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Unformatted text preview: Optical Communication Components and Subsystems ECE 565 Spring 2005 Homework 1 Solutions 1. Calculate the frequency and photon energy of optical communication systems operating at 0.88, 1.3 and 1.55 m. This is best done with a short program. Solution: The frequency of a photon is related to its energy through the relation E = h , where h is Planks constant, h =6.6262 x 10-34 Js. Also, = c / , where c is the speed of light in vacuum c = 2.9979x10 8 m/s, and is the photons wavelength. ( m ) (Hz) E ( J ) E ( eV ) 0.88 3.4067 x10 14 0.2257 x10-18 1.4089 1.3 2.3061 x10 14 0.1528 x10-18 0.9537 1.55 1.9341 x10 14 0.1282 x10-18 0.7999 Matlab code: h=6.6262e-34; c=2.9979e8; lambda=[0.88 1.3 1.55]*1.0e-6; nu = c ./lambda; E=h*nu; Eev = E/1.6022e-19; 2. Calculate the distances over which the optical power is attenuated by a factor of 10 for fibers with losses 0.2 dB/km and 2000 dB/km. Write a program to compute this distance for any loss quoted in dB/km. Solution: If we launch an optical power of P over distance L (in km), and assume that the power at the destination P , then the loss coefficient in dBm/km is = - L-1 10 log 10 ( P / P ). In this problem, P / P = 0.1....
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