Mobile Radio Channels

8/12/2019 Mobile Radio Channels

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3. The Mobile Radio Channels

One of the distinguishing features of mobile communications is the channelbehavior, of which most is unwelcome. Well look at these :

1.path loss analogous to inverse square law, but worst, much worse.Inverse fourth power ??

2.Shadowing variation due to obstacles blocking the path (makeshandoff difficult problem).

3. fading tens of dB and phase reversal in a fraction of a second. Very

challenging, very destructive.

We will elaborate on fading :

- two phenomena; fading in time, dispersion (delay spread, frequency

selective fading). Keep them separate in your mind !

- differences between mobile and base

- fading spectra, fade rate, delay profile.

3.1 Path Loss

References :

1.Steele, R, Mobile Radio Communications, Pentech Press, 1992 ;TK6570; [1.2.5,2.7]


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2.Lee, William C. Y., Mobile Communications Engineering : Theory andApplications, McGraw Hill, 1998, 2

ndEdition; TK6570; [3.1-3.4]

Free space inverse square law is optimistic. Even in the absence of localscatterers and obstacles, there are large geometry objects with reflections.

Consider the following simple propagation model. Let s(t) be thetransmitted complex envelop of the transmitted bandpass signal

{ }2( ) Re ( ) cj f ts t s t e =! . The received bandpass signal (ignoring free space path

loss for the time being) can be written as :

( ){ }

( ){ }

( ){ }

2 ( / )

( )

( , ) Re /

Re /

Re /

c

c

c

j f t x c

j t x

j tjx

r t x s t x c e

s t x c e

s t x c e e

=

=

=

!

(plane wave)

wherexis the distance from the transmitter to the receiver (in the direction

of propagation), cis the speed of light, 2c cf = is the carrier frequency in

rad/s, 2 / = , and / cc f= is the wavelength.

Let ah and mh be the height of the transmitting and receiving antennae

respectively, and dthe horizontal distance between them.

mh

ah

d


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Any phase difference between the direct and reflected rays leads to partial

cancellation in addition to the21/d free space loss.

Direct path length :2

2 2 2 2 12( ) 1 ( ) / 1 a md a m a m h hx d h h d h h d d

d

! "# $= + = + +% &' (% &) *+ ,

Reflected path length :

2

2 2 12

( ) 1 a mr a mh h

x d h h dd

! "+# $= + + +% &' (% &) *+ ,

Differential path length : 2 /r d a mx x x h h d = =

Sum of arrivals (note inverse square and a reflection coefficient of 1) :

( ) ( )

( ) ( )

1 1( ) / /

1

/ 1

d r

d

jx jx

d r

jx j x

d

sum t s t x c e s t x c ed d

s t x c e ed

=

since ( ) ( )/ /d rs t x c s t x c (narrowband processes, time scale >> /x c )

Received power is proportional to :

( )2 2

12 2

2 2

2

2

2

1( ) ( / ) 1 cos( )

2 for 1 (far away)2

4

d

o

a mo

E sum t E s t x c xd

P xx

d

h hP

d

# $ # $= ) * ) *

=

! "= % &

+ ,

"


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that is the inverse fourth power of the distance.

Example : Consider the case 30 , 5 , ' / a mh h d d = = = . The plot of

the normalized power as a function of 'd is as shown below. There are2 /mh maxima and the farthest one is at ' 4( / )( / )a md h h = . The inverse

fourth power phenomenon kicks in after the last peak.

Reality is not a plane earth obstacles, terrain variations, etc introducediffraction, reflections. However, the exponent is usually between 3 and 4

when averaged over a large ensemble of real configurations [M. Hata,

Empirical Forms for Propagation Loss in Land Mobile Radio, IEEE

Trans. On Vehicular Technology, Aug 1980]

Consequences of faster than inverse square

bad news

- need a powerful transmitter for reasonable range- accentuated near-far problem


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- n increase in transmit power doesnt give a proportionalincrease in coverage area.

