Fading Statistical description of the wireless channel · Ghassan Dahman \ Fredrik Tufvesson Department of Electrical and Information Technology Lund University, Sweden Channel Modelling

$Page 1: Fading Statistical description of the wireless channel · Ghassan Dahman \ Fredrik Tufvesson Department of Electrical and Information Technology Lund University, Sweden Channel Modelling$
Ghassan Dahman \ Fredrik TufvessonDepartment of Electrical and Information Technology

Lund University, Sweden

Channel Modelling – ETIM10Lecture no:

2013-01-28 Fredrik Tufvesson - ETIM10 1

3

FadingStatistical description of the

wireless channel


Contents

• Why statistical description• Large scale fading• Fading margin• Small scale fading

– without dominant component– with dominant component

• Statistical models• Measurement example


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”


The WSSUS modelAssumptions

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 differentcontributions (incoming waves), the following is assumed:

The channel is Wide-Sense Stationary (WSS), meaningthat the time correlation of the channel is invariant over time.

The channel is built up by Uncorrelated Scatterers (US), meaningthat contributions with different delays are uncorrelated.


What is large scale and small scale?


Large-scale fadingBasic principle

d

Received power

PositionA B C C

A

B

C

D


Large-scale fadingLog-normal distribution

A normal distributionin the dB domain.

2| 0|

| 2||

1 exp22

dB dBdB

F dBF dB

L Lpdf L

dBDeterministic meanvalue of path loss, L0|dB

|dBpdf L


Large-scale fadingExample• What is the probability that the small scale averaged amplitude

will be 10 dB above the mean if the large scale fading can be described as log-normal with a standard deviation of 5 dB?

dB

|dBpdf L

What if the standard deviation is 10 dB instead?

02.021010Pr 0 QQLLPdB

dBout

0L


The Q(.)-functionUpper-tail probabilities

4.265 0.000014.107 0.000024.013 0.000033.944 0.000043.891 0.000053.846 0.000063.808 0.000073.775 0.000083.746 0.000093.719 0.000103.540 0.000203.432 0.000303.353 0.000403.291 0.000503.239 0.000603.195 0.000703.156 0.000803.121 0.00090

3.090 0.001002.878 0.002002.748 0.003002.652 0.004002.576 0.005002.512 0.006002.457 0.007002.409 0.008002.366 0.009002.326 0.010002.054 0.020001.881 0.030001.751 0.040001.645 0.050001.555 0.060001.476 0.070001.405 0.080001.341 0.09000

1.282 0.100000.842 0.200000.524 0.300000.253 0.400000.000 0.50000

x Q(x) x Q(x) x Q(x)


If these are consideredrandom and independent,we should get a normaldistribution in thedB domain.

Large-scale fading, why log-normal?

12

N

Many diffraction points adding extra attenuation to the pathloss.

Total pathloss:

1 2tot NL L d| 1| 2| ||tot dB dB dB N dBdB

L L d


Small-scale fadingTwo waves

Wave 2

Wave 1

ETX(t)=A cos(2 fct)

ERX1(t)=A1 cos(2 fct-2 / *d1)k0=2 /

E=E1 exp(-jkod1)

ERX2(t)=A2 cos(2 fct-2 / *d2)E=E2 exp(-jkod2) E1

E2


Small-scale fadingTwo waves

Wave 1 + Wave 2

Wave 2

Wave 1


Small-scale fadingDoppler shifts

c

rv

0f f

Frequency of received signal:

0 cosrvfc

where the doppler shift isReceiving antenna moves withspeed vr at an angle relativeto the propagation directionof the incoming wave, whichhas frequency f0.

The maximal Doppler shift is

max 0vfc


Small-scale fadingDoppler shifts

• f0=5.2 109 Hz, v=5 km/h, (1.4 m/s) 24 Hz• f0=900 106 Hz, v=110 km/h, (30.6 m/s) 92 Hz

max 0vfc

How large is the maximum Doppler frequency at pedestrian speeds for 5.2 GHz WLAN and at highway speeds using GSM 900?


