Convert the continuous time received signal into a vector
without loss of information (or performance).

r
t
=
s
m
t+
N
t
r
t
s
m
t
N
t

(1)
r
t
=∑
n
=1N
s
m
n
ψ
n
t+∑
n
=1N
η
n
ψ
n
t+
N
t
˜
r
t
n
1
N
s
m
n
ψ
n
t
n
1
N
η
n
ψ
n
t
N
t
˜

(2)
r
t
=∑
n
=1N(
s
m
n
+
η
n
)
ψ
n
t+
N
t
˜
r
t
n
1
N
s
m
n
η
n
ψ
n
t
N
t
˜

(3)
r
t
=∑
n
=1N
r
n
ψ
n
t+
N
t
˜
r
t
n
1
N
r
n
ψ
n
t
N
t
˜

(4)
The noise projection coefficients
η
n
η
n
's are zero mean, Gaussian random variables and
are mutually independent if
N
t
N
t
is a white Gaussian process.

μ
η
n=E
η
n
=E∫0T
N
t
ψ
n
td
t
μ
η
n
η
n
t
0
T
N
t
ψ
n
t

(5)
μ
η
n=∫0TE
N
t
ψ
n
td
t
=0
μ
η
n
t
0
T
N
t
ψ
n
t
0

(6)
E
η
k
η
n
¯=E∫0T
N
t
ψ
k
td
t
∫0T
N
t
′
¯
ψ
k
t
′
¯d
t
′
=∫0T∫0T
N
t
N
t
′
¯
ψ
k
t
ψ
n
t
′
dtd
t
′
η
k
η
n
t
0
T
N
t
ψ
k
t
t
′
0
T
N
t
′
ψ
k
t
′
t
′
0
T
t
0
T
N
t
N
t
′
ψ
k
t
ψ
n
t
′

(7)
E
η
k
η
n
¯=∫0T∫0T
R
N
t−
t
′
ψ
k
t
ψ
n
¯d
t
d
t
′
η
k
η
n
t
′
0
T
t
0
T
R
N
t
t
′
ψ
k
t
ψ
n
t
′

(8)
E
η
k
η
n
¯=
N
0
2∫0T∫0Tδt−
t
′
ψ
k
t
ψ
n
t
′
¯d
t
d
t
′
η
k
η
n
N
0
2
t
′
0
T
t
0
T
δ
t
t
′
ψ
k
t
ψ
n
t
′

(9)
E
η
k
η
n
¯=
N
0
2∫0T
ψ
k
t
ψ
n
t¯d
t
=
N
0
2
δ
k
n
={
N
0
2 if k=n0 if k≠n
η
k
η
n
N
0
2
t
0
T
ψ
k
t
ψ
n
t
N
0
2
δ
k
n
N
0
2
k
n
0
k
n

(10)
η
k
η
k
's are uncorrelated and since they are Gaussian they are
also independent. Therefore,

η
k
≃Gaussian0
N
0
2
η
k
Gaussian
0
N
0
2
and

R
η
kn=
N
0
2
δ
k
n
R
η
k
n
N
0
2
δ
k
n
The
r
n
r
n
's, the projection of the received signal
r
t
r
t
onto the orthonormal bases
ψ
n
t
ψ
n
t
's, are independent from the residual noise
process
N
t
˜
N
t
˜
.

The residual noise
N
t
˜
N
t
˜
is irrelevant to the decision process on
r
t
r
t
.

Recall
r
n
=
s
m
n
+
η
n
r
n
s
m
n
η
n
, given
s
m
t
s
m
t
was transmitted. Therefore,

μ
r
n=E
s
m
n
+
η
n
=
s
m
n
μ
r
n
s
m
n
η
n
s
m
n

(11)
Var
r
n
=Var
η
n
=
N
0
2
Var
r
n
Var
η
n
N
0
2

(12)
The correlation between

r
n
r
n
and

N
t
˜
N
t
˜
E
N
t
˜
r
n
¯=E(
N
t
−∑
k
=1N
η
k
ψ
k
t)
s
m
n
+
η
n
¯
N
t
˜
r
n
N
t
k
1
N
η
k
ψ
k
t
s
m
n
η
n

(13)
E
N
t
˜
r
n
¯=E
N
t
−∑
k
=1N
η
k
ψ
k
t
s
m
n
+E
η
k
η
n
¯−∑
k
=1NE
η
k
η
n
¯
ψ
k
t
N
t
˜
r
n
N
t
k
1
N
η
k
ψ
k
t
s
m
n
η
k
η
n
k
1
N
η
k
η
n
ψ
k
t

(14)
E
N
t
˜
r
n
¯=E
N
t
∫0T
N
t
′
¯
ψ
n
t
′
¯d
t
′
−∑
k
=1N
N
0
2
δ
k
n
ψ
k
t
N
t
˜
r
n
N
t
t
′
0
T
N
t
′
ψ
n
t
′
k
1
N
N
0
2
δ
k
n
ψ
k
t

(15)
E
N
t
˜
r
n
¯=∫0T
N
0
2δt−
t
′
ψ
n
t
′
d
t
′
−
N
0
2
ψ
n
t
N
t
˜
r
n
t
′
0
T
N
0
2
δ
t
t
′
ψ
n
t
′
N
0
2
ψ
n
t

(16)
E
N
t
˜
r
n
¯=
N
0
2
ψ
n
t−
N
0
2
ψ
n
t=0
N
t
˜
r
n
N
0
2
ψ
n
t
N
0
2
ψ
n
t
0

(17)
Since both

N
t
˜
N
t
˜
and

r
n
r
n
are Gaussian then

N
t
˜
N
t
˜
and

r
n
r
n
are also independent.

The conjecture is to ignore
N
t
˜
N
t
˜
and extract information from
r
1
r
2
…
r
N
r
1
r
2
…
r
N
.
Knowing the vector
r
r
we can reconstruct the relevant part of random process
r
t
r
t
for
0≤t≤T
0
t
T

r
t
=
s
m
t+
N
t
=∑
n
=1N
r
n
ψ
n
t+
N
t
˜
r
t
s
m
t
N
t
n
1
N
r
n
ψ
n
t
N
t
˜

(18)
Once the received signal has been converted to a vector, the
correct transmitted signal must be detected based upon
observations of the input vector. Detection is covered elsewhere.