Edge of coverage defined by specific received power level, so

1 1 2 2/ /k kP r P r = . This means

2 2/

2 1 2 1 2 1/ ( / ) ( / ) k

A A r r P P= =

Good news

- coverage areas are relatively sharply defined which is good forcellular layout

3.2 Shadowing

Hills, large buildings cause variations in the received power indicated bypath loss. These obstacles prevent the existence of a direct line of sight path

between the transmitter and the receiver. When the receiver is in a shadow,

the signal strength will be weak.

Shadowing can also be caused by foliage attenuation and precipitation.This is especially true when transmitting at super high frequency (tens of

GHz).

Shadowing is particular problematic for inbound (mobile to base) signalswhere the transmitted power is low.

/ k

r tP P r=


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One remedy is to use auxiliary (or satellite) antennas, usually for inboundtransmission. Why not outbound (i.e. simulcast)? Simultaneous

transmission on same frequency from different sites gives destructive

interference in some places if exactly the same frequency; beats if not

exactly the same; garbling or ISI if envelopes not matched in time.

Shadowing variation is usually log-normal. That means if you take the dBequivalent of the received signal strength, it will be a Gaussian random

variable. A typical variation is 6-8 dB.

3.3 Slow Fading

Here we examine the fine structure of the received signal for mobilechannels. Discussion is limited to the pdf (time variations in a later

section), so it applies to very slow fading (large , slow or motionless

mobile) so that the fade rate df " modulation bandwidth.

3.3.1 Fading Mechanism


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Receive superposition of several reflections, each with own path length

from transmitter ix and complex reflection coefficient ia (magnitude and

phase change).

Average path length :1

1 N

i

i

x x

N == -

Differential path length : i ix x x =

Received signal :

{ }

{ }

( )

( )

( , ) Re ( / )

Re ( / )

Re ( [ ] / )

c i

i c

i c

j t x

i i

i

j x j t

i i

i

j x j t x

i i

i

r t x a s t x c e

a s t x c e e

a s t x x c e e

# $= ) *

# $=

) *# $= + ) *

-

-

-

!

Assume that modulation s(t) changes slowly enough to be unaffected by

differential delays /ix c , i.e. / 1iB x c " . Then

( )

( )

( , ) Re ( / )

Re ( / )

i c

c

j x j t x

i

i

j t x

r t x a e s t x c e

g s t x c e

# $! " ' (% &

+ ,) *

# $= ) *

-!

where g is a complex gain, constant at that point in space. As vehicle

moves, set of different path lengths change, get time varying g(t).

The last equation depicts the phenomenon of flat fading. We examine nextthe first order statistics of g, the complex gain.


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3.3.2 Gaussian Model Flat Fading

Examine the first order statistics of g, the complex gain.

Apply central limit theorem to real and imaginary part ofij x

i I Q

i

g a e g jg

= = +- and it becomes Gaussian :

2

2 2

1 | |( ) exp

2 2g

gp g

! "= % &

+ ,where

( )2

2 2 2 21 12 2g I QE g E g E g # $ # $ # $ = = +) * ) *) *

If you plot the pdf ( )gp g as a function of Ig and Qg , you see circular

symmetry.

In polar co-ordinate,j

I Qg g jg re

= + = , where | |r g= , 0 r < and

( )arctan / Q Ig g = , . Can show that

Ig

Qg

constant prob.contour


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2 2/ 2

, 2

1( , )

2

r

r

rp r e

! "= % &

+ ,Consequently

2 2/ 2

2( ) 0rr

rp r e r

=

(Rayleigh)1

( )2

p

=

Clearly r, are independent.

The received power is proportional to 2z r= , which is exponentiallydistributed. Follows from change of variables in Rayleigh, or from

2 2 2

I Qr g g= + (sum of independent squared Gaussian is2 and 2 with 2

degrees of freedom is exponential with mean22 ). In other word

2/ 2

2

1( ) 0

2

z

zp z e z

=

How good is the Gaussian approximation ? As shown in the diagrambelow, even 6 to 10 components is a close approximation for

/ 2r r = , except for the tails.