Small-scale fadingTwo waves with Doppler

0 2 20

2exp( cos( ) )cvE E jk d f tc

Wave 2

Wave 1

ETX(t)=A cos(2 fct)

0 1 10

1exp( cos( ) )cvE E jk d f tc

v

E1E2

The two components have different Doppler shifts!The Doppler shifts will cause a random frequency modulation


Small-scale fadingMany incoming waves

1 1,r 2 2,r

3 3,r4 4,r

,r

1 1 2 2 3 3 4 4exp exp exp exp expr j r j r j r j r j

1r1

2r 2

3r

3

4r4

r

Many incoming waves withindependent amplitudes

and phases

Add them up as phasors


Small-scale fadingMany incoming waves

1r1

2r 2

3r

3

4r4

r

Re and Im components aresums of many independentequally distributed components

Re(r) and Im(r) are independent

The phase of r has a uniform distribution

2Re( ) (0, )r N


Small-scale fadingRayleigh fading

No dominant component(no line-of-sight)

2D Gaussian(zero mean)

Tap distribution

2

2 2exp2

r rpdf r

Amplitude distribution

Rayleigh

0 1 2 30

0.2

0.4

0.6

0.8

r a

No line-of-sightcomponent

TX RXX

Im a Re a


Small-scale fadingRayleigh fading

minr

min 2min

min 20

Pr 1 expr

rms

rr r pdf r drr

Probability that the amplitudeis below some threshold rmin:

0 r

Rayleigh distribution

2rmsr

2

2 2exp2

r rpdf r


Small-scale fadingRayleigh fading – outage probability

• What is the probability that we will receive an amplitude 20 dB below the rrms?

• What is the probability that we will receive an amplitude below rrms?

2min

min 2Pr 1 exp 1 exp( 0.01) 0.01rms

rr rr

2min

min 2Pr 1 exp 1 exp( 1) 0.63rms

rr rr


Small-scale fadingRayleigh fading – fading margin• To Ensure that we in most cases receive enough power we

transmit extra power – fading margin

2

2min

rmsrMr

2

| 10 2min

10log rmsdB

rMr

minr0 r2rmsr


Small-scale fadingRayleigh fading – fading margin

How many dB fading margin, against Rayleigh fading, do we need toobtain an outage probability of 1%?

2min

min 2Pr 1 exprms

rr rr

1% 0.01

Some manipulation gives2

min21 0.01 exp

rms

rr

2min

2ln 0.99rms

rr

2min

2 ln 0.99 0.01rms

rr

2

2min

1/ 0.01 100rmsrMr

| 20dBM


Small-scale fadingRayleigh fading – signal and interference• Both the desired signal and the

interference undergo fading• For a single user inteferer and

Rayleigh fading:

• where is the mean signal to interference radio

2

2 2 2

2( )( )

SIRrpdf rr

2

2 2( ) 1

( )SIR

rcdf rr

22 221

-20 -15 -10 -5 0 5 10 15 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

SIR(dB)

pdf

pdf for 10 dB mean signal to interference ratio


Small-scale fadingRayleigh 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?

2

minmin 2 2

min

10Pr 1 1 0.09(10 1)( )

rr rr


Small-scale fadingone dominating componentIn case of Line-of-Sight (LOS) one component dominates.• Assume it is aligned with the real axis

• The recieved amplitude has now a Ricean distribution instead of a Rayleigh– The fluctuations are smaller– 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

2Re( ) ( , )r N A 2Im( ) (0, )r N

2

2

Power in LOS componentPower in random components 2

Ak


Small-scale fadingRice fading

A dominant component(line of sight)

2D Gaussian(non-zero mean)

Tap distribution

A

Line-of-sight (LOS)component with

amplitude A.

2 2

02 2 2exp2

r r A rApdf r I

Amplitude distribution

Rice

r

0 1 2 30

0.5

1

1.5

2

2.5k = 30

k = 10

k = 0

TX RX

Im a Re a


Small-scale fadingRice fading, phase distribution

The distribution of the phase is dependent on the K-factor


Small-scale fadingNakagami 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 2 3 4 5}


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:

2

22

20log( )24

20

20 1( )2 ln(10) 2

F

r

F

rpdf r e e

log-normal sdt

log-normal mean

small-scale std for complex components


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.


Some special cases

• Rayleigh fading

• Rice fading, K=0

Rice fading with K=0 becomes Rayleigh

• Nakagami, m=1

Nakagami with m=1 becomes Rayleigh

2

2 2exp2

r rpdf r

Example, shadowing from people


Two persons communicating with each other using PDAs, signal sometimes blocked by persons moving randomly

Example, shadowing from people

0 2 4 6 8 10-40

-30

-20

-10

0

10

Time [s]

Rec

eive

d po

wer

[dB

]


line-of-sight

obstructed LOS

Documents

Fading Statistical description of the wireless channel · Ghassan Dahman \ Fredrik Tufvesson Department of Electrical and Information Technology Lund University, Sweden Channel Modelling