Question 4 of 7: Digital fibre-link design — dispersion limit, attenuation limit and repeater spacing

Question 4: Digital fibre-link design — dispersion limit, attenuation limit and repeater spacing (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 40 km multimode digital link, fully specified by the component list above.

Find. The dispersion-limited and attenuation-limited repeater spacings, the resulting number of repeaters and the length of each link, and the number of guided modes.

Approach. Fix the line rate from the sampling and coding, then solve two independent limits — a time-domain one (intermodal spread must stay below half a bit period) and a power-domain one (the received power must exceed the photons-per-bit sensitivity after all losses and the margin) — and let the smaller one govern the span.

1. Line rate. Nyquist sampling of a 100 kHz baseband needs 200 kSa/s, and each sample becomes 9 bits: $$B = 2 f_{b}\times\text{bits} = 2(100\ \text{kHz})(9) = 1.8\ \text{Mb/s}$$ The detector’s 1 GHz bandwidth is three orders of magnitude larger than this, so the receiver never limits the design — the fibre does.
2. Allowed pulse spreading. For NRZ signalling the bit period is $1/B$, and the paper allows half of it: $$\delta\tau_{\text{allow}} = \frac{0.5}{B} = \frac{0.5}{1.8\times10^{6}} = 277.8\ \text{ns}$$
3. Dispersion-limited span. The fibre is multimode step-index and the source linewidth is not given, so intermodal delay dominates: $$\frac{\delta\tau}{L} = \frac{n_{1}-n_{2}}{c} = \frac{1.500-1.495}{2.998\times10^{8}} = 1.668\times10^{-11}\ \text{s/m} = 16.68\ \text{ns/km}$$ Dividing the allowance by this rate, $$L_{\text{disp}} = \frac{277.8\ \text{ns}}{16.68\ \text{ns/km}}$$ $$\boxed{L_{\text{disp}} = 16.7\ \text{km}}$$
4. Receiver sensitivity in dBm. A photon at 900 nm carries $h\nu = hc/\lambda = 2.207\times10^{-19}\ \text{J}$ (1.378 eV), and the detector needs 400 of them in each of $B$ bits per second: $$P_{rx} = N_{p}\,h\nu\,B = (400)(2.207\times10^{-19})(1.8\times10^{6}) = 1.589\times10^{-10}\ \text{W}$$ $$P_{rx}[\text{dBm}] = 10\log_{10}\!\left(\frac{1.589\times10^{-10}}{10^{-3}}\right) = -68.0\ \text{dBm}$$
5. Attenuation-limited span. The launch is $P_{tx} = 10\ \text{mW} = +10.0\ \text{dBm}$, so the total optical budget is $$P_{tx} - P_{rx} = 10.0 - (-68.0) = 78.0\ \text{dB}$$ Taking off the two coupling losses and the margin leaves what the fibre may absorb: $$78.0 - (3+2) - 5 = 68.0\ \text{dB} \qquad\Rightarrow\qquad L_{\text{att}} = \frac{68.0\ \text{dB}}{3.5\ \text{dB/km}}$$ $$\boxed{L_{\text{att}} = 19.4\ \text{km}}$$
6. Which limit governs, and how many repeaters. Since $L_{\text{disp}} = 16.7\ \text{km}$ is shorter than $L_{\text{att}} = 19.4\ \text{km}$, the system is dispersion limited and no span may exceed 16.7 km. The number of spans needed to cover 40 km is $$N_{\text{span}} = \left\lceil\frac{40}{16.66}\right\rceil = \lceil 2.40\rceil = 3$$ $$\boxed{N_{\text{rep}} = N_{\text{span}} - 1 = 2\ \text{repeaters}}$$ Splitting the route evenly (which is what keeps every span inside both limits with equal headroom) gives $$\boxed{L_{\text{link}} = \frac{40}{3} = 13.3\ \text{km per span}}$$ Each 13.3 km span suffers $13.3\times3.5 = 46.7\ \text{dB}$ of fibre loss against a 68.0 dB allowance, so it clears the power budget with about 21 dB to spare — confirming that dispersion, not loss, set the answer.
7. Mode count. The numerical aperture follows from the two indices, $$\mathrm{NA} = \sqrt{n_{1}^{2}-n_{2}^{2}} = \sqrt{1.500^{2}-1.495^{2}} = 0.1224$$ and with $a = 50\ \mu\text{m}$ at 900 nm, $$V = \frac{2\pi a\,\mathrm{NA}}{\lambda} = 42.7 \qquad\Rightarrow\qquad \boxed{M \approx \frac{V^{2}}{2} = 913\ \text{modes}}$$