22-Elec-B10 Electro-Optical Engineering · December 2017
Question 7 of 7: Graded-index versus step-index fibre — total dispersion and bandwidth–distance product
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Elec-B10, Electro-Optical Engineering —
National Exams, December 2017. Three hours, closed book; one 8.5 in × 11 in
double-sided sheet of hand-written notes and a Casio or Sharp approved calculator are
permitted. Seven questions, all of equal value (20 marks each); the rubric says
any five questions constitute a complete paper. All seven are solved below,
because the set is a study resource rather than a timed sitting.
Reference texts.
G. Keiser, Optical Fiber Communications, 4th ed. — fibre modes,
dispersion, link power and rise-time budgets, photodetectors.
J. M. Senior, Optical Fiber Communications: Principles and Practice, 3rd ed.
— numerical aperture, intermodal/intramodal dispersion, receiver noise.
B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. —
laser rate equations, photon lifetime, electro-optic modulators.
J. Wilson and J. F. B. Hawkes, Optoelectronics: An Introduction, 3rd ed.
— pin and avalanche photodiodes, Pockels cells.
Physical constants are the ones printed on page 1 of the examination paper:
$c = 2.998\times10^{8}\ \text{m/s}$, $q = 1.602\times10^{-19}\ \text{C}$,
$h = 6.626\times10^{-34}\ \text{J}\cdot\text{s}$,
$k = 1.381\times10^{-23}\ \text{J/K}$,
$\varepsilon_{0} = 8.854\times10^{-12}\ \text{F/m}$, and for the modulator crystal
$\varepsilon_{r}(\text{LiNbO}_3) = 32$, $n_{o} = 2.30$, $r = 30\ \text{pm/V}$.
Check: two book-keeping notes on the source paper.
The page-5 marking scheme lists five sub-parts, (a)–(e), for Questions 5 and 7, but
the printed questions on pages 3 and 4 carry only four sub-parts each; the mark totals
still come to 20, so the extra row is a marking-sheet artefact and every printed sub-part
is answered here. Question 6 also carries the hand-added annotation
“rework this one” next to its heading — an examiner’s note on the
printed paper, not part of the question. It is solved exactly as printed.
Question 7: Graded-index versus step-index fibre — total dispersion and
bandwidth–distance product (20 marks)
Given. One set of fibre materials and dimensions, used twice —
once with an optimum near-parabolic index profile and once as a step index.
Given data — Question 7
Quantity
Symbol
Value
Core diameter (radius)
$2a$ ($a$)
30 µm (15 µm)
Axial core index
$n_{1}$
1.474
Cladding index
$n_{2}$
1.453
Source wavelength
$\lambda$
1300 nm
Source spectral width (FWHM)
$\sigma_{\lambda}$
3 nm
Material dispersion coefficient
$D_{\text{mat}}$
−5 ps km$^{-1}$ nm$^{-1}$
Find. The total dispersion per kilometre of the graded-index fibre, its
bandwidth–distance product, the same product for the step-index version, and a
physical explanation of the difference.
Why the profile matters. In the step-index fibre every ray travels at the same speed, so the longer zig-zag path arrives late. In the graded-index fibre the off-axis ray spends its extra path length in lower-index (faster) material, and the group delays nearly cancel.
Approach. Compute the relative index difference, then the two
independent contributions — chromatic (material) dispersion driven by the source
linewidth, and intermodal dispersion driven by the profile — and combine them in
quadrature. The bandwidth–distance product follows from the r.m.s. criterion
$B = 0.2/\sigma$.
Relative index difference. Both fibres share the same materials, so
$$\Delta = \frac{n_{1}-n_{2}}{n_{1}} = \frac{1.474-1.453}{1.474}
= 1.4247\times10^{-2}$$
a 1.42 % step — large, as befits a multimode fibre designed to capture plenty of
light.
