Page 372 - Fiber Optic Communications Fund

Performance Analysis 353

and
(√ )
P het = 1 erfc E av . (8.137)
b 2 hf

The corresponding expressions for homodyne detection are

hf
homo
N = , (8.138)
0 2
(√ )
P homo = 1 erfc 2E av . (8.139)
b 2 hf

Comparing Eqs. (8.137) and (8.139), we find that the homodyne receiver has a 3-dB advantage over the
heterodyne receiver when the shot noise is dominant [3]–[5]. In other words, to reach a fixed error probability,
the received signal power in the case of a heterodyne receiver should be twice that of a homodyne receiver.
This can be explained as follows. In the case of the heterodyne receiver, the signal is modulated by a carrier at
a frequency . Since the average energy of the signal s(t) cos ( t +Δ) is half of that of the transmitted
IF
IF
signal s(t), the heterodyne receivers have a 3-dB disadvantage over the homodyne receivers. However, for
long-haul systems, the noise due to in-line amplifiers is dominant and, hence, the performances of homodyne
and heterodyne receivers are roughly the same.
Sine E ∕hf is the mean number of received signal photons N , Eqs. (8.137) and (8.139) may be rewritten as
av s
1 √
het
P = erfc( N ), (8.140)
b s
2
√
P homo = 1 erfc( 2N ). (8.141)
b 2 s
√
1
−9
For example, when = 1 and N = 9, P homo = erfc( 18)= 10 , in agreement with Eq. (7.63). When N =
s
s
b
2
18, P het = 10 −9 [4].
b
8.4.2 OOK: Synchronous Detection
The received signal in the absence of noise can be written as
{
= I (t) for ‘1’
1
I(t) (8.142)
= I (t) for ‘0’
0
I (t)= s (t) cos ( t), j = 0, 1, (8.143)
j
IF
j
with s (t)= 0. Here Δ is set to zero for simplicity. Proceeding as in Section 8.4.1, we find
0
(√ )
1 het
P = erfc . (8.144)
b
2 4