wireless communications

1

EE 463 - Wireless CommunicationsCommunications

Dr. Ahmed Iyanda SulymanAssociate Professor

Electrical Engineering Department Electrical Engineering Department King Saud University

Chapter 4: Mobile radio propagation: Small-scale fading

1. Introduction: small-scale fading & multipath

2. Impulse response model of a multipath channel

3. Multipath channel parameters

4. Types of small-scale fading

5. Statistical models

2

6. Fading channel Simulations

2

Chapter 4: Mobile radio propagation: Small-scale fading

Direct path or line of sight (LOS) wave

Reflectedwave

Diffractedwave

3

Mobile userBase station

Scatteredwave

Small-scale fading (or simply fading) describes the rapid fluctuations of a radio signal over a short period of time (few seconds) or a short travel distance (few

1 – Introductions

( ) (wavelength).

Path 1

Path 2Path 3

4

The radio waves from the transmitter arrive at the mobile from different directions, each with different amplitude and propagation delays (multipath comp.).

Path 4

3

1 – Introductions

• Multipath components may combine (vectorial addition) either constructively or destructively at the mobiles, y y ,and may thus cause the signal received by the mobile to distort or fade (small-scale fading).

Factors causing small-scale fading:

Multipath propagation: presence of reflectors, scatterers, etc., in the environment.

Speed of the mobile: movement of transmitter or receiver or both cause frequency shift in the transmitted signal (called Doppler shift)

5

transmitted signal (called Doppler shift). Speed of surrounding object: movements of

surrounding objects induce time-varying Doppler shift. Transmission bandwidth of the signal: If the

transmitted radio signal has bandwidth greater than the bandwidth of the multipath channel.

Doppler shift (revision):

Consider a mobile moving at a constant velocity v

1 – Introductions

Consider a mobile moving at a constant velocity, v, along a path segment of length d between points Aand B, while it receives signals from a remote source S.

The phase shift in the received signal due to the difference in path length as the mobile moves from points A to B is given by:

)(22 dL

6

)cos(2

)cos(

tv

4

Source, S

Illustration: Doppler Shift

L )cos(2

)cos(22

tv

dL

• Phase shift between A and B:

7

d

v

)cos(2

1

v

tfDoppler

A B

• Frequency shift between A and B(Doppler frequency):

Doppler shift (revision):

The apparent change in frequency (Doppler shift or

1 – Introductions

The apparent change in frequency (Doppler shift or Doppler frequency) is.

Positive Doppler shift (apparent received frequency is increased): receiver moves towards transmitter.

Negative Doppler shift (apparent received frequency d d) f

)cos(2

1

v

tf D

8

is decreased): receiver moves away from transmitter.

If source transmits fc, the received frequency is fc ± fD

5

Example 4.1:

1 – Introductions

Consider a transmitter which radiates a sinusoidal carrier frequency of 1850MHz. For a vehicle moving at 28m/s, compute the received carrier frequency if the mobile is moving:

(a) directly toward the transmitter.

(b) directly away from the transmitter.

(c) in a direction perpendicular to the direction of

9

(c) in a direction perpendicular to the direction of arrival of the transmitted signal.

Modeling the impulse response of wireless channels allows numerical performance evaluation of different

2 – Impulse response model of a multipath channel

pmobile communication systems.

For a mobile at a fixed position, d, the received signal can be expressed as.

d

v

Spatial position

10

p

For a causal system ( ),

dtdhxtdhtxtdy ),()(),(*)(),(

0 ,0)( tth

t

dtdhxtdy ),()(),(

6

Assume the mobile moves at a constant velocity v, then d = vt, thus.


or

where x(t) = transmitted bandpass signal (modulated signal at

d

v

Spatial position

t

dtvthxtvty ,)(),( ),(*)( ,)()( thtxdthxtyt

11

where x(t) = transmitted bandpass signal (modulated signal at carrier frequency fc)

y(t)=received signal, h (t -) = impulse response of the multi-path radio channel,

t= time variable, =channel multipath delay for fixed t.

The impulse response of a multipath channel can be expressed as.


1Np

ak(t, ) is the amplitude of the kth multipath component at time t.

k(t) is the excess delay of the kth multipath component at time t.

k(t,) is the phase shift of the kth multipath component, which is a function of the delay and time t

1

0

))(()),(exp(),(),(N

kkkk ttjtath

12

is a function of the delay and time t. N is the number of paths. If the channel is assumed to be time invariant, or stationary

over a small-scale time or distance interval (quasi-static fading), then

1

0

)()exp(),(N

kkkk jath

7

),( tht4

Illustration: time-varying impulse response

t2

t3

4

10 2 3

10 2 3

1

02222 ))(()),(exp(),(),( E.g.

