Page 393 - Fiber Optic Communications Fund

374 Fiber Optic Communications

Using Eqs. (8.290), (8.291), and (8.293), we obtain
∞ ∞
P(0|1)= p (i )di p (i )di
∫ 1 1 1 ∫ 1 0 0
0 i 1
√
[ ] ⎛ 2 ⎞
2
∞ RE + i i RE ∞ ( )
1
1 1 1 ⎜ 1 ⎟ i 0
= exp − I di 1 exp − di 0
(2 ) ∫ 0 2 2 0 ⎜ 2 ⎟ ∫ 2 2
2 2
⎜ ⎟ i 1
⎝ ⎠
√
[ ] ⎛ 2 ⎞
2
∞ RE + 2i i RE
1
1 1 1 ⎜ 1 ⎟
= exp − I di . (8.295)
1
2 ∫ 0 2 2 0 ⎜ 2 ⎟
2
⎜ ⎟
⎝ ⎠
2
′
Let i = 2i and x = RE ∕2. Now, Eq. (8.295) becomes
1
1 1
√
[ ] ⎛ ′ ⎞
∞ ′ i x
1 2x + i 1 ⎜ 1 ⎟ ′
P(0|1)= exp − I di
4 ∫ 0 2 2 0 ⎜ ⎜ 2 ⎟ 1
2
⎟
⎝ ⎠
√
[ ] ⎛ ′ ⎞
∞ ′ i x
1 − x x + i 1 ⎜ 1 ⎟ ′
= e 2 2 exp − I ⎟ di 1
4 2 ∫ 0 2 2 0 ⎜ 2
⎜ ⎟
⎝ ⎠
∞
x
1 −
= e 2 2 p (i )di 1
1 1
2 ∫ 0
( )
1 E 1
= exp − . (8.296)
2 2
ASE
For FSK, E = E = E . Using Eq. (8.268), Eq. (8.296) may be rewritten as
1
av
0
( DD )
1
P(0|1)= exp − . (8.297)
2 2
Case (ii): bit ‘0’ transmitted. Owing to the symmetry of the problem, P(0|1) is same as P(1|0). The error
probability is
( DD )
1 1
P = [P(0|1)+ P(1|0)] = exp − (8.298)
b
2 2 2
Note that this error probability is the same as that given by Eq. (8.251) for asynchronous detection if we
replace het by 2 DD . If we ignore shot noise, we see that two expressions are identical.
8.5.3 DPSK
In the case of PSK, the information is transmitted as the absolute phase of the complex field envelope s(t).
But in the case of DPSK, the information is transmitted as the phase of the field envelope relative to the
previous bit. To estimate the absolute phase of the transmitted PSK signal, a reference is required at the
receiver. This reference is provided by the local oscillator whose phase should be synchronized with that of