Cryptography Reference
In-Depth Information
signal
s
1
(
t
)
, is therefore Gaussian, with mean
√
E
b
and variance
N
0
.Thus
probability
Peb
1
is equal to:
exp
−
√
E
b
)
2
2
N
0
dz
0
1
√
2
πN
0
(
z
Peb
1
=
−
−∞
Introducing the complementary error function,
Peb
1
is written:
2
erfc
E
b
Peb
1
=
1
2
N
0
It is easy to check that the error probability conditionally to the emission of
signal
s
2
(
t
)
is identical to the error probability conditionally to the emission of
signal
s
1
(
t
)
and thus, we obtain:
2
erfc
E
b
1
Peb
=
(2.91)
2
N
0
If we compare the performance of 2-FSK modulation to that of 2-PSK we note
that the former requires 3 dB more signal to noise ratio to obtain the same
performance as the latter.
Continuous phase frequency shift keying - CPFSK
For continuous phase frequency shift keying, the modulated signal has the ex-
pression:
S
(
t
)=
A
cos(2
πf
0
t
+
φ
(
t
))
(2.92)
where the phase
φ
(
t
)
,ontheinterval
[
iT,
(
i
+1)
T
[
,isequalto:
i
φ
(
t
)=2
πh
a
n
q
(
t
−
nT
)+
θ
i−L
(2.93)
n
=
i−L
+1
with:
i−L
θ
i−L
=
πh
a
n
n
=
−∞
h
is the modulation index and the symbols
a
i
are
M
-ary in the general case.
They take their values in the alphabet
{±
1
,
±
3
,
···
,
±
(2
p
+1)
,
···
,
±
(
M
;
M
=2
m
.
If the modulation index
h
=
m/p
where
m
and
p
are relatively prime integers,
phase
θ
i−L
takes its values in the following sets:
−
1)
}
0
,
(
p−
1)
πm
p
πm
p
2
πm
p
θ
i−L
∈
,
,
···
,
if
m
even
(2.94)
0
,
(2
p−
1)
πm
p
πm
p
2
πm
p
θ
i−L
∈
,
,
···
,
if
m
odd
