Problem Solutions for Chapter 12

12-1. We need to evaluate Pin using Eq. (12-11). Here Fc = 0.20,

CT = 0.05, Fi = 0.10, P0 = 0.5 mW, and A0 = e −2.3( 3) / 10 = 0.933

Values of Pin as a function of N are given in the table below. Pin in

Pin (mW )
dBm is found from the relationship Pin(dBm) = 10 log
1 mW

N Pin(nW) Pin(dBm) N Pin(nW) Pin(dBm)

2 387 -34.1 8 5.0 -53.0

3 188 -37.3 9 2.4 -56.2

4 91 -40.4 10 1.2 -59.2

5 44.2 -43.5 11 0.6 -62.2

6 21.4 -46.7 12 0.3 -65.5

7 10.4 -49.8

(b) Using the values in the above table, the operating margin for 8 stations is

-53 dBm - (-58 dBm) = 5 dB

(c) To have a 6-dB power margin, we can transmit over at most seven stations.

The dynamic range with N = 7 is found from Eq. (12-13):

DR = −10(N − 2) log [.933(.8) (.95) (.9)] = −50 log (0.485) = 15.7 dBm
2 2

12-2. (a) Including a power margin, we have from Eq. (12-16)

PS − PR − power m arg in = L excess + α(2L) + 2L c + 10log N

Thus

0 – (-38 dBm) – 6 dB = 3 dB + (0.3 dB/km)2(2 km) + 2(1.0 dB) + 10 log N

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 so that 10 log N = 25.8. This yields N = 380.1, so that 380 stations can be

attached.

(b) For a receiver sensitivity of –32 dBm, one can attach 95 stations.

12-3. (b) Let the star coupler be located in the ceiling in the wire room, as shown in the

figure below.

Wire
room
A B C D

For any row we need seven wires running from the end of the row of offices to

each individual office. Thus, in any row we need to have (1+2+3+4+5+6+7)x15 ft

= 420 ft of optical fiber to connect the offices. From the wiring closet to the

second row of offices (row B), we need 8(10 + 15) ft = 200 ft; from the wiring

closet to the third row of offices (row C), we need 8(10 + 30) ft = 320 ft; and from

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 the wiring closet to the fourth row of offices (row D), we need 8(20 + 45) ft = 520

ft of cable. For the 28 offices we also need 28x7 ft = 196 ft for wall risers.

Therefore for each floor we have the following cable needs:

(1) 4 x 420 ft for row runs

(2) 200 + 320 + 520 ft = 1040 for row connections

(3) 196 ft for wall risers

Thus, the total per floor = 2916 ft

Total cable in the building: 2x9 ft risers + 2916 ft x 2 floors = 5850 ft

12-4. Consider the following figure:

d
N

(a) For a bus configuration:

Cable length = N rows×(M-1)stations/row + (N-1) row interconnects

= N(M-1)d + (N-1)d = (MN-1)d

(b) The ring is similar to the bus, except that we need to close the loop with one

cable of length d. Therefore the cable length = MNd

(c) In this problem we consider the case where we need individual cables run

from the star to each station. Then the cable length is

L = cables run along the M vertical rows + cables run along the N horizontal

rows:

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 N −1 M −1
N(N − 1) M(M − 1) MN
= Md ∑ i + Nd ∑j= M
2
d+N
2
d =
2
(M + N − 2)d
i =1 j=1

12-5. (a) Let the star be located at the relative position (m,n). Then

 m−1 M− m n−1 N−n 

L = N ∑ j + N ∑ j + M∑ i + M ∑ i d
 j=1 j =1 i =1

i =1 

 m(m − 1) (M − m)(M − m + 1)  n(n − 1) (N − n)(N − n + 1)  

= N  + + M + d
  2 2   2 2 

MN
=  (M + N + 2) − Nm(M − m + 1) − Mn(N − n + 1) d
 2 

(b) When the star coupler is located in one corner of the grid, then

m = n= 1, so that the expression in (a) becomes

MN MN
L=  (M + N + 2) − NM − MN  d = (M + N − 2)d
 2  2

(c) To find the shortest distance, we differentiate the expression for L given in (a)

with respect to m and n, and set the result equal to zero:

dL M +1
= N(m - 1 - M) + Nm = 0 so that m=
dm 2
Similarly

dL N +1
= M(n - 1 - N) + nM = 0 yields n=
dn 2
Thus for the shortest cable runs the star should be located in the center of the grid.