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The distribution of the resultant of Nunit vectors with random relativephase. The x-component of each point represents ( )Pr /a N y> where ais

the magnitude of the sum and y is the corresponding value on the y-axis.

Mobile satellite with a line of sight component (LOS) is better modeled byRice fading. In this case, the complex gain g becomes

s dg g g= +

2 212

| | 2 | |s s sg K P g K = . = =

2 212 d dE g P# $ = =

) *sg

dg

g


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where sg is the specular (LOS) component and dg is the diffuse

component.

K represents the power ratio in the specular and diffuse components. The

total power is

2 2 2 21 1 1

2 2 2| | | | | | (1 )s dP E g g E g K # $ # $= = + = +) * ) * .

Since ghas a non zero mean of sg , it means2| |z g= is non central

2 with

2 degree of freedom, or the pdf of 2| |r z g= = is Ricean :

2

2 2

2( ) exp

2r o

r r r K p r K I

! "! "= % &% & % &

+ , + ,

and phase [see also Eqn 5-2-55 of Proakis]

( )2

cos ( )1( ) 1 4 cos( ) [1 ( 2 cos )]2

K Kp e K e Q K

= +


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Note that2 itself varies with the lognormal distribution determined by

shadowing. The Kfactor depends on terrain.

3.4 Fading Autocorrelation Function and Power Spectrum

We need second order statistics of the complex gain process if we are todescribe its time variation. Its a Gaussian process, so thats all we need. If

second order statistics are constant in time, than it is WSS.

3.4.1 Power Spectrum and Autocorrelation at Mobile

References:

1.Jakes, William C. Jr., Microwave Mobile Communications, Wiley,1974; TK6570; [Chapter 1].

2. Lee [7.1]

Consider the case we transmit an unmodulated carrier and a movingreceiver. Consequences :

1.The received complex envelop is the fading gain g. Since thereceiver is moving, gchanges with time, i.e. a random process g(t).

2.When there is a relative motion between the transmitter andreceiver, there exists a Doppler frequency.

How do we relate the two ?


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Assume the mobile is traveling through a set of scatterers at a velocity of V.

(Top View)

The scatterer at an angle of to the direction of motion introduces a

Doppler shift of

cos( ) cos( )dV

f f

= =

to the unmodulated carrier, where /df V = is the maximum Dopplerfrequency.

The Doppler shift is the largest (i.e. df ) when the mobile is traveling

towards the scatterer and the smallest (i.e. df ) when away from the

scatterer

Conclude :

- the transmission of an unmodulated tone produces multiple tones withdifferent frequencies because of the scatterers and the relative motion.


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- The power distribution, as a function of the Doppler frequency shift, isthe power spectral density of the received complex envelop, which as

mentioned earlier, is the channel complex gain process g(t).

- The inverse Fourier transform of the PSD gives the autocorrelation

function.

Note :

1.Both and contribute to the same Doppler shift.

2.Since cos( )df f = ,21 ( / )

d ddf f f f d =

3.Let ( )p be the (raw) received power density coming from the

direction . For isotropic scattering, this is a constant.

4. ( )G be the antenna gain pattern; assume the antenna is pointing at

an angle of with respect to the direction of motion. For a vertical

polarized antenna,( )G

is a constant.

V


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Based on the above, we can conclude that the power arriving in dat is

proportional to2 ( ) ( )G p d . So power in df at f is

{ }2 22

( ) ( ) ( ) ( ) ( )

1 ( / )g

d d

dfS f df G p G p

f f f

+

#

For isotropic scattering and vertical polarized antenna,

2

2 2| |

( )

0 otherwise

d

g d

f fS f f f

/


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*12

2 2

2

( ) ( ) ( ) IFT ( )

(2 ) 2

2

g g

o d o

o

R E g t g t S f

VJ f J

xJ

# $ # $= = ) *) *

! "= = % &

+ ,

! "=

% &+ ,

A plot of the autocorrelation function versus the normalized

(to the wavelength) distance of travel

From the autocorrelation function plot, we see that points separated by 0.4

are uncorrelated; however oscillations die slowly, so, in principle,

correlation lasts for many wavelengths

Exercise: Write a computer program to simulate Rayleigh fading with aJakes spectrum.

