 1) min ) + 016-01- Fredrik Tufvesson - ETIM10 4
one dominating component In case of Line-of-Sight (LOS) one component dominates. Assume it is aligned with the real axis Re( r) N( Aσ, ) The recieved amplitude has now a Ricean distribution instead of a Rayleigh The fluctuations are smaller Im( r) N(0, σ ) The phase is dominated by the LOS component In real cases the mean propagation loss is often smaller due to the LOS The ratio between the power of the LOS component and the diffuse components is called Ricean K-factor Power in LOS component A k = = Power in random components σ 016-01- Fredrik Tufvesson - ETIM10 5
Rice fading Tap distribution D Gaussian (non-zero mean) A Im( a ) Re( a) A dominant component (line of sight) r = α Amplitude distribution.5 1.5 1 0.5 Rice k = 30 k = 10 k = 0 TX RX 0 0 1 3 Line-of-sight (LOS) component with amplitude A. ( ) pdf r r exp r + A ra = I 0 σ σ σ 016-01- Fredrik Tufvesson - ETIM10 6
Rice fading, phase distribution The distribution of the phase is dependent on the K-factor 016-01- Fredrik Tufvesson - ETIM10 7
Nakagami distribution In many cases the received signal can not be described as a pure LOS + diffuse components The Nakagami distribution is often used in such cases with m it is possible to adjust the dominating power increasing m {1 3 4 5} 016-01- Fredrik Tufvesson - ETIM10 8
Both small-scale and large-scale fading Large-scale fading - lognormal fading gives a certain mean Small scale fading Rayleigh distributed given a certain mean The two fading processes adds up in a db-scale Suzuki distribution: log-normal mean 4σ pdf () r e e = 0 π r πr 0 1 σ σσ π ln(10) F 0log( σ ) µ σ log-normal sdt small-scale std for complex components F 016-01- Fredrik Tufvesson - ETIM10 9
Both small-scale and large-scale fading An alternative is to add fading terms (fading margins) independently Pessimistic approach, but used quite often The communication link can be OK if there is a fading dip for the small scale fading but the log-normal fading have a rather small attenuation. 016-01- Fredrik Tufvesson - ETIM10 30
Some special cases Rayleigh fading pdf r ( ) Rice fading, K=0 r r = exp σ σ Rice fading with K=0 becomes Rayleigh Nakagami, m=1 Nakagami with m=1 becomes Rayleigh 016-01- Fredrik Tufvesson - ETIM10 31
Example, shadowing from people Two persons communicating with each other using PDAs, signal sometimes blocked by persons moving randomly 016-01- Fredrik Tufvesson - ETIM10 3
Example, shadowing from people 10 line-of-sight 0 Received power [db] -10-0 -30-40 0 4 6 8 10 Time [s] obstructed LOS 016-01- Fredrik Tufvesson - ETIM10 33