Performance of binary Non-

orthogonal signals in Random

Phase channels

YOGESH SINGH

Performance of Binary non-

orthogonal signals• For binary case, the probability of error can be simplified

much further as:

• The receiver compares are still given by eq. 5.94 and 5.94b of Simon.

• Multiplying Zci and Zsi, i=0,1, by , we have:

νc0 = Zc0 = 2Es/No+uc0

……. (5.110)

νs0 = zs0 = us0

0,1ifor , and where and 22

10 zzzz sicisici

NEs 02

NE s

0

2

NE s

0

2

2

Cnt'd…

assuming m0 was transmitted. The Gaussian noise variables,

are zero-mean, unit variance with

E{uc0 uc1}=E{us0 us1}=(2Es/N0)α, E{uc0 us1}= -E{uc1 us0} =

(2Es/N0)β.

The probability of error, assuming message m0 is transmitted, is

equal to the probability that exceeds or

…..(5.112)

Which requires the computation of the joint pdf .

General quadratic form in complex-valued RV by,

…..(5.113)

uuuu sccand

11s00, ,

1v

1v

),(10

yxfvv

)(**

3

*

3

2

2

2

11

xyAyxAyAxA kkkkkk

L

k

A

dxdyyxPEx

vvvvs fmmP ),()/()/(0

100010

3

Cnt'd…

where A1, A2, A3 are constants and {xk,yk} are pairs of correlated

complex-valued Gaussian random variables.

Consequently the variables and

are Rician and independent and the probability of error reduces to

the probability that one Rician variable exceeds another

independent Rician variable. Defining the vector x as,

…..(5.114)

then the mean and covariance ,

…..(5.115)

vv swcww

2

0,

2

0,0, vv swcww

2

1,

2

1,1,

) ( 1100 vvvvx scsc

TT

0

1

)(

)(

0

)(

0

2

2

2

2

0

0

0

NE

ms

x

NE

NE

NE

s

s

s

4

Cnt'd…

And also,

…..(5.116)

Due to the symmetry of the problem, it is easily shown that

Ps(E/m0)=Ps(E/m1)=Ps(E). Hence, the error probability is

given by,

…..(5.117)

Where,

10

01

10

01

0

2

NE s

x

)0()( QxP x

T

sPE

5

Cnt'd…

1000

0100

0010

0001

Q …..(5.119)

6

Cnt'd…

In order to whiten the vector x, we need to perform a linear

nonsingular transformation Ƭ on x which diagonalizes ,

xw=Ƭx …..(5.120)

One such transformation is given by,

…..(5.121)

Where,

k= …..(5.122)

x

k

k

k

k

kk

s

10

01

10

01

)1(2

EN0

22

1

7

Cnt'd…

Also,

…..(5.123)

…..(5.124)

Hence, the probability of error becomes,

.....(5.125)

Where are unit variance uncorrelated

Gaussian (and hence independent) RV with mean

xxNE

xxQx w

T

w

s

w

T

wx

T

T

x

QkQ

x

T

w

0

1 21

2

1,

2

1,

2

0,

2

0,

00

swcwswcw

w

T

wx

T

s

P

QPPE xxQxP

1,1,0,0, and ,,

swcwswcw

8

Cnt'd…

…..(5.126)

The optimum receiver compares the pdf of is,

…..(5.127)

…..(5.128)

k

k

kk

kkxEE

NENE

NENE

x

s

s

s

s

w

0

0

0

0

2

2

0

12

12

iww ,

2

0, and

iw,

)1(

)1(

2exp

0

2

1

0

2

0

0

22

0,

k

k

where

NE

s

NE

s

sIs

mf

s

s

i

i

iw

9

Cnt'd…

Hence, the probability of error becomes,

……(5.129)

Which is similar to (5.112) except the variables are now

independent. The Rician RV(s) R0 and R1, the probability that R1

exceeds R0 is given by,

P=Prob (R0< R1)

…...(5.130)

…..(5.131)

dydxyx

PE

x

wws

ff

mP

ww

1,0

0,

00,1,

)(

and 1,0, ww

2

1

2

0

2

0

2

1

2

0

2

1

02

1

2

0

2

0

and

2exp,

ss

I

ba

where

abba

baQ

10

Cnt'd…

In our case 1 and,

…..(5.132)

With . Hence, the probability of error for two

nonorthogonal signals is given by,

…..(5.133)

2

0

2

1

2

00

2

00

112

)1(2

112

)1(2

NE

NE

NE

NE

ss

ss

kb

ka

j

NE

INE

NE

NE

P

ss

ss

sQE

22 0

0

0

2

0

2

0

exp2

1

112

.112

11

Cnt'd…

which is identical to 1-Ps(C) as given by (5.109) in M-ary

equicorrelated signals. It is clear that orthogonal signals (i.e. ρ=0)

minimizes Ps(E). In this case, Ps(E) reduces to,

…..(5.134)

But since , then

…..(5.135)

Consider the cross-correlation between two FSK signals at

frequencies f0 and f1. then the coss-correlation between the

corresponding complex baseband signals is,

NE

NE

NE

P

x

NE

NE

P

sss

s

ss

s

E

xQ

QE

222

2

000

2

00

exp2

1exp

2

1exp

2exp,0

exp2

1,0

12

Cnt'd…

T

T

f

T

Tdt

ff

T

dtf

Tf

T

f

f

ef

fe

eEeEE

tjT

tj

tj

s

Ttj

s

s

1,0

1,0

2

1,0

1,0

0

)(2

2

0

2

sin

sin1

222

1

1,010

10

13