Channel Modelling –ETI 085 Contents 2 · 2009. 10. 28. · 2009-10-29 Fredrik Tufvesson -ETI 085...

Channel Modelling – ETI 085gLecture no: 22

Statistical description of theStatistical description of the wireless channel

Fredrik TufvessonDepartment of Electrical and Information Technology

Lund University, [email protected]

2009-10-29 Fredrik Tufvesson - ETI 085 1

Contents

• Why statistical description• Small scale fading

– without dominant componentith d i t t– with dominant component

• Doppler effectsT l d i ti• Temporal description


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 b h l l ”above the level y”


What is large scale and small scale?


The WSSUS modelAssumptionsAssumptions

A very common wide-band channel model is the WSSUS-modelA very common wide band channel model is the WSSUS model.

Recalling that the channel is composed of a number of differentcontributions (incoming waves) the following is assumed:contributions (incoming waves), the following is assumed:

The channel is Wide-Sense Stationary (WSS), meaningth t th ti l ti f th h l i i i t tithat the time correlation of the channel is invariant over time.(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 different delaysare uncorrelated.)


Small-scale fadingTwo wavesTwo waves

Wave 1ERX1(t)=A1 cos(2πfct-2π/λ*d1)

k0=2π/λk0=2π/λE=E1 exp(-jkod1)

ETX(t)=A cos(2πfct)

W 2

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

E2

Wave 2


Small-scale fadingTwo wavesTwo waves

Wave 1

Wave 1 + Wave 2

W 2Wave 2


Small-scale fadingDoppler shiftsDoppler shifts

rvθ Frequency of received signal:

cθ

0f f ν= +

where the doppler shift is

( )0 cosrvfc

ν θ= −

where the doppler shift isReceiving antenna moves withspeed vr at an angle θ relative

cto the propagation directionof the incoming wave, whichhas frequency f0.

The maximal Doppler shift isq y 0

max 0vfc

ν =


Small-scale fadingDoppler shiftsDoppler shifts

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

max 0vfc

ν =

• f0=5.2 109 Hz, v=5 km/h, (1.4 m/s) 24 Hzf =900 106 Hz v=110 km/h (30 6 m/s) 92 Hz• f0=900 106 Hz, v=110 km/h, (30.6 m/s) 92 Hz


Small-scale fadingTwo waves with DopplerTwo waves with Doppler

Wave 1

v0 1 1

0

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

θ= − −

ETX(t)=A cos(2πfct) v

0 2 20


θ= − −

W 2

E1E2

Wave 2

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


The Doppler shifts will cause a random frequency modulation

Small-scale fadingMany incoming wavesMany incoming waves

Many incoming waves with Add them up as phasorsindependent amplitudes

and phases

1 1,r φ 2 2,r φ

r2r 2φ

3φ

φ

4 4,r φ

1r1φ3r

r

φ

3 3,r φ4 4,φ

,r φ 4r4φ

( ) ( ) ( ) ( ) ( )1 1 2 2 3 3 4 4exp exp exp exp expr j r j r j r j r jφ φ φ φ φ= + + +


Small-scale fadingMany incoming wavesMany incoming waves

Re and Im components aree a d co po e ts a esums of many independentequally distributed components

r2r 2φ

3φ

φ

2Re( ) (0, )r N σ∈1r

1φ3r

r

φRe(r) and Im(r) are independent

4r4φThe phase of r has a uniform distribution


Small-scale fadingRayleigh fadingRayleigh fading

No dominant component TX RXX(no line-of-sight)

2D GaussianTap distribution Amplitude distribution

2D Gaussian(zero mean) Rayleigh

0.8

0.4

0.6

r a=( )Im a ( )Re a

0 1 2 30

0.2

( )2

2 2exp2

r rpdf rσ σ

⎛ ⎞= −⎜ ⎟

⎝ ⎠No line-of-sight

component


σ σ⎝ ⎠component

Small-scale fadingRayleigh fadingRayleigh fading

Rayleigh distribution

( )2

2 2expr rpdf r⎛ ⎞

= −⎜ ⎟( ) 2 2exp2

pdf rσ σ⎜ ⎟

⎝ ⎠

minr0 r2rmsr σ=

⎛ ⎞

Probability that the amplitudeis below some threshold rmin:

( ) ( )min 2

minmin 2

0

Pr 1 expr

rms

rr r pdf r drr

⎛ ⎞< = = − −⎜ ⎟

⎝ ⎠∫


0 rms⎝ ⎠

Small-scale fadingRayleigh fading – outage probabilityRayleigh fading outage probability