As expected from the autocorrelation function, signal magnitude is quasi-periodic in space, with a dip every / 2 to or so.

From the complex envelope trajectory, we see that fades are associated

with rapid phase change and rapid fractional amplitude change.


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0 2 4 6 8 10 12 14 16 18 20-60

-50

-40

-30

-20

-10

0

10

20

x/lambda

|g(x)|indB

Magnitude of g(x) in dB scale. The horizontal line denotes the

root-mean value in dB.

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Real[g(x)]

Imag[g(x)]

Complex envelop trajectory. The point near (0,0) corresponds to

the deep fade at 6x = in the amplitude plot. Separation in spacebetween successive circles is /10 .


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In summary, the received signal in a flat fading channel is

( ) ( ) ( )r t g t s t =

where( )g t

is a zero mean complex Gaussian process with anautocorrelation function of

2( ) (2 )g o dR J f = .

If the transmitted complex envelop is ( )( ) exp 2 os t j f t = , the received

complex envelop is ( )( ) ( )exp 2 or t g t j f t = . In other word, ( )g t is the

frequency response of the channel at time tand frequency of . Since ( )g t is

actually independent of of , the instantaneous frequency response at any

time tis flat, hence the term flat fading; see the figure below.


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3.4.2 Autocorrelation at the Base

The channel is reciprocal in an electromagnetic sense. This means theautocorrelation function of the complex channel gain is the same at the

base as it is at the mobile.

The spatial correlation at the base, however, is considerably different - farless decorrelation when base is moved than when mobile moves. Base

station antennas need much wider spacing for diversity (at mobile, only

half a wavelength or so is enough for decorrelation).

Source of difference is geometry : mobile is surrounded by scatterers, basehas signal from narrow spatial angle.

Jakes derives a detailed model based on a ring of scatterers about themobile, none near the base. It agrees with experiment only in a general

sense, and it is immediately thrown off by scatterers near the base.

Use of antenna arrays at the base is currently a very active research topic,and better models for correlation among the antennas are needed.


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3.5 Multipath Spread and Correlation Bandwidth

References : Lee [1.5, 1.6], Steele [Chapter 2].

Returning to Section 3.3, we now examine the case of large differentialdelays amongst the different paths. By large, we mean the differential

delays are large with respect to signal variations in time.

Rewrite the received signal as

{ }

( )

( ) Re ( / )

Re ( )

Re ( )

c i

i c

c

j t x

i i

i

j x j t

i i

i

j t

i i

i

r t a s t x c e

a s t e e

h s t e

# $= ) *

# $= ) *

# $! "= ' (% &

+ ,) *

--

-

!

where /i ix c = is the delay associated with the i-th path, andi c ij x j

i i ih a e a e

= = .

The complex envelop of the received signal is

( ) ( )i ii

r t h s t = -

implying a (baseband equivalent) channel with an impulse response of

( ) ( )i i

i

h t h t = -and a frequency response of


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2( ) i

j f

i

i

H f h e

= -

Since we assume the differential delays are large with respect to signal

variations in time, this means ( )H f has significant variation across the

signal band (frequency selective).

Alternative description for the channel :

( ) ( )i ii

h h = -

is defined as the response of the channel at the present moment due to animpulse applied second earlier. This means when the input to the channelis ( )s t , the output is

( ) ( ) ( ) ( )i ii

r t h s t d h s t = = -3

Multipath spread (or delay spread), d , is the range of i over which thereis significant energy, that is significant values for

2| |ih .

Correlation bandwidth (coherence bandwidth) is the frequency separationover which there is a significant change in ( )H f . It is the width of

*12

( ) ( )H f H f f df 3 or*1

2( ) ( )E H f H f f# $ ) * (loose definition).

A rough approximation is: coherence bandwidth 1/ d# , since fine structure

width in one domain (frequency) is roughlythe reciprocal of the width in

the other (time). Little point in being more precise.