12-6. (a) For a star network, one cannot reuse wavelengths. Thus, since each node must

be connected to N – 1 other nodes through a central point, we need N – 1

wavelengths.

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 For a bus network, these equations can easily be verified by drawing sample

diagrams with several even or odd stations.

For a ring network, each node must be connected to N – 1 other nodes. Without

wavelength reuse one thus needs N(N – 1) wavelengths. However, since each

wavelength can be used twice in the network, the number of wavelengths needed

is N(N-1)/2.

12-7. From Tables 12-4 and 12-5, we have the following:

OC-48 output for 40-km links: –5 to 0 dBm; α = 0.5 dB/km; PR = -18 dBm

OC-48 output for 80-km links: –2 to +3 dBm; α = 0.3 dB/km; PR = -27 dBm
The margin is found from: Margin = (Ps − PR ) − αL − 2L c

(a) Minimum power at 40 km:

Margin = [-2 – (-27)] – 0.5(40) –2(1.5) = +2 dB

(b) Maximum power at 40 km:

Margin = [0 – (-27)] – 0.5(40) –2(1.5) = +4 dB

(c) Minimum power at 80 km:

Margin = [-2 – (-27)] – 0.3(80) –2(1.5) = -2 dB

(d) Maximum power at 80 km:

Margin = [3 – (-27)] – 0.3(80) –2(1.5) = +3 dB

12-8. Expanding Table 12-6:

P = 10
P1 / 10
# of λs P1(dBm) (mW) Ptotal(mW) Ptotal(dBm)

1 17 50 50 17

2 14 25 50 17

3 12.2 16.6 49.8 17

4 11 12.6 50.4 17

5 10 10 50 17

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 6 9.2 8.3 49.9 17

7 8.5 7.1 49.6 17

8 8.0 6.3 50.4 17

12-9. See Figure 20 of ANSI T1.105.01-95.

12-10. See Figure 21 of ANSI T1.105.01-95.

12-11. The following wavelengths can be added and dropped at the three other nodes:

Node 2: add/drop wavelengths 3, 5, and 6

Node 3: add/drop wavelengths 1, 2, and 3

Node 4: add/drop wavelengths 1, 4, and 5

12-12. (b) From Eq. (12-18) we have

N λ = kpk +1 = 2(3)3 = 54

(c) From Eq. (12-20) we have

2(3)2 (3 − 1)(6 − 1) − 4(32 − 1)

H= = 2.17
2(3 − 1)[2(32 ) − 1]

(d) From Eq. (12-21) we have

2(3)
3

C= = 8.27
2.17
12-13. See Hluchyj and Karol, Ref. 25, Fig. 6, p. 1391 (Journal of Lightwave

Technology, Oct. 1991).

12-14. From Ref. 25:

In general, for a (p,k) ShuffleNet, the following spanning tree for assigning fixed

routes to packets generated by any given user can be obtained:

h Number of users h hops away from the source

1 p

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 2 p2

k–1 pk-1

k Pk - 1

k+1 Pk - p

k+2 Pk - p2

2k – 1 Pk - pk-1

Summing these up results in Eq. (12-20).

12-15. See Li and Lee (Ref 40) for details.

