European Journal of Scientific Research - EuroJournals

830 Taba Mohamed Tahar, S. Femmame and D. Mossadeg
⎡
π
0
n
exp j
⎣ 2 n=
−∞
= −α
nα n−1a
0,
n−2
⎦
−1
a = j
, n ⎢ ∑α n⎥
= α na
0,
n
⎡
π
a = j
j
, n α n ⎢ ∑α = n⎥
α na
0,
n
j exp 1 −2
2 n=
−∞
n
⎤
⎤
⎣ ⎦
In other words, the GMSK signal can be approximated with almost no error by the sum of two
QAM signals with pulse shapes h0(t) and h1(t). These two pulses in the linear approximation [3], are
shown in figure 1.
For the case L = 4, BT=0.3,
h0 ( t)
= β ( t − 4T
) β ( t − 3T
) β ( t − 2T
) β ( t − T )
0 ≤ t≤ 5T
(9)
h 1(
t)
= β ( t − T)
β ( t − 2T
) β ( t − 4T
) β ( t + T)
0 ≤ t ≤ 3T
With,
(10)
⎧sin[
πh
−πhψ
( t)]
⎪
,
sin( πh)
⎪
β ( t) = ⎨β
( −t),
⎪
0,
⎪
⎩
With h=0.5 β(t) becomes:
tε
[ 0,
LT )
tε
( LT , 0]
t ≥ 0
(11)
π ⎧cos(
2 g(
t))
⎪
β ( t) = ⎨β
( −t),
⎪
⎩0,
tε
[ 0,
LT )
1 t1Q
( σ t1)
−t2Q
( σ t2)
ψ(
t)
= +
−...
tε
( LT , 0]
2 2Tp
2 2
2 2
exp( −σ
/ 2)
t exp( σ / 2)
t ≥ 0
1 − − t2
−
2Tσ
2π
(12)
Tp
Tp
t1 = t − 2 , t 2 t + 2 , σ =
Tp: sampling period.
2πB = ln 2
Figure 1: Represents two pulse shapes in the GMSK linear approximation. The power in h1(t) is 0.48% of the
power in h0(t).
h 1(t)
Because the majority (99.5%) of signal energy in GMSK signal s(t) is contained in the first
pulse approximation h0(t) figure 1, we can further simplify s(t) into a single QAM transmission
∞
= ∑ ( t − nT ), a =
0
n j na
n
n 1
n=
−∞
s( t)
a h
α (13)
−
It can be noted that the approximation error may be viewed as an additive interference.
Therefore, even in noiseless channels, the maximum signal-to-noise ratio (SNR) of this approximation
h 0(t)
p