performance of Binary non coherent signals in random phase channel

1. Performance of binary Non-
orthogonal signals in Random
Phase channels
YOGESH SINGH

2. 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,andwhereand
22
10
 zzzz sicisici
NEs 02
N
Es
0
2
N
Es
0
2
2

3. 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 scc
and 11s00
,,
 1v
 1v
),(
10
yxf vv 
)(
**
3
*
3
2
2
2
1
1
xyAyxAyAxA kkkkkk
L
k
A  
dxdyyxPE
x
vvvvs fmmP ),()/()/(
0
100010 
 



3

4. 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
N
Em s
x
N
E
N
E
N
E
s
s
s
4

5. 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




 N
Es
x
)0()(  QxP x
T
s
PE
5

6. Cnt'd…















1000
0100
0010
0001
Q …..(5.119)
6

7. 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
E
N0





 22
1 
7

8. 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
   xx
N
ExxQx 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

9. Cnt'd…
…..(5.126)
The optimum receiver compares the pdf of is,
…..(5.127)
…..(5.128)
   
   
 
  












 





k
k
kk
kk
xEE
NE
NE
NE
NE
x
s
s
s
s
w
0
0
0
0
2
2
0
12
12
 iww ,
2
0,
and  iw,
   
)1(
)1(
2
exp
0
2
1
0
2
0
0
22
0,
k
k
where
N
Es
N
Es
sI
s
mf
s
s
i
i
i
w



























 
 


9

10. 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
2
exp,









 



ss
I
ba
where
ab
ba
baQ
10

11. 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
11
2
)1(
2
11
2
)1(
2
N
E
N
E
N
E
N
E
ss
ss
kb
ka
 j
 












































N
E
I
N
E
N
E
N
E
P
ss
ss
s
QE
22 0
0
0
2
0
2
0
exp
2
1
11
2
.11
2
11

12. 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,
 
   
  













































N
E
N
E
N
E
P
x
N
E
N
E
P
sss
s
ss
s
E
xQ
QE
222
2
000
2
00
exp
2
1
exp
2
1
exp
2exp,0
exp
2
1
,0
12

13. Cnt'd…
 
 
 
 T
T
f
T
T
dtff
T
dtfTfT
f
f
ef
f
e
eEeE
E
tj
T
tj
tj
s
T
tj
s
s
1,0
1,0
2
1,0
1,0
0
)(2
2
0
2
sin
sin1
22
2
1
1,010
10















13