ber in fading environment

Chapter 10: Signaling Over Fading Channels

A First Course in Digital Communications

Ha H. Nguyen and E. Shwedyk

February 2009

A First Course in Digital Communications 1/45


Introduction

We have considered that the transmitted signal is onlydegraded by AWGN and it might be subjected to filtering.

There are communication channels (e.g., mobile wirelesschannels) where the received signal is not subjected to aknown transformation or filtering.

Typically the gain and/or phase of a digitally modulatedtransmitted signal is not known precisely at the receiver.

It is common to model these parameters as random.

Shall consider channel models where the amplitude and/orphase of the received signal is random.



Demodulation with Random Amplitude

r(t) = as(t) + w(t),

where a is a random variable with known pdf fa(a).

1r

2r

E

1T

Ea

0 0R T 1R

2Ea

0D 1D

1r

2r

E

1T

Ea

0T 1R

0D 1D

0R

E Ea

1r0

2r

21 rr =

0D

1D

1T

0T

Ea

1R

0R

E

E

Ea



Performance of BPSK and BFSK

P[error] = Q

(a

2EbN0

)(antipodal),

P[error] = Q

(a

EbN0

)(orthogonal).

E{P[error]} =

0Q

(a

2EbN0

)fa(a)da (antipodal),

E{P[error]} =

0Q

(a

EbN0

)fa(a)da (orthogonal).



Optimum Demodulation of BASK

Optimum receiver is determined by the maximum likelihoodratio:

fr1(r1|1T )fr1(r1|0T )

1D

R0D

1.

fr1(r1|0T ) is N (0, N0/2), while

fr1(r1|1T ) =

0fr1(r1|1T ,a = a)fa(a)da

= E {fr1(r1|1T ,a = a)} .

Need to know fa(a) to proceed further.

In general the threshold (and hence the decision regions) is abalance between the different regions given by the values thata takes on weighted by the probability that a takes on thesevalues.



M -ary Demodulation with Random Amplitude

If all the signal points lie at distance ofEs from the origin (i.e.,

equal energy), then the optimum decision regions are invariant toany scaling by a, provided that a 0.The matched-filter or correlation receiver structure is still optimum,one does not even need to know fa(a).

The error performance, however, depends crucially on a and fa(a).

1r

2r

1=a

0 0

2r

1r

0.5=a



Demodulation with Random Phase

Phase uncertainty can be modeled as a uniform random variable (over

[0, 2] or [, ]). It does not change the energy of the received signal.

E

)( of locus 2 tsR

1T2( )s t1 1( ) ( )Rs t s t=

2( ) sin(2 )Q cb

t f tT

pi=

0 0R T

2 ( ) 1R Rs t

( )2

cos(2 )I

c

b

t

f tT

pi

=

E

1 2locus of ( ), ( )R Rs t s t

!

"

( )I t1T2( )s t1( )s t

( )Q t

0T

2 ( ) 1R Rs t

1 ( ) 0R Rs t

E

#

$% &

'()

1, ( )I t1( )s t

1, ( )Q t

0T

1 ( ) 0R Rs t

E

2, ( )I t2 ( )s t

2, ( )Q t

1T

2 ( ) 1R Rs t



Optimum Receiver for Noncoherent BASK

r(t) =

{w(t), if 0T

E

2Tb

cos(2fct ) + w(t), if 1T .

rI =

{wI , if 0T

E cos + wI , if 1T, rQ =

{wQ, if 0T

E sin + wQ, if 1T,

where wI and wQ are statistically independent N (0, N0/2).f(rI , rQ|0T ) = f(rI |0T )f(rQ|0T ) = 1piN0 exp

( r

2I+r

2Q

N0

).

f(rI , rQ|1T ) = 2pi0

1

N0exp

(rI

E cos

)2+(rQ

E sin

)2N0

1

2d

=1

N0e

r2I+r2Q

N0 eEN0 I0

(2E

N0

r2I + r

2Q

).

e EN0 I0

(2E

N0

r2I + r

2Q

)1D

R0D

1 r2I + r

2Q

1D

R0D

N0

2EI10

(eEN0

).



Decision Regions of BASK with Random Phase

E

)( of locus 2 tsR

1T

( )Q t

0 0R T

2 ( ) 1R Rs t

( )I t

1 1D

0 0D

02 2 100 e2

EN

I Q hN

r r T IE

* ++ = = , -. /

)(tr 2( ) cos(2 )I cb

t f tT

pi=

2( ) sin(2 )Q cb

t f tT

pi=

( )( 1)

db

b

kT

k T

t

0

( )( 1)

db

b

kT

k T

t

1

bt kT=

bt kT=

Ir

Qr

( ) 23456768371

0

D

h

D

T>

0, present bit equals previous bitIf 0, present bit is complement of previous bit

k

k

r

r

r

s

t

But the above method can lead to error propagation!