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Example : Two path model : 1 2( ) ( ) ( )dh h h = + . The transfer

function of the channel is2

1 2( ) dj fH f h h e

= + . Assume 1h and 2h are

drawn from an ensemble, the frequency correlation function becomes

*1

2

222 2

1 2

1

( ) ( ) ( )

1 d

h

j f

S f E H f H f f

e

# $ = ) *! "

= +% &+ ,

Caution: ( )hS f is NOT the transform of autocorrelation of gain.

Now a general model is

2( ) ( ) j fH f h e d = 3and

( )( )2 * 2 ( )

* 2 2 ( )

* 2 2 ( )

1( ) ( ) ( )

2

1 ( ) ( )

21 [ ( ) ( )]

2

j f j f f s

h

j f j f f s

j f j f f s

S f E h e d h s e ds

E h h s e e d ds

E h h s e e d ds

# $ =) *

# $= ) *

=

3 3

3 3

3 3

If we assume uncorrelated scattering gains at different delays are

uncorrelated (wide-sense stationary uncorrelated scatterers WSSUS),

*12

( ) ( ) ( ) ( )E h h s G s # $ = ) *

where ( )G is the delay power profile. Then

2( ) ( ) j fh

S f G e d = 3 : frequency correlation function


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Root mean square (RMS) delay spread is a common measure

2

2( ) ( )

; = ( )( )

rms

G dG d

G d

=3

33

If the mobile moves, then the path length, hence phases change, so

coefficients ih in impulse response are time varying. Now it is

( , ) ( ) ( )i ii

h t h t = -

It describes the response, seen at time t, to an impulse seconds earlier.The tap-delay line model for the channel is shown below.

( ) ( , ) ( ) ( ) ( )i ii

r t h t s t d h t s t = = -3

Finally, lets take the transform in the input/output relation

( )r t

( )s t

( )r t

3( )h t2 ( )h t1( )h t1( )h t

1 3 2 2 1


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( )2 2

2

( ) ( , ) ( )

( ) ( , )

j ft j f

j ft

r t h t S f e df e d

S f H f t e df

=

=

3 3

3where

2( , ) ( , ) j fH f t h t e d = 3

is called the time variant transfer functionof the channel.

Example : If ( ) ( )oS f f f = , then ( )2

( ) , oj f t

or t H f t e

= . So ( ),oH f t must

be the frequency response of the channel at time t and frequency of . A

sample ( , )H f t is shown below.


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At this point, want to find out if ( , )h t is still Gaussian for any t and .Look at the complex envelop of the received signal again.

( ) ( ) ( )c ij

i i i i

i i

r t h s t a e s t = = - - (any given point in space)

If the time variation of s(t) is comparable to or less than the individual

delays (i.e very wideband), then a collection of separable paths, no

opportunity for central limit theorem. With lower resolving power, can

aggregate paths into bins { }: ( 1)kT k k = + and rewrite r(t) as :

( ) ( )kk

r t h s t k = -where

c i

i k

j

k i

T

h a e

=-

are still more or less Gaussian, but with less plausibility than flat fading. So

channel becomes less Gaussian as the resolving power (bandwidth) of the

signal increases.

In general, the gain kh is time dependent because of motion of mobile, just

like in the flat fading case.

Given that ( )kh t are plausibly Gaussian for any t, the time variant transfer

function

2( , ) ( ) j kfkk

H f t h t e = -

can be viewed as a Gaussian process inf, with autocorrelation function


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*12

* 2 ( )12

( ) ( , ) ( , )

( ) ( )

h

j kf f f

k

k

S f E H f t H f f t

E h t h t e +

# $ = ) *

# $= ) *-- $ $

$

$

It is reasonable to assume that complex gains in paths with resolvabledelays are independent, i.e.