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

( )2

mini 2Pr 1 exp 1 exp( 0.01) 0.01rr r

⎛ ⎞< = − − = − − ≈⎜ ⎟( )min 2Pr 1 exp 1 exp( 0.01) 0.01

rms

r rr

< ⎜ ⎟⎝ ⎠

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

( )2

minmin 2Pr 1 exp 1 exp( 1) 0.63

rms

rr rr

⎛ ⎞< = − − = − − ≈⎜ ⎟

⎝ ⎠⎝ ⎠


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

transmit extra power fading margintransmit extra power – fading margin

22

2min

rmsrMr

=2r⎛ ⎞

| 10 2min

10 log rmsdB

rMr

⎛ ⎞= ⎜ ⎟

⎝ ⎠

minr0 r2rmsr σ=


Small-scale fadingRayleigh fading – fading marginRayleigh fading – fading margin

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

( )2

minP 1 r⎛ ⎞⎜ ⎟ 1% 0 01( ) min

min 2Pr 1 exprms

r rr

< = − −⎜ ⎟⎝ ⎠

1% 0.01= =

Some manipulation gives2

min21 0.01 exp r⎛ ⎞

− = −⎜ ⎟ ( )2

min2ln 0.99 r

⇒ = −2prmsr

⎜ ⎟⎝ ⎠

( ) 2rmsr

( )2

min l 0 99 0 01r 2

1/ 0 01 100rmsrM( )min2 ln 0.99 0.01

rmsr⇒ = − = 2

min

1/ 0.01 100rmsMr

⇒ = = =

20M⇒ =


| 20dBM⇒ =

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

interference undergo fading 0.5interference undergo fading• For a single user inteferer and

Rayleigh fading: 0.35

0.4

0.45

2

2 2 2

2( )( )

SIRrpdf rr

σ

σ=

+0.15

0.2

0.25

0.3

pdf

2

2 2( ) 1

( )SIR

rcdf rr

σ

σ= −

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

0.05

0.1

0.15

• where is the mean signal 22 22

σσσ

=

SIR(dB)

pdf for 10 dB mean signal to interference ratio

to interference radio1σ


Small-scale fadingRayleigh fading – signal and interferenceRayleigh 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 theif the mean SIR is 10 dB when both the desired signal and the interferer experience Rayleigh fading?

210( ) min

min 2 2min

10Pr 1 1 0.09(10 1)( )

rr rr

σ

σ< = − = − ≈

++


Small-scale fadingone dominating componentone dominating componentIn case of Line-of-Sight (LOS) one component dominates.

A it i li d ith th l i• Assume it is aligned with the real axis2Re( ) ( , )r N A σ∈ 2Im( ) (0, )r N σ∈

• 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 p p g

to the LOS• The ratio between the power of the LOS component and

the diffuse components is called Ricean K-factorthe diffuse components is called Ricean K-factor2

2

Power in LOS componentPower in random components 2

Akσ

= =


Small-scale fadingRice fadingRice fading

A dominant component(line of sight) TX RX(line of sight)

2D Gaussian(non-zero mean)

Tap distribution Amplitude distribution

Rice( )

A

1.5

2

2.5k = 30

k = 10

r α=0.5

1 k = 0( )Im a ( )Re a

0 1 2 30

Line-of-sight (LOS)component with

amplitude A.

( )2 2

02 2 2exp2

r r A rApdf r Iσ σ σ

⎛ ⎞+ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠


p

Small-scale fadingRice fading phase distributionRice fading, phase distribution

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


Small-scale fadingNakagami distributionNakagami distribution• In many cases the received signal can not be described as a pure

LOS + diffuse componentsLOS + diffuse components• The Nakagami distribution is often used in such cases

1.4

1.6

1.8

2

0.4

0.6

0.8

1

1.2

pdf

0 0.5 1 1.5 2 2.5 30

0.2

0.4

signal amplitude

increasing m

with m it is possible to adjust the dominating power

g{1 2 3 4 5}


Some special cases

• Rayleigh fading2⎛ ⎞( )2

2 2exp2

r rpdf rσ σ

⎛ ⎞= −⎜ ⎟

⎝ ⎠• Rice fading, K=0

Rice fading with K=0 becomes Rayleigh

• Nakagami, m=1

N k i ith 1 b R l i h


Nakagami with m=1 becomes Rayleigh

Measurement example

• Measurement in the lab• Center frequency 3.2 GHz• Measurement bandwidth 200 MHz, 201 frequency points• 60 measurement positions, spaced 1 cm apart• Measured with a vector network analyzer


