Question

A delay difference of ‘x’ns between the slowest and fastest fiber with a core refractive index of 1.5 and refractive index difference of 2%. Estimate the following
The length of the optical fiber link
RMS pulse broadening due to intermodal dispersion link
Bandwidth –length product for the maximum bit rate.

Answer 1

Given:

$\Delta=0.02;\\ \delta T_b=x\text{ ns};\\ n=1.5.$

Solution:

1) The length of the optical fiber link assuming the delay difference of ‘x’ ns:

$$x\cdot10^{-9}=\delta T_b\approx\frac{Ln\Delta}{c}\rightarrow L=10x\text{ m}.$$

2) RMS pulse broadening due to intermodal dispersion link

$$\sigma_s\approx\frac{Ln\Delta}{2\sqrt3 c}=2.889x\cdot10^{-10}\text{ s}.$$

3) Bandwidth–length product for the maximum bit rate.

$$B_{T\text{max}}\cdot L=\frac{0.2L}{\sigma_s}=692.3\cdot10^7\text{ Hz}\cdot\text{km}.$$