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# hw8_sol - We also use α f =1dB/km α con = 2dB M s = 6 dB...

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Optical Communication Components and Subsystems ECE 565 Spring 2005 Homework 8 Solutions Problem 5.1. We use equation (5.1.1): P n = P T C {(1- δ )(1-C} n-1 . Generally, the insertion loss factor δ is (p in -p out )/p out = p in /p out –1. Thus, if p in /p out = -1 dB, then 1- δ =10 -0.1 = 0.794. Also, C=0.1 in this case. Use P T =1 mW to calculate: P 8 = 9.5 µ W, P 9 = 6.8 µ W and P 10 = 4.9 µ W. Problem 5.2. As in Problem 5.1, P n = P T C {(1- δ )(1-C} n-1 . In this case we want P n = 100 nW. Here, C=0.05, and 1- δ =10 -0.05 = 0.8913. Thus, N can be solved for as 38.4; taking the integer part yields N=38. Problem 5.4. We use the power-budget equation s splice con f rec tr M L P P + + + + = α α α . Now 1 . 0 = tr P mW= -10dBm and 100 = rec P nW = -40 dBm.
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Unformatted text preview: We also use α f =1dB/km, α con = 2dB, M s = 6 dB, α splice = 0.2 └ L/2 ┘ dB. Thus, we solve for L and obtain L=20 km. Problem 5.9. From Eq.(5.2.11) and Eq.(5.2.12) in text, we have 2 2 2 mod 2 2 rec GVD la tr r T T T T T + + + = . Now, T tr = 10 ns and T rec = 15 ns. Also for a graded-index fiber, T modal ≈ (n 1 ∆ 2 /8c)L= 0.60 ns. Also, T GVD = |D|L ∆λ = 40 ns. Thus, T r = 43.9 ns. From Eq.(5.2.10), we know that T r should be < 0.35/B = 7 ns for RZ modulation and T r should be < 0.7/B = 14 ns for NRZ modulation. Hence, none of these formats can be supported using the current system....
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