Transfer function

-35

-45

-40

-55

-50

p (d

B)

70

-65

-60

eque

ncy

res

-80

-75

-70

Fre

0 0 2 0 4 0 6 0 8 1 1 2 1 4 1 6 1 8 2-90

-85


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 108Frequency (Hz)

Transfer function, all positions

40

60

-50

-40

esp

(dB

)

80

-70

-60

Freq

uenc

y re

1 52

-90

-80F

0

0.2

0.4

00.5

1

1.5

x 108

( )Frequency (Hz)


0Position (m)Frequency (Hz)

Impulse response

-50

-60

-80

-70

p (d

B)

-90

Impu

lse

resp

110

-100

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-120

-110


x 10-7Delay (s)

Impulse response, all positions

-60

-40

)

-80

-60

se re

sp (d

B)

0

0.2

-100 Positi

Impu

ls

0.4

0.6

00.5

11.5

2

-120

on (m)


2

x 10-7 Delay (s)

Probability density function

180 d

140

160measuredtheoretical

100

120

80

100

pdf

40

60

0 0 002 0 004 0 006 0 008 0 01 0 0120

20


0 0.002 0.004 0.006 0.008 0.01 0.012amplitude

Cumulative distribution function

1

0.8

0.9measuredtheoretical

0.6

0.7

0.4

0.5cdf

0.2

0.3

0 0.002 0.004 0.006 0.008 0.01 0.0120

0.1


amplitude

Small-scale fadingDoppler spectraDoppler spectra

Wave 1

v0 1 1

0


θ= − −

ETX(t)=A cos(2πfct) v

0 2 20


θ= − −

W 2

E1E2

Wave 2

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


The Doppler shifts will cause a random frequency modulation

Small-scale fadingDoppler spectrumDoppler spectrum

Incoming waves from several directionsSpectrum of received signal

when a f0 Hz signal is transmitted.

Incoming waves from several directions(relative to movement or RX)

RX movement

11

2

2

3 4

RX 4

3

0 maxf ν− 0 maxf ν+0fAll waves of equal strength inthis example, for simplicity.


Small-scale fadingDoppler spectrumDoppler spectrum

Isotropic uncorrelatedscattering

Time correlaion

1RX movement

0.5

1

RX

0

U if i iU l t d

0 0.5 1 1.5 2-0.5

max tν ΔUniform incomingpower distribution

(isotropic)

Uncorrelatedamplitudesand phases


Small-scale fadingDoppler spectrumDoppler spectrum• Time correlation – how static is the channel?

F if tt i i t th ti l ti f th R• For a uniform scattering environment the time correlation of the Re-and Im-components is described by a “zeroth-order Bessel function of the first kind”

1

Th ti l ti f th lit d i

( ) ( ) ( ){ } ( )*0 max2t E a t a t t J tρ πνΔ = + Δ ∝ Δ

0.5

1

• The time correlation for the amplitude is

( ) ( )20 2t J tρ πνΔ ∝ Δ

0

( ) ( )0 max2t J tρ πνΔ ∝ Δ0 0.5 1 1.5 2

-0.5

max tν Δ


Small-scale fadingThe Doppler spectrumThe Doppler spectrum

Uncorrelated scatteringDoppler spectrumat center frequency f

( )0DS fν −

Uncorrelated scatteringwith uniform angular distribution

D l t b F i

at center frequency f0.

Doppler spectrum by Fouriertransformation of the timecorrelation of the signal:

( ) ( ) 2jS dπν τ− ΔΔ Δ∫0 maxf ν− 0 maxf ν+0f

( ) ( ) 2

1

jDS e dπν τν ρ τ τΔ= Δ Δ

∝

∫

2 2maxπ ν ν−

for ν ν ν− < <What does this mean in practice?


for max maxν ν ν< <

Small-scale fadingFading dipsFading dips

Wh t b t th l th d th fReceived amplitude [dB]

r

What about the length and the frequencyof fading dips ?

*rrmsr

|dBMImportant for system designblock lengthcode design

*rg

choice of modulation

TimeTime


Small-scale fadingStatistics of fading dipsStatistics of fading dips

Frequency of the fading dips Length of fading dipsFrequency of the fading dips(normalized dips/second)

Length of fading dips(normalized dip-length)

These curves are for Rayleigh fading and isotropic


These curves are for Rayleigh fading and isotropicuncorrelated scattering (Jakes’ doppler spectrum).

Channel Modelling –ETI 085 Contents 2 · 2009. 10. 28. · 2009-10-29 Fredrik Tufvesson -ETI 085...

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