The bit-error probability for a correlation receiver with an
antipodal signal set (
Figure 1) can be found as follows:
P
e
=Pr
m
̂
≠m=Pr
b
̂
≠b=
π
0
Pr
r
1
<γ|m=1+
π
1
Pr
r
1
≥γ|m=2=
π
0
∫-∞γf
r
1
s
1
trdr+
π
1
∫γ∞f
r
1
s
2
trdr
P
e
m
̂
m
b
̂
b
π
0
m
1
r
1
γ
π
1
m
2
r
1
γ
π
0
r
γ
f
r
1
s
1
t
r
π
1
r
γ
f
r
1
s
2
t
r
(1)
if
π
0
=
π
1
=1/2
π
0
π
1
12
,
then the optimum threshold is
γ=0
γ
0
.
f
r
1
|
s
1
tr=
E
s
N
0
2
f
|
r
1
s
1
t
r
E
s
N
0
2
(2)
f
r
1
|
s
2
tr=-
E
s
N
0
2
f
|
r
1
s
2
t
r
E
s
N
0
2
(3)
If the two symbols are equally likely to be transmitted then
π
0
=
π
1
=1/2
π
0
π
1
12
and if the threshold is set to zero, then
P
e
=1/2∫-∞012π
N
0
2ⅇ-|r-
E
s
|2
N
0
dr+1/2∫0∞12π
N
0
2ⅇ-|r+
E
s
|2
N
0
dr
P
e
12
r
0
1
2
π
N
0
2
r
E
s
2
N
0
12
r
0
1
2
π
N
0
2
r
E
s
2
N
0
(4)
P
e
=1/2∫-∞-2
E
s
N
0
12πⅇ-|
r
′
|22d
r
′
+1/2∫2
E
s
N
0
∞12πⅇ-|
r
″
|22d
r
″
P
e
12
r
′
2
E
s
N
0
1
2
π
r
′
2
2
12
r
″
2
E
s
N
0
1
2
π
r
″
2
2
(5)
with
r
′
=r-
E
s
N
0
2
r
′
r
E
s
N
0
2
and
r
″
=r+
E
s
N
0
2
r
″
r
E
s
N
0
2
P
e
=12Q2
E
s
N
0
+12Q2
E
s
N
0
=Q2
E
s
N
0
P
e
1
2
Q
2
E
s
N
0
1
2
Q
2
E
s
N
0
Q
2
E
s
N
0
(6)
where
Qb=∫b∞12πⅇ-x22dx
Q
b
x
b
1
2
π
x
2
2
.
Note that
P
e
=Q
d
1
2
2
N
0
P
e
Q
d
1
2
2
N
0
(7)
where
d
1
2
=2
E
s
=∥s1-s2∥2
d
1
2
2
E
s
s
1
s
2
2
is the Euclidean distance between the two constellation points
(
Figure 2).
This is exactly the same bit-error probability as for the matched
filter case.
A similar bit-error analysis for matched filters can be found
here. For the bit-error analysis
for correlation receivers with an orthogonal signal set, refer
here.