Wireless Communication Channels Lecture 3: Fading

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Wireless Communication ChannelsLecture 3: Fading

EITN85, FREDRIK TUFVESSON

ELECTRICAL AND INFORMATION TECHNOLOGY

Wireless Communication Channels 2

Fading – Statistical description of thewireless channel

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

– without dominant component– with dominant component

• Statistical models• Measurement example

2016-01-22VT 2018

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VT 2018 Wireless Communication Channels 3

Why statistical description?

• Unknown environment• Complicated environment• Can not describe everything in detail• Large fluctuations

• Need a statistical measure since we can not describeevery point everywhere

“There is a x% probability that theamplitude/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 remainthe 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 overtime. (Contributions with different Doppler frequency areuncorrelated.)

The channel is built up by Uncorrelated Scatterers (US),meaning that the frequency correlation of the channels isinvariant over frequency. (Contributions with differentdelays are uncorrelated.)

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What is large scale and small scale?


Large-scale fadingBasic principle

d

Received power

PositionA B C C

A

B

C

D

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Large-scale fadingLog-normal distribution

A normal distributionin the dB domain.

( ) ( )2| 0|

| 2||

1 exp22

dB dBdB

F dBF dB

L Lpdf L

sps

æ ö-ç ÷= -ç ÷è ø

dBDeterministic meanvalue of path loss, L0|dB

( )|dBpdf L


Large-scale fading - Example

What is the probability that the small scale averaged amplitude willbe 10 dB below the mean if the large scale fading can be describedas log-normal with a standard deviation of 5 dB?

dB

( )|dBpdf L

What if the standard deviation is 10 dB instead?

{ } ( ) 02.021010Pr 0 »=÷÷ø

öççè

æ=+³= QQLLP

dBdBout s

0L

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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?

1a2a

Na

Many diffraction points adding extra attenuation to thepathloss.This is, however, only one of several possible explanations.

Total pathloss:( ) 1 2tot NL L d a a a= ´ ´ ´ ´K

( )| 1| 2| ||tot dB dB dB N dBdBL L d a a a= + + + +K

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Small-scale fadingTwo waves

Wave 2

Wave 1

ETX(t)=A cos(2pfct)

ERX1(t)=A1 cos(2pfct-2p/l*d1)k0=2p/lE=E1 exp(-jkod1)

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

E2


Small-scale fadingTwo waves

Wave 1 + Wave 2

Wave 2

Wave 1

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Small-scale fadingDoppler shifts

c

rvq

0f f n= +

Frequency of received signal:

( )0 cosrvfc

n q= -

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

The maximal Doppler shift is

max 0vfc

n =


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

n =

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

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Small-scale fadingTwo waves with Doppler

0 2 20

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

q= - -

Wave 2

Wave 1

ETX(t)=A cos(2pfct)

0 1 10

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

q= - -

v

E1E2

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


Small-scale fadingMany incoming waves

1 1,r f 2 2,r f

3 3,r f4 4,r f

,r f

( ) ( ) ( ) ( ) ( )1 1 2 2 3 3 4 4exp exp exp exp expr j r j r j r j r jf f f f f= + + +

1r1f

2r 2f

3r

3f

4r4f

r

f

Many incoming waves withindependent amplitudesand phases

Add them up as phasors

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Small-scale fadingMany incoming waves

1r1f

2r 2f

3r

3f

4r4f

r

f

Re and Im components aresums of many independentequally distributedcomponents

Re(r) and Im(r) areindependent

The phase of r has a uniformdistribution

2Re( ) (0, )r N sÎ


Small-scale fadingRayleigh fading

No dominant component(non line-of-sight)

2D Gaussian(zero mean)

