Question 1 of 6: Group Index, Dispersion and a Step-Index Fibre

Question 1: Group Index, Dispersion and a Step-Index Fibre (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. A single glass of wavelength-dependent index $n(\lambda)$, in which phase and group velocities are $v_p = c/n$ and $v_g = c/N$; and, for part (b), a step-index fibre carrying 1350 nm light with the data collected below.

Find. The group-index identity and the second-order dispersion coefficient in terms of $N$; then the modal character and mode count of the fibre, and the greatest length it can span at 100 Mbit/s in RZ format.

Approach. Differentiate the propagation constant $\beta = n\omega/c$ once to reach the group index and twice to reach $\beta_2$, converting $\omega$-derivatives to $\lambda$-derivatives; then size the fibre with the normalised frequency $V$, and set the total pulse spread against the RZ rise-time budget $0.35/B$.

1. Write the propagation constant and differentiate once. For a plane wave in a medium of index $n$, $$\beta = k_0 n = \frac{n\,\omega}{c}.$$ The group velocity is $v_g = d\omega/d\beta$, so its reciprocal is $$\frac{1}{v_g} = \frac{d\beta}{d\omega} = \frac{1}{c}\left(n + \omega\,\frac{dn}{d\omega}\right).$$ Comparing this with the definition $v_g = c/N$ identifies the group index as $N = n + \omega\,dn/d\omega$.
2. Convert the frequency derivative to a wavelength derivative. Since $\omega = 2\pi c/\lambda$, differentiating gives $d\omega/d\lambda = -2\pi c/\lambda^{2}$, so $$\omega\,\frac{dn}{d\omega} = \frac{2\pi c}{\lambda}\cdot\left(-\frac{\lambda^{2}}{2\pi c}\right) \frac{dn}{d\lambda} = -\lambda\,\frac{dn}{d\lambda}.$$ Substituting into the result of Step 1 gives the required identity, $$\boxed{\;N = n - \lambda\,\frac{dn}{d\lambda}\;}$$ which says that the group index exceeds the phase index wherever the glass is normally dispersive ($dn/d\lambda \lt 0$).
3. Differentiate a second time for the dispersion parameter. The first-order coefficient is the group delay per unit length, $\beta_1 = d\beta/d\omega = N/c$. Differentiating once more and converting to wavelength in the same way, $$\beta_2 = \frac{d^{2}\beta}{d\omega^{2}} = \frac{1}{c}\frac{dN}{d\omega} = \frac{1}{c}\frac{dN}{d\lambda}\cdot\frac{d\lambda}{d\omega}, \qquad \frac{d\lambda}{d\omega} = -\frac{\lambda^{2}}{2\pi c},$$ so that $$\boxed{\;\frac{d^{2}\beta}{d\omega^{2}} = -\frac{\lambda^{2}}{2\pi c^{2}}\,\frac{dN}{d\lambda}\;}$$ Because $dN/d\lambda = -\lambda\,d^{2}n/d\lambda^{2}$ follows directly from the identity of Step 2, the same coefficient may be written $\beta_2 = \lambda^{3}/(2\pi c^{2})\cdot d^{2}n/d\lambda^{2}$, and the engineering dispersion parameter is $D = d\beta_1/d\lambda = (1/c)\,dN/d\lambda = -2\pi c\,\beta_2/\lambda^{2}$.
4. Size the fibre: numerical aperture and V-number. With $a = 10$ µm, $$\mathrm{NA} = \sqrt{n_1^{2}-n_2^{2}} = \sqrt{1.465^{2}-1.462^{2}} = 0.0937,$$ $$V = \frac{2\pi a}{\lambda}\,\mathrm{NA} = \frac{2\pi(10\times10^{-6})(0.0937)}{1350\times10^{-9}} = 4.36 .$$ Since $V = 4.36$ exceeds the single-mode cut-off of 2.405, the fibre is multimode at 1350 nm. For a step-index guide the supported mode count is $$M \approx \frac{V^{2}}{2} = \frac{4.36^{2}}{2} = \boxed{\;9.5 \approx 9\ \text{to}\ 10\ \text{modes}\;}$$
5. Collect the pulse-spreading mechanisms. Two act at once. The chromatic (intramodal) term is stated directly: $$\frac{\Delta t_{\text{intra}}}{L} = D\,\Delta\lambda = \left(20\ \tfrac{\text{ps}}{\text{nm}\cdot\text{km}}\right)(15\ \text{nm}) = 300\ \text{ps/km} = 0.300\ \text{ns/km}.$$ The intermodal term is the delay difference between the axial and the highest-order ray of the previous figure, evaluated with the group indices the question supplies: $$\frac{\Delta t_{\text{inter}}}{L} = \frac{N_1(n_1-n_2)}{c\,n_2} = \frac{1.474\,(0.003)}{(2.998\times10^{8})(1.462)} = 1.009\times10^{-11}\ \text{s/m} = 10.09\ \text{ns/km}.$$ This is the reason the group indices were given, and it dwarfs the chromatic term by a factor of thirty-four.
6. Add the mechanisms in quadrature. Because the two broadening processes are statistically independent, $$\frac{\Delta t_{\text{tot}}}{L} = \sqrt{\left(10.09\right)^{2}+\left(0.300\right)^{2}} = 10.09\ \text{ns/km}.$$ To three figures the total is indistinguishable from the intermodal term alone — in a step-index multimode fibre the mode delay is the whole story.
7. Apply the RZ rise-time budget. A return-to-zero pulse occupies half a bit slot, so the accepted engineering allowance is that the received pulse may spread to 35 % of the bit period: $$\Delta t_{\max} = \frac{0.35}{B} = \frac{0.35}{100\times10^{6}} = 3.5\ \text{ns}.$$ Dividing by the spread per kilometre, $$L_{\max} = \frac{3.5\ \text{ns}}{10.09\ \text{ns/km}} = \boxed{\;0.35\ \text{km}\ \ (\approx 350\ \text{m})\;}$$ Had the chromatic term acted alone the reach would have been $3.5/0.300 = 11.7$ km; the mode delay costs a factor of thirty-three in distance. A bandwidth–distance product of roughly $100\ \text{Mbit/s}\times0.35\ \text{km} \approx 35$ Mbit·km/s is exactly what a step-index multimode fibre of this small index contrast delivers.