N

kkkk ttjtath

13

t1

1

0

0 20k

Cellular operators develop channel impulse response for different environments to allow prior planning and design. ~ typical entry-level duties given wireless Engineers

2 – Impulse response Measurements

~ typical entry level duties given wireless Engineers

Small-scale Multipath channel Measurements techniques:

Small scale channel state for an environment is typically

d

v

Spatial position

14

Small-scale channel state for an environment is typically recorded in the form of average power delay profiles, defined as P() Ek[|h(tk, )|2 ], where Ek[] denotes ensemble average over samples taken at different times tk.

Different channel sounding techniques for estimating the power delay profiles of wireless systems are:

8

Direct RF pulse system:transmits a repetitive pulse ofwidth Tb , and uses a receiver


cf TxRF

b ,with a wide bandpass filter tocollect all multipath signalreceived from each of thesepulses and take the ensembleaverage.

Advantage: low complexity (easily implemented).

Disadvantage: it is subject

Pulse Generator

Tb

Trep

Pulse width = Tb, Pulse period = Trep

BPFEnvelope

DetectorAmpl.

Oscillos-

cope

RF

15

Disadvantage: it is subject to interference and noise in the environment during the test which may not represent the actual channel being measured.

Spread Spectrum based channelsounding: rather than transmittingordinary pulses, the transmitted

l i thi h d d


cf TxRF

pulses in this approach are encodedusing binary PN (pseudo-noise)sequence, which spreads the pulseover a wide band causing noise &interference rejections.

Advantage: moderate complexity, and noise/interference rejection.

Processing gain of the SS system allows much lower power than the

PN sequence

Generator

Sequence length

RF PN sequence

Generator

BPF

16

allows much lower power than the direct RF pulse system.

Widely used for indoor and outdoor channel sounding (3G and 4G).

Disadvantage: instantaneous measurements (real time) are not made, but over PN sequence length

BPF

(wide)Ampl.

Oscillos-

cope

Envelope

Detector

BPF

(narrow)

9

Frequency domain channelsounding: this method measureschannel impulse response inf d i d


frequency-domain and useInverse Discrete Fourier Transform(IDFT) to convert from frequencyto time domain measurements.

The S parameter measured infrequency domain is proportionalto H (j)= FT{h(t)}.

Advantage: provides amplitude & phase information of the time-

RF

Inverse

DFT processor

S parameter

Test set

Vector Network Analyzer

with

Swept Freq. Oscillator

RF

)( jX)()(21 jHjS

)(

)()(

jX

jYjH

1

)( jYPort1 Port2

17

phase information of the time-domain channel (complex).

Disadvantage: hard wired sync. needed between transmitter & receiver, making it useful only for very close measurement.

Used in experiments on indoor channel measurements.

)}({1)( jHFTth

Quantitative multipath channel parameters have been developed in order to compare different multipath channels and for receiver design purposes.

3 – Parameters of mobile multipath channel

g p p

Time dispersion parameters:

Mean Excess delay:

1

0

2

1

0

2

N

kk

N

kkk

122

N

kk

Pt (t)

t

Mobile

channel

maxTT

Channel span (max. excess delay)

t

Prec (t)

18

RMS delay spread: where

Maximum excess delay max (x dB): Is the time delay during which multipath energy falls to x dB below the maximum.

These parameters are obtained directly from power delay profile.

22

1

0

2

02N

kk

k

10

Example 4.3 (Ex5.4 in Ref1)


Compute the RMS delay spread for the power delay profile of a multipath channel shown below. What is the maximum symbol rate that can be transmitted through the channel without needing an Equalizer?.

)(recP

0 dB 0 dB

19

Ans: = 0.5sec, and Rs = 200k symb./sec. For BPSK, Rb=Rs.

0

1s

Coherence bandwidth, Bc.


Is the range of frequencies over which the channel can be considered “flat” (i.e., the channel’s frequency response stays correlated).

Bc is inversely proportional to the RMS delay spread, but its exact value depends on how it is defined: → If the coherence bandwidth is defined as the bandwidth

over which the frequency correlation function is above 0.9, then 1

B

20

0.9, then

→ If the coherence bandwidth is defined as the bandwidth over which the frequency correlation function is above 0.5, then

50cB

5

1cB

11

Example 4.4 (Ex 5.5 in Ref1)

Calculate the mean excess delay and the rms delay spread for the


Calculate the mean excess delay, and the rms delay spread for themultipath power delay profile shown below.

