Answer to Question #214424 in Optics for Soham Samanta

Question

A multimode step index fiber has a material dispersion parameter of 250 ps 

nm−1km−l at 1.2 μm. Estimate the pulse broadening for a 10 kilometer long fiber 

with an LED source of spectral width 50 nm.

Answer 1

s

elative refractive index ∆ = 2 % = 0.02

Core refractive index n₁ = 1.48

Link length L = 5 km = 5000 m

Maximum optical bandwidth $BW = 3 MHz = 3\times 10^6 \;Hz$

Material dispersion parameter M = 80 ps nm^-1 km^-1

Spectral width due to waveguide dispersion $σ_λ = 1$

The expression of the rms pulse broadening due to chromatic dispersion τ:

$τ ≈ \frac{L(NA)^2}{2n_{1}c}$

NA is the numerical aperture, n₁ is core refractive index

c is the velocity of light $3\times 10^8\; m/s$

The expression for numerical aperture NA:

$NA= n_1\sqrt{2∆}$

∆ is the relative refractive index

$τ ≈ \frac{L(n_1\sqrt{2∆})^2}{2n_1c}$

$τ ≈ \frac{5000(1.48\sqrt{2\times0.02})^2}{2\times1.48\times 3\times10^8}$

$τ ≈ 0.493 \;μs$

The expression to calculate the rms pulse broadening per km

$\frac{τ}{L} = \frac{0.493\; μs}{5\; km} = 98.6\; \frac{ns}{km}$

The expression to calculate spectral width of source:

$σ_m = σ_λ\times M\times L$

$σ_m = 1\times 80\times 5 = 0.4 \; \frac{ns}{km}$