Page 405 - Fiber Optic Communications Fund

386 Fiber Optic Communications

Since the filter is not matched to x(t), Eq. (8.35) can not be used to find max . Instead, we use Eqs. (8.27) and
(8.28) to calculate P . The Fourier transform of Eq. (8.365) is
b
̃ s() ̃ s( − 2 )e −i2Δ ̃ s( + 2 )e i2Δ
IF
IF
̃ x()= + + . (8.366)
2 4 4
The transfer function of the filter matched to s(t) is
∗
∗
H ()=[̃s ()− ̃s ()] exp (iT )
s
b
0
1
∗
= 2̃s () exp (iT ). (8.367)
1 b
Substituting Eqs. (8.366) and (8.367) into Eq. (8.6), we find
∞
1 ̃ s () ∗
1
u (T )= ⋅ 2̃s ()d, (8.368)
1 b 2 ∫ −∞ 2 1
where we have ignored the overlap between the frequency components at ± 2 and since the bandwidth
IF
of s(t) is assumed to be much smaller than . From Parseval’s relations, it follows that
IF
u (T )= E =−u (T ). (8.369)
b
0
1
b
1
Next, let us consider the noise propagation. Let the noise before the demodulator be
n (t)= n ASE (t)+ n (t), (8.370)
d
het
n ASE = n cos + n cQ sin , (8.371)
cI
n = n cos + n dQ sin , (8.372)
dI
d
= t +Δ, (8.373)
IF
n (t)= n het,I cos + n het,Q sin , (8.374)
het
n (t)= n + n , (8.375)
het,I cI dI
n (t)= n + n . (8.376)
het,Q cQ dQ
After the synchronous demodulator, the noise is
n(t)= n (t) cos
het
2
= n het,I (t)cos + n het,Q (t) sin cos
n het,I (t) n het,I (t) cos 2 + n het,Q sin 2
= + , (8.377)
2 2
̃ n het,I ()
̃ n()= + terms at 2 . (8.378)
IF
2
The components of ̃n() around 2 are removed by the low-pass filter, H () placed just before the decision.
IF
s
Hence, we ignore these terms and obtain
2
⟨|̃n het,I ()| ⟩
2
⟨|̃n()| ⟩ = (8.379)
4