Estimate the 50% coherence bandwidth of the channel.Would this channel be suitable for AMPS or GSM cellular service

without the use of an Equalizer ?.

Ans: =4.38sec, = 1.37sec, and

)(recP

0 dB

21

Bc (50%)= 146KHz.

Bc>30kHz used in AMPS (no Equalizer required)

But Bc<200kHz used in GSM (Equalizer required for GSM).

-20 dB

01s

-10 dB

2s 5s

Doppler spread (BD) and coherence time (TC) ,.

D l d d h b d idth B d ib th


Delay spread, στ, and coherence bandwidth, BC, describe the time dispersion of the channel in a short time window. However, they do not give information about the time-varying nature of the channel.

Doppler spread BD and coherence time TC describe the effect of time-varying nature of the channel on the received signal.

Doppler spread BD is a measure of the spectral broadening caused by the time rate of change of the mobile radio

22

y gchannel:

If the baseband signal bandwidth is much greater than BD, then the effects of the Doppler spread are negligible.

v

fB DD max

12

Coherence time is a statistical measure of time duration over which the channel impulse response is essentially invariant:


p p y

If is defined as the time over which the time correlation function is above 0.5, then

Example 4.5

A measurement team traveling at 50 m/s uses a 900MHz

max

11

DDc fB

T

max16

9

Dc f

T

23

gcarrier to estimate the small-scale propagation parameter for an urban environment: (a) What is the Doppler spread BD for the mobile channel, (b) What is the coherence time Tc for the mobile channel.

Ans: BD =(3x50)Hz, Tc = 3/(50x16) sec.

Depending on the relation between the signal parameters (such as bandwidth, symbol period, etc) and the propagation

4 – Types of small-scale fading

( , y p , ) p p gchannel parameters (such as RMS delay spread and Doppler spread), transmitted signals will undergo different types of small-scale fading.

Basically there are two fading mechanisms, one independent of the other:

Small-scale fading based on multipath delay spread

24

Small-scale fading based on Doppler spread

13

Small-scale fading based on multipath delay spread:


Time dispersion of the multipath causes the transmitted signal to undergo either flat or frequency selective fading;

Flat fading:Signal BW (Bs) < Channel BW (Bc)Symbol period (Ts) > Delay spread (στ)

[note: if one is satisfied, the other will be satisfied as well]

25

Frequency selective fading:Signal BW > Channel BWSymbol period < Delay spread

Illustration of flat fading:

)(tx ),( th)(ty

0 sT

)(tx ),( th

0 max t0 maxsT

)(*),()( txthty

t

sTmax

X(f) H(f) Y(f)= H(f).X(f)

26

cff

cff

cff

X(f) H(f) Y(f) H(f).X(f)

sc BB

14

Illustration of frequency selective fading:

)(tx ),( th)(ty

0 sT

)(tx ),( th

0 max t0 maxsT

)(ty

t

sTmax

X(f) H(f)

27

cff

cff

cff

X(f) H(f) Y(f)sc BB

Small-scale fading based on Doppler spread:

F di i d t D l d B ( l ll d


Frequency dispersion due to Doppler spread BD, (also called time selective fading since ), leads to signal distortion (either slow or fast fading):

Fast fading:Symbol period (Ts) > Coherence time (Tc)Large Doppler spread

Channel variations faster than base-band signal variations

Sl f di

Dc B

T1

28

Slow fading:Symbol period < Coherence timeSmall Doppler spread

Channel variations slower than base-band signal variations

15

Rayleigh fading model:

Baseband received signal can be written as y(t)= yI(t)+ jyQ(t).

5 – Statistical Models

Q

The I-Q (In-Phase and Quad.) components of the received signal follow zero-mean Gaussian distribution. Thus the envelope of the received signal can be modeled as a zero-mean Rayleigh random variable, when LOS is absent.

Prob. density function (PDF):

0.4

0.5

0.6

0.7)(rp

10|)(|,)2

exp()(2

2

2 tyr

rrrp

29

The cum. distribution function (cdf),is the probability that received signal envelope does not exceed a value R:

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

0

0.1

0.2

0.3

r

2

2

0 2exp1 )(Pr

R

drrpRrR

Rayleigh fading model:

Baseband received signal power is given b ( ) | ( )|2


by Prec(t)=|y(t)|2. Thus the received signal power for a

Rayleigh fading model, is exponentially distributed with mean E[Prec(t)]=2σ2.

