Question

Imagine that you have three graded index fiber cables which are connected to each other in series. The refractive indices of the core/clad for each fiber is 1.57/1.50, 1.68/1.46 and 1.77/1.68, and the lengths of the fibers are 5, 8, and 2 km, respectively. What are the time dispersion, maximum bit-rate and frequency bandwidth of each fiber?

Answer 1

Let the refractive index of core be n₁, refractive index of cladding be n₂ and length of the fiber be L. c is the velocity of light in free space

The time dispersion is given by t=Ln₁(n₁-n₂)/cn₂

The maximum bit-rate is given by B=1/2t

And the frequency bandwidth is given by W=cn_2²/[2n₁²L(n₁-n₂)]

For 1st graded index fiber:-

n₁= 1.57 and n₂=1.5(given)

L=5km=5000m

Thus, t=5000*1.57(0.07)/(3*10⁸*1.5)=122.11 *10^-8sec

B=1/(2*122.11 *10^-8)=4.09*10⁵ bits/sec

W=3*10⁸*1.5²/[2*1.57²*5000(0.07)]=3.91*10⁵ Hz

For 2nd graded index fiber:-

n₁=1.68 and n₂=1.46

L=8 km=8000 m

Thus, t=8000*1.68(0.22)/(3*10⁸*1.46)=675.07 *10^-8sec

B=1/(2*675.07 *10^-8)=7.41*10⁴ bits/sec

W=3*10⁸*1.46²/[2*1.68²*8000(0.22)]=6.44*10⁴ Hz

For 3rd graded index fiber:-

n₁=1.77 and n₂=1.68

L=2 km=2000 m

Thus, t=2000*1.77(0.09)/(3*10⁸*1.68)=63.21 *10^-8sec

B=1/(2*63.21 *10^-8)=7.9*10⁵ bits/sec

W=3*10⁸*1.68²/[2*1.77²*2000(0.09)]=75.07*10⁴ Hz