Differential BPSK Modulation and Demodulation

0T : no phase change,1T : phase change.

The decision rule is:

rk

1D

R0D

0.

which is independent of the previous decision.

Since DBPSK is orthogonal signaling, the error analysis fornoncoherent BFSK therefore applies to DBPSK:

P [error]DBPSK =1

2eEb/N0 .

The only difference is rather than Eb joules/bit, the energy inDBPSK becomes 2Eb. This is because the received signalover two bit intervals is used to make a decision.



DBPSK shows about 1 dB degradation over coherent BPSK.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15106

105

104

103

102

101

100

SNR per bit (Eb/N0) in dB

P[err

or]

Coherent BPSK

Noncoherent DBPSK

Coherent BASK & BFSKNoncoherent BFSK

Noncoherent BASKwith threshold of 0.5E1/2

Noncoherent BASKwith the optimum threshold



Detection over Fading Channels

Fading channel model arises when there are multiple transmission paths

from the transmitter to the receiver.

uv

w x

y

v

z{

|

}~w} x}

~

uv

w

v

}x

}

v

v

}x x~}w x

~ww

y



Rayleigh Fading Channel Model

Consider the transmitted signal sT (t) = s(t) cos(2fct), where

s(t) = Eb

2Tb

over the bit interval with bit rate rb fc(lowpass signal).

The received signal is:

r(t) =j

rj(t) =j

s(t tj)j cos(2fc(t tj))

s(t)j

j cos (2fct 2fctj) = s(t)j

j cos (2fct j)

where j represents the attenuation and tj the delay along the jthpath, which are random variables. Also because s(t) is lowpass, weapproximate s(t) s(t tj).Since tj 1/fc, the random phase j lies in the range [0, 2). Now

r(t) = s(t)

j

j cosj

cos(2fct) +

j

j sinj

sin(2fct)

.



Rayleigh Fading Channel Model

nF,I =(

j j cosj

)and nF,Q =

(j j sinj

)have the following

moments:

E {nF,I} =j

E{j}E{cosj} = 0, E {nF,Q} =j

E{j}E{sinj} = 0,

E{n2F,I

}=j

E{2j

}E{cos2 j

}=

2F2,

E{n2F,Q

}=j

E{2j

}E{sin2 j

}=

2F2,

E {nF,InF,Q} = E

j

j cosjk

k sink

=j

k

E {jk}E {cosj sink} =0

= 0,

Since the number of multipaths is large, the central limit theorem says

that nF,I , nF,Q are Gaussian random variables.



nF,I and nF,Q are statistically independent Gaussian randomvariables, zero-mean, variance 2F /2:

fnI ,nQ(nI , nQ) = fnI (nI)fnQ(nQ) = N(0,2F2

)N(0,2F2

).

The received signal is therefore:

r(t) = s(t) [nF,I cos(2fct) + nF,Q sin(2fct)]

= s(t) [ cos(2fct )] ,where =

n2F,I + n

2F,Q, = tan

1(

nF,Q

nF,I

)and

f() =1

2(uniform),

f() =2

2Fexp

{

2

2F

}u() (Rayleigh).

The term Rayleigh fading comes from the envelope distribution.

The phase of the received signal severely degraded but that theamplitude is affected as well: The incoming signals add not onlyconstructively but also destructively.



Noncoherent Demodulation of BFSK in Rayleigh Fading

s(t) =

Eb

2Tb

cos(2f1t), if 0T Eb

2Tb

cos(2f2t), if 1T ,

r(t) =

Eb

2Tb cos(2f1t ) + w(t), if 0T

Eb

2Tb cos(2f2t ) + w(t), if 1T

.

=

EbnF,I

2

Tbcos(2f1t) 1,I(t)

+EbnF,Q

2

Tbsin(2f1t) 1,Q(t)

+w(t), 0T,

EbnF,I

2

Tbcos(2f2t) 2,I(t)

+EbnF,Q

2

Tbsin(2f2t) 2,Q(t)

+w(t), 1T,

The transmitted signal lies entirely within the signal space spanned by

1,I(t), 1,Q(t), 2,I(t) and 2,Q(t).



0T

r1,I =EbnF,I + w1,I

r1,Q =EbnF,Q + w2,Q

r2,I = w2,I

r2,Q = w2,Q

1T

r1,I = w1,I

r1,Q = w2,Q

r2,I =EbnF,I + w2,I

r2,Q =EbnF,Q + w2,Q

w1,I , w1,Q, w2,I , w2,Q are due to thermal noise, are Gaussian,statistically independent, zero-mean, and variance N0/2.

nF,I and nF,Q, are also Gaussian, statistically independent,zero-mean and variance 2F /2.

The sufficient statistics are Gaussian, statistically independent,zero-mean, with a variance of either N0/2 or Eb

2F /2 +N0/2,

depending on whether a 0T or 1T .

