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# hw4_sol - Optical Communication Components and Subsystems...

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Optical Communication Components and Subsystems ECE 565 Spring 2005 Homework 4 Solutions Problem 2.16. Received power in dm = 10log 10 0.3x10 -3 = -35.2288 dBm Losses: Fiber loss = (50)(0.5)=25 dBm Connector loss = (2)(1)=2 dBm Number of sections =50/5; Number of tips = 11; Number of spliced ends = 11 –2=9; Splice loss = (9)(0.2)=1.8 dBm Total loss = 28.8 dBm Power budget equation (in dBm): Launched power - total loss Receive power Or minimum launched power = 28.8 dBm -35.2288 dBm = -6.4288 dBm = 0.2276 mW. Problem 2.17. Assuming that fiber loss is negligible, Eq. (2.6.14) becomes: φ NL = γ P in L. We can calculate γ using γ = λ π eff A n / 2 2 . Thus, with the given parameters, γ = 2.1 W -1 /km. With P in = 6dBm and by setting φ NL = 2 π , we obtain L=745 km. We need to minimize (T 1 /T 0 ) in Eq.(2.4.17) with respect to z. For, simplicity, we equivalently minimize (T 1 /T 0 ) 2 by differentiating and setting the derivative to zero. (Note that the coefficient of z 2 in Eq.(2.4.17) is always positive and the critical point of the above quadratic equation is therefore a minimum.) d(T 1 /T 0 ) 2 /dz = (d/dz)[(1+C β 2 z/ T 0 2 ) 2 + ( β 2 z/ T 0 2 ) 2 ] = 2(C β

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hw4_sol - Optical Communication Components and Subsystems...

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