r
t
⇒(r=
r
1
r
2
T=sm+η)
r
t
r
r
1
r
2
s
m
η
(see Figure 1)
Decide
s
1
t
s
1
t
was transmitted if
r
1
≥
r
2
r
1
r
2
.
P
e
=Pr
m
̂
≠m=Pr
b
̂
≠b
P
e
m
̂
m
b
̂
b
(1)
P
e
=1/2Prr∈
R
2
|
s
1
t
transmitted
+1/2Prr∈
R
1
|
s
2
t
transmitted
=1/2∫
R
2
∫f
r
s
1
t
rd
r
1
d
r
2
+1/2∫
R
1
∫f
r
s
2
t
rd
r
1
d
r
2
=1/2∫
R
2
∫12π
N
0
2e−|
r
1
−
E
s
|2
N
0
1π
N
0
e−|
r
2
|2
N
0
d
r
1
d
r
2
+1/2∫
R
1
∫12π
N
0
2e−|
r
1
|2
N
0
1π
N
0
e−|
r
2
−
E
s
|2
N
0
d
r
1
d
r
2
P
e
12
s
1
t
transmitted
r
R
2
12
s
2
t
transmitted
r
R
1
12
r
2
R
2
r
1
f
r
s
1
t
r
12
r
2
R
1
r
1
f
r
s
2
t
r
12
r
2
R
2
r
1
1
2
N
0
2
r
1
E
s
2
N
0
1
N
0
r
2
2
N
0
12
r
2
R
1
r
1
1
2
N
0
2
r
1
2
N
0
1
N
0
r
2
E
s
2
N
0
(2)
Alternatively, if
s
1
t
s
1
t
is transmitted we decide on the wrong signal if
r
2
>
r
1
r
2
r
1
or
η
2
>
η
1
+
E
s
η
2
η
1
E
s
or when
η
2
−
η
1
>
E
s
η
2
η
1
E
s
.
P
e
=1/2∫
E
s
∞12π
N
0
e−
η
′
22
N
0
d
η
′
+1/2Pr
r
1
≥
r
2
|
s
2
t
transmitted
=Q
E
s
N
0
P
e
12
η
′
E
s
1
2
N
0
η
′
2
2
N
0
12
s
2
t
transmitted
r
1
r
2
Q
E
s
N
0
(3)
Note that the distance between
s
1
s
1
and
s
2
s
2
is
d
1
2
=2
E
s
d
1
2
2
E
s
. The average bit error probability
P
e
=Q
d
1
2
2
N
0
P
e
Q
d
1
2
2
N
0
as we had for the
antipodal
case. Note also that the bit-error probability is the
same as for the
matched filter
receiver.