Cristian Parra Alvial Elo 352: Comunicaciones por fibra óptica Profesor: Ricardo Olivares Fecha: 10/10/2017
Problema 2.1 A multimode fiber with a 50-µm core diameter is designed to limit the t he intermodal dispersion to 10 ns/km. What is the numerical aperture of this fiber? What is the limiting bit rate for transmission over 10 km at 0.88 µm? Use 1.45 for the refractive index of the cladding.
Δ = ⋅ ≈ Δ = 1010 − 1 0 ⋅ 1 0 = 1000 ⋅3⋅10 + = 1,453 = = 0,0933 Δ = = 2,06⋅10− < ⋅ ⋅Δ < 10
Reemplazando los valores obtenidos en la figura de mérito se obtiene el bit rate para el e l enlace:
Problema 2.5 A single-mode single-mode fiber has an index step n1 −n2 = 0.005. Calculate the core radius if the fiber has a cutoff wavelength of 1 µm. Estimate the spot size (FWHM) of the fiber mode and the fraction of the mode power inside the core when this fiber is used at 1.3 µm. Use n1 = 1.45.
= 1 = 1,445 N = = = 0,12031 = 2 ⋅ V = 2,405 = 3,18 = 1,3 = 2 ⋅3,⋅1,3,31818 ⋅ = 1,849849
Como opera en monomodo la frecuencia normalizada de cort e
está en el límite:
Se tiene que el radio es:
Para el caso de
Se tiene que la frecuencia normalizada de corte es:
Extrayendo información del gráfico de flujo de potencias se encuentra que:
= 0,66
Problema 2.6 A 1.55-µm unchirped Gaussian pulse of 100-ps width (FWHM) is launched into a single-mode fiber. Calculate its FWHM after 50 km if the fiber has a dispersion of 16 ps/(km-nm). Neglect the source spectral width.
= 2 ln2 ⋅ = √ = = 6 ⋅ 10− () = 11 + 2 + 1 + 2 + 1 + + 4√ 2 = 0 = 0 () = 2 2 = ( ) ⋅ ⇒ = 2 1550 ⋅ 23⋅ 155010 = 20,3939 = 16 = 44,1246 Como
y
Se tiene que
Del enunciado se tiene que C=0 y se desprecia el ancho espectral de la fuente ( la longitud de onda está lejos de por lo tanto Se tiene que:
Finalmente reemplazando los valores para obtener , se tiene:
) y además
Problema 2.10 Estimate the limiting bit rate for a 60-km single-mode fiber link at 1.3- and 1.55- µm wavelengths assuming transform-limited, transform-limited, 50-ps (FWHM) ( FWHM) input pulses. Assume that β2 = 0 and
= .. and
and 0 at 1.3- and 1.55-µm 1 .55-µm wavelengths, respectively. respectively. Also assume
that Vω Vω <<1
= 2 = 21,233 2 l n 2 () = 11 + 2 + 1 + 2 + 1 + + 4√ 2 = 20 20 = 0 ≪ 1 () = 1 + 2 = 35,34 < 41 < 7,07 = 0,1 = 0 ≪ 1 () = 1 + 4√ 2 = 21,233 < 11,77 •
Para
•
Para
,
,
y
y
Problema 2.11 A 0.88-µm communication system transmits data over a 10-km single-mode fiber by using 10-ns (FWHM) pulses. Determine the maximum bit rate if the LED has a spectral FWHM of 30 nm. Use D = −80 ps/(km-nm). ps/(km -nm).
10 = 2 = 2 ln2 2 2 ln22 = 4,246 = 4246 = ( ) 23⋅ 1 0 = 0,88⋅10− ⋅30⋅10− = 7,302302 ⋅ 10 [ ] = 2 = 620107,54 ≫ 1 8 0 0 880 = 2 = 23⋅10 = 32,32,86 () = 11 + 2 + 1 + 2 + 1 + + 4√ 2 = 0 = 0 () = 1 + 1 + 2 = 24,36767 4 < 1 ⇒ < 41 < 10,259 Se tiene que
y
, por lo tanto:
Reemplazando los valores en la formula, se tiene finalmente:
Para calcular el máximo bit rate se considera lo siguiente:
Luego:
Problema 2.13 Repeat Problem 2.12 for the case c ase of a single-mode semiconductor laser for which V ω << 1 and show that the bit rate is limited by B(|β B(| β3|L)1/3 < 0.324. What is the limiting bit rate for L = 100 km if β3 = 0.1 ps3/km?
() = 11 + 2 + 1 + 2 + 1 + + 4√ 2 =0 = 0 = 0 () = 1 + 4√ 2 = + 4√ 22 = |3|44 = i = 32 (4) 4≤ 1 ||| ≤ 0,324 = 100, = 0,1 ≤ 150,388b b De los datos del problema se tiene:
Por lo tanto, la ecuación nos queda de la siguiente manera:
El valor mínimo para se consigue para
Usando
, y está dado por:
, se tiene:
Finalmente, para los valores dados:
Se calcula el bit rate:
Problema2.15 A 1.55-µm optical communication system operating at 5 Gb/s is using Gaussian pulses of width 100 ps (FWHM) chirped such that C = −6. What is the dispersion-limited dispersion -limited maximum fiber length? How much will it change if the pulses were unchirped? Neglect laser linewidth and assume that β2 = −20 ps2/km.
= 2020 = 0 = 0 100 100 p p = 2 = 2 ln2 2 21ln2 = 42,466pp 4 < 1 → < 4 ⋅ 5 → < 5050 () = 11 + 2 + 1 + 2 + 1 + + 4√ 2 () = 11 + 2 + 2 = 6 = 5,32 =0 = 112,07676
•
Para
•
Para
, y reemplazando los valores en la ecuación resultante se obtiene:
, y reemplazando los valores en la e cuación resultante se obtiene:
Problema 2.16
A 1.3-µm lightwave system uses a 50-km fiber link and requires at least 0.3 µW at the receiver. The fiber loss is 0.5 dB/km. Fiber is spliced every 5 km and has two connectors of 1-dB loss at both ends. Splice loss is only 0.2 dB. Determine the minimum power that must be launched into the fiber
Prec = 0,0003 003mW mW = 35, 233dbm dbm = 0,2 = (5 1)⋅ = (505 1) ⋅ 0,2 = 1,8 ≥ (50 ⋅0,5 [])2⋅1 1,8 ≥ 35,23 ≥ 6,433 = 0,2275 275
