Answer to Question #132933 in Optics for Shreyasi Chatterjee

Question

1. A multimode step index fiber has a relative refractive index difference of 2% and core refractive index of 1.48. The maximum optical bandwidth that may be obtained with a particular source on a 5km link is 3Mhz. Determine the rms pulse broadening per kilometre resulting from chromatic dispersion mechanisms. Assuming waveguide dispersion to be negligible, estimate the rms spectral width of the source used, if the material dispersion parameter for the fiber at the operating wavelength is 80 ps nm-1 km-1.

2. A 1km length multimode step index fiber has a spectral width of 50nm with intermodal and intramodal dispersion of 10 ns/km and 100 ps/km respectively. Calculate the total dispersion and maximum bit rate.

Answer 1

(1) Relative 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:

$σ_m = σ_λ\times M\times L$

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

Answer: 98.6 ns/km, 0.4 ns/km

(2) Total dispersion $∆t = \sqrt{∆t_{modal}^2+ ∆t_{ch}^2}$

Intermodal or modal dispersion $∆t_{modal} = L \times 10 \;ns/km = 1 \;km\times 10 \;ns/km = 10 \;ns$

Intramodal or chromatic dispersion $∆t_{ch} = L \times 100 \; ps/km \times 50\; nm = 5000 \;ps = 5\; ns$

$∆t = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.18\; ns$

Number of bits per second can be determined as

$B_{max}=\frac{1}{∆t_{modal}\times L} = \frac{1}{ 10\; ns/km*1\;km} = \frac{1}{10\times 10^{-9}\;s} = 100\; Mbit/s\; km$

Answer: 11.18 ns, 100 Mbit/s km