12-16. The following is one possible solution:

(a) Wavelength 1 for path A-1-2-5-6-F

(b) Wavelength 1 for path B-2-3-C

(c) Wavelength 2 for the partial path B-2-5 and Wavelength 1 for path 5-6-F

(d) Wavelength 2 for path G-7-8-5-6-F

(e) Wavelength 2 for the partial path A-1-4 and Wavelength 1 for path 4-7-G

12-17. See Figure 4 of Barry and Humblet (Ref. 42).

12-18. See Shibata, Braun, and Waarts (Ref. 67).

(a) The following nine 3rd-order waves are generated due to FWM:

ν113 = 2(ν2 - ∆ν) – (ν2 + ∆ν) = ν2 - 3∆ν

ν112 = 2(ν2 - ∆ν) – ν2 = ν2 - 2∆ν

ν123 = (ν2 - ∆ν) + ν2 – (ν2 + ∆ν) = ν2 - 2∆ν

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ν223 = 2ν2 – (ν2 + ∆ν) = ν2 - ∆ν = ν1

ν132 = (ν2 - ∆ν) + (ν2 + ∆ν) – ν2 = ν2

ν221 = 2ν2 – (ν2 - ∆ν) = ν2 + ∆ν = ν3

ν231 = ν2 + (ν2 + ∆ν) – (ν2 - ∆ν) = ν2 + 2∆ν

ν331 = 2(ν2 + ∆ν) – (ν2 - ∆ν) = ν2 + 3∆ν

ν332 = 2(ν2 + ∆ν) – ν2 = ν2 + 2∆ν

(b) In this case the nine 3rd-order waves are:

ν113 = 2(ν2 - ∆ν) – (ν2 + 1.5∆ν) = ν2 – 1.5∆ν

ν112 = 2(ν2 - ∆ν) – ν2 = ν2 - 2∆ν

ν123 = (ν2 - ∆ν) + ν2 – (ν2 + 1.5∆ν) = ν2 – 2.5∆ν

ν223 = 2ν2 – (ν2 + 1.5∆ν) = ν2 – 1.5∆ν

ν132 = (ν2 - ∆ν) + (ν2 + 1.5∆ν) – ν2 = ν2 + 0.5∆ν

ν221 = 2ν2 – (ν2 - ∆ν) = ν2 + ∆ν

ν231 = ν2 + (ν2 + 1.5∆ν) – (ν2 - ∆ν) = ν2 + 2.5∆ν

ν331 = 2(ν2 + 1.5∆ν) – (ν2 - ∆ν) = ν2 + 4∆ν

ν332 = 2(ν2 + 1.5∆ν) – ν2 = ν2 + 3∆ν

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12-19. Plot: from Figure 2 of Y. Jaouën, J-M. P. Delavaux, and D. Barbier, “Repeaterless

bidirectional 4x2.5-Gb/s WDM fiber transmission experiment,” Optical Fiber

Technology, vol. 3, p. 239-245, July 1997.

12-20. (a) From Eq. (12-50) the peak power is

A eff λ3 D
2
 1.7627 
Ppeak = 2 = 11.0 mW
2π  n 2 c Ts

(b) From Eq. (12-49) the dispersion length is

Ldisp = 43 km

(c) From Eq. (12-51) the soliton period is

π
L period = L = 67.5 km
2 disp
(d) From Eq. (12-50) the peak power for 30-ps pulses is

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  1.7627 
2
A eff λ3 D
Ppeak = = 3.1 mW
 2π  n 2 c Ts2

12-21. Soliton system design.

12-22. Soliton system cost model.

12-23. (a) From the given equation, Lcoll = 80 km.

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(b) From the given condition, L amp ≤ L = 40 km
2 coll

12-24. From the equation and conditions given in Prob. 12-23, we have that

Ts 20 ps
∆λ max = = = 2 nm
DL amp [0.4 ps /(nm ⋅ km)](25 km)

Thus 2.0/0.4 = 5 wavelength channels can be accommodated.

12-25 Plot from Figure 3 of Ref. 103.

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