We now consider the example of detecting a binary signal after
it has passed through a channel which adds noise. The
transmitted signal is typically as shown in (a) of Figure 1.

In order to reduce the channel noise, the receiver will
include a lowpass filter. The aim of the filter is to reduce
the noise as much as possible without reducing the peak values
of the signal significantly. A good filter for this has a
half-sine impulse response of the form:

ht={π2
Tb
sinπt
Tb
if 0≤t≤
Tb
0 otherwise
h
t
2
Tb
t
Tb
0
t
Tb
0

(1)
Where

Tb
Tb
= bit period.

This filter will convert the rectangular data bits into
sinusoidally shaped pulses as shown in (b) of Figure 1 and it Will also
convert wide bandwidth channel noise into the form shown in
(c) of
Figure 1. Bandlimited noise of
this form will usually have an approximately Gaussian pdf.

Because this filter has an impulse response limited to just
one bit period and has unit gain at zero frequency (the area
under
ht
h
t
is unity), the signal values at the center of each
bit period at the detector will still be
±1
±
1
. If we choose to sample each bit at the detector at
this optimal mid point, the pdfs of the signal plus noise at
the detector will be shown in Figure 2.

Let the filtered data signal be
Dt
D
t
and the filtered noise be
Ut
U
t
, then the detector signal is

Rt=Dt+Ut
R
t
D
t
U
t

(2)
If we assume that

+1
+
1
and

-1-1 bits are
equiprobable and the noise is a symmetric zero-mean process,
the optimum detector threshold is clearly midway between these
two states, i.e. at zero. The probability of error when the
data =

+1
+
1
is then given by:

Prerror|
D=+1
=PrRt<0|
D=+1
=
FU
-1=∫−∞-1
fU
udu
D
+
1
error
D
+
1
R
t
0
FU
-1
u
-1
fU
u

(3)
where

FU
F
U
and

fU
f
U
are the cdf and pdf of

UU. This is the shaded area in

Figure 2.

Similarly the probability of error when the data =
-1-1 is then given by:

Prerror|
D=-1
=PrRt>0|
D=-1
=1−
FU
+1=∫1∞
fU
udu
D
-1
error
D
-1
R
t
0
1
FU
+
1
u
1
fU
u

(4)
Hence the overall probability of error is:

Prerror=Prerror|
D=+1
PrD=+1+Prerror|
D=-1
PrD=-1=∫−∞-1
fU
uduPrD=+1+∫1∞
fU
uduPrD=-1
error
D
+
1
error
D
+
1
D
-1
error
D
-1
u
-1
fU
u
D
+
1
u
1
fU
u
D
-1

(5)
since

fU
fU
is symmetric about zero

Prerror=∫1∞
fU
udu(PrD=+1+PrD=-1)=∫1∞
fU
udu
error
u
1
fU
u
D
+
1
D
-1
u
1
fU
u
To be a little more general and to account for signal
attenuation over the channel, we shall assume that the signal
values at the detector are

±
v0
±
v0
(rather than

±1
±
1
) and that the filtered noise at the detector has a
zero-mean Gaussian pdf with variance

σ2
σ
2
:

fU
u=12πσ2e−u22σ2
fU
u
1
2
σ
2
u
2
2
σ
2

(6)
and so

Prerror=∫
v0
∞
fU
udu=∫
v0
σ∞Q
v0
σdu
error
u
v0
fU
u
u
v0
σ
fU
σ
u
σ
Q
v0
σ

(7)
where

Qx=12π∫x∞e−u22du
Q
x
1
2
u
x
u
2
2

(8)
This integral has no analytic solution, but a good
approximation to it exists ans is discussed in some detail in ?????????????

(Reference).

From Equation 7 we may obtain the
probability of error in the
binary detector, which is ofter expressed as the bit
error rate or BER. For
example, if
Prerror=2×103
error
2
10
3
, this would often be expressed as a bit error rate
of
2×103
2
10
3
, or alternatively as 1 error in 500 bits (on
average).

The argument (
v0
σ
v0
σ
) in Equation 7 is the
signal-to-noise voltage ratio (SNR) at
the detector, and the BER rapidly diminishes with increasing
SNR (see ??????????? (Reference)).