Prob. density function (PDF) of received power:

0|)(|,)2

exp(2

1)(

222 tyrx

xpr

t

rec

P

P

30

Note that E[Prec(t)] is the received signal power based on path loss and shadowing alone, while the pdf above models the random variations around E[Prec(t)], based on small-scale fading (see illustrations).

)log(d

16

Example 4.6

Gi th t th i d i l lit d i l


Given that the received signal amplitude r, over a wireless channel, is a Rayleigh distributed random variable. Compute the mean value, and the variance of r. Calculate also the median value, what does this signify?.

Ans:

0 2

)(][drrrprErmean

]2[])[()var( 222meanmeanmean rrrrErrEr

2

31

Median value represents the value of r for which the cdf is 0.5

Solve (for Rayleigh ch.)

22

0

2 4292.0)2

2(2

)( drrpr

, )(5.00

medianr

drrp 177.1 medianr

Example 4.6b

C id h l ith R l i h f di l d


Consider a channel with Rayleigh fading envelope and average received power E[Prec(t)]=20dBm. Find the probability that the received power is below 10dBm.

Ans: note that E[Prec(t)]=20dBm=100mW. We want to find the prob. that r 2 < 10dBm =10mW.

,095.0)100

exp(100

1]10Pr[

102 dx

xr

32

,)100

p(100

][0

17

Ricean fading model:

Baseband received signal can be written as y(t)= yI(t)+ jyQ(t).


Q

The I-Q (In-Phase and Quad.) components of the received signal follow Gaussian distribution with E[y(t)]=A , when LOS is present. Thus the envelope of the received signal can be modeled as a Ricean random variable with E[y(t)]=A, when LOS is present.

Prob. density function (PDF):

0,)()2

)(exp()(

202

2

2

r

ArI

Arrrp

)(rp

0.4

0.5

0.6

1,1 A

33

A= amplitude of the LOS signal, I0 =Bessel function of the first kind ,

of order 0. Becomes a Rayleigh fading model

if A=0 (no LOS). Rice factor: dB

00.5 1 1.5 2 2.5 3 3.5 4 4.5

0

0.1

0.2

0.3

r

2

2

2log10

A

K

Ricean fading model:

The average received power in Ricean fading model is given by


Where 2σ2 is the average power in the non-LOS multipath components, and A2 is the power in the LOS component.

Thus the Rice factor K is the ratio of power in the LOS to the power in the non-LOS components.

22

0

22 2)(][

AdrrprrEPrec

)2( 222

22 PA

34

Note:

Using these relations, the Ricean pdf can also be written as:

)1/(2 and,)1(

.)1

2(

)2(2 2

2

222

KP

K

PK

A

AKKA rec

rec

0,))1(

2())1(

exp)1(2

)( 0

2

r

P

KKrI

P

rKK

P

Krrp

recrecrec

18

Nakagami fading model:

Gives more generalized fading statistics than Rayleigh and Ricean


g g y gmodels. Practical measurements show that Nakagami distribution models signal received over mobile channels, more accurately.

The Nakagami pdf is given by

is the gamma function, and

0

1 exp dyyym m

0 ,5.0,exp)(

2)(

2)12(

rmP

mrr

P

m

mrp

rec

m

m

rec

][ 2rEPrec

35

m is called the Nakagami fading parameter (usually m ≥0.5) m =1, Nakagami fading becomes Rayleigh fading model. m ≥2, Less severe fading can be modeled. As m → ∞, non-fading or AWGN channel, can be modeled. For m=(K+1)2/(2K+1), Ricean fading with factor K can be modeled.

Nakagami fading model:

Thus the Nakagami distribution can model Rayleigh, Ricean, AWGN


g y g , ,(no fading), and more general fading in more or less severer than Rayleigh and Ricean. Thus measurements in different environments can fit the Nakagami distribution by choosing appropriate m.

The power distribution for Nakagami fading, obtained by making change of variable x=r2 in above pdf, is given by

050)(1

mxxm mm

36

0 ,5.0,exp)(

)(

xmPmP

xprecrec

19

5 – Level Crossing and Average Fade durations

• Two important fading parameters are the average # of level crossings or fades and the duration of these gfades.

