Cryptography Reference
In-Depth Information
Considering the modulated signal on a frequency carrier, its PSD is given by
expression (2.13):
γ
(
f
)=
1
f
0
)+
1
4
γ
e
(
f
+
f
0
)
In conclusion, the modulated signal on a frequency carrier uses the
f
0
−
4
γ
e
(
f
−
(1+
α
)
/
2
T
to
f
0
+(1+
α
)
/
2
T
band, that is, a bandwidth
2
W
=(1+
α
)
/T
or again
(1+
α
)
R
m
where
R
m
is the symbol rate. The spectral eciency of the M-QAM modulation
expressed in bit/s/Hz is then equal to:
D
2
W
R
m
log
2
(
M
)
(1 +
α
)
R
m
log
2
(
M
)
(1 +
α
)
η
=
=
=
(2.143)
where
D
is the bit rate of the transmission expressed in bit/s.
The spectral eciency increases as a function of the number of states
M
of the modulation but the performance of the modulation, in terms of error
probability, decreases as a function of this parameter
M
.
2.3.4 Expression of the error probability in presence of
Nyquist filtering
Let us determine the bit error probability provided by the source by considering,
for example, a 4-PSK modulation. In this case, each symbol
a
i
(respectively
b
i
)
transmits a bit
d
i
every
T
seconds. The error probability on the data
d
i
is
therefore identical to the error probability on the symbols
a
i
or
b
i
.L tus
calculate, for example, the error probability on symbol
a
i
.
The output of the reception filter of the in-phase component at sampling
time
t
0
+
nT
is equal to:
y
c
(
t
0
+
nT
)=
Aa
n
p
(
t
0
)+
b
c
(
t
0
+
nT
)
(2.144)
where
b
c
(
t
0
+
nT
)
is the real part of the noise
b
(
t
0
+
nT
)
.
Assuming the data
d
n
iid
, the error probability on the symbols
a
n
has the
expression:
Pe
a
n
=
2
Pr
{
y
c
(
t
0
+
nT
)
>
0
|
a
n
=
−
1
}
(2.145)
1
2
Pr
{
y
c
(
t
0
+
nT
)
<
0
|
a
n
=+1
}
Since
y
c
(
t
0
+
nT
)
is Gaussian, with mean
Aa
n
p
(
t
0
)
and variance
σ
b
c
, the error
probability
Pe
a
n
is equal to:
Pe
a
n
=
1
2
erfc
Ap
(
t
0
)
σ
b
c
√
2
(2.146)
The variance
σ
b
c
of the noise
b
c
(
t
0
+
nT
)
is equal to:
+
∞
+
∞
2
df
=
N
0
σ
b
c
=
N
0
|
G
r
(
f
)
|
p
(
t
0
)
CS
α
(
f
)
df
=
N
0
p
(
t
0
)
(2.147)
−∞
−∞