Material (chromatic) dispersion. This depends only on the source and
the glass, so it is identical for both fibres:
$$\frac{\sigma_{\text{mat}}}{L} = |D_{\text{mat}}|\,\sigma_{\lambda}
= (5\ \text{ps}\,\text{km}^{-1}\text{nm}^{-1})(3\ \text{nm})
= 15\ \text{ps/km}$$
The sign of $D_{\text{mat}}$ tells us which spectral components run ahead (here the fibre is
in the normal-dispersion regime at 1300 nm), but only the magnitude enters a pulse-spreading
budget.
Intermodal dispersion, optimum graded profile. For a near-parabolic
($\alpha \approx 2$) profile the r.m.s. modal delay spread is
$$\frac{\sigma_{\text{mod}}}{L} = \frac{n_{1}\Delta^{2}}{20\sqrt{3}\,c}
= \frac{(1.474)(1.4247\times10^{-2})^{2}}{20\sqrt{3}\,(2.998\times10^{8})}$$
$$\boxed{\frac{\sigma_{\text{mod}}}{L} = 28.8\ \text{ps/km}}$$
The key feature is the $\Delta^{2}$: grading the profile turns a first-order dependence on
the index difference into a second-order one.
Total dispersion for the graded-index fibre. The two mechanisms are
statistically independent, so they add in quadrature:
$$\frac{\sigma_{T}}{L} = \sqrt{\left(\frac{\sigma_{\text{mat}}}{L}\right)^{2}
+ \left(\frac{\sigma_{\text{mod}}}{L}\right)^{2}}
= \sqrt{15.0^{2} + 28.8^{2}}$$
$$\boxed{\frac{\sigma_{T}}{L} = 32.5\ \text{ps/km}}$$
Modal dispersion is still the larger term, but only by a factor of about two — the
grading has brought the two contributions into the same league, which is the sign of a
well-matched design.
Step-index version of the same fibre. Replacing the graded profile with
a step, the r.m.s. modal spread becomes first order in $\Delta$:
$$\frac{\sigma_{\text{mod}}}{L} = \frac{n_{1}\Delta}{2\sqrt{3}\,c}
= \frac{(1.474)(1.4247\times10^{-2})}{2\sqrt{3}\,(2.998\times10^{8})}
= 20.2\ \text{ns/km}$$
This is roughly 1350 times larger than the 15 ps/km of material dispersion, so the
quadrature sum is indistinguishable from the modal term alone,
$\sigma_{T}/L = 20.2\ \text{ns/km}$, and
$$B\!\cdot\!L = \frac{0.2}{20.22\times10^{-9}\ \text{s/km}}$$
$$\boxed{B\!\cdot\!L = 9.9\ \text{Mb}\,\text{s}^{-1}\,\text{km}}$$
Comparison. The graded-index fibre outperforms its step-index twin by
$$\frac{6.16\times10^{9}}{9.89\times10^{6}} \approx 623\times$$
Physically, the step-index fibre gives every guided ray the same phase velocity $c/n_{1}$,
so a ray bouncing at the critical angle simply travels a longer geometric path and arrives
late by $n_{1}\Delta L/c$. In the graded fibre the index falls away from the axis, so the
off-axis ray spends most of its longer path in faster material; to first order the
extra path length and the higher speed cancel, and the residual spread is second order in
$\Delta$. That is the entire content of the $\Delta \to \Delta^{2}$ change between steps 3
and 6, and it is why every multimode fibre made for data transmission is graded. Note also
that the graded fibre’s figure would improve no further by grading alone: at 6.2
Gb s$^{-1}$ km the material term is already a third of the budget, so the next
design move is a narrower-linewidth source.
Check: dispersion conventions. The r.m.s.
formulations used here are Senior’s ($\sigma_{\text{step}} = n_{1}\Delta L/(2\sqrt{3}c)$,
$\sigma_{\text{graded}} = n_{1}\Delta^{2}L/(20\sqrt{3}c)$, $B = 0.2/\sigma$). Texts that work
with the maximum delay difference $\delta\tau$ instead (with
$B \approx 1/2\delta\tau$) obtain numerically different products — typically a factor
of two to three — though the graded-to-step ratio, which is what part (d)
asks about, is essentially the same. The convention is stated so the arithmetic can be
followed.
