Digital Signal Processing Reference
In-Depth Information
f
[
n
0
(
T
)
]
1
0
VT
τ
VT
−
n
o
(
T
)
τ
Fig. 3.8
PDF of noise sample
e
−
n
o
(
T
)
/
2
σ
2
0
f
[
n
o
(
T
)]
=
2
(3.12)
0
πσ
VT
τ
Since, at the end of bit interval,
S
0
(
T
)
=−
(for logic 0 representation), the
VT
τ
error will occur if
n
0
(
. Hence, the probability of error in this case is given
by the marked area (0) shown in Fig.
3.8
.
Hence,
T
) >
∞
P
e
=
f
[
n
o
(
T
)]
dn
o
(
T
)
VT
τ
(3.13)
∞
e
−
n
o
(
T
)
/
2
σ
0
=
2
dn
o
(
T
)
2
0
πσ
VT
τ
/
√
2
0
, we get
Substituting
x
=
n
o
(
T
)
σ
0
and putting the value of
σ
∞
1
2
×
2
π
e
−
x
2
dx
P
e
=
V
T
η
erfc
V
T
η
1
2
×
=
(3.14)
erfc
V
2
T
η
2
1
2
×
=
erfc
E
S
η
1
2
1
2
×
=
V
2
T
where, Signal Energy
=
E
S
=
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