Tap distribution

( )2

2 2exp2

r rpdf rs s

æ ö= -ç ÷

è ø

Amplitude distributionRayleigh

0 1 2 30

0.2

0.4

0.6

0.8

r a=

No line-of-sightcomponent

TX RXX

( )Im a ( )Re a

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Small-scale fadingRayleigh fading

minr

( ) ( )min 2

minmin 2

0

Pr 1 expr

rms

rr r pdf r drr

æ ö< = = - -ç ÷

è øò

Probability that the amplitudeis below some threshold rmin:

0 r

Rayleigh distribution

2rmsr s=

( )2

2 2exp2

r rpdf rs s

æ ö= -ç ÷

è ø


Small-scale fadingRayleigh fading – outage probability

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

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

( )2

minmin 2Pr 1 exp 1 exp( 0.01) 0.01

rms

rr rr

æ ö< = - - = - - »ç ÷

è ø

( )2

minmin 2Pr 1 exp 1 exp( 1) 0.63

rms

rr rr

æ ö< = - - = - - »ç ÷

è ø

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Small-scale fadingRayleigh fading – fading margin

To Ensure that we in most cases receive enough power we transmitextra power – fading margin

2

2min

rmsrMr

=2

| 10 2min

10log rmsdB

rMr

æ ö= ç ÷

è ø

minr0 r2rmsr s=


Small-scale fadingRayleigh fading – fading margin

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

( )2

minmin 2Pr 1 exp

rms

rr rr

æ ö< = - -ç ÷

è ø1% 0.01= =

Some manipulation gives2

min21 0.01 exp

rms

rr

æ ö- = -ç ÷

è ø( )

2min

2ln 0.99rms

rr

Þ = -

( )2

min2 ln 0.99 0.01

rms

rr

Þ = - =2

2min

1/ 0.01 100rmsrMr

Þ = = =

| 20dBMÞ =

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Small-scale fadingRayleigh fading – signal and interference

Both the desired signal and theinterference undergo fading

For a single user inteferer andRayleigh fading:

where is the mean signal to

interference ratio

2

2 2 2

2( )( )

SIRrpdf rr

s

s=

+2

2 2( ) 1

( )SIR

rcdf rr

s

s= -

+

22 22

1

ss

s=

-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 tointerference ratio


Small-scale fadingRayleigh fading – signal and interference

What is the probability that the instantaneous SIR will be below 0 dB ifthe mean SIR is 10 dB when both the desired signal and the interfererexperience Rayleigh fading?

( )2

minmin 2 2

min

10Pr 1 1 0.09(10 1)( )

rr rr

s

s< = - = - »

++

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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 distributioninstead 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 dueto the LOS

• The ratio between the power of the LOS component and thediffuse components is called Ricean K-factor

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

2

2Power in LOS component

Power in random components 2Aks

= =


Small-scale fadingRice fading

A dominant component(line of sight)

2D Gaussian(non-zero mean)

Tap distribution

A

Line-of-sight (LOS)component withamplitude A.

( )2 2

02 2 2exp2

r r A rApdf r Is s s

æ ö+ æ ö= -ç ÷ ç ÷è øè ø

Amplitude distributionRice

r a=

0 1 2 30

0.5

1

1.5

2

2.5k = 30k = 10

k = 0

TX RX

( )Im a ( )Re a

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Small-scale fadingRice fading, phase distribution

The distribution of thephase is dependent onthe K-factor


Small-scale fadingNakagami distribution

In many cases the received signal can not be described as a pureLOS + 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}

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Both small-scale and large-scale fadingLarge-scale fading - lognormal fading gives a certain meanSmall scale fading – Rayleigh distributed given a certainmeanThe 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 es mp

ssps ss p

-¥ --= ò

log-normal std

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 havea rather small attenuation.

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Some special casesRayleigh 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 rs s

æ ö= -ç ÷

è ø

Example, shadowing from people


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

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Example, shadowing from people

0 2 4 6 8 10-40

-30

-20

-10

0

10

Time [s]

Rec

eive

dpo

wer

[dB

]


line-of-sight

obstructed LOS

Wireless Communication Channels Lecture 3: Fading

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