Level crossing rate (LCR): the expected rate at which the Rayleigh fading envelope crosses a specified level R, when envelope is in a positive going direction.(i.e. r1 – r2 = +ve).

The number of level crossing per second is given by

)exp(2)( 2

frdrRprN

37

Where , and is the joint pdf of at r =R. Also , where Rrms is the local rms amplitude of the fading envelope [Ref 5].

Note also: where P0 is a target received power (receiver sensitivity).

)exp(2),(0

max DR frdrRprN

rr of slope theis ),( rRp rr and rmsRR /

recPP /0


• Example: Rayleigh fading simulation, fc=900MHz, and Mobile Speed = 120km/hrp

l id2/

38

ncorrelatio-de2/

Exercise: at threshold level -20dB about RMS, estimate NR for this simulation.

20


• Example (5.7 in Ref1): For a Rayleigh fading signal, compute the positive-going level crossing rate for =1, p p g g g when the maximum Doppler frequency (fm) is 20Hz. What is the maximum velocity of the mobile for this Doppler frequency if the carrier frequency is 900MHz?.

Ans:

crossings per sec. The maximum velocity of the mobile can be obtained

using the Doppler relation, .

44.18)1exp()1)(20(2 2 RN

/max

vf D

39

Thus v =20Hz(1/3m)=6.66m/s=24 km/hr.max


Average fade duration (AFD): is the average period of time for which the received signal is below a of time for which the received signal is below a specified level R. For a Rayleigh fading signal, this is given by

The average fade duration helps to determine the likely number of signaling bits that may be lost when fading occurs

2

}1){exp(]Pr[

1

max

2

DRAFD

fRr

N

40

fading occurs.

21


• Example (5.9 in Ref1): (a) Find the average fade durationfor a threshold level of =0.707 when the Doppler frequencyi 20H F bi di it l d l ti ith d t t f 50is 20Hz. For a binary digital modulation with data rate of 50bps, is this Rayleigh fading scenario slow or fast?. (b) What isthe average number of bit errors per second for the givendata rate for the case = 0.1 [assuming that a bit erroroccurs whenever any portion of a bit encounters fading]?.

Ans: (a)

For Rb=50bps, Tb= 20ms, which is > . Thus the signalundergoes fast Rayleigh fading.

ms 3.182)20)(707.0(

1)707.0exp( 2

FD

AFD

41

u de goes ast ay e g ad g(b) For = 0.1, we have . This is less thanthe duration of one bit. Therefore, only one bit on average willbe lost when a fading event occur.

For = 0.1, crossings per second. Thus there aretotal of 5 bits in error per sec., resulting in BER=(5/50)=0.1

ms2sec002.0 AFD

96.4RN

Frequency Flat /Freq selective fading models:

Two-ray frequency-selective fading model: Models

5 – Fading Channel Simulations

the effect of multipath delay spread as well as fading.

Delay

Input(transmittedsignal)

Output(receivedsignal)

11 exp j

42

The impulse response of the channel model is

y 22 exp j

tjtjth exp exp, 2211

22

Two-ray Rayleigh fading model:

α1 and α2 are independent and Rayleigh distributed.


Input(transmitted

Output(received

11 exp j

p y g

θ1 and θ2 are independent and uniformly distributed over [0,2π].

τ is the time delay between the two rays.

Setting α = 0, a flat Rayleigh fading channel is obtained.

By varying τ, it is

43

Delay signal) signal)

22 exp j

y y gpossible to create a wide range of frequency selective fading effects.

N-ray Rayleigh fading model:

Example 4.7


p

Develop an (N+1) -ray mobile channel simulator as shown below:

Input(transmittedsignal)

Output(receivedsignal)

11 exp j

1

44

Delay 1 22 exp j

Delay N NN j exp

1

N

23

Clarke’s model for flat fading (freq domain):

While stat. models (Rayleigh, Ricean, etc.) directly predict received signal envelope in time domain Clarke’s model predicts


received signal envelope in time domain, Clarke’s model predicts the power spectrum of the received signal first in frequency domain. This is then used to produce a time-domain waveform.

The model is widely used to model narrow-band channel in the IS-54 (US digital cellular system).

Given that the source, S, transmits a continuous wave signal of frequency fc.

Source, S (Base Station)

45

Then the instantaneous frequency of the received signal component arriving at an angle θ at the mobile is:

v(mobile)

cDc fffv

f

coscosmax

Clarke’s model for flat fading:

Thus,

ff


Clarke’s model uses statistical characteristic of electromagnetic waves to show that the received power spectral density (PSD), S(f), is proportional to [Ref3].