*12

( ) ( )0 otherwise

k

k

G kE h t h t

=/# $ =1) *

2$

$

with

( ) ( )kk

G G k = -

being the delay power profile of the channel. Consequently

2 ( )( ) j k fh k

k

S f G e

= -

is simply the Fourier transform of the delay power profile, evaluated at f

A more general correlation function is :

*12

* 2 ( )12

2 ( )( )

( , ) ( , ) ( , )

( ) ( )

( )

( ) ( )

h

j kf f f

k

k

j k f

k G

k

G h

S f t E H f t H f f t t

E h t h t t e

G t e

t S f

+

# $ = ) *

# $= ) *

=

=

--

-

$ $

$

$

where


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*12

212

( ) ( )( )

| ( ) |

k k

G

k

E h t h t tt

E h t

# $ ) * =

# $) *

is the normalized autocorrelation function for any delay.

Conclusion : uncorrelated scattering is a convenient model since the joint

time-frequency correlation function is separable.

3.6 More on Flat Fading

As shown earlier, in the absence of a LOS component, fading in a mobileradio channel can be modeled as a zero mean complex Gaussian process

( )g t with an autocorrelation of2( ) (2 )g o dR J f = (the corresponding

fading spectrum is ( )2 2 2( )g dS f f f = , df f ). We now examine the

characteristics of this fading model.

3.6.1 Some important probability density functions

References:

1. S.O. Rice Mathematical Analysis of Random Noise, BSTJ Vol. 23, pp. 282-332,July 44, and BSTJ Vol. 24, pp. 46-156, Jan 45.

2. S.O. Rice, Statistical Properties of s Sine Wave Plus Random Noise, BSTJ, Vol.

27, pp. 109-157, Jan 48.

3. W.B. Davenport, W.L. Root, An Introduction to the Theory of Random Signalsand Noise, McGraw Hill, 58.


28/403-28

The probability densities functions given below are general in the sensethat they are applicable to any fading spectrum ( )gS f with even symmetry.

Exercise : Consider the fading process

2( ) ( )o

j f th t e g t

=

, where g(t) is afading process with a Jakes spectrum. Show that the PSD of h(t) is not

symmetrical aboutf = 0.

Note : the above represents the scenario where there is a frequency offset

between the oscillators in the transmitter and receiver, in addition to fading.

First order( )

( ) ( ) ( ) ( ) j tI Qg t g t jg t a t e

= + = :

2

2 2

1 | |( ) exp

2 2g

gp g

! "= % &

+ ,

2 2/ 2

, 2

1( , )

2

a

a

ap a e

! "= % &

+ ,

Derivative fading process( )( ) ( ) ( ) ( ) ( ) j td I Qdtg t g t g t jg t b t e

= = + =

% % %:

2

2 2

1 | |( ) exp

2 2g

gp g

! "= % &

+ ,%

%%

where

( )

22 2

2 rmsf = (variance of derivative process)

2

2( )

( )

g

rms

g

f S f dff

S f df =3

3 (mean square Doppler spread)


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Proof:

The derivative process ( )g t%

can be obtained by passing g(t) through a

linear filter with frequency response 2j f . The output process is

Gaussian and has a PSD of ( )2

( ) 2 ( )g gS f f S f =%

Exercise : Show that 2 2/ 2rms d f f= for the Jakes spectrum.

Joint pdf of g(t)and ( )g t%

:

( )( )

2 2

, 2 22 21 | | | |( , ) exp

2 22 2g g g gp g g

! "# $= +% &' (

) *+ ,%

%

%

Proof:

The processes g(t) and ( )g t%

are jointly Gaussian. Their cross-correlation

can be written as :

* **1 12 2

0

* *

0

0

( ) ( )( ) ( ) ( ) lim ( )

1 ( ) ( ) ( ) ( ) lim

2

( ) ( ) lim ( )

ggt

t

g g

gt

g t g t t R E g t g t E g tt

g t g t g t g t t E

t

R R tR

t

# $ # $= = ' () *) *

# $ = ' (

) *

+ / 4= = 1 5

2 6

%%

%

Therefore [ ]12 ( ) ( ) (0)gE g t g t R= %%

. For Jakes spectrum, the derivative ofthe autocorrelation function at 0 = is 0.

Conclude : g(t) and ( )g t%

are independent when evaluated at the same

time instants. Therefore , ( , ) ( ) ( )g g g gp g g p g p g=% %% %

, which is basically

what the equations says.


