ECE-T603 Homewrok 1 Solution

Spring, AY 05-06

1. (a) Since λ = c/f ,

∆λ ≈ −c

f2∆f = −

λ2

c∆f

(b) For the λ = 1.29 and 1.33µm lasers, use D(λ) ≈ S(λ − λZD) to get the dispersionparameters and then compute the BL product according to BL < 1/(D∆λ). For theλ = 1.31µm = λZD laser, consider the higher-order dispersion and compute the BLproduct according to BL < 1/[S(∆λ)2].

2. The basic formula to use is

L =Pin − Pout

α

where the powers are measured in dB. Here, Pin = 10 log 3 dBm and Pout = −20 dBm, andα = 0.4 and 0.2 for the two wavelengths.

3.

4. For dispersion-limited distance of the three systems is determined by(i)

L =2c

n1∆2B(1)

and n1 = 1.46, ∆ = 0.01.(ii)

L =1

|D|σλB(2)

and D = 1 ps/(km-nm), σλ = 2 nm.(iii)

L ==1

Sσ2λB

(3)

and S = 0.05 ps/(km-nm2), σλ = 2 nm.

The corresponding Distance versus Bit-Rate curves are shown in figure below

5. (a) Ptr = 1.5 mW = 1.76 dBmαf = 0.4 db/km (A typical value for SMF at 1.3 µm not given in the problem)αsplice = 0.1 dB/kmαcon = 0.2 dBPre = −30 db/km for 622 Mb/s and −38 dB/km for 155 Mb/sPmargin = 7 dBFor 622 Mb/s system, 1.76 − (−30) = (0.4 + 0.1)L + 0.2 + 7, hence the distance shouldnot exceed L = 48.72 km. For 155 Mb/s system, the result should be L = 64.72 km.

1

107

108

109

1010

1011

1012

0

100

200

300

400

500

600

Bit Rate

Dis

tanc

e (k

m)

0.85, MMF−GI, Power−limited1.30, SMF, Power−limited1.55, DSF, Power−limited0.85, MMF−GI, Dispersiona−limited1.30, SMF, Dispersion−limited1.55, DSF, Dispersion−limited

Figure 1: Distance versus bit-rate curves

(b) If NRZ is used, Tsys ≤ 0.70/BTre = 0.2 nsTtr = 0.25 nsFor SMF Tf = TGV D = DL∆λ = 0.002L nsAccording to T 2

sys = T 2tr +T 2

f +T 2re, for 622 Mb/s system, Tsys ≤ 0.70/622×106 = 1.1254

ns, then (1.1254)2 = (.25)2 +(.002L)2 +(.2)2 results in L ≤ 539.45 km. Similarly, for 155Mb/s system, L ≤ 2252.37 km.

(c) Judging from (a) and (b), both system are loss limited. For 622 Mb/s system the maxi-mum distance should not exceed 48.72 km. For 155 Mb/s system the maximum distanceshould not exceed 64.72 km.

(d) The extinction ratio will course the power penalty of δex = 10 log(

1+rex

1−rex

)

= 0.8715

dB, then the maximum distance will change by ∆L = 0.8715/(0.1 + 0.4) = 1.743 km.For 622Mb/s system, the L = 48.72-1.743 = 46.98 km. For 155Mb/s system, the L =64.72-1.743=62.98 km.

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