Page 400 - Fiber Optic Communications Fund

Performance Analysis 381

Comparing Eqs. (8.327) and (8.328), we find that PSK and DPSK have a similar exponential dependence on
when is large. Fig. 8.29 shows the performance of PSK with homodyne receiver and DPSK with direct
−3
detection receiver. To reach a BER of 10 , the DPSK requires roughly 1.2 dB more power than the PSK
−9
when the noise power is kept constant. However, to reach a BER of 10 , the DPSK requires only 0.5 dB
more power than the PSK.

8.6 Additional Examples

Example 8.4
An optical signal passes through the single-mode fiber with dispersion coefficient and length L. The trans-
2
mitted field envelope is
[ ]
⎧ (t−T B ∕2) 2
⎪±A exp − for 0 < t < T
x (t)= 2T 2 B (8.329)
tr ⎨ 0
⎪ 0 otherwise.
⎩
Find the filter matched to the received signal. Ignore and . Assume T << T .
3
1
0
B
Solution:
Since T ≪ T , we approximate that the signal is Gaussian in the range [−∞, ∞] since T << T :
0
B
0
b
x (t)= s (t − T ∕2)., (8.330)
B
tr
i
( )
t 2
s (t)=±A exp − (8.331)
i 2
2T
0
From Eq. (2.153), the Fourier transform of the signal s (t) after the fiber transmission is
0
̃ s ()= ̃s ()H (, L)
0
f
i
( 2 )
A
=± exp − , (8.332)
a 4b 2
1 1
= − i2 L (8.333)
2
b 2 a 2
1
T . (8.334)
0
a = √
2
The received signal is
x (t)= s (t − T ∕2), (8.335)
o o B
x ()= ̃s ()e iT b ∕2 . (8.336)
o o
Let the received signal corresponding to bit ‘1’ and bit ‘0’ be x (t) and x (t), respectively:
o,0
o,1
[ ]
A 2 iT b
̃ x ()=−̃x ()= exp − + . (8.337)
o,1 o,0 2
a 4b 2