30/403-30

Exercise : Show that [ ]12 ( ) ( ) 0E g t g t =%

is true for any ( )gS f with even

symmetry.

Joint PDF, polar co-ordinate :

Since ,j j j j

g ae g be ae j ae = = = +

%% %, we have

2 2 2 2 2g b a a= = +

%% %

This information can be used (see [Rice]) to show that

2 2 2 2 2

, , , 2 2 2 2 2

1( , , , ) exp

4 2a a

a a a ap a a

/ 4! "+0 0= +1 5% &

0 0+ ,2 6%%

% %%%

where a range from 0 to , range from to , and both a%

and %

range from to .

Because of the form of this joint pdf, the marginal pdfs can be easily

obtained.

Why are these probability density functions of importance? Well, they aredirectly related to performance. Will examine two here first :

1. random FM2. fade rate and fade duration

More to come later.

3.6.2 Random FM


31/40


32/403-32

This is a long-tailed pdf where2 2 3( ) 2 / rmsp f %

% %for large

%. So even

though ( )t is bounded, its derivative is not.

It is quite non-Gaussian

Example: A 4800bR = bps FSK modem, modulation index h=1/2, detected

with discriminator and no post-filter, transmit at 800 MHz, vehicle at 100

km/hr, what is the irreducible error rate ?

- A modulation index of means ( )( ) exp 2s t j f t = where

/ 2 1200bf hR = = .

- Doppler frequency : / / 74d cf v vf c= = = Hz.

- Make an error if random FM exceeds signal. If the negative frequencywas sent, then the irreducible error probability is

[ ]

( )

( )

2

2

212

Pr 2 ( )

1 1 1

2 / 1

1 1 1 for Jakes spectrum

2 / 1

eo

f

rms

d

P f p d

f f

f f

= > =

! "% &= % &

++ ,

! "% &= % &

+

+ ,

3 %

% % %

When / 1df f


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3.6.3 Frequency and Duration of Fades

The big questions in relating fading to system designs :

- what type of burst error correcting code ?

- and is burst error correction useful, anyway ?- length of forward error correcting code (FEC) ?- depth of interleaving to break up bursts ?- in antenna switched diversity, how much delay is tolerable ?

Answers to some can be provided by analysis. Others through

measurements and/or simulations.

Preliminary observations

- Let2

( ) ( )z t g t= . As shown in Section 3.3.2,2/ 2 2

( ) (2 )z

zp z e = . Thus

[ ]2/(2 )

Pr 1 oz

oz z e

=

- fades are of varying depths, and instantaneous system performance

(like bit error probability, output signal-to-noise ratio) may berelatively soft function of depth.

- But we will think in terms of threshold, since useful for codedsystems, antenna switching etc., and some applicability to uncoded

systems

- Define threshold level with respect to the mean (rms) level. Then atlower levels :

1. less time spent in fade state2. fewer fades3.shorter fades


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

- 5 0

- 4 0

- 3 0

- 2 0

- 1 0

0

1 0

2 0

x / l a m b d a

|g(x)|indB

Fade Rate:

- Refer to [Lee 2.7, 6.4], [Rice 1948]

- This is a level crossing rate (LCR). For a given amplitude threshold A,the number of upward crossings per second is

,( ) ( , )a ao

n A a p A a da

= 3 %% % %

where , ( , )a ap a a%%

is the joint pdf of the fading envelop ( )a g z= = and

its derivative. The proof below is due to [Rice] :

Amplitude of

fading gain

a adt + %

a

A

t dt+tTime


35/403-35

- In any time range t, t+dt, there is an upward crossing if

A a a dt %

and 0a >%

Equivalently, we need

A a dt a A %

Therefore

[ ] ,0

,

0

Pr upward crossing in ( , )

( , )

A

a a

a a A adt

a a

a

dt da p a a da

dt da a p A a

= =

=

=

=

3 3

3

%

% %

%

%

% %

% % %

This means the expected number of crossings in any dt, hence the

crossing rate is

,( ) ( , )a ao

n A a p A a da

= 3 %% % %

Note that the lower limit of integration is zero because we are interested

in positive crossings.

