The data is impressed upon the carrier frequency. Therefore,
the M M different signals are
s
m
t=A
P
T
tcos2π
f
c
t+2π(m−1)Δft+
θ
m
s
m
t
A
P
T
t
2
f
c
t
2
m
1
Δ
f
t
θ
m
(1)
for
m∈12…M
m
1
2
…
M
The M M different signals have
M M different carrier frequencies
with possibly different phase angles since the generators of
these carrier signals may be different. The carriers are
f
2
=
f
c
+Δf
f
2
f
c
Δ
f
f
M
=
f
c
−MΔf
f
M
f
c
M
1
Δ
f
Thus, the
MM signals may be
designed to be orthogonal to each other.
〈
s
m
,
s
n
〉=∫0TA2cos2π
f
c
t+2π(m−1)Δft+
θ
m
cos2π
f
c
t+2π(n−1)Δft+
θ
n
dt=A22∫0Tcos4π
f
c
t+2π(n+m−2)Δft+
θ
m
+
θ
n
dt+A22∫0Tcos2π(m−n)Δft+
θ
m
−
θ
n
dt=A22sin4π
f
c
T+2π(n+m−2)ΔfT+
θ
m
+
θ
n
−sin
θ
m
+
θ
n
4π
f
c
+2π(n+m−2)Δf+A22(sin2π(m−n)ΔfT+
θ
m
−
θ
n
2π(m−n)Δf−sin
θ
m
−
θ
n
2π(m−n)Δf)
s
m
s
n
t
0
T
A
2
2
f
c
t
2
m
1
Δ
f
t
θ
m
2
f
c
t
2
n
1
Δ
f
t
θ
n
A
2
2
t
0
T
4
f
c
t
2
n
m
2
Δ
f
t
θ
m
θ
n
A
2
2
t
0
T
2
m
n
Δ
f
t
θ
m
θ
n
A
2
2
4
f
c
T
2
n
m
2
Δ
f
T
θ
m
θ
n
θ
m
θ
n
4
f
c
2
n
m
2
Δ
f
A
2
2
2
m
n
Δ
f
T
θ
m
θ
n
2
m
n
Δ
f
θ
m
θ
n
2
m
n
Δ
f
(3)
If
2
f
c
T+(n+m−2)ΔfT
2
f
c
T
n
m
2
Δ
f
T
is an integer, and if
(m−n)ΔfT
m
n
Δ
f
T
is also an integer, then
〈
S
m
,
S
n
〉=0
S
m
S
n
0
if
ΔfT
Δ
f
T
is an integer,
then
〈
s
m
,
s
n
〉≃0
s
m
s
n
0
when
f
c
f
c
is much larger than
1T
1
T
.
In case
∀m,
θ
m
=0:
θ
m
=0
m
θ
m
0
〈
s
m
,
s
n
〉≃A2T2sinc2(m−n)ΔfT
s
m
s
n
A
2
T
2
sinc
2
m
n
Δ
f
T
(4)
Therefore, the frequency spacing could be as small as
Δf=12T
Δ
f
1
2
T
since
sincx=0
sinc
x
0
if
x=±1
x
±
1
or
±2
±
2
.
If the signals are designed to be orthogonal then the average
probability of error for binary FSK with optimum receiver is
P
‐
e
=Q
E
s
N
0
P
‐
e
Q
E
s
N
0
(5)
in AWGN.
Note that
sincx
sinc
x
takes its minimum value not at
x=±1
x
±
1
but at
±1.4
±
1.4
and the minimum value is -0.216-0.216.
Therefore if
Δf=0.7T
Δ
f
0.7
T
then
P
‐
e
=Q1.216
E
s
N
0
P
‐
e
Q
1.216
E
s
N
0
(6)
which is a gain of
10×log1.216≃0.85dθ
10
1.216
0.85
d
θ
over orthogonal FSK.