Computing the likelihood ratio gives the following decision rule:

r22,I + r22,Q

1D

R0D

r21,I + r21,Q,



Equivalently the decision rule can be expressed as:

r22,I + r

22,Q

1D

R0D

r21,I + r

21,Q.

which is identical to that for noncoherent BFSK!

bt kT=

11

2cos(2 ), 0( )

0, elsewhere

bb

f t t Th t T

pi

=

bt kT=

1

0

0

D

D

>

0 andR() > 0, the pdf is

fy(y) =yN2 1ey/(2

2)

2N2 N

(N2

) u(y),where (x) =

0 t

x1etdt = (x 1)! for x integer.A First Course in Digital Communications 36/45


Error Performance of BFSK with Diversity

Define 1 =4N

j=2N+1 r2j and 0 =

2Nj=1 r

2j . The decision rule is:

1

1D

R0D

0.

P [error] = P [error|0T ] =

0f(0|0T )

[ 0

f(1|0T )d1]d0.

f(1|0T ) and f(0|0T ) are chi-square distributions:

f(1|0T ) = N11 e

1/(22w)

2N2Nw (N)u(1), f(0|0T ) =

N10 e

0/(22t )

2N2Nt (N)u(0).

It can be shown that

P [error] =

Nj=1

(2t2w

)Nj1(

1 +2t2w

)2Nj (2N j)(N)(N j + 1) .A First Course in Digital Communications 37/45


Define T = E

b2F /N0 as the averaged SNR per transmission.

Recognize that2t2w

= 1 + T and (x) = (x 1)! for integerx. Then

P [error] =1

(2 + T )N

N1k=0

(N 1 + k

k

)(1 + T2 + T

)k.

For large values of SNR, 1 + T 2 + T T and

P [error] 1(T )N

N1k=0

(N 1 + k

k

)=

1

(T )N

(2N 1N

).

The error performance now decays inversely with the N thpower of the received SNR.

The exponent N of the SNR is generally referred to as thediversity order of the modulation scheme.



Compared to no diversity, there is a significant improvement in

performance with diversity.

0 5 10 15 20 25 30 35106

105

104

103

102

101

100

Received SNR per transmission (dB)

P[err

or]

N=1 (No diversity)

N=2

N=4

N=6

N=8Noncoherent FSK

in random phase only



Optimum Diversity

As the diversity order N increases the error performance improves.

This improvement comes at the expense of a reduced data rate inthe case of time diversity.

If the transmitters power or equivalently the energy expended perinformation bit is constrained to Eb joules then increasing N doesnot necessarily lead to a better error performance.

With increased N we increase the probability of avoiding a deepfade, at the same time the energy, E

b, of each transmission isreduced. Therefore the SNR of each transmission is reduced whichin turn increases the error probability.

There is an optimum value for the diversity order N at each level oferror probability. An empirical relationship is:

Nopt = Ke10 log10 T .

where K is some constant.



P [error] versus the averaged received SNR per bit, 10 log10

(Eb

2F

N0

).

0 5 10 15 20 25 30 35106

105

104

103

102

101

100

Received SNR per bit (dB)

P[err

or]

N=1 (No diversity)

N=2

N=4

N=6

N=8

Noncoherent FSKin random phase only



Determining The Optimum Diversity Order

0 5 10 15 20104

103

102

101

100

Diversity order N (number of transmissions per bit)

P[err

or]

SNR per bit ranges from 2 to 16 dB in 1dB steps

(a)

2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

Average received SNR per bit (dB)

Opt

imum

div

ersit

y or

der

N opt

(b)



Central Limit Theorem

Central limit theorem states that under certain generalconditions the sum of n statistically independent continuousrandom variables has a pdf that approaches a Gaussian pdf asn increases.

Let x =n

i=1 xi. where the xis are statistically independentrandom variables with mean, E{xi} = mi, and variance,E{(xi mi)2

}= 2i . Then x is a random variable with

mean mx =n

i=1 mi, variance 2x=n

i=1 2i and a pdf of

fx(x) = fx1(x) fx2(x) fxn(x).By the central limit theorem fx(x) approaches a Gaussian pdfas n increases, i.e.,

fx(x) 12x

e (xmx)

2

22x .



Example 1: fxi(xi) are Zero-Mean Uniform

8 6 4 2 0 2 4 6 80

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

(a)

x

f x(x)

n=2

n=4

n=6

n=10

Actual distributionGaussian fit



Example 2: fxi(xi) are Laplacian

5 0 5 10 15 20 250

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4(b)

x

f x(x)

n=6

n=2

n=4

n=10

Actual distributionGaussian fit


Chapter 10: Signaling Over Fading ChannelsIntroductionDemodulation with Random AmplitudeDemodulation with Random PhaseDetection with Random Amplitude and Random Phase: Rayleigh Fading ChannelDiversityCentral Limit Theorem

ber in fading environment

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