- The joint pdf , ( , )a ap a a%%

is

, , , ,

2 2

2 22 2

( , ) ( , , , )

1 exp

22

a a a ap a a p a a d d

a a a

= =

=

/ 4! "0 0= +1 5% &

0 0+ ,2 6

3 3 %% %%

% %% %

%

Substitution gives


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2

2

22( ) exp

22

2 for Jakes spectrumRd

A An A

f R e

# $= ' (

) *

=

where

22

AR

=

is the crossing threshold normalized to the RMS envelop level.

- The figure below shows the normalized crossing rate (with respect to the

Doppler frequency) as a function of the normalized threshold levelR.

-40 -35 -30 -25 -20 -15 -10 -5 0 5 1010

-4

10-3

10-2

10-1

100

101

20*log10(R)

n(A)/fd

Normalized zero crossing (fade) rate versus normalized threshold level.


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It is observed that

a. a 6 dB decrease in level (Rcut in half) means cutting the fade(zero crossing) rate by half,

b.fade rate proportional to df ,

c.maximum rate at 3R = dB

Fade Duration

- Refer to [Lee 6.5], [Jakes 1.3.5]

- Easy, now that we have the other relations

2Pr / 2 ( ) ( )a R T R n R# $ =

) *Hence

2 2

2

1 1( )

2 22

R R

Rd dd

e e RT R

f R ff R e

= =

- Plot of normalized fade duration ( ( )df T R ) vs R is shown in the plot

below.

1. fade duration is proportional to amplitude level forRequal 5 dB andbelow

2.a 6 dB decrease in level means fades are half as long (and half asfrequent, as we saw earlier)

3. fade duration inversely proportional to df


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-40 -35 -30 -25 -20 -15 -10 -5 0 5 1010

-3

10-2

10-1

100

101

10

2

103

104

20*log10(R)

T(R)*fd

Normalized fade duration versus normalized threshold

- It is shown in [Arnold and Boltman, Interfade Interval Statistics of aRayleigh distributed wave, IEEE Trans. Communications, Sept. 83]

that intervals between fades are approximately exponentially distributed

for 15R < dB or so.

The same two authors in [Boltman and Arnold, Fade distribution

statistics of Rayleigh distributed wave, IEEE Trans. Communications,

March 82] show that the fade duration is not exponential

Example : Consider a mobile data system. What does a 3 dB increase intransmit power buy in terms of error burst statistics ?

Solutions :


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Since 2/ 2R A = , increase 2 by 3 dB (i.e doubling it) effectively reduces

R to 0.71 times its old value. Since for small R, the fade duration is

approximately proportional toR, so bursts are now 71% as long.

Example : Have a mobile data system operating at 1 GHz, vehicle speed108 km/hr, data rate 9600 bps using FSK with discriminator detection,

average error probability (BEP) is 310 . Estimate (roughly) burst duration

and frequency, both in bits. Use the following expressions for the

instantaneous and the average BEPs :

2/ 41( )

2

a

eP a e

= (instantaneous BEP)

2

0

1( ) ( )

2e e a

a

P P a p a da

=

= =+

3 (average BEP)

where 2 is the variance of the underlying complex fading process. For this

question, it also respresents the average SNR in the system.

Solutions :

- From the BEP expressions, we can deduce that 2 1000 or 30 dB.

- Major approximation: bursts defined by instantaneous BEP > 0.1. This

means2.5373A =

is the threshold.

- The normalized threshold is2/ 2 0.0032R A = = or approximately 25

dB.


40/40

- From equations (or graphs), the normalized fade rate and normal fadeduration are approximately 0.14 and 0.022 respectively.

- Since / 100df V = = Hz, we have ( ) 14n A = deep fades/s and

( ) 224T A s=

- Bit duration is 1/ 9600 104T s= = . So average fade duration is 2 bits

- Inter-fade interval is 9600/n(A) = 686 bits.

Exercise: Repeat last exercise but with a velocity of 50 km/hr.

Mobile Radio Channels

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