Part (c) — the sources of dispersion. Dispersion is any mechanism that makes different parts of the launched energy arrive at different times, and in a fibre four mechanisms are conventionally distinguished.

Intermodal (modal) dispersion exists only in multimode fibre. Each guided mode has its own axial phase velocity, so a pulse launched into many modes arrives as a superposition of differently delayed replicas. In a step-index guide the spread is set by the index contrast, as the calculation above showed; grading the profile parabolically equalises the path delays and reduces the spread by two to three orders of magnitude, which is why graded-index fibre replaced step-index fibre for local-area links. Single-mode fibre removes the mechanism entirely.

Material dispersion arises because the silica refractive index itself varies with wavelength, so the several nanometres of source linewidth travel at slightly different group velocities. This is the term governed by $d^{2}n/d\lambda^{2}$ in Step 3; it passes through zero near 1310 nm in pure silica, which is why that window was the first to be exploited for long-haul systems.

Waveguide dispersion arises because the fraction of modal power carried in the cladding — and therefore the effective index seen by the mode — is itself wavelength dependent. It is small compared with material dispersion in multimode fibre, but in single-mode fibre it is comparable and of opposite sign, and deliberately shaping the index profile to exploit that cancellation produces dispersion-shifted and dispersion-flattened fibres. Taken together, material and waveguide dispersion constitute chromatic (intramodal) dispersion, the 20 ps/(nm·km) figure quoted in part (b).

Polarisation-mode dispersion and profile dispersion complete the list. Residual core ellipticity and stress birefringence give the two polarisation states of the fundamental mode slightly different group delays, accumulating as a random-walk term in $\text{ps}/\sqrt{\text{km}}$ that limits 10 Gbit/s and faster single-mode systems; profile dispersion is the weak second-order effect of the index contrast $\Delta$ itself being wavelength dependent.