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

The delay difference is

$$\delta T_b\approx\frac{Ln_1\Delta}{c}=\frac{L\cdot1.5\cdot0.02}{2.998\cdot10^6}=L\cdot10^{-10}\text{ s},$$

where L is the length of the optical link, therefore, if $\delta T_b=x\cdot10^{-9},$

$$L=10x\text{ m}$$

if x is measured in nanoseconds.

Find the RMS pulse broadening due to intermodal dispersion link:

$$\sigma_s\approx\frac{Ln_1\Delta}{2\sqrt3 c}=\frac{10x\cdot1.5\cdot0.02}{2\sqrt3 \cdot2.998\cdot10^8}=2.889x\cdot10^{-10}\text{ s}.$$

Prior to find the bandwidth-length product, calculate the maximum bit rate:

$$B_{T\text{max}}=\frac{0.2}{\sigma_s}=\frac{692.3}{x}\cdot10^6\text{ Hz}.$$

The bandwidth-length product is

$$B_{T\text{max}}\cdot L=\frac{692.3}{x}\cdot10^6\cdot10x=692.3\cdot10^7\text{ Hz}\cdot\text{km}.$$