"Introduction". In: Fiber-Optic Communication Systems

496 CHAPTER 10. COHERENT LIGHTWAVE SYSTEMS

branches of the dual filter receiver, where both of them include noise currents through
Eq. (10.4.12). Consider the case in which 1 bits are received in the upper branch. The
currentIis then given by Eq. (10.4.12) and follows a Rice distribution withIp=I 1
in Eq. (10.4.13). On the other hand,I′consists only of noise and its distribution is
obtained by settingIp=0 in Eq. (10.4.13). An error is made whenI′>I, as the signal
is then below the decision level, resulting in

P( 0 / 1 )=

∫∞

0

p(I,I 1 )

[∫∞

I

p(I′, 0 )dI′

]

dI, (10.4.20)

where the inner integral provides the error probability for a fixed value ofIand the
outer integral sums it over all possible values ofI. The probabilityP( 1 / 0 )can be
obtained similarly. In fact,P( 1 / 0 )=P( 0 / 1 )because of the symmetric nature of a
dual-filter receiver.
The integral in Eq. (10.4.20) can be evaluated analytically. By using Eq. (10.4.13)
in the inner integral withIp=0, it is easy to verify that

∫∞

I

p(I′, 0 )dI′=exp

(

−

I^2

2 σ^2

)

. (10.4.21)

By using Eqs. (10.4.14), (10.4.20), and (10.4.21) withP( 1 / 0 )=P( 0 / 1 ), the BER is
given by

BER=

∫∞

0

I

σ^2

exp

(

−

I^2 +I 12

2 σ^2

)

I 0

(

I 1 I

σ^2

)

exp

(

−

I^2

2 σ^2

)

dI, (10.4.22)

wherep(I,Ip)was substituted from Eq. (10.4.13). By introducing the variablex=

√

2 I,

Eq. (10.4.22) can be written as

BER=

1

2

exp

(

−

I^2

4 σ^2

)∫∞

0

x

σ^2

exp

(

−

x^2 +I 12 / 2

2 σ^2

)

I 0

(

I 1 x

σ^2

√

2

)

dx. (10.4.23)

The integrand in Eq. (10.4.23) is justp(x,I 1 /

√

2 )and the integral must be 1. The BER
is thus simply given by

BER=^12 exp(−I 12 / 4 σ^2 )=^12 exp(−SNR/ 4 ). (10.4.24)

By using SNR= 2 ηNpfrom Eq. (10.1.14), we obtain the final result

BER=^12 exp(−ηNp/ 2 ), (10.4.25)

which should be compared with Eq. (10.4.11) obtained for the case of synchronous
FSK heterodyne receivers. Figure 10.7 compares the BER in the two cases. Just as
in the ASK case, the BER is larger for asynchronous demodulation. However, the
difference is small, and the receiver sensitivity is degraded by only about 0.5 dB com-
pared with the synchronous case. If we assume thatη=1,Np=40 at a BER of 10−^9
(Np=36 in the synchronous case).N ̄palso equals 40, since the same number of pho-
tons are received during 1 and 0 bits. Similar to the synchronous case,N ̄pis the same
for both the ASK and FSK formats.