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Unformatted text preview:  Delta are used in calculating the cutoﬀ wavelength. Problem 3 From the assumptions, we can derive the pulse spread and the group velocity dispersion • Δ τ = 1 . 732 ps • β 2 = 0 . 1732 ps 2 /km 1. The pulse output width is equal to 3.6 ps 2. The pulse output width is equal to 2.2 ps 1 Problem 4 From the course, the pulse broadening relation is T z T = s ± 1 + κβ 2 z T 2 ² 2 + ± β 2 z T 2 ² 2 This equation rewrites: T z T = v u u t ± 1κz L D ² 2 + ± z L 2 D ² 2 1. From the diﬀerentiation of the above equation, we yield to z min = κ 1 + κ 2 L D For κ = 4, we obtain z min = 0 . 154 L D 2. The pulse width equals that of an unchirped pulse if bigg (1κz L D ² 2 + ± z L 2 D ² 2 = 1 + ± z L 2 D ² 2 For κ = 4, we obtain z min = 0 . 5 L D 2...
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This note was uploaded on 09/01/2010 for the course ECE 6543 taught by Professor Boussert during the Spring '09 term at Georgia Tech.
 Spring '09
 Boussert

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