Therefore,

max

1cosD

c

f

ff

f

1 1cos

c

f

ff

ff

fS Doppler power Spectrum

fcfc -fDmax fc +fDmax

46

i.e.,

Ref1, pages 214-218, show that K = 1.5/fDmax .

2

max

max

1

D

cD

f

fffff

fS

2

max

1

)(

D

c

f

ff

KfS

24

Simulation of Clarke’s fading model using Doppler filter:

Baseband received signal: y(t)= yI(t)+ jyQ(t), and r (t)=|y(t)|.


Q

Generate two independent, random complex Gaussian source.

Then use spectral filter defined by S(f) in the previous slide, to shape the random signals in frequency domain.

Time domain waveform of the resulting fading generated can be produced by using an inverse fast Fourier transform (IFFT) at the last stage of the simulator.

Baseband BasebandMixer

47

Gaussian

Noise Source

Doppler

Filter

Baseband

Gaussian

Noise Source

Baseband

Doppler

Filter

Ind

epen

den

t

Cos(2fct)

Sin(2fct)

Mixer

To IFFT

Note: for baseband waveform simulations, mixers will be omitted.

Simulation of Clarke’s fading model using Doppler filter:

Ref 4 provide steps for computer program to implements this:


1. Specify # of frequency points N to represent , and the . Usually N is to power of 2, i.e. N=2k , where k=1,2,3,…

2. Calculate frequency spacing between adjacent spectral lines as f=2fDmax /(N-1). Time duration of the fading waveform is T=1/f.

3. Generate complex Gaussian random variables for each of the N/2 positive frequency component of the noise source.

4 C t t th ti f t f th i

)( fS

maxDf

48

4. Construct the negative frequency components of the noise source by conjugating positive frequency values & assigning these at the negative frequency values.

5. Multiply the in-phase and quadrature noise sources by the fading spectrum . (Freq domain operation ends here))( fS

25


Simulation of Clarke’s fading model:

6. Perform an IFFT on the resulting frequency 6. Perform an IFFT on the resulting frequency domain signals from the I-Q parts to get two N-length time series, & add the squares of each signal point in time to create an N-point time series.

7. Take the square root of the sum obtained in step 6 to obtain an N-point time series of a simulated Rayleigh fading signal, to model the expression: r (t)=|y(t)|=sqrt(yI(t)2+ yQ(t)2), with proper Doppler

d d ti l ti

49

spread and time correlation.

By making a freq component dominant in amplitude within , and at f=0, the fading is changed from Rayleigh to Ricean.

)( fS


Block diagram:


50

26


Example: Fig 5.15 in [Ref 1, 6] – typical Rayleigh fading envelope at 900MHz and 120km /hr


at 900MHz, and v =120km /hr.

51

Other models: Jake’s model [Ref 5], Young and Beaulieu’s method for computational efficient Rayleigh fading sim.[Ref7].

Jakes’ fading model [5]:

Clarke’s model (time domain): defines the complex channel gain for non LOS frequency flat and 2 D isotropic scattering


gain, for non-LOS, frequency flat, and 2-D isotropic scattering assumptions as [3]

N= number of multipaths, nUniform(-,), and nUniform(-,) are the phase and amplitudes of the nth multipath component.

Jakes’ model: the high degree of randomness in equation above

N

n

tfj nnDeN

th1

])cos(2[max

2)(

52

is not desirable for efficient simulation. Thus, Jakes proposed the following sum of sinusoid model to simulate h(t), & is widely used.

M

nDDI t

M

nf

M

ntfth

1max ]

24

2cos[2cos)

2cos(2)2cos(2)(

max

M

nDQ t

M

nf

M

nth

1

]24

2cos[2cos)

2sin(2)(

max

27

Jake’s fading model [5]:


0 4

0.6se

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08-0.2

0

0.2

0.4

Time

Imp

ulse

Res

pon

s

1

1.5x 10

-3

53

Other recent models: Zheng et al. [8], Young and Beaulieu’s method for computational efficient Rayleigh fading sim.[Ref7].

-800 -600 -400 -200 0 200 400 600 8000

0.5

Frequency

PS

D

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Ed., Prentice HallPTR, Upper Saddle River, NJ 07458, USA.2 A Goldsmith “Wireless Communications ” Cambridge University Press New2. A. Goldsmith, Wireless Communications, Cambridge University Press, New

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