Question 7 of 7: Optimal Graded-Index Fiber — Dispersion and Bandwidth–Length Product

Question 7: Optimal Graded-Index Fiber — Dispersion and Bandwidth–Length Product (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) total dispersion per km (material + modal, optimal graded-index profile), (b) the BL product for the graded-index fiber, (c) the BL product for an equivalent step-index fiber, (d) the physical reason for the difference.

Approach. Material dispersion follows directly from the given coefficient and source linewidth. For an optimal graded-index profile, modal (intermodal) dispersion is suppressed from $\propto\Delta$ (step index) to $\propto\Delta^2$ (a well-known result of the self-focusing ray behaviour in a near-parabolic profile); the two independent broadening mechanisms combine in quadrature to give the total dispersion, and the familiar rule-of-thumb $B\!\cdot\!L\!\cdot\!\sigma\approx0.2$ converts a total rms spread per km into an estimated bandwidth–length product.

1. Part (a) — total dispersion. The material-dispersion contribution is $$\frac{\Delta\tau_{mat}}{L}=|D_{mat}|\,\Delta\lambda=(5)(3)=15\ \text{ps/km}.$$ The index difference is $\Delta=(n_1-n_2)/n_1=(1.474-1.453)/1.474=0.01425$. For an optimal-profile graded-index fiber the intermodal spread is $$\frac{\Delta\tau_{modal}}{L}=\frac{n_1\Delta^2}{20\sqrt3\,c} =\frac{(1.474)(0.01425)^2}{20\sqrt3(2.998\times10^8)}\times1000\times10^{12}\ \text{ps/km} =28.8\ \text{ps/km}.$$ Combining the two independent mechanisms in quadrature, $$\frac{\Delta\tau_{tot}}{L}=\sqrt{15^2+28.8^2}=\boxed{32.5\ \text{ps/km}}.$$
2. Part (b) — BL product, graded-index. Using the standard NRZ rule-of-thumb $B\!\cdot\!L\approx0.2/\sigma_{tot}$ (per unit length), $$B\!\cdot\!L=\frac{0.2}{32.5\times10^{-12}\ \text{s/km}}=\boxed{6.16\ \text{Gb/s}\cdot\text{km}}.$$
3. Part (c) — BL product, equivalent step-index fiber. For a step-index profile with the same $n_1$, $n_2$ (hence the same $\Delta$), the uncorrected ray-theory modal dispersion is $$\frac{\Delta\tau_{modal,step}}{L}=\frac{n_1\Delta}{c} =\frac{(1.474)(0.01425)}{2.998\times10^8}\times1000\times10^{12}\ \text{ps/km} \approx7.0\times10^4\ \text{ps/km}=70.0\ \text{ns/km},$$ so overwhelmingly dominant that the total is essentially unchanged by the material term: $\Delta\tau_{tot,step}/L\approx70.0$ ns/km. The corresponding bandwidth–length product is $$B\!\cdot\!L=\frac{0.2}{70.0\times10^{-9}\ \text{s/km}}=\boxed{2.86\ \text{Mb/s}\cdot\text{km}}.$$
4. Part (d) — comparison. Grading the index parabolically improves the BL product by a factor of roughly $6.16\times10^3/2.86\approx\boxed{2150\times}$ over the step-index fiber with identical core size and index contrast. The physical reason is the ray picture in the figure above: in a step-index core every ray travels at the same local speed $c/n_1$, so the more obliquely a ray zig-zags, the longer its path and the later it arrives — the spread scales directly with $\Delta$. In the optimal graded-index profile the index (and hence the local speed $c/n(r)$) increases smoothly away from the axis, so the more oblique, longer-path rays spend more of their journey in faster, lower-index material near the cladding, very nearly compensating their extra path length; the residual mismatch is only second order in $\Delta$, which is exactly the $\Delta^